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3,300 | aa41481c-6ddd-11ea-8a43-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-state-of-the-diatomic-chlorine-element | 0 | start physical_unit 7 9 oxidation_state none qc_end substance 7 9 qc_end end | [{"type":"physical unit","value":"Oxidation state [OF] diatomic chlorine element"}] | [{"type":"physical unit","value":"0"}] | [{"type":"substance name","value":"Diatomic chlorine element"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation state of the diatomic chlorine element?</h1> | null | 0 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All neutral substances that contain atoms of only one element have an oxidation number of zero.</p>
<p>Other examples include helium, sodium, and dihydrogen <mathjax>#("H"_2")#</mathjax>.</p></div>
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<div class="markdown"><p>The oxydation state of diatomic chlorine, <mathjax>#"Cl"_2"#</mathjax>, is zero.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All neutral substances that contain atoms of only one element have an oxidation number of zero.</p>
<p>Other examples include helium, sodium, and dihydrogen <mathjax>#("H"_2")#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation state of the diatomic chlorine element?</h1>
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<div class="markdown"><p>The oxydation state of diatomic chlorine, <mathjax>#"Cl"_2"#</mathjax>, is zero.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All neutral substances that contain atoms of only one element have an oxidation number of zero.</p>
<p>Other examples include helium, sodium, and dihydrogen <mathjax>#("H"_2")#</mathjax>.</p></div>
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</article> | What is the oxidation state of the diatomic chlorine element? | null |
3,301 | ab067162-6ddd-11ea-a831-ccda262736ce | https://socratic.org/questions/if-given-sn-s-cl-2-g-sncl-2-s-deltah-325-kj-and-sncl-2-s-cl-2-g-sncl-4-l-deltah- | -511 kJ | start physical_unit 25 25 deltah kj qc_end chemical_equation 26 31 qc_end end | [{"type":"physical unit","value":"DeltaH [OF] the reaction [IN] kJ"}] | [{"type":"physical unit","value":"-511 kJ"}] | [{"type":"chemical equation","value":"Sn(s) + Cl2(g) -> SnCl2(s) deltaH = −325 kJ"},{"type":"chemical equation","value":"SnCl2(s) + Cl2(g) -> SnCl4(l), deltaH = −186 kJ"},{"type":"chemical equation","value":"Sn(s) + 2 Cl2(g) -> SnCl4(l)"}] | <h1 class="questionTitle" itemprop="name">If given: #Sn(s) + Cl_2(g) -> SnCl_2(s)#, #DeltaH = -325 kJ# and #SnCl_2(s) + Cl_2(g) -> SnCl_4(l)#, #DeltaH = -186 kJ#, what is #DeltaH# for this reaction? #Sn(s) + 2Cl_2(g) -> SnCl_4(l)#</h1> | null | -511 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First of all, write out the equations</p>
<p><mathjax>#Sn(s) + Cl_"2"(g) -> SnCl_"2"(s)#</mathjax> <mathjax>#DeltaH = -325kJ#</mathjax></p>
<p><mathjax>#SnCl_"2"(s) + Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = -186kJ#</mathjax></p>
<p>_- - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - </p>
<p><mathjax>#Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = ???#</mathjax></p>
<p>Next, we will rework the two equations on top. What we will do determines the answer for the problem. First, we will flip around equations (if necessary) so that the products of the equation or reactants of the top equations will be the same products and reactants as the very bottom equation.</p>
<p>Next, we will change the coefficients of the top two equations to match those of the bottom equation. Let's work it out below:</p>
<p><mathjax>#Sn(s) + Cl_"2"(g) -> SnCl_"2"(s)#</mathjax> <mathjax>#DeltaH = -325kJ#</mathjax></p>
<p><mathjax>#SnCl_"2"(s) + Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = -186kJ#</mathjax></p>
<p>_- - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - </p>
<p><mathjax>#Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = ???#</mathjax></p>
<p>It seems in this equation, however, the coefficients are set correctly and the equations do not need to be flipped. Now we will do the last step, which is canceling out unneeded products/reactants to get an equation identical to the bottom equation. the "x" is attached to those products/reactants that are canceled out</p>
<p><mathjax>#Sn(s) + Cl_"2"(g) ->#</mathjax> x(<mathjax>#SnCl_"2"(s))#</mathjax> <mathjax>#DeltaH = -325kJ#</mathjax></p>
<p>x(<mathjax>#SnCl_"2"(s)) + Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = -186kJ#</mathjax></p>
<p>_- - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - </p>
<p><mathjax>#Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = ???#</mathjax></p>
<p>=</p>
<p><mathjax>#Sn(s) + Cl_"2"(g) -> )#</mathjax> <mathjax>#DeltaH = -325kJ#</mathjax></p>
<p><mathjax># Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = -186kJ#</mathjax></p>
<p>_- - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - </p>
<p><mathjax>#Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = ???#</mathjax></p>
<p>Now, when we add the two equations together to equal the bottom equation, we also add the <mathjax>#DeltaH#</mathjax> for each equation together</p>
<pre><code> #DeltaH = -325kJ#
</code></pre>
<p><mathjax># Sn(s) + Cl_"2"(g) + Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <br/>
<mathjax>#DeltaH = -186kJ + -325kJ#</mathjax></p>
<p>_- - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - </p>
<p><mathjax>#Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = ???#</mathjax></p>
<p>Now combine like terms to finish the problem</p>
<p><mathjax># Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = -511kJ#</mathjax></p>
<p>_- - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - </p>
<p><mathjax>#Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = ???#</mathjax></p>
<p>See how the two equations are now exactly the same? Well, so are the <mathjax>#DeltaH#</mathjax> values!. So, now your final answer is:</p>
<p><mathjax># Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = -511kJ#</mathjax></p>
<p>Final answer: <mathjax>#-511kJ#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Use Hess's law to simplify</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First of all, write out the equations</p>
<p><mathjax>#Sn(s) + Cl_"2"(g) -> SnCl_"2"(s)#</mathjax> <mathjax>#DeltaH = -325kJ#</mathjax></p>
<p><mathjax>#SnCl_"2"(s) + Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = -186kJ#</mathjax></p>
<p>_- - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - </p>
<p><mathjax>#Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = ???#</mathjax></p>
<p>Next, we will rework the two equations on top. What we will do determines the answer for the problem. First, we will flip around equations (if necessary) so that the products of the equation or reactants of the top equations will be the same products and reactants as the very bottom equation.</p>
<p>Next, we will change the coefficients of the top two equations to match those of the bottom equation. Let's work it out below:</p>
<p><mathjax>#Sn(s) + Cl_"2"(g) -> SnCl_"2"(s)#</mathjax> <mathjax>#DeltaH = -325kJ#</mathjax></p>
<p><mathjax>#SnCl_"2"(s) + Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = -186kJ#</mathjax></p>
<p>_- - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - </p>
<p><mathjax>#Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = ???#</mathjax></p>
<p>It seems in this equation, however, the coefficients are set correctly and the equations do not need to be flipped. Now we will do the last step, which is canceling out unneeded products/reactants to get an equation identical to the bottom equation. the "x" is attached to those products/reactants that are canceled out</p>
<p><mathjax>#Sn(s) + Cl_"2"(g) ->#</mathjax> x(<mathjax>#SnCl_"2"(s))#</mathjax> <mathjax>#DeltaH = -325kJ#</mathjax></p>
<p>x(<mathjax>#SnCl_"2"(s)) + Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = -186kJ#</mathjax></p>
<p>_- - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - </p>
<p><mathjax>#Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = ???#</mathjax></p>
<p>=</p>
<p><mathjax>#Sn(s) + Cl_"2"(g) -> )#</mathjax> <mathjax>#DeltaH = -325kJ#</mathjax></p>
<p><mathjax># Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = -186kJ#</mathjax></p>
<p>_- - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - </p>
<p><mathjax>#Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = ???#</mathjax></p>
<p>Now, when we add the two equations together to equal the bottom equation, we also add the <mathjax>#DeltaH#</mathjax> for each equation together</p>
<pre><code> #DeltaH = -325kJ#
</code></pre>
<p><mathjax># Sn(s) + Cl_"2"(g) + Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <br/>
<mathjax>#DeltaH = -186kJ + -325kJ#</mathjax></p>
<p>_- - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - </p>
<p><mathjax>#Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = ???#</mathjax></p>
<p>Now combine like terms to finish the problem</p>
<p><mathjax># Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = -511kJ#</mathjax></p>
<p>_- - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - </p>
<p><mathjax>#Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = ???#</mathjax></p>
<p>See how the two equations are now exactly the same? Well, so are the <mathjax>#DeltaH#</mathjax> values!. So, now your final answer is:</p>
<p><mathjax># Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = -511kJ#</mathjax></p>
<p>Final answer: <mathjax>#-511kJ#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If given: #Sn(s) + Cl_2(g) -> SnCl_2(s)#, #DeltaH = -325 kJ# and #SnCl_2(s) + Cl_2(g) -> SnCl_4(l)#, #DeltaH = -186 kJ#, what is #DeltaH# for this reaction? #Sn(s) + 2Cl_2(g) -> SnCl_4(l)#</h1>
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<div class="markdown"><p>Use Hess's law to simplify</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First of all, write out the equations</p>
<p><mathjax>#Sn(s) + Cl_"2"(g) -> SnCl_"2"(s)#</mathjax> <mathjax>#DeltaH = -325kJ#</mathjax></p>
<p><mathjax>#SnCl_"2"(s) + Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = -186kJ#</mathjax></p>
<p>_- - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - </p>
<p><mathjax>#Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = ???#</mathjax></p>
<p>Next, we will rework the two equations on top. What we will do determines the answer for the problem. First, we will flip around equations (if necessary) so that the products of the equation or reactants of the top equations will be the same products and reactants as the very bottom equation.</p>
<p>Next, we will change the coefficients of the top two equations to match those of the bottom equation. Let's work it out below:</p>
<p><mathjax>#Sn(s) + Cl_"2"(g) -> SnCl_"2"(s)#</mathjax> <mathjax>#DeltaH = -325kJ#</mathjax></p>
<p><mathjax>#SnCl_"2"(s) + Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = -186kJ#</mathjax></p>
<p>_- - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - </p>
<p><mathjax>#Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = ???#</mathjax></p>
<p>It seems in this equation, however, the coefficients are set correctly and the equations do not need to be flipped. Now we will do the last step, which is canceling out unneeded products/reactants to get an equation identical to the bottom equation. the "x" is attached to those products/reactants that are canceled out</p>
<p><mathjax>#Sn(s) + Cl_"2"(g) ->#</mathjax> x(<mathjax>#SnCl_"2"(s))#</mathjax> <mathjax>#DeltaH = -325kJ#</mathjax></p>
<p>x(<mathjax>#SnCl_"2"(s)) + Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = -186kJ#</mathjax></p>
<p>_- - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - </p>
<p><mathjax>#Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = ???#</mathjax></p>
<p>=</p>
<p><mathjax>#Sn(s) + Cl_"2"(g) -> )#</mathjax> <mathjax>#DeltaH = -325kJ#</mathjax></p>
<p><mathjax># Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = -186kJ#</mathjax></p>
<p>_- - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - </p>
<p><mathjax>#Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = ???#</mathjax></p>
<p>Now, when we add the two equations together to equal the bottom equation, we also add the <mathjax>#DeltaH#</mathjax> for each equation together</p>
<pre><code> #DeltaH = -325kJ#
</code></pre>
<p><mathjax># Sn(s) + Cl_"2"(g) + Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <br/>
<mathjax>#DeltaH = -186kJ + -325kJ#</mathjax></p>
<p>_- - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - </p>
<p><mathjax>#Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = ???#</mathjax></p>
<p>Now combine like terms to finish the problem</p>
<p><mathjax># Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = -511kJ#</mathjax></p>
<p>_- - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - </p>
<p><mathjax>#Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = ???#</mathjax></p>
<p>See how the two equations are now exactly the same? Well, so are the <mathjax>#DeltaH#</mathjax> values!. So, now your final answer is:</p>
<p><mathjax># Sn(s) + 2Cl_"2"(g) -> SnCl_"4"(l)#</mathjax> <mathjax>#DeltaH = -511kJ#</mathjax></p>
<p>Final answer: <mathjax>#-511kJ#</mathjax></p></div>
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Michael
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<div class="markdown"><p><mathjax>#sf(DeltaH=-511color(white)(x)kJ)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/thermochemistry/hess-law">Hess' Law</a> states that the overall <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of a chemical reaction is independent of the route taken. This is a consequence of The Conservation of Energy.</p>
<p>I have set up a Hess Cycle:</p>
<p><img alt="MFDocs" src="https://useruploads.socratic.org/Sd6CpldOSrOBvrOUJr53_Tin%20Hess.jpg"/> </p>
<p>Applying Hess' Law you can say that the enthalpy change of the <mathjax>#sf(color(red)(red))#</mathjax> route is equal to the enthalpy change of the <mathjax>#sf(color(blue)(blue))#</mathjax> route. This is because the arrows start and finish in the same place.</p>
<p>We can say:</p>
<p><mathjax>#sf(DeltaH=-325-186=-511color(white)(x)kJ)#</mathjax></p>
<p>For the reaction <mathjax>#sf(Sn_((s))+2Cl_(2(g))rarrSnCl_(4(l))#</mathjax></p></div>
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</article> | If given: #Sn(s) + Cl_2(g) -> SnCl_2(s)#, #DeltaH = -325 kJ# and #SnCl_2(s) + Cl_2(g) -> SnCl_4(l)#, #DeltaH = -186 kJ#, what is #DeltaH# for this reaction? #Sn(s) + 2Cl_2(g) -> SnCl_4(l)# | null |
3,302 | aadf95ae-6ddd-11ea-bf35-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-an-aqueous-solution-containing-22-5-grams-of-glucose-in- | 4.90 M | start physical_unit 6 7 molarity mol/l qc_end physical_unit 12 12 9 10 mass qc_end physical_unit 7 7 14 15 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] aqueous solution [IN] M"}] | [{"type":"physical unit","value":"4.90 M"}] | [{"type":"physical unit","value":"Mass [OF] glucose [=] \\pu{22.5 grams}"},{"type":"physical unit","value":"Volume [OF] glucose solution [=] \\pu{25.5 mL}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of an aqueous solution containing 22.5 grams of glucose in 25.5 mL of solution?</h1> | null | 4.90 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by calculating the number of moles of glucose present in <mathjax>#"25.5 mL"#</mathjax> of solution. </p>
<p>To do that, use the <strong>molar mass</strong> of glucose</p>
<blockquote>
<p><mathjax>#22.5 color(red)(cancel(color(black)("g"))) * "1 mole glucose"/(180.156color(red)(cancel(color(black)("g")))) = "0.1249 moles glucose"#</mathjax></p>
</blockquote>
<p>Now, in order to find the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the solution, you need to determine the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present in <mathjax>#"1 L" = 10^3#</mathjax> <mathjax>#"mL"#</mathjax> of solution. </p>
<p>The fact that you're dealing with a <em>solution</em>, which, as you know, is a <strong>heterogeneous mixture</strong>, i.e. it has the same composition throughout, you can use the known composition as a conversion factor to find the number of moles of solute present in <mathjax>#10^3#</mathjax> <mathjax>#"mL"#</mathjax> of solution.</p>
<blockquote>
<p><mathjax>#10^3 color(red)(cancel(color(black)("mL solution"))) * "0.1249 moles glucose"/(25.5color(red)(cancel(color(black)("mL solution")))) = "4.898 moles glucose"#</mathjax></p>
</blockquote>
<p>You can thus say that the molarity of the solution is equal to</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("molarity = 4.90 mol L"^(-1))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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<div class="markdown"><p><mathjax>#"4.90 mol L"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by calculating the number of moles of glucose present in <mathjax>#"25.5 mL"#</mathjax> of solution. </p>
<p>To do that, use the <strong>molar mass</strong> of glucose</p>
<blockquote>
<p><mathjax>#22.5 color(red)(cancel(color(black)("g"))) * "1 mole glucose"/(180.156color(red)(cancel(color(black)("g")))) = "0.1249 moles glucose"#</mathjax></p>
</blockquote>
<p>Now, in order to find the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the solution, you need to determine the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present in <mathjax>#"1 L" = 10^3#</mathjax> <mathjax>#"mL"#</mathjax> of solution. </p>
<p>The fact that you're dealing with a <em>solution</em>, which, as you know, is a <strong>heterogeneous mixture</strong>, i.e. it has the same composition throughout, you can use the known composition as a conversion factor to find the number of moles of solute present in <mathjax>#10^3#</mathjax> <mathjax>#"mL"#</mathjax> of solution.</p>
<blockquote>
<p><mathjax>#10^3 color(red)(cancel(color(black)("mL solution"))) * "0.1249 moles glucose"/(25.5color(red)(cancel(color(black)("mL solution")))) = "4.898 moles glucose"#</mathjax></p>
</blockquote>
<p>You can thus say that the molarity of the solution is equal to</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("molarity = 4.90 mol L"^(-1))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of an aqueous solution containing 22.5 grams of glucose in 25.5 mL of solution?</h1>
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<div class="markdown"><p><mathjax>#"4.90 mol L"^(-1)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by calculating the number of moles of glucose present in <mathjax>#"25.5 mL"#</mathjax> of solution. </p>
<p>To do that, use the <strong>molar mass</strong> of glucose</p>
<blockquote>
<p><mathjax>#22.5 color(red)(cancel(color(black)("g"))) * "1 mole glucose"/(180.156color(red)(cancel(color(black)("g")))) = "0.1249 moles glucose"#</mathjax></p>
</blockquote>
<p>Now, in order to find the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the solution, you need to determine the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present in <mathjax>#"1 L" = 10^3#</mathjax> <mathjax>#"mL"#</mathjax> of solution. </p>
<p>The fact that you're dealing with a <em>solution</em>, which, as you know, is a <strong>heterogeneous mixture</strong>, i.e. it has the same composition throughout, you can use the known composition as a conversion factor to find the number of moles of solute present in <mathjax>#10^3#</mathjax> <mathjax>#"mL"#</mathjax> of solution.</p>
<blockquote>
<p><mathjax>#10^3 color(red)(cancel(color(black)("mL solution"))) * "0.1249 moles glucose"/(25.5color(red)(cancel(color(black)("mL solution")))) = "4.898 moles glucose"#</mathjax></p>
</blockquote>
<p>You can thus say that the molarity of the solution is equal to</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("molarity = 4.90 mol L"^(-1))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | What is the molarity of an aqueous solution containing 22.5 grams of glucose in 25.5 mL of solution? | null |
3,303 | a9af2a8d-6ddd-11ea-b72e-ccda262736ce | https://socratic.org/questions/59aa5b2e11ef6b7f52b247b2 | 1/2 N2(g) + 3/2 H2(g) -> NH3(g) | start chemical_equation qc_end substance 7 7 qc_end substance 10 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reduction"}] | [{"type":"chemical equation","value":"1/2 N2(g) + 3/2 H2(g) -> NH3(g)"}] | [{"type":"substance name","value":"Nitrogen"},{"type":"substance name","value":"Ammonia"}] | <h1 class="questionTitle" itemprop="name">How do you represent the reduction of nitrogen to give ammonia?</h1> | null | 1/2 N2(g) + 3/2 H2(g) -> NH3(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Simple, ain't it? Several biological processes achieve this reduction. And yet this reaction is extraordinarily difficult to do, and it really has only been in the last 10-20 years or so that chemists have developed convincing model systems that demonstrate this reduction in the laboratory.....</p>
<p>Industrially this reaction is catalyzed on metal surfaces, that can reduce the activation energies associated with cleavage of the <mathjax>#N-=N#</mathjax> triple bond. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#1/2N_2(g) + 3/2H_2(g) rarr NH_3(g)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Simple, ain't it? Several biological processes achieve this reduction. And yet this reaction is extraordinarily difficult to do, and it really has only been in the last 10-20 years or so that chemists have developed convincing model systems that demonstrate this reduction in the laboratory.....</p>
<p>Industrially this reaction is catalyzed on metal surfaces, that can reduce the activation energies associated with cleavage of the <mathjax>#N-=N#</mathjax> triple bond. </p></div>
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<h1 class="questionTitle" itemprop="name">How do you represent the reduction of nitrogen to give ammonia?</h1>
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anor277
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<span class="dateCreated" datetime="2017-09-02T08:57:30" itemprop="dateCreated">
Sep 2, 2017
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<div class="markdown"><p><mathjax>#1/2N_2(g) + 3/2H_2(g) rarr NH_3(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Simple, ain't it? Several biological processes achieve this reduction. And yet this reaction is extraordinarily difficult to do, and it really has only been in the last 10-20 years or so that chemists have developed convincing model systems that demonstrate this reduction in the laboratory.....</p>
<p>Industrially this reaction is catalyzed on metal surfaces, that can reduce the activation energies associated with cleavage of the <mathjax>#N-=N#</mathjax> triple bond. </p></div>
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</article> | How do you represent the reduction of nitrogen to give ammonia? | null |
3,304 | aa89ac18-6ddd-11ea-83f4-ccda262736ce | https://socratic.org/questions/3-761x10-1-l-of-an-ideal-gas-is-in-a-balloon-at-1-000x10-0-atm-the-weather-chang | 7.52 × 10^(-1) atm | start physical_unit 6 7 pressure atm qc_end physical_unit 6 7 0 3 volume qc_end physical_unit 6 7 13 16 pressure qc_end physical_unit 6 7 28 31 volume qc_end c_other constant_temperature qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the ideal gas [IN] atm"}] | [{"type":"physical unit","value":"7.52 × 10^(-1) atm"}] | [{"type":"physical unit","value":"Volume1 [OF] the ideal gas [=] \\pu{3.761 × 10^1 L}"},{"type":"physical unit","value":"Pressure1 [OF] the ideal gas [=] \\pu{1.000 × 10^0 atm}"},{"type":"physical unit","value":"Volume2 [OF] the ideal gas [=] \\pu{5.00 × 10^1 L}"},{"type":"other","value":"ConstantTemperature"}] | <h1 class="questionTitle" itemprop="name">#3.761xx10^1color(white)(.)"L"# of an ideal gas is in a balloon at #1.000xx10^0 color(white)(.)"atm"#. The weather changes, and the volume of the balloon changes to #5.00xx10^1 "L"#. What is the new pressure, assuming no change in temperature? </h1> | null | 7.52 × 10^(-1) atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a></strong> , which states that the volume of a gas held at constant amount and temperature, varies inversely with the pressure. This means that if the pressure increases, the volume decreases, and vice-versa. The equation for this law is:</p>
<p><mathjax>#"P_1V_1=P_2V_2#</mathjax></p>
<p>where <mathjax>#P#</mathjax> is pressure and <mathjax>#V#</mathjax>is volume.</p>
<p><strong>Organize the information:</strong></p>
<p><strong>Known</strong><br/>
<mathjax>#P_1=1.000xx10^0"atm"#</mathjax><br/>
<mathjax>#V_1=3.761xx10^1 "L"#</mathjax><br/>
<mathjax>#V_2=5.00xx10^1"L"#</mathjax></p>
<p><strong>Unknown:</strong> <mathjax>#P_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#P_2#</mathjax>, substitute the known values into the equation and solve.</p>
<p><mathjax>#P_2=(P_1V_1)/V_2#</mathjax></p>
<p><mathjax>#P_2=(1.000xx10^0"atm"xx3.761xx10^1color(red)cancel(color(black)"L"))/(5.00xx10^1color(red)cancel(color(black)"L"))=7.52xx10^(-1)"atm"#</mathjax> rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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<div class="markdown"><p>The new pressure is <mathjax>#7.52xx10^(-1)color(white)(.)"atm"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a></strong> , which states that the volume of a gas held at constant amount and temperature, varies inversely with the pressure. This means that if the pressure increases, the volume decreases, and vice-versa. The equation for this law is:</p>
<p><mathjax>#"P_1V_1=P_2V_2#</mathjax></p>
<p>where <mathjax>#P#</mathjax> is pressure and <mathjax>#V#</mathjax>is volume.</p>
<p><strong>Organize the information:</strong></p>
<p><strong>Known</strong><br/>
<mathjax>#P_1=1.000xx10^0"atm"#</mathjax><br/>
<mathjax>#V_1=3.761xx10^1 "L"#</mathjax><br/>
<mathjax>#V_2=5.00xx10^1"L"#</mathjax></p>
<p><strong>Unknown:</strong> <mathjax>#P_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#P_2#</mathjax>, substitute the known values into the equation and solve.</p>
<p><mathjax>#P_2=(P_1V_1)/V_2#</mathjax></p>
<p><mathjax>#P_2=(1.000xx10^0"atm"xx3.761xx10^1color(red)cancel(color(black)"L"))/(5.00xx10^1color(red)cancel(color(black)"L"))=7.52xx10^(-1)"atm"#</mathjax> rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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<h1 class="questionTitle" itemprop="name">#3.761xx10^1color(white)(.)"L"# of an ideal gas is in a balloon at #1.000xx10^0 color(white)(.)"atm"#. The weather changes, and the volume of the balloon changes to #5.00xx10^1 "L"#. What is the new pressure, assuming no change in temperature? </h1>
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<div class="markdown"><p>The new pressure is <mathjax>#7.52xx10^(-1)color(white)(.)"atm"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a></strong> , which states that the volume of a gas held at constant amount and temperature, varies inversely with the pressure. This means that if the pressure increases, the volume decreases, and vice-versa. The equation for this law is:</p>
<p><mathjax>#"P_1V_1=P_2V_2#</mathjax></p>
<p>where <mathjax>#P#</mathjax> is pressure and <mathjax>#V#</mathjax>is volume.</p>
<p><strong>Organize the information:</strong></p>
<p><strong>Known</strong><br/>
<mathjax>#P_1=1.000xx10^0"atm"#</mathjax><br/>
<mathjax>#V_1=3.761xx10^1 "L"#</mathjax><br/>
<mathjax>#V_2=5.00xx10^1"L"#</mathjax></p>
<p><strong>Unknown:</strong> <mathjax>#P_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#P_2#</mathjax>, substitute the known values into the equation and solve.</p>
<p><mathjax>#P_2=(P_1V_1)/V_2#</mathjax></p>
<p><mathjax>#P_2=(1.000xx10^0"atm"xx3.761xx10^1color(red)cancel(color(black)"L"))/(5.00xx10^1color(red)cancel(color(black)"L"))=7.52xx10^(-1)"atm"#</mathjax> rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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</article> | #3.761xx10^1color(white)(.)"L"# of an ideal gas is in a balloon at #1.000xx10^0 color(white)(.)"atm"#. The weather changes, and the volume of the balloon changes to #5.00xx10^1 "L"#. What is the new pressure, assuming no change in temperature? | null |
3,305 | a9f3ccc0-6ddd-11ea-a352-ccda262736ce | https://socratic.org/questions/the-chemical-formula-of-ilmenite-is-fetio-3-what-mass-of-titanium-can-be-obtaine | 157.73 grams | start physical_unit 10 10 mass g qc_end physical_unit 4 4 15 16 mass qc_end chemical_equation 6 6 qc_end end | [{"type":"physical unit","value":"mass [OF] titanium [IN] grams"}] | [{"type":"physical unit","value":"157.73 grams"}] | [{"type":"physical unit","value":"Mass [OF] ilmenite [=] \\pu{500.0 g}"},{"type":"chemical equation","value":"FeTiO3"}] | <h1 class="questionTitle" itemprop="name">The chemical formula of ilmenite is #FeTiO_3#. What mass of titanium can be obtained from 500.0 g of ilmenite?</h1> | null | 157.73 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here will be to determine how many grams of titanium, <mathjax>#"Ti"#</mathjax>, are present in <strong>one mole</strong> of ilmenite by using the <strong>molar masses</strong> of ilmenite and of titanium. </p>
<p>Ilmenite's <strong>molar mass</strong> is equal to <mathjax>#"151.71 g mol"^(-1)#</mathjax>. This tells you that <strong>every mole</strong> of ilmenite will have a mass of <mathjax>#"151.71 g"#</mathjax>. </p>
<p>As you can see by inspecting the compound's chemical formula, <strong>one mole</strong> of ilmenite will also contain </p>
<blockquote>
<ul>
<li><em><strong>one mole</strong> of iron</em>, <mathjax>#1 xx "Fe"#</mathjax></li>
<li><em><strong>one mole</strong> of titanium</em>, <mathjax>#1 xx "Ti"#</mathjax></li>
<li><em><strong>three moles</strong> of oxygen</em>, <mathjax>#3 xx "O"#</mathjax></li>
</ul>
</blockquote>
<p>Since titanium has a <strong>molar mass</strong> of <mathjax>#"47.857 g mol"^(-1)#</mathjax>, you can say that <mathjax>#"151.71 g"#</mathjax> of ilmenite will contain <mathjax>#"47.857 g"#</mathjax> of titanium. </p>
<p>So, if you get <mathjax>#"47.857 g"#</mathjax> of titanium <strong>for every</strong> <mathjax>#"151.71 g"#</mathjax> of ilmenite, it follows that <mathjax>#"500.0 g"#</mathjax> of ilmenite will contain </p>
<blockquote>
<p><mathjax>#500.0 color(red)(cancel(color(black)("g ilmenite"))) * "47.857 g Ti"/(151.71 color(red)(cancel(color(black)("g ilmenite")))) = color(green)(|bar(ul(color(white)(a/a)"157.7 g Ti"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to four <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p>
<p>Notice that it's more <em>practical</em> to determine the <strong><a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> of the compound because it makes the ratio of titanium to ilmenite easier to work with. </p>
<blockquote>
<p><mathjax>#"% Ti" = (47.857 color(red)(cancel(color(black)("g"))))/(151.71color(red)(cancel(color(black)("g")))) xx 100 = 31.545%#</mathjax></p>
</blockquote>
<p>This means that you get <mathjax>#"31.545 g"#</mathjax> of titanium <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of ilmenite. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"157.7 g Ti"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here will be to determine how many grams of titanium, <mathjax>#"Ti"#</mathjax>, are present in <strong>one mole</strong> of ilmenite by using the <strong>molar masses</strong> of ilmenite and of titanium. </p>
<p>Ilmenite's <strong>molar mass</strong> is equal to <mathjax>#"151.71 g mol"^(-1)#</mathjax>. This tells you that <strong>every mole</strong> of ilmenite will have a mass of <mathjax>#"151.71 g"#</mathjax>. </p>
<p>As you can see by inspecting the compound's chemical formula, <strong>one mole</strong> of ilmenite will also contain </p>
<blockquote>
<ul>
<li><em><strong>one mole</strong> of iron</em>, <mathjax>#1 xx "Fe"#</mathjax></li>
<li><em><strong>one mole</strong> of titanium</em>, <mathjax>#1 xx "Ti"#</mathjax></li>
<li><em><strong>three moles</strong> of oxygen</em>, <mathjax>#3 xx "O"#</mathjax></li>
</ul>
</blockquote>
<p>Since titanium has a <strong>molar mass</strong> of <mathjax>#"47.857 g mol"^(-1)#</mathjax>, you can say that <mathjax>#"151.71 g"#</mathjax> of ilmenite will contain <mathjax>#"47.857 g"#</mathjax> of titanium. </p>
<p>So, if you get <mathjax>#"47.857 g"#</mathjax> of titanium <strong>for every</strong> <mathjax>#"151.71 g"#</mathjax> of ilmenite, it follows that <mathjax>#"500.0 g"#</mathjax> of ilmenite will contain </p>
<blockquote>
<p><mathjax>#500.0 color(red)(cancel(color(black)("g ilmenite"))) * "47.857 g Ti"/(151.71 color(red)(cancel(color(black)("g ilmenite")))) = color(green)(|bar(ul(color(white)(a/a)"157.7 g Ti"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to four <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p>
<p>Notice that it's more <em>practical</em> to determine the <strong><a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> of the compound because it makes the ratio of titanium to ilmenite easier to work with. </p>
<blockquote>
<p><mathjax>#"% Ti" = (47.857 color(red)(cancel(color(black)("g"))))/(151.71color(red)(cancel(color(black)("g")))) xx 100 = 31.545%#</mathjax></p>
</blockquote>
<p>This means that you get <mathjax>#"31.545 g"#</mathjax> of titanium <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of ilmenite. </p></div>
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<h1 class="questionTitle" itemprop="name">The chemical formula of ilmenite is #FeTiO_3#. What mass of titanium can be obtained from 500.0 g of ilmenite?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"157.7 g Ti"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here will be to determine how many grams of titanium, <mathjax>#"Ti"#</mathjax>, are present in <strong>one mole</strong> of ilmenite by using the <strong>molar masses</strong> of ilmenite and of titanium. </p>
<p>Ilmenite's <strong>molar mass</strong> is equal to <mathjax>#"151.71 g mol"^(-1)#</mathjax>. This tells you that <strong>every mole</strong> of ilmenite will have a mass of <mathjax>#"151.71 g"#</mathjax>. </p>
<p>As you can see by inspecting the compound's chemical formula, <strong>one mole</strong> of ilmenite will also contain </p>
<blockquote>
<ul>
<li><em><strong>one mole</strong> of iron</em>, <mathjax>#1 xx "Fe"#</mathjax></li>
<li><em><strong>one mole</strong> of titanium</em>, <mathjax>#1 xx "Ti"#</mathjax></li>
<li><em><strong>three moles</strong> of oxygen</em>, <mathjax>#3 xx "O"#</mathjax></li>
</ul>
</blockquote>
<p>Since titanium has a <strong>molar mass</strong> of <mathjax>#"47.857 g mol"^(-1)#</mathjax>, you can say that <mathjax>#"151.71 g"#</mathjax> of ilmenite will contain <mathjax>#"47.857 g"#</mathjax> of titanium. </p>
<p>So, if you get <mathjax>#"47.857 g"#</mathjax> of titanium <strong>for every</strong> <mathjax>#"151.71 g"#</mathjax> of ilmenite, it follows that <mathjax>#"500.0 g"#</mathjax> of ilmenite will contain </p>
<blockquote>
<p><mathjax>#500.0 color(red)(cancel(color(black)("g ilmenite"))) * "47.857 g Ti"/(151.71 color(red)(cancel(color(black)("g ilmenite")))) = color(green)(|bar(ul(color(white)(a/a)"157.7 g Ti"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to four <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p>
<p>Notice that it's more <em>practical</em> to determine the <strong><a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> of the compound because it makes the ratio of titanium to ilmenite easier to work with. </p>
<blockquote>
<p><mathjax>#"% Ti" = (47.857 color(red)(cancel(color(black)("g"))))/(151.71color(red)(cancel(color(black)("g")))) xx 100 = 31.545%#</mathjax></p>
</blockquote>
<p>This means that you get <mathjax>#"31.545 g"#</mathjax> of titanium <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of ilmenite. </p></div>
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</article> | The chemical formula of ilmenite is #FeTiO_3#. What mass of titanium can be obtained from 500.0 g of ilmenite? | null |
3,306 | aa16508c-6ddd-11ea-8111-ccda262736ce | https://socratic.org/questions/what-is-the-percent-yield-of-o-2-if-10-2-g-of-o-2-is-produced-from-the-decomposi | 67.5% | start physical_unit 6 6 percent_yield none qc_end physical_unit 6 6 8 9 mass qc_end physical_unit 21 21 18 19 mass qc_end end | [{"type":"physical unit","value":"Percent yield [OF] O2"}] | [{"type":"physical unit","value":"67.5%"}] | [{"type":"physical unit","value":"Mass [OF] O2 [=] \\pu{10.2 g}"},{"type":"physical unit","value":"Mass [OF] H2O [=] \\pu{17.0 g}"}] | <h1 class="questionTitle" itemprop="name">What is the percent yield of #O_2# if 10.2 g of #O_2# is produced from the decomposition of 17.0 g of #H_2O#?</h1> | null | 67.5% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="http://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a> can be calculated by:</p>
<p><mathjax>#%Yield=("Actual Yield")/("Theoretical Yield")xx100%#</mathjax></p>
<p>The actual yield of oxygen is given and it is <mathjax>#10.2g#</mathjax>.</p>
<p>We will need to find the theoretical yield.</p>
<p>The decomposition reaction of water can be written as:</p>
<p><mathjax>#2H_2O(l)->2H_2(g)+O_2(g)#</mathjax></p>
<p>To find the theoretical yield of oxygen we will use dimensional analysis:</p>
<p><mathjax>#?g O_2=underbrace(17.0gH_2Oxx(1molH_2O)/(18.0gH_2O))_(color(blue)("g to mol"))xxunderbrace((1molO_2)/(2molH_2O))_color(green)("molar ratio")xxunderbrace((32.0gO_2)/(1molO_2))_(color(red)("mol to g"))=15.1gO_2#</mathjax></p>
<p>Thus, the theoretical yield is equal to <mathjax>#15.1g#</mathjax>.</p>
<p>The percent yield is then: <mathjax>#%Yield=(10.2)/(15.1)xx100%=67.5%#</mathjax></p></div>
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<div>
<div class="markdown"><p><mathjax>#%Yield=67.5%#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="http://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a> can be calculated by:</p>
<p><mathjax>#%Yield=("Actual Yield")/("Theoretical Yield")xx100%#</mathjax></p>
<p>The actual yield of oxygen is given and it is <mathjax>#10.2g#</mathjax>.</p>
<p>We will need to find the theoretical yield.</p>
<p>The decomposition reaction of water can be written as:</p>
<p><mathjax>#2H_2O(l)->2H_2(g)+O_2(g)#</mathjax></p>
<p>To find the theoretical yield of oxygen we will use dimensional analysis:</p>
<p><mathjax>#?g O_2=underbrace(17.0gH_2Oxx(1molH_2O)/(18.0gH_2O))_(color(blue)("g to mol"))xxunderbrace((1molO_2)/(2molH_2O))_color(green)("molar ratio")xxunderbrace((32.0gO_2)/(1molO_2))_(color(red)("mol to g"))=15.1gO_2#</mathjax></p>
<p>Thus, the theoretical yield is equal to <mathjax>#15.1g#</mathjax>.</p>
<p>The percent yield is then: <mathjax>#%Yield=(10.2)/(15.1)xx100%=67.5%#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the percent yield of #O_2# if 10.2 g of #O_2# is produced from the decomposition of 17.0 g of #H_2O#?</h1>
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<div class="markdown"><p><mathjax>#%Yield=67.5%#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="http://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a> can be calculated by:</p>
<p><mathjax>#%Yield=("Actual Yield")/("Theoretical Yield")xx100%#</mathjax></p>
<p>The actual yield of oxygen is given and it is <mathjax>#10.2g#</mathjax>.</p>
<p>We will need to find the theoretical yield.</p>
<p>The decomposition reaction of water can be written as:</p>
<p><mathjax>#2H_2O(l)->2H_2(g)+O_2(g)#</mathjax></p>
<p>To find the theoretical yield of oxygen we will use dimensional analysis:</p>
<p><mathjax>#?g O_2=underbrace(17.0gH_2Oxx(1molH_2O)/(18.0gH_2O))_(color(blue)("g to mol"))xxunderbrace((1molO_2)/(2molH_2O))_color(green)("molar ratio")xxunderbrace((32.0gO_2)/(1molO_2))_(color(red)("mol to g"))=15.1gO_2#</mathjax></p>
<p>Thus, the theoretical yield is equal to <mathjax>#15.1g#</mathjax>.</p>
<p>The percent yield is then: <mathjax>#%Yield=(10.2)/(15.1)xx100%=67.5%#</mathjax></p></div>
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</article> | What is the percent yield of #O_2# if 10.2 g of #O_2# is produced from the decomposition of 17.0 g of #H_2O#? | null |
3,307 | acdfb554-6ddd-11ea-ae9b-ccda262736ce | https://socratic.org/questions/the-pressure-exerted-on-800ml-of-a-gas-is-decreased-from-300-kpa-to-198-kpa-what | 1.21 L | start physical_unit 23 24 volume l qc_end physical_unit 23 24 4 5 volume qc_end physical_unit 23 24 12 13 pressure qc_end physical_unit 23 24 15 16 pressure qc_end c_other constant_temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"1.21 L"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{800 mL}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{300 kPa}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{198 kPa}"},{"type":"other","value":"ConstantTemperature"}] | <h1 class="questionTitle" itemprop="name">The pressure exerted on 800mL of a gas is decreased from 300 kPa to 198 kPa. What is the new volume of the gas if the temp remains constant? </h1> | null | 1.21 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At constant temperature, <mathjax>#PV=k#</mathjax>, for a fixed amount (mass) of gas. This is old <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a>:</p>
<p>Since <mathjax>#PV=k#</mathjax>, <mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p><mathjax>#V_2=(P_1V_1)/P_2=(300*kPaxx800*mL)/(198*kPa)=??*mL#</mathjax>. A good question in that you don't have to use a calculator.........</p></div>
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<div class="markdown"><p><mathjax>#V_2=1200*mL........#</mathjax></p></div>
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</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At constant temperature, <mathjax>#PV=k#</mathjax>, for a fixed amount (mass) of gas. This is old <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a>:</p>
<p>Since <mathjax>#PV=k#</mathjax>, <mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p><mathjax>#V_2=(P_1V_1)/P_2=(300*kPaxx800*mL)/(198*kPa)=??*mL#</mathjax>. A good question in that you don't have to use a calculator.........</p></div>
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<h1 class="questionTitle" itemprop="name">The pressure exerted on 800mL of a gas is decreased from 300 kPa to 198 kPa. What is the new volume of the gas if the temp remains constant? </h1>
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May 12, 2017
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<div class="markdown"><p><mathjax>#V_2=1200*mL........#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>At constant temperature, <mathjax>#PV=k#</mathjax>, for a fixed amount (mass) of gas. This is old <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a>:</p>
<p>Since <mathjax>#PV=k#</mathjax>, <mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p><mathjax>#V_2=(P_1V_1)/P_2=(300*kPaxx800*mL)/(198*kPa)=??*mL#</mathjax>. A good question in that you don't have to use a calculator.........</p></div>
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</article> | The pressure exerted on 800mL of a gas is decreased from 300 kPa to 198 kPa. What is the new volume of the gas if the temp remains constant? | null |
3,308 | a8e49e4c-6ddd-11ea-87e9-ccda262736ce | https://socratic.org/questions/for-the-reaction-of-ammonia-with-molecular-oxygen-forming-nitrogen-monoxide-and- | 12.0 g | start physical_unit 21 21 theoretical_yield g qc_end physical_unit 25 25 23 24 mass qc_end physical_unit 30 30 28 29 mass qc_end substance 12 12 qc_end end | [{"type":"physical unit","value":"Theoretical yield [OF] NO [IN] g"}] | [{"type":"physical unit","value":"12.0 g"}] | [{"type":"physical unit","value":"Mass [OF] NH3 [=] \\pu{17.0 g}"},{"type":"physical unit","value":"Mass [OF] O2 [=] \\pu{16.0 g}"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">For the reaction of ammonia with molecular oxygen, forming nitrogen monoxide and water. How do you calculate the theoretical yield of #NO# when 17.0g #NH_3# reacts with 16.0g #O_2#?</h1> | null | 12.0 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given stoichiometric oxygen, ammonia gives an equimolar quantity of <mathjax>#NO#</mathjax>. </p>
<p><mathjax>#"Moles of ammonia"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(17.0*g)/(17.03*g)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#1*mol#</mathjax></p>
<p><mathjax>#"Moles of dioxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(16.0*g)/(32.03*g)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#0.50*mol#</mathjax></p>
<p>Clearly, dioxygen is in deficiency, and only <mathjax>#4/5xx0.5*mol=0.40*mol#</mathjax> <mathjax>#NH_3#</mathjax> will react. </p>
<p>And thus only <mathjax>#0.40*mol#</mathjax> <mathjax>#NO#</mathjax> will be produced, i.e. <mathjax>#0.40*molxx30.01*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#12.0*g#</mathjax> <mathjax>#NO#</mathjax>.</p></div>
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<div class="markdown"><p>Well first you write a stoichiometrically balanced equation:</p>
<p><mathjax>#2NH_3(aq) + 5/2O_2(g) rarr 2NO(g) +3H_2O(l)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Given stoichiometric oxygen, ammonia gives an equimolar quantity of <mathjax>#NO#</mathjax>. </p>
<p><mathjax>#"Moles of ammonia"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(17.0*g)/(17.03*g)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#1*mol#</mathjax></p>
<p><mathjax>#"Moles of dioxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(16.0*g)/(32.03*g)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#0.50*mol#</mathjax></p>
<p>Clearly, dioxygen is in deficiency, and only <mathjax>#4/5xx0.5*mol=0.40*mol#</mathjax> <mathjax>#NH_3#</mathjax> will react. </p>
<p>And thus only <mathjax>#0.40*mol#</mathjax> <mathjax>#NO#</mathjax> will be produced, i.e. <mathjax>#0.40*molxx30.01*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#12.0*g#</mathjax> <mathjax>#NO#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">For the reaction of ammonia with molecular oxygen, forming nitrogen monoxide and water. How do you calculate the theoretical yield of #NO# when 17.0g #NH_3# reacts with 16.0g #O_2#?</h1>
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<div class="markdown"><p>Well first you write a stoichiometrically balanced equation:</p>
<p><mathjax>#2NH_3(aq) + 5/2O_2(g) rarr 2NO(g) +3H_2O(l)#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Given stoichiometric oxygen, ammonia gives an equimolar quantity of <mathjax>#NO#</mathjax>. </p>
<p><mathjax>#"Moles of ammonia"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(17.0*g)/(17.03*g)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#1*mol#</mathjax></p>
<p><mathjax>#"Moles of dioxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(16.0*g)/(32.03*g)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#0.50*mol#</mathjax></p>
<p>Clearly, dioxygen is in deficiency, and only <mathjax>#4/5xx0.5*mol=0.40*mol#</mathjax> <mathjax>#NH_3#</mathjax> will react. </p>
<p>And thus only <mathjax>#0.40*mol#</mathjax> <mathjax>#NO#</mathjax> will be produced, i.e. <mathjax>#0.40*molxx30.01*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#12.0*g#</mathjax> <mathjax>#NO#</mathjax>.</p></div>
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</article> | For the reaction of ammonia with molecular oxygen, forming nitrogen monoxide and water. How do you calculate the theoretical yield of #NO# when 17.0g #NH_3# reacts with 16.0g #O_2#? | null |
3,309 | a9aedb90-6ddd-11ea-812e-ccda262736ce | https://socratic.org/questions/the-concentration-of-co-2-in-a-can-of-soda-is-approximately-0-045-m-if-all-of-th | 0.36 liters | start physical_unit 31 32 volume l qc_end physical_unit 8 8 20 21 mass qc_end c_other STP qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Volume [OF] CO2 gas [IN] liters"}] | [{"type":"physical unit","value":"0.36 liters"}] | [{"type":"physical unit","value":"Concentration [OF] CO2 in soda [=] \\pu{0.045 M}"},{"type":"physical unit","value":"Mass [OF] soda [=] \\pu{12 ounce}"},{"type":"other","value":"STP"},{"type":"other","value":"all of the CO2 in soda evolves from solution."}] | <h1 class="questionTitle" itemprop="name">The concentration of #CO_2# in a can of soda is approximately 0.045 M. If all of the #CO_2# in a 12 ounce soda evolves from solution, how many liters would the #CO_2# gas take up at STP?</h1> | null | 0.36 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to figure out how many <strong>moles</strong> of carbon dioxide, <mathjax>#"CO"_2#</mathjax>, are initially dissolved in the soda by using the <em>volume</em> and <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> of the solution. </p>
<p>As you know, a <strong>molarity</strong> of <mathjax>#"0.045 M"#</mathjax> means that <strong>every liter</strong> of solution, which in your case is soda, will contain <mathjax>#0.045#</mathjax> <strong>moles</strong> of carbon dioxide. </p>
<p>This means that your </p>
<blockquote>
<p><mathjax>#12 color(red)(cancel(color(black)("oz"))) * (29.57 color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("oz")))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.35484 L"#</mathjax></p>
</blockquote>
<p>sample of soda will contain </p>
<blockquote>
<p><mathjax>#0.35484 color(red)(cancel(color(black)("L soda"))) * "0.045 moles CO"_2/(1color(red)(cancel(color(black)("L soda")))) = "0.01597 moles CO"_2#</mathjax></p>
</blockquote>
<p>Now, you are told that <strong>all of the moles</strong> of carbon dioxide evolve from the solution. </p>
<p>As you know, <strong>STP conditions</strong>, which are defined as a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>, are characterized by the fact that <strong>one mole</strong> of any ideal gas occupies exactly <mathjax>#"22.7 L"#</mathjax> <mathjax>#->#</mathjax> this is known as the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a></strong>. </p>
<p>This means that your sample of carbon dioxide will occupy</p>
<blockquote>
<p><mathjax>#0.01597 color(red)(cancel(color(black)("moles CO"_2))) * "22.7 L"/(1color(red)(cancel(color(black)("mole CO"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.36 L")color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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<div class="markdown"><p><mathjax>#"0.36 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to figure out how many <strong>moles</strong> of carbon dioxide, <mathjax>#"CO"_2#</mathjax>, are initially dissolved in the soda by using the <em>volume</em> and <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> of the solution. </p>
<p>As you know, a <strong>molarity</strong> of <mathjax>#"0.045 M"#</mathjax> means that <strong>every liter</strong> of solution, which in your case is soda, will contain <mathjax>#0.045#</mathjax> <strong>moles</strong> of carbon dioxide. </p>
<p>This means that your </p>
<blockquote>
<p><mathjax>#12 color(red)(cancel(color(black)("oz"))) * (29.57 color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("oz")))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.35484 L"#</mathjax></p>
</blockquote>
<p>sample of soda will contain </p>
<blockquote>
<p><mathjax>#0.35484 color(red)(cancel(color(black)("L soda"))) * "0.045 moles CO"_2/(1color(red)(cancel(color(black)("L soda")))) = "0.01597 moles CO"_2#</mathjax></p>
</blockquote>
<p>Now, you are told that <strong>all of the moles</strong> of carbon dioxide evolve from the solution. </p>
<p>As you know, <strong>STP conditions</strong>, which are defined as a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>, are characterized by the fact that <strong>one mole</strong> of any ideal gas occupies exactly <mathjax>#"22.7 L"#</mathjax> <mathjax>#->#</mathjax> this is known as the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a></strong>. </p>
<p>This means that your sample of carbon dioxide will occupy</p>
<blockquote>
<p><mathjax>#0.01597 color(red)(cancel(color(black)("moles CO"_2))) * "22.7 L"/(1color(red)(cancel(color(black)("mole CO"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.36 L")color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">The concentration of #CO_2# in a can of soda is approximately 0.045 M. If all of the #CO_2# in a 12 ounce soda evolves from solution, how many liters would the #CO_2# gas take up at STP?</h1>
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<div class="markdown"><p><mathjax>#"0.36 L"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The idea here is that you need to figure out how many <strong>moles</strong> of carbon dioxide, <mathjax>#"CO"_2#</mathjax>, are initially dissolved in the soda by using the <em>volume</em> and <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> of the solution. </p>
<p>As you know, a <strong>molarity</strong> of <mathjax>#"0.045 M"#</mathjax> means that <strong>every liter</strong> of solution, which in your case is soda, will contain <mathjax>#0.045#</mathjax> <strong>moles</strong> of carbon dioxide. </p>
<p>This means that your </p>
<blockquote>
<p><mathjax>#12 color(red)(cancel(color(black)("oz"))) * (29.57 color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("oz")))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.35484 L"#</mathjax></p>
</blockquote>
<p>sample of soda will contain </p>
<blockquote>
<p><mathjax>#0.35484 color(red)(cancel(color(black)("L soda"))) * "0.045 moles CO"_2/(1color(red)(cancel(color(black)("L soda")))) = "0.01597 moles CO"_2#</mathjax></p>
</blockquote>
<p>Now, you are told that <strong>all of the moles</strong> of carbon dioxide evolve from the solution. </p>
<p>As you know, <strong>STP conditions</strong>, which are defined as a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>, are characterized by the fact that <strong>one mole</strong> of any ideal gas occupies exactly <mathjax>#"22.7 L"#</mathjax> <mathjax>#->#</mathjax> this is known as the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a></strong>. </p>
<p>This means that your sample of carbon dioxide will occupy</p>
<blockquote>
<p><mathjax>#0.01597 color(red)(cancel(color(black)("moles CO"_2))) * "22.7 L"/(1color(red)(cancel(color(black)("mole CO"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.36 L")color(white)(a/a)|)))#</mathjax></p>
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</article> | The concentration of #CO_2# in a can of soda is approximately 0.045 M. If all of the #CO_2# in a 12 ounce soda evolves from solution, how many liters would the #CO_2# gas take up at STP? | null |
3,310 | ab050154-6ddd-11ea-8dbb-ccda262736ce | https://socratic.org/questions/what-is-the-total-mass-of-oxygen-in-1-00-mole-of-al-2-cro-4-3 | 192.00 g | start physical_unit 6 6 mass g qc_end physical_unit 11 11 8 9 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] oxygen [IN] g"}] | [{"type":"physical unit","value":"192.00 g"}] | [{"type":"physical unit","value":"Mole [OF] Al2(CrO4)3 [=] \\pu{1.00 mole}"}] | <h1 class="questionTitle" itemprop="name">What is the total mass of oxygen in 1.00 mole of #Al_2(CrO_4)_3#?</h1> | null | 192.00 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Here, I speak of oxygen ATOMS (and not oxygen molecules). Clearly, there are 4 moles of oxygen atoms in one mole of chromate ion. We have a molar quantity of aluminum chromate, and thus there is 12-fold molar quantity of oxygen ATOMS: i.e. a mass of <mathjax>#12#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#16.00*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#192.0#</mathjax> <mathjax>#g#</mathjax>.</p></div>
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<div class="markdown"><p>In <mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#Al_2(CrO_4)_3#</mathjax> there are <mathjax>#12#</mathjax> <mathjax>#"mol oxygen atoms"#</mathjax>, or <mathjax>#192#</mathjax> <mathjax>#g#</mathjax>.</p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Here, I speak of oxygen ATOMS (and not oxygen molecules). Clearly, there are 4 moles of oxygen atoms in one mole of chromate ion. We have a molar quantity of aluminum chromate, and thus there is 12-fold molar quantity of oxygen ATOMS: i.e. a mass of <mathjax>#12#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#16.00*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#192.0#</mathjax> <mathjax>#g#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the total mass of oxygen in 1.00 mole of #Al_2(CrO_4)_3#?</h1>
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<div class="markdown"><p>In <mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#Al_2(CrO_4)_3#</mathjax> there are <mathjax>#12#</mathjax> <mathjax>#"mol oxygen atoms"#</mathjax>, or <mathjax>#192#</mathjax> <mathjax>#g#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Here, I speak of oxygen ATOMS (and not oxygen molecules). Clearly, there are 4 moles of oxygen atoms in one mole of chromate ion. We have a molar quantity of aluminum chromate, and thus there is 12-fold molar quantity of oxygen ATOMS: i.e. a mass of <mathjax>#12#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#16.00*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#192.0#</mathjax> <mathjax>#g#</mathjax>.</p></div>
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</article> | What is the total mass of oxygen in 1.00 mole of #Al_2(CrO_4)_3#? | null |
3,311 | aa4dfc30-6ddd-11ea-8df9-ccda262736ce | https://socratic.org/questions/how-many-o-2-molecules-are-needed-to-react-with-7-87-g-of-s | 1.48 × 10^23 | start physical_unit 2 3 number none qc_end physical_unit 12 12 9 10 mass qc_end end | [{"type":"physical unit","value":"Number [OF] O2 molecules"}] | [{"type":"physical unit","value":"1.48 × 10^23"}] | [{"type":"physical unit","value":"Mass [OF] S [=] \\pu{7.87 g}"}] | <h1 class="questionTitle" itemprop="name">How many #O_2# molecules are needed to react with 7.87 g of S?</h1> | null | 1.48 × 10^23 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#S + O_2(g) rarr SO_2(g)#</mathjax></p>
<p>Moles of <mathjax>#S#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(7.87*g)/(32.06*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.245*mol#</mathjax></p>
<p>And thus, by reaction <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, we need <mathjax>#0.245*mol#</mathjax> dioxygen gas.</p>
<p>So we need, <mathjax>#0.245*molxx6.022xx10^23*mol^-1" oxygen molecules"#</mathjax>.</p>
<p>How many oxygen atoms does this represent?</p></div>
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<div class="markdown"><p>Wee assume formation of the <mathjax>#S(IV)#</mathjax> oxide, <mathjax>#SO_2(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#S + O_2(g) rarr SO_2(g)#</mathjax></p>
<p>Moles of <mathjax>#S#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(7.87*g)/(32.06*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.245*mol#</mathjax></p>
<p>And thus, by reaction <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, we need <mathjax>#0.245*mol#</mathjax> dioxygen gas.</p>
<p>So we need, <mathjax>#0.245*molxx6.022xx10^23*mol^-1" oxygen molecules"#</mathjax>.</p>
<p>How many oxygen atoms does this represent?</p></div>
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<h1 class="questionTitle" itemprop="name">How many #O_2# molecules are needed to react with 7.87 g of S?</h1>
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<div class="markdown"><p>Wee assume formation of the <mathjax>#S(IV)#</mathjax> oxide, <mathjax>#SO_2(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#S + O_2(g) rarr SO_2(g)#</mathjax></p>
<p>Moles of <mathjax>#S#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(7.87*g)/(32.06*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.245*mol#</mathjax></p>
<p>And thus, by reaction <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, we need <mathjax>#0.245*mol#</mathjax> dioxygen gas.</p>
<p>So we need, <mathjax>#0.245*molxx6.022xx10^23*mol^-1" oxygen molecules"#</mathjax>.</p>
<p>How many oxygen atoms does this represent?</p></div>
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</article> | How many #O_2# molecules are needed to react with 7.87 g of S? | null |
3,312 | a8386a2e-6ddd-11ea-9e8c-ccda262736ce | https://socratic.org/questions/what-mass-of-oxygen-is-needed-for-the-complete-combustion-of-8-90-10-3-g-of-meth | 0.04 g | start physical_unit 3 3 mass g qc_end c_other OTHER qc_end physical_unit 16 16 11 14 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] oxygen [IN] g"}] | [{"type":"physical unit","value":"0.04 g"}] | [{"type":"other","value":"Complete combustion."},{"type":"physical unit","value":"Mass [OF] methane [=] \\pu{8.90 × 10^(−3) g}"}] | <h1 class="questionTitle" itemprop="name">What mass of oxygen is needed for the complete combustion of #8.90 * 10^-3# #g# of methane?</h1> | null | 0.04 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of methane"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(8.90xx10^-3*g)/(16.04*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#"moles"#</mathjax></p>
<p>Given the stoichiometric equation, quite clearly, we need twice this molar quantity of dioxygen gas for complete combustion:</p>
<p><mathjax>#2xx(8.90xx10^-3*g)/(16.04*g*mol^-1)xx32.00*g*mol^-1" dioxygen gas"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#??g#</mathjax> <mathjax>#"dioxygen gas"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(l)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of methane"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(8.90xx10^-3*g)/(16.04*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#"moles"#</mathjax></p>
<p>Given the stoichiometric equation, quite clearly, we need twice this molar quantity of dioxygen gas for complete combustion:</p>
<p><mathjax>#2xx(8.90xx10^-3*g)/(16.04*g*mol^-1)xx32.00*g*mol^-1" dioxygen gas"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#??g#</mathjax> <mathjax>#"dioxygen gas"#</mathjax>.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What mass of oxygen is needed for the complete combustion of #8.90 * 10^-3# #g# of methane?</h1>
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anor277
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<div class="markdown"><p><mathjax>#CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(l)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of methane"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(8.90xx10^-3*g)/(16.04*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#"moles"#</mathjax></p>
<p>Given the stoichiometric equation, quite clearly, we need twice this molar quantity of dioxygen gas for complete combustion:</p>
<p><mathjax>#2xx(8.90xx10^-3*g)/(16.04*g*mol^-1)xx32.00*g*mol^-1" dioxygen gas"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#??g#</mathjax> <mathjax>#"dioxygen gas"#</mathjax>.</p></div>
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</article> | What mass of oxygen is needed for the complete combustion of #8.90 * 10^-3# #g# of methane? | null |
3,313 | abe604be-6ddd-11ea-9166-ccda262736ce | https://socratic.org/questions/when-the-temperature-of-a-rigid-hollow-sphere-containing-685-l-of-helium-gas-is- | 250.61 moles | start physical_unit 12 13 mole mol qc_end physical_unit 12 13 9 10 volume qc_end physical_unit 12 13 17 18 temperature qc_end physical_unit 12 13 25 28 pressure qc_end end | [{"type":"physical unit","value":"Mole [OF] helium gas [IN] moles"}] | [{"type":"physical unit","value":"250.61 moles"}] | [{"type":"physical unit","value":"Volume [OF] helium gas [=] \\pu{685 L}"},{"type":"physical unit","value":"Temperature [OF] helium gas [=] \\pu{621 K}"},{"type":"physical unit","value":"Pressure [OF] helium gas [=] \\pu{1.89 × 10^3 kPa}"}] | <h1 class="questionTitle" itemprop="name">When the temperature of a rigid hollow sphere containing 685 L of helium gas is held at 621 K, the pressure of the gas is #1.89*10^3# kPa. How many moles of helium does the sphere contain?</h1> | null | 250.61 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find the <em>number of moles</em> of gas present in that sample of helium under those conditions for pressure and temperature, you must use the <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong> equation</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the <em>number of moles of gas</em><br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given as <mathjax>#0.0821("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the <strong>absolute temperature</strong> of the gas</p>
<p>Now, before plugging in your values into the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation, make sure that the <strong>units</strong> given to you for pressure, temperature, and volume <strong>match</strong> the units used in the expression of the universal gas constant. </p>
<p>As you can see, <mathjax>#R#</mathjax> uses <em>atm</em> as the unit for pressure. The problem gives you the pressure of the gas expressed in <em>kPa</em>. This means that you're going to have to convert the pressure from <em>kPa</em> to <em>atm</em> by using the conversion factor</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 atm" = 1.01325 * 10^3"kPa")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The units for volume and temperature match those used by <mathjax>#R#</mathjax>, so rearrange the ideal gas law equation to solve for <mathjax>#n#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies n = (PV)/(RT)#</mathjax></p>
</blockquote>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#n = ((1.89 * 10^3color(red)(cancel(color(black)("atm"))))/(1.01325 * 10^2color(red)(cancel(color(black)("atm")))) * 685color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 621color(red)(cancel(color(black)("K")))) = "250.61 moles"#</mathjax></p>
</blockquote>
<p>Rounded to three <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the answer will be</p>
<blockquote>
<p><mathjax>#n = color(green)(|bar(ul(color(white)(a/a)"251 moles" color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"251 moles"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find the <em>number of moles</em> of gas present in that sample of helium under those conditions for pressure and temperature, you must use the <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong> equation</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the <em>number of moles of gas</em><br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given as <mathjax>#0.0821("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the <strong>absolute temperature</strong> of the gas</p>
<p>Now, before plugging in your values into the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation, make sure that the <strong>units</strong> given to you for pressure, temperature, and volume <strong>match</strong> the units used in the expression of the universal gas constant. </p>
<p>As you can see, <mathjax>#R#</mathjax> uses <em>atm</em> as the unit for pressure. The problem gives you the pressure of the gas expressed in <em>kPa</em>. This means that you're going to have to convert the pressure from <em>kPa</em> to <em>atm</em> by using the conversion factor</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 atm" = 1.01325 * 10^3"kPa")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The units for volume and temperature match those used by <mathjax>#R#</mathjax>, so rearrange the ideal gas law equation to solve for <mathjax>#n#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies n = (PV)/(RT)#</mathjax></p>
</blockquote>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#n = ((1.89 * 10^3color(red)(cancel(color(black)("atm"))))/(1.01325 * 10^2color(red)(cancel(color(black)("atm")))) * 685color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 621color(red)(cancel(color(black)("K")))) = "250.61 moles"#</mathjax></p>
</blockquote>
<p>Rounded to three <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the answer will be</p>
<blockquote>
<p><mathjax>#n = color(green)(|bar(ul(color(white)(a/a)"251 moles" color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">When the temperature of a rigid hollow sphere containing 685 L of helium gas is held at 621 K, the pressure of the gas is #1.89*10^3# kPa. How many moles of helium does the sphere contain?</h1>
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Stefan V.
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Mar 8, 2016
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<div class="markdown"><p><mathjax>#"251 moles"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find the <em>number of moles</em> of gas present in that sample of helium under those conditions for pressure and temperature, you must use the <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong> equation</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the <em>number of moles of gas</em><br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given as <mathjax>#0.0821("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the <strong>absolute temperature</strong> of the gas</p>
<p>Now, before plugging in your values into the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation, make sure that the <strong>units</strong> given to you for pressure, temperature, and volume <strong>match</strong> the units used in the expression of the universal gas constant. </p>
<p>As you can see, <mathjax>#R#</mathjax> uses <em>atm</em> as the unit for pressure. The problem gives you the pressure of the gas expressed in <em>kPa</em>. This means that you're going to have to convert the pressure from <em>kPa</em> to <em>atm</em> by using the conversion factor</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 atm" = 1.01325 * 10^3"kPa")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The units for volume and temperature match those used by <mathjax>#R#</mathjax>, so rearrange the ideal gas law equation to solve for <mathjax>#n#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies n = (PV)/(RT)#</mathjax></p>
</blockquote>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#n = ((1.89 * 10^3color(red)(cancel(color(black)("atm"))))/(1.01325 * 10^2color(red)(cancel(color(black)("atm")))) * 685color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 621color(red)(cancel(color(black)("K")))) = "250.61 moles"#</mathjax></p>
</blockquote>
<p>Rounded to three <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the answer will be</p>
<blockquote>
<p><mathjax>#n = color(green)(|bar(ul(color(white)(a/a)"251 moles" color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</article> | When the temperature of a rigid hollow sphere containing 685 L of helium gas is held at 621 K, the pressure of the gas is #1.89*10^3# kPa. How many moles of helium does the sphere contain? | null |
3,314 | a8c89194-6ddd-11ea-9b12-ccda262736ce | https://socratic.org/questions/a-nitrogen-and-oxygen-containing-molecule-is-decomposed-into-its-elements-it-is- | N2O5 | start chemical_formula qc_end physical_unit 1 1 16 17 mass qc_end physical_unit 3 3 21 22 mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"N2O5"}] | [{"type":"physical unit","value":"Mass [OF] nitrogen [=] \\pu{14 g}"},{"type":"physical unit","value":"Mass [OF] oxygen [=] \\pu{40 g}"}] | <h1 class="questionTitle" itemprop="name">A nitrogen and oxygen containing molecule is decomposed into its elements. It is found to contain 14 g of nitrogen and 40 g of oxygen. What is the empirical formula of the compound?</h1> | null | N2O5 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find a compound's <a href="http://socratic.org/chemistry/the-mole-concept/empirical-and-molecular-formulas">empirical formula</a> you must find the <strong>smallest whole number ratio</strong> that exists between the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> that are a part of this compound. </p>
<p>You know that your unknown compound contains <mathjax>#"14 g"#</mathjax> of nitrogen and <mathjax>#"40 g"#</mathjax> of oxygen. </p>
<p>Your first goal here will be to use the <strong>molar masses</strong> of these two elements to figure out how many <em>moles</em> of each you get in this sample. </p>
<blockquote>
<p><mathjax>#"For N: " 14 color(red)(cancel(color(black)("g"))) * "1 mole N"/(14.007color(red)(cancel(color(black)("g")))) = "0.9995 moles N"#</mathjax></p>
<p><mathjax>#"For O: " 40 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994 color(red)(cancel(color(black)("g")))) = "2.500 moles O"#</mathjax></p>
</blockquote>
<p>In order to find the <em>mole ratio</em> that exists between these two elements in the compound, divide both values by the <em>smallest</em> one </p>
<blockquote>
<p><mathjax>#"For N: " (0.9995 color(red)(cancel(color(black)("moles"))))/(0.9995color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For O: " (2.500 color(red)(cancel(color(black)("moles"))))/(0.9995color(red)(cancel(color(black)("moles")))) = 2.501 ~~ 2.5#</mathjax></p>
</blockquote>
<p>In order to correctly determine the empirical formula of the compound, you need the <strong>smallest whole number ratio</strong> that exists between these two elements. </p>
<p>To get that, simply multiply both values by <mathjax>#2#</mathjax>. This will get you </p>
<blockquote>
<p><mathjax>#("N"_1"O"_2.5)_2 <=> "N"_2"O"_5#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"N"_2"O"_5#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find a compound's <a href="http://socratic.org/chemistry/the-mole-concept/empirical-and-molecular-formulas">empirical formula</a> you must find the <strong>smallest whole number ratio</strong> that exists between the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> that are a part of this compound. </p>
<p>You know that your unknown compound contains <mathjax>#"14 g"#</mathjax> of nitrogen and <mathjax>#"40 g"#</mathjax> of oxygen. </p>
<p>Your first goal here will be to use the <strong>molar masses</strong> of these two elements to figure out how many <em>moles</em> of each you get in this sample. </p>
<blockquote>
<p><mathjax>#"For N: " 14 color(red)(cancel(color(black)("g"))) * "1 mole N"/(14.007color(red)(cancel(color(black)("g")))) = "0.9995 moles N"#</mathjax></p>
<p><mathjax>#"For O: " 40 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994 color(red)(cancel(color(black)("g")))) = "2.500 moles O"#</mathjax></p>
</blockquote>
<p>In order to find the <em>mole ratio</em> that exists between these two elements in the compound, divide both values by the <em>smallest</em> one </p>
<blockquote>
<p><mathjax>#"For N: " (0.9995 color(red)(cancel(color(black)("moles"))))/(0.9995color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For O: " (2.500 color(red)(cancel(color(black)("moles"))))/(0.9995color(red)(cancel(color(black)("moles")))) = 2.501 ~~ 2.5#</mathjax></p>
</blockquote>
<p>In order to correctly determine the empirical formula of the compound, you need the <strong>smallest whole number ratio</strong> that exists between these two elements. </p>
<p>To get that, simply multiply both values by <mathjax>#2#</mathjax>. This will get you </p>
<blockquote>
<p><mathjax>#("N"_1"O"_2.5)_2 <=> "N"_2"O"_5#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A nitrogen and oxygen containing molecule is decomposed into its elements. It is found to contain 14 g of nitrogen and 40 g of oxygen. What is the empirical formula of the compound?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-01-27T23:47:55" itemprop="dateCreated">
Jan 27, 2016
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<div class="markdown"><p><mathjax>#"N"_2"O"_5#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find a compound's <a href="http://socratic.org/chemistry/the-mole-concept/empirical-and-molecular-formulas">empirical formula</a> you must find the <strong>smallest whole number ratio</strong> that exists between the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> that are a part of this compound. </p>
<p>You know that your unknown compound contains <mathjax>#"14 g"#</mathjax> of nitrogen and <mathjax>#"40 g"#</mathjax> of oxygen. </p>
<p>Your first goal here will be to use the <strong>molar masses</strong> of these two elements to figure out how many <em>moles</em> of each you get in this sample. </p>
<blockquote>
<p><mathjax>#"For N: " 14 color(red)(cancel(color(black)("g"))) * "1 mole N"/(14.007color(red)(cancel(color(black)("g")))) = "0.9995 moles N"#</mathjax></p>
<p><mathjax>#"For O: " 40 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994 color(red)(cancel(color(black)("g")))) = "2.500 moles O"#</mathjax></p>
</blockquote>
<p>In order to find the <em>mole ratio</em> that exists between these two elements in the compound, divide both values by the <em>smallest</em> one </p>
<blockquote>
<p><mathjax>#"For N: " (0.9995 color(red)(cancel(color(black)("moles"))))/(0.9995color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For O: " (2.500 color(red)(cancel(color(black)("moles"))))/(0.9995color(red)(cancel(color(black)("moles")))) = 2.501 ~~ 2.5#</mathjax></p>
</blockquote>
<p>In order to correctly determine the empirical formula of the compound, you need the <strong>smallest whole number ratio</strong> that exists between these two elements. </p>
<p>To get that, simply multiply both values by <mathjax>#2#</mathjax>. This will get you </p>
<blockquote>
<p><mathjax>#("N"_1"O"_2.5)_2 <=> "N"_2"O"_5#</mathjax></p>
</blockquote></div>
</div>
</div>
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</article> | A nitrogen and oxygen containing molecule is decomposed into its elements. It is found to contain 14 g of nitrogen and 40 g of oxygen. What is the empirical formula of the compound? | null |
3,315 | a9a58d2c-6ddd-11ea-a610-ccda262736ce | https://socratic.org/questions/a-gas-with-a-volume-of-4-0-l-at-90-0-kpa-expands-until-the-pressure-drops-to-20- | 18.00 L | start physical_unit 1 1 volume l qc_end physical_unit 1 1 6 7 volume qc_end physical_unit 1 1 9 10 pressure qc_end physical_unit 1 1 17 18 pressure qc_end c_other constant_temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"18.00 L"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{4.0 L}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{90.0 kPa}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{20.0 kPa}"},{"type":"other","value":"ConstantTemperature"}] | <h1 class="questionTitle" itemprop="name"> A gas with a volume of 4.0 L at 90.0 kPa expands until the pressure drops to 20.0 kPa. What is its new volume if the temperature doesn't change?</h1> | null | 18.00 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Before doing any calculations, try to predict what you <em>expect</em> to see happen to the volume of the sample. </p>
<p>The problem tells you that the <em>temperature</em> of the gas <strong>remains unchanged</strong>. No mention is made about the <em>number of moles</em> of gas present in the sample, which means that you can assume it to be <strong>constant</strong> as well. </p>
<p>As you know, temperature is actually a measure of the <strong>average kinetic energy</strong> of the gas molecules. <a href="http://socratic.org/chemistry/the-behavior-of-gases/gas-pressure">Gas pressure</a> is a result of the <em>frequency</em> and <em>intensity</em> of the collisions that take place between the gas molecules and the walls of the container. </p>
<p>If a pressure of <mathjax>#"90.0 kPa"#</mathjax> corresponds to a given average kinetic energy of the molecules, i.e. a given <em>temperature</em>, then <strong>decreasing</strong> would result in an <strong>increase</strong> in volume. </p>
<p>Think about it like this - the average kinetic energy of the gas molecules remains <strong>unchanged</strong>, which means that in order to produce a <strong>lower pressure</strong> they must hit the walls of the container <strong>less often</strong>. </p>
<p>The molecules are moving with the same average kinetic energy, which is why they must hit the walls of the container <em>less often</em>. </p>
<p>This can only be achieved by an <strong>increase in volume</strong>. </p>
<p>So, when temperature and number of moles are kept constant, pressure and volume have an <strong>inverse relationship</strong> - this is known as <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong>.</p>
<p><img alt="https://saylordotorg.github.io/text_introductory-chemistry/s10-03-gas-laws.html" src="https://useruploads.socratic.org/GXXZJVdCQFeIotXgQswi_72ab7ad8a1fdc1f4fdadd93d52db9294.jpg"/> </p>
<p>When pressure <strong>increases</strong>, volume <strong>decreases</strong>, and when pressure <strong>decreases</strong>, volume <strong>increases</strong>. </p>
<p>This means that you can expect the volume of the gas to <strong>increase</strong>. </p>
<p>Mathematically, you can write this as</p>
<blockquote>
<p><mathjax>#color(blue)(P_1V_1 = P_2V_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax> - the pressure and volume of the gas at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax> - the pressure and volume of the gas at a final state</p>
<p>Rearrange and solve for <mathjax>#V_2#</mathjax>, the new volume of the gas</p>
<blockquote>
<p><mathjax>#V_2 = P_1/P_2 * V_1#</mathjax></p>
<p><mathjax>#V_2 = (90.0 color(red)(cancel(color(black)("kPa"))))/(20.0color(red)(cancel(color(black)("kPa")))) * "4.0 L" = color(green)("18 L")#</mathjax></p>
</blockquote>
<p>As predicted, the volume of the gas <strong>increased</strong> as a result of the <em>decrease</em> in pressure. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"18 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Before doing any calculations, try to predict what you <em>expect</em> to see happen to the volume of the sample. </p>
<p>The problem tells you that the <em>temperature</em> of the gas <strong>remains unchanged</strong>. No mention is made about the <em>number of moles</em> of gas present in the sample, which means that you can assume it to be <strong>constant</strong> as well. </p>
<p>As you know, temperature is actually a measure of the <strong>average kinetic energy</strong> of the gas molecules. <a href="http://socratic.org/chemistry/the-behavior-of-gases/gas-pressure">Gas pressure</a> is a result of the <em>frequency</em> and <em>intensity</em> of the collisions that take place between the gas molecules and the walls of the container. </p>
<p>If a pressure of <mathjax>#"90.0 kPa"#</mathjax> corresponds to a given average kinetic energy of the molecules, i.e. a given <em>temperature</em>, then <strong>decreasing</strong> would result in an <strong>increase</strong> in volume. </p>
<p>Think about it like this - the average kinetic energy of the gas molecules remains <strong>unchanged</strong>, which means that in order to produce a <strong>lower pressure</strong> they must hit the walls of the container <strong>less often</strong>. </p>
<p>The molecules are moving with the same average kinetic energy, which is why they must hit the walls of the container <em>less often</em>. </p>
<p>This can only be achieved by an <strong>increase in volume</strong>. </p>
<p>So, when temperature and number of moles are kept constant, pressure and volume have an <strong>inverse relationship</strong> - this is known as <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong>.</p>
<p><img alt="https://saylordotorg.github.io/text_introductory-chemistry/s10-03-gas-laws.html" src="https://useruploads.socratic.org/GXXZJVdCQFeIotXgQswi_72ab7ad8a1fdc1f4fdadd93d52db9294.jpg"/> </p>
<p>When pressure <strong>increases</strong>, volume <strong>decreases</strong>, and when pressure <strong>decreases</strong>, volume <strong>increases</strong>. </p>
<p>This means that you can expect the volume of the gas to <strong>increase</strong>. </p>
<p>Mathematically, you can write this as</p>
<blockquote>
<p><mathjax>#color(blue)(P_1V_1 = P_2V_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax> - the pressure and volume of the gas at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax> - the pressure and volume of the gas at a final state</p>
<p>Rearrange and solve for <mathjax>#V_2#</mathjax>, the new volume of the gas</p>
<blockquote>
<p><mathjax>#V_2 = P_1/P_2 * V_1#</mathjax></p>
<p><mathjax>#V_2 = (90.0 color(red)(cancel(color(black)("kPa"))))/(20.0color(red)(cancel(color(black)("kPa")))) * "4.0 L" = color(green)("18 L")#</mathjax></p>
</blockquote>
<p>As predicted, the volume of the gas <strong>increased</strong> as a result of the <em>decrease</em> in pressure. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name"> A gas with a volume of 4.0 L at 90.0 kPa expands until the pressure drops to 20.0 kPa. What is its new volume if the temperature doesn't change?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-03-02T00:01:04" itemprop="dateCreated">
Mar 2, 2016
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<div class="markdown"><p><mathjax>#"18 L"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Before doing any calculations, try to predict what you <em>expect</em> to see happen to the volume of the sample. </p>
<p>The problem tells you that the <em>temperature</em> of the gas <strong>remains unchanged</strong>. No mention is made about the <em>number of moles</em> of gas present in the sample, which means that you can assume it to be <strong>constant</strong> as well. </p>
<p>As you know, temperature is actually a measure of the <strong>average kinetic energy</strong> of the gas molecules. <a href="http://socratic.org/chemistry/the-behavior-of-gases/gas-pressure">Gas pressure</a> is a result of the <em>frequency</em> and <em>intensity</em> of the collisions that take place between the gas molecules and the walls of the container. </p>
<p>If a pressure of <mathjax>#"90.0 kPa"#</mathjax> corresponds to a given average kinetic energy of the molecules, i.e. a given <em>temperature</em>, then <strong>decreasing</strong> would result in an <strong>increase</strong> in volume. </p>
<p>Think about it like this - the average kinetic energy of the gas molecules remains <strong>unchanged</strong>, which means that in order to produce a <strong>lower pressure</strong> they must hit the walls of the container <strong>less often</strong>. </p>
<p>The molecules are moving with the same average kinetic energy, which is why they must hit the walls of the container <em>less often</em>. </p>
<p>This can only be achieved by an <strong>increase in volume</strong>. </p>
<p>So, when temperature and number of moles are kept constant, pressure and volume have an <strong>inverse relationship</strong> - this is known as <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong>.</p>
<p><img alt="https://saylordotorg.github.io/text_introductory-chemistry/s10-03-gas-laws.html" src="https://useruploads.socratic.org/GXXZJVdCQFeIotXgQswi_72ab7ad8a1fdc1f4fdadd93d52db9294.jpg"/> </p>
<p>When pressure <strong>increases</strong>, volume <strong>decreases</strong>, and when pressure <strong>decreases</strong>, volume <strong>increases</strong>. </p>
<p>This means that you can expect the volume of the gas to <strong>increase</strong>. </p>
<p>Mathematically, you can write this as</p>
<blockquote>
<p><mathjax>#color(blue)(P_1V_1 = P_2V_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax> - the pressure and volume of the gas at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax> - the pressure and volume of the gas at a final state</p>
<p>Rearrange and solve for <mathjax>#V_2#</mathjax>, the new volume of the gas</p>
<blockquote>
<p><mathjax>#V_2 = P_1/P_2 * V_1#</mathjax></p>
<p><mathjax>#V_2 = (90.0 color(red)(cancel(color(black)("kPa"))))/(20.0color(red)(cancel(color(black)("kPa")))) * "4.0 L" = color(green)("18 L")#</mathjax></p>
</blockquote>
<p>As predicted, the volume of the gas <strong>increased</strong> as a result of the <em>decrease</em> in pressure. </p></div>
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</article> | A gas with a volume of 4.0 L at 90.0 kPa expands until the pressure drops to 20.0 kPa. What is its new volume if the temperature doesn't change? | null |
3,316 | ad1d6230-6ddd-11ea-828f-ccda262736ce | https://socratic.org/questions/59102b0011ef6b54f49b0d51 | 400.00 mmHg | start physical_unit 26 27 pressure mmhg qc_end physical_unit 26 27 7 8 pressure qc_end physical_unit 26 27 1 2 volume qc_end physical_unit 26 27 17 18 volume qc_end c_other constant_temperature qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the gas [IN] mmHg"}] | [{"type":"physical unit","value":"400.00 mmHg"}] | [{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{800 mmHg}"},{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{2.5 L}"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{5.0 L}"},{"type":"other","value":"ConstantTemperature"}] | <h1 class="questionTitle" itemprop="name">A #2.5*L# volume of gas at #800*mm*Hg# pressure, is expanded with constant temperature to a #5.0*L# volume. What is the resultant pressure of the gas?</h1> | null | 400.00 mmHg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And so <mathjax>#P_2=(P_1V_1)/V_2=(800*mm*Hgxx0.250*L)/(0.500*L)=#</mathjax></p>
<p><mathjax>#400*mm*Hg#</mathjax>.</p>
<p>Just to ADD that you DO NOT measure a pressure in <mathjax>#mm*Hg#</mathjax> OVER <mathjax>#1*atm#</mathjax> (<mathjax>#1*atm-=760*mm*Hg#</mathjax>). This is an inappropriate question posed by someone who has NEVER used a mercury manometer, and certainly NEVER cleaned up a mercury spill. </p></div>
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<div class="markdown"><p>Well <mathjax>#P_1V_1=P_2V_2...........#</mathjax> given constant <mathjax>#T#</mathjax>, which is old <mathjax>#"Boyle's Law".................#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And so <mathjax>#P_2=(P_1V_1)/V_2=(800*mm*Hgxx0.250*L)/(0.500*L)=#</mathjax></p>
<p><mathjax>#400*mm*Hg#</mathjax>.</p>
<p>Just to ADD that you DO NOT measure a pressure in <mathjax>#mm*Hg#</mathjax> OVER <mathjax>#1*atm#</mathjax> (<mathjax>#1*atm-=760*mm*Hg#</mathjax>). This is an inappropriate question posed by someone who has NEVER used a mercury manometer, and certainly NEVER cleaned up a mercury spill. </p></div>
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<h1 class="questionTitle" itemprop="name">A #2.5*L# volume of gas at #800*mm*Hg# pressure, is expanded with constant temperature to a #5.0*L# volume. What is the resultant pressure of the gas?</h1>
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<div class="markdown"><p>Well <mathjax>#P_1V_1=P_2V_2...........#</mathjax> given constant <mathjax>#T#</mathjax>, which is old <mathjax>#"Boyle's Law".................#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And so <mathjax>#P_2=(P_1V_1)/V_2=(800*mm*Hgxx0.250*L)/(0.500*L)=#</mathjax></p>
<p><mathjax>#400*mm*Hg#</mathjax>.</p>
<p>Just to ADD that you DO NOT measure a pressure in <mathjax>#mm*Hg#</mathjax> OVER <mathjax>#1*atm#</mathjax> (<mathjax>#1*atm-=760*mm*Hg#</mathjax>). This is an inappropriate question posed by someone who has NEVER used a mercury manometer, and certainly NEVER cleaned up a mercury spill. </p></div>
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</article> | A #2.5*L# volume of gas at #800*mm*Hg# pressure, is expanded with constant temperature to a #5.0*L# volume. What is the resultant pressure of the gas? | null |
3,317 | abaf918a-6ddd-11ea-b0ad-ccda262736ce | https://socratic.org/questions/59456d59b72cff1628486322 | 33.5 mL | start physical_unit 5 7 volume ml qc_end physical_unit 5 7 3 4 molarity qc_end physical_unit 19 20 17 18 molarity qc_end physical_unit 7 7 14 15 volume qc_end end | [{"type":"physical unit","value":"Volume [OF] sodium sulfide solution [IN] mL"}] | [{"type":"physical unit","value":"33.5 mL"}] | [{"type":"physical unit","value":"Molarity [OF] sodium sulfide solution [=] \\pu{0.230 mol/L}"},{"type":"physical unit","value":"Molarity [OF] silver nitrate solution [=] \\pu{0.513 mol/L}"},{"type":"physical unit","value":"Volume [OF] silver nitrate solution [=] \\pu{30.00 mL}"}] | <h1 class="questionTitle" itemprop="name">What volume of 0.230 mol/L sodium sulfide solution do I need to react with 30.00 ml of 0.513 mol/L silver nitrate?</h1> | null | 33.5 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step1. Write the balanced chemical equation</strong></p>
<p><mathjax>#"Na"_2"S" + "2AgNO"_3 → "2NaNO"_3 + "Ag"_2"S"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the moles of <mathjax>#"AgNO"_3#</mathjax></strong></p>
<p><mathjax>#"Moles of AgNO"_3 = "0.030 00" color(red)(cancel(color(black)("L AgNO"_3))) × "0.513 mol AgNO"_3/(1 color(red)(cancel(color(black)("L AgNO"_3)))) = "0.015 39 mol AgNO"_3#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the moles of <mathjax>#"Na"_2"S"#</mathjax></strong></p>
<p><mathjax>#"Moles of Na"_2"S" = "0.015 39" color(red)(cancel(color(black)("mol AgNO"_3))) × ("1 mol Na"_2"S")/(2 color(red)(cancel(color(black)("mol AgNO"_3)))) = "0.007 695 mol Na"_2"S"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4. Calculate the volume of <mathjax>#"Na"_2"S"#</mathjax></strong></p>
<p><mathjax>#"Vol. of Na"_2"S" = "0.007 695" color(red)(cancel(color(black)("mol Na"_2"S"))) ×("1 L Na"_2"S")/(0.230 color(red)(cancel(color(black)("mol Na"_2"S")))) = "0.0335 L Na"_2"S" = "33.5 mL Na"_2"S"#</mathjax></p></div>
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<div class="markdown"><p>You will need 33.5 mL of <mathjax>#"Na"_2"S""#</mathjax> solution.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step1. Write the balanced chemical equation</strong></p>
<p><mathjax>#"Na"_2"S" + "2AgNO"_3 → "2NaNO"_3 + "Ag"_2"S"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the moles of <mathjax>#"AgNO"_3#</mathjax></strong></p>
<p><mathjax>#"Moles of AgNO"_3 = "0.030 00" color(red)(cancel(color(black)("L AgNO"_3))) × "0.513 mol AgNO"_3/(1 color(red)(cancel(color(black)("L AgNO"_3)))) = "0.015 39 mol AgNO"_3#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the moles of <mathjax>#"Na"_2"S"#</mathjax></strong></p>
<p><mathjax>#"Moles of Na"_2"S" = "0.015 39" color(red)(cancel(color(black)("mol AgNO"_3))) × ("1 mol Na"_2"S")/(2 color(red)(cancel(color(black)("mol AgNO"_3)))) = "0.007 695 mol Na"_2"S"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4. Calculate the volume of <mathjax>#"Na"_2"S"#</mathjax></strong></p>
<p><mathjax>#"Vol. of Na"_2"S" = "0.007 695" color(red)(cancel(color(black)("mol Na"_2"S"))) ×("1 L Na"_2"S")/(0.230 color(red)(cancel(color(black)("mol Na"_2"S")))) = "0.0335 L Na"_2"S" = "33.5 mL Na"_2"S"#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What volume of 0.230 mol/L sodium sulfide solution do I need to react with 30.00 ml of 0.513 mol/L silver nitrate?</h1>
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<div class="markdown"><p>You will need 33.5 mL of <mathjax>#"Na"_2"S""#</mathjax> solution.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step1. Write the balanced chemical equation</strong></p>
<p><mathjax>#"Na"_2"S" + "2AgNO"_3 → "2NaNO"_3 + "Ag"_2"S"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the moles of <mathjax>#"AgNO"_3#</mathjax></strong></p>
<p><mathjax>#"Moles of AgNO"_3 = "0.030 00" color(red)(cancel(color(black)("L AgNO"_3))) × "0.513 mol AgNO"_3/(1 color(red)(cancel(color(black)("L AgNO"_3)))) = "0.015 39 mol AgNO"_3#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the moles of <mathjax>#"Na"_2"S"#</mathjax></strong></p>
<p><mathjax>#"Moles of Na"_2"S" = "0.015 39" color(red)(cancel(color(black)("mol AgNO"_3))) × ("1 mol Na"_2"S")/(2 color(red)(cancel(color(black)("mol AgNO"_3)))) = "0.007 695 mol Na"_2"S"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4. Calculate the volume of <mathjax>#"Na"_2"S"#</mathjax></strong></p>
<p><mathjax>#"Vol. of Na"_2"S" = "0.007 695" color(red)(cancel(color(black)("mol Na"_2"S"))) ×("1 L Na"_2"S")/(0.230 color(red)(cancel(color(black)("mol Na"_2"S")))) = "0.0335 L Na"_2"S" = "33.5 mL Na"_2"S"#</mathjax></p></div>
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</article> | What volume of 0.230 mol/L sodium sulfide solution do I need to react with 30.00 ml of 0.513 mol/L silver nitrate? | null |
3,318 | ac09a6fe-6ddd-11ea-8dc2-ccda262736ce | https://socratic.org/questions/what-is-the-formal-charge-of-n-in-nh-4 | +1 | start physical_unit 6 6 charge none qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Formal charge [OF] N"}] | [{"type":"physical unit","value":"+1"}] | [{"type":"chemical equation","value":"NH4+"}] | <h1 class="questionTitle" itemprop="name">What is the formal charge of N in #NH_4^+#?</h1> | null | +1 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Consider the ammonia molecule, <mathjax>#H_3N:#</mathjax>, which is formally neutral Why so? Because it has 2 inner shell electrons, it has a half share of the 6 electrons that comprise the <mathjax>#3xxN-H#</mathjax> bonds, and it gets the full contribution from the lone pair of electrons: <mathjax>#2+3+2=7e^-#</mathjax>. This electronic charge balances the 7 positively charged protons present in the nitrogen nucleus. In ammonium, it still gets the 2 inner core electrons, but only half of the <mathjax>#4xxN-H#</mathjax> electrons, i.e. 6 electrons in total, and thus a formal positive charge. </p>
<p>This can also be seen in the acid-base reaction:</p>
<p><mathjax>#NH_3(aq) + H_2O rightleftharpoonsNH_4^+ + HO^-#</mathjax>.</p>
<p>Charge is conserved as always.</p>
<p>In liquid ammonia, the amide ion, <mathjax>#NH_2^-#</mathjax>, is known, as well as the imide ion, <mathjax>#NH^(2-)#</mathjax>. By means of the formalisms above, can you tell me where the formal charge lies?</p></div>
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<div class="markdown"><p>The formal charge of <mathjax>#N#</mathjax> is <mathjax>#+1#</mathjax>, i.e. it is the charge of the ion. </p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Consider the ammonia molecule, <mathjax>#H_3N:#</mathjax>, which is formally neutral Why so? Because it has 2 inner shell electrons, it has a half share of the 6 electrons that comprise the <mathjax>#3xxN-H#</mathjax> bonds, and it gets the full contribution from the lone pair of electrons: <mathjax>#2+3+2=7e^-#</mathjax>. This electronic charge balances the 7 positively charged protons present in the nitrogen nucleus. In ammonium, it still gets the 2 inner core electrons, but only half of the <mathjax>#4xxN-H#</mathjax> electrons, i.e. 6 electrons in total, and thus a formal positive charge. </p>
<p>This can also be seen in the acid-base reaction:</p>
<p><mathjax>#NH_3(aq) + H_2O rightleftharpoonsNH_4^+ + HO^-#</mathjax>.</p>
<p>Charge is conserved as always.</p>
<p>In liquid ammonia, the amide ion, <mathjax>#NH_2^-#</mathjax>, is known, as well as the imide ion, <mathjax>#NH^(2-)#</mathjax>. By means of the formalisms above, can you tell me where the formal charge lies?</p></div>
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<h1 class="questionTitle" itemprop="name">What is the formal charge of N in #NH_4^+#?</h1>
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<div class="markdown"><p>The formal charge of <mathjax>#N#</mathjax> is <mathjax>#+1#</mathjax>, i.e. it is the charge of the ion. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Consider the ammonia molecule, <mathjax>#H_3N:#</mathjax>, which is formally neutral Why so? Because it has 2 inner shell electrons, it has a half share of the 6 electrons that comprise the <mathjax>#3xxN-H#</mathjax> bonds, and it gets the full contribution from the lone pair of electrons: <mathjax>#2+3+2=7e^-#</mathjax>. This electronic charge balances the 7 positively charged protons present in the nitrogen nucleus. In ammonium, it still gets the 2 inner core electrons, but only half of the <mathjax>#4xxN-H#</mathjax> electrons, i.e. 6 electrons in total, and thus a formal positive charge. </p>
<p>This can also be seen in the acid-base reaction:</p>
<p><mathjax>#NH_3(aq) + H_2O rightleftharpoonsNH_4^+ + HO^-#</mathjax>.</p>
<p>Charge is conserved as always.</p>
<p>In liquid ammonia, the amide ion, <mathjax>#NH_2^-#</mathjax>, is known, as well as the imide ion, <mathjax>#NH^(2-)#</mathjax>. By means of the formalisms above, can you tell me where the formal charge lies?</p></div>
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</article> | What is the formal charge of N in #NH_4^+#? | null |
3,319 | aa3833e6-6ddd-11ea-985f-ccda262736ce | https://socratic.org/questions/for-the-reaction-2h-2o-137-kcal-2h-2-g-o-2-g-how-many-kcal-are-needed-to-form-2- | 274 kcal | start physical_unit 1 2 energy kcal qc_end chemical_equation 3 12 qc_end end | [{"type":"physical unit","value":"Needed energy [OF] the reaction [IN] kcal"}] | [{"type":"physical unit","value":"274 kcal"}] | [{"type":"physical unit","value":"Mole [OF] O2(g) [=] \\pu{2 moles}"},{"type":"chemical equation","value":"2 H2O + 137 kcal -> 2 H2(g) + O2(g)"}] | <h1 class="questionTitle" itemprop="name">For the reaction: #2H_2O + 137 kcal -> 2H_2(g) + O_2(g)# how many kcal are needed to form 2.00 moles #O_2(g)#?</h1> | null | 274 kcal | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by taking a look at the <strong>thermochemical equation</strong> given to you</p>
<blockquote>
<p><mathjax>#2"H"_ 2"O"_ ((l)) + "137 kcal" -> 2"H"_ (2(g)) + "O"_ (2(g))#</mathjax></p>
</blockquote>
<p>Notice that this thermochemical equation describes an <strong>endothermic reaction</strong>. You can say that because you have a <em>heat term</em> added to the <strong>reactants' side</strong>.</p>
<p>This essentially tells you that heat is <strong>needed</strong> in order for the reaction to take place. This can also be shown by using the <em><a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of reaction</em>, <mathjax>#DeltaH_"rxn"#</mathjax>, which in this case is equal to </p>
<blockquote>
<p><mathjax>#DeltaH_"rxn" = +"137 kcal"#</mathjax></p>
<blockquote>
<p><em>The enthalpy change of reaction is <strong>positive</strong> because the reaction is <strong>endothermic</strong></em></p>
</blockquote>
</blockquote>
<p>Now, take a look at the <strong>stoichiometric coefficients</strong>. You can say that it takes <mathjax>#"137 kcal"#</mathjax> of heat in order to get <mathjax>#2#</mathjax> <strong>moles</strong> of water to produce <mathjax>#2#</mathjax> <strong>moles</strong> of hydrogen gas and <mathjax>#1#</mathjax> <strong>mole</strong> of oxygen gas. </p>
<p>The problem wants you to figure out how much heat is needed in order for the reaction to produce <mathjax>#2.00#</mathjax> <strong>moles</strong> of oxygen gas. </p>
<p>You can use the fact that <mathjax>#"137 kcal"#</mathjax> of heat are needed in order to produce <mathjax>#1#</mathjax> <strong>mole</strong> of oxygen gas to say that</p>
<blockquote>
<p><mathjax>#2.00 color(red)(cancel(color(black)("moles O"_2))) * overbrace("137 kcal"/(1color(red)(cancel(color(black)("mole O"_2)))))^(color(blue)("given by the thermochemical equation")) = color(darkgreen)(ul(color(black)("274 kcal")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"274 kcal"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by taking a look at the <strong>thermochemical equation</strong> given to you</p>
<blockquote>
<p><mathjax>#2"H"_ 2"O"_ ((l)) + "137 kcal" -> 2"H"_ (2(g)) + "O"_ (2(g))#</mathjax></p>
</blockquote>
<p>Notice that this thermochemical equation describes an <strong>endothermic reaction</strong>. You can say that because you have a <em>heat term</em> added to the <strong>reactants' side</strong>.</p>
<p>This essentially tells you that heat is <strong>needed</strong> in order for the reaction to take place. This can also be shown by using the <em><a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of reaction</em>, <mathjax>#DeltaH_"rxn"#</mathjax>, which in this case is equal to </p>
<blockquote>
<p><mathjax>#DeltaH_"rxn" = +"137 kcal"#</mathjax></p>
<blockquote>
<p><em>The enthalpy change of reaction is <strong>positive</strong> because the reaction is <strong>endothermic</strong></em></p>
</blockquote>
</blockquote>
<p>Now, take a look at the <strong>stoichiometric coefficients</strong>. You can say that it takes <mathjax>#"137 kcal"#</mathjax> of heat in order to get <mathjax>#2#</mathjax> <strong>moles</strong> of water to produce <mathjax>#2#</mathjax> <strong>moles</strong> of hydrogen gas and <mathjax>#1#</mathjax> <strong>mole</strong> of oxygen gas. </p>
<p>The problem wants you to figure out how much heat is needed in order for the reaction to produce <mathjax>#2.00#</mathjax> <strong>moles</strong> of oxygen gas. </p>
<p>You can use the fact that <mathjax>#"137 kcal"#</mathjax> of heat are needed in order to produce <mathjax>#1#</mathjax> <strong>mole</strong> of oxygen gas to say that</p>
<blockquote>
<p><mathjax>#2.00 color(red)(cancel(color(black)("moles O"_2))) * overbrace("137 kcal"/(1color(red)(cancel(color(black)("mole O"_2)))))^(color(blue)("given by the thermochemical equation")) = color(darkgreen)(ul(color(black)("274 kcal")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">For the reaction: #2H_2O + 137 kcal -> 2H_2(g) + O_2(g)# how many kcal are needed to form 2.00 moles #O_2(g)#?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"274 kcal"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by taking a look at the <strong>thermochemical equation</strong> given to you</p>
<blockquote>
<p><mathjax>#2"H"_ 2"O"_ ((l)) + "137 kcal" -> 2"H"_ (2(g)) + "O"_ (2(g))#</mathjax></p>
</blockquote>
<p>Notice that this thermochemical equation describes an <strong>endothermic reaction</strong>. You can say that because you have a <em>heat term</em> added to the <strong>reactants' side</strong>.</p>
<p>This essentially tells you that heat is <strong>needed</strong> in order for the reaction to take place. This can also be shown by using the <em><a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of reaction</em>, <mathjax>#DeltaH_"rxn"#</mathjax>, which in this case is equal to </p>
<blockquote>
<p><mathjax>#DeltaH_"rxn" = +"137 kcal"#</mathjax></p>
<blockquote>
<p><em>The enthalpy change of reaction is <strong>positive</strong> because the reaction is <strong>endothermic</strong></em></p>
</blockquote>
</blockquote>
<p>Now, take a look at the <strong>stoichiometric coefficients</strong>. You can say that it takes <mathjax>#"137 kcal"#</mathjax> of heat in order to get <mathjax>#2#</mathjax> <strong>moles</strong> of water to produce <mathjax>#2#</mathjax> <strong>moles</strong> of hydrogen gas and <mathjax>#1#</mathjax> <strong>mole</strong> of oxygen gas. </p>
<p>The problem wants you to figure out how much heat is needed in order for the reaction to produce <mathjax>#2.00#</mathjax> <strong>moles</strong> of oxygen gas. </p>
<p>You can use the fact that <mathjax>#"137 kcal"#</mathjax> of heat are needed in order to produce <mathjax>#1#</mathjax> <strong>mole</strong> of oxygen gas to say that</p>
<blockquote>
<p><mathjax>#2.00 color(red)(cancel(color(black)("moles O"_2))) * overbrace("137 kcal"/(1color(red)(cancel(color(black)("mole O"_2)))))^(color(blue)("given by the thermochemical equation")) = color(darkgreen)(ul(color(black)("274 kcal")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | For the reaction: #2H_2O + 137 kcal -> 2H_2(g) + O_2(g)# how many kcal are needed to form 2.00 moles #O_2(g)#? | null |
3,320 | a92b3a22-6ddd-11ea-908d-ccda262736ce | https://socratic.org/questions/a-compound-is-found-be-36-5-na-25-3-s-and-38-0-o-what-is-its-empirical-formula | Na2SO3 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"Na2SO3"}] | [{"type":"physical unit","value":"Percentage [OF] Na in the compound [=] \\pu{36.5%}"},{"type":"physical unit","value":"Percentage [OF] S in the compound [=] \\pu{25.3%}"},{"type":"physical unit","value":"Percentage [OF] O in the compound [=] \\pu{38.0%}"}] | <h1 class="questionTitle" itemprop="name">A compound is found be 36.5% #Na#, 25.3% #S#, and 38.0% #O#. What is its empirical formula?</h1> | null | Na2SO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To solve for an empirical formula, when given the percent compositions, first, divide each <a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> by the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of THAT element.</p>
<ol>
<li><mathjax>#(36.5% Na)/(22.99 g/(mol) Na#</mathjax> <mathjax>#(25.3% S)/(32.07 g/(mol) S#</mathjax> <mathjax>#(38.0% O)/(16 g/(mol) O#</mathjax> <br/>
(considering that this is not oxygen gas). For these atomic masses, I rounded to the nearest hundredth.</li>
</ol>
<p>You get: <mathjax>#1.587#</mathjax> <mathjax>#0.788#</mathjax> <mathjax>#2.375#</mathjax> (in corresponding order to the above equations- these values don't have units).</p>
<ol>
<li>Then, you divide each of these values by the SMALLEST of these values.<br/>
<mathjax>#1.587/0.788#</mathjax> <mathjax>#0.788/0.788#</mathjax> <mathjax>#2.375/0.788#</mathjax> <br/>
Then, still corresponding to Na, S, and O respectively, we get:<br/>
<mathjax>#2.013#</mathjax> mol Na= <mathjax>#2#</mathjax> mol Na<br/>
<mathjax>#1#</mathjax> mol S<br/>
<mathjax>#3.013#</mathjax> mol O= <mathjax>#3#</mathjax> mol O <br/>
(after rounding the answers)</li>
</ol>
<p>Therefore, the empirical formula is <mathjax>#Na_2SO_3#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#Na_2SO_3#</mathjax> </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To solve for an empirical formula, when given the percent compositions, first, divide each <a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> by the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of THAT element.</p>
<ol>
<li><mathjax>#(36.5% Na)/(22.99 g/(mol) Na#</mathjax> <mathjax>#(25.3% S)/(32.07 g/(mol) S#</mathjax> <mathjax>#(38.0% O)/(16 g/(mol) O#</mathjax> <br/>
(considering that this is not oxygen gas). For these atomic masses, I rounded to the nearest hundredth.</li>
</ol>
<p>You get: <mathjax>#1.587#</mathjax> <mathjax>#0.788#</mathjax> <mathjax>#2.375#</mathjax> (in corresponding order to the above equations- these values don't have units).</p>
<ol>
<li>Then, you divide each of these values by the SMALLEST of these values.<br/>
<mathjax>#1.587/0.788#</mathjax> <mathjax>#0.788/0.788#</mathjax> <mathjax>#2.375/0.788#</mathjax> <br/>
Then, still corresponding to Na, S, and O respectively, we get:<br/>
<mathjax>#2.013#</mathjax> mol Na= <mathjax>#2#</mathjax> mol Na<br/>
<mathjax>#1#</mathjax> mol S<br/>
<mathjax>#3.013#</mathjax> mol O= <mathjax>#3#</mathjax> mol O <br/>
(after rounding the answers)</li>
</ol>
<p>Therefore, the empirical formula is <mathjax>#Na_2SO_3#</mathjax></p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">A compound is found be 36.5% #Na#, 25.3% #S#, and 38.0% #O#. What is its empirical formula?</h1>
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May 19, 2016
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<div class="markdown"><p><mathjax>#Na_2SO_3#</mathjax> </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To solve for an empirical formula, when given the percent compositions, first, divide each <a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> by the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of THAT element.</p>
<ol>
<li><mathjax>#(36.5% Na)/(22.99 g/(mol) Na#</mathjax> <mathjax>#(25.3% S)/(32.07 g/(mol) S#</mathjax> <mathjax>#(38.0% O)/(16 g/(mol) O#</mathjax> <br/>
(considering that this is not oxygen gas). For these atomic masses, I rounded to the nearest hundredth.</li>
</ol>
<p>You get: <mathjax>#1.587#</mathjax> <mathjax>#0.788#</mathjax> <mathjax>#2.375#</mathjax> (in corresponding order to the above equations- these values don't have units).</p>
<ol>
<li>Then, you divide each of these values by the SMALLEST of these values.<br/>
<mathjax>#1.587/0.788#</mathjax> <mathjax>#0.788/0.788#</mathjax> <mathjax>#2.375/0.788#</mathjax> <br/>
Then, still corresponding to Na, S, and O respectively, we get:<br/>
<mathjax>#2.013#</mathjax> mol Na= <mathjax>#2#</mathjax> mol Na<br/>
<mathjax>#1#</mathjax> mol S<br/>
<mathjax>#3.013#</mathjax> mol O= <mathjax>#3#</mathjax> mol O <br/>
(after rounding the answers)</li>
</ol>
<p>Therefore, the empirical formula is <mathjax>#Na_2SO_3#</mathjax></p></div>
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</article> | A compound is found be 36.5% #Na#, 25.3% #S#, and 38.0% #O#. What is its empirical formula? | null |
3,321 | ad07a130-6ddd-11ea-8c8f-ccda262736ce | https://socratic.org/questions/how-many-grams-are-contained-in-2-00-mol-of-neon | 40.40 grams | start physical_unit 9 9 mass g qc_end physical_unit 9 9 6 7 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] neon [IN] grams"}] | [{"type":"physical unit","value":"40.40 grams"}] | [{"type":"physical unit","value":"Mole [OF] neon [=] \\pu{2.00 mol}"}] | <h1 class="questionTitle" itemprop="name">How many grams are contained in 2.00 mol of neon?</h1> | null | 40.40 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Multiply the given moles times the molar mass of neon, <mathjax>#"20.180 g/mol"#</mathjax> (periodic table).</p>
<p><mathjax>#2.00cancel"mol Ne"xx(20.180"g Ne")/(1cancel"mol Ne")="40.4 g Ne"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"2.00 mol Ne"#</mathjax> has a mass of <mathjax>#"40.4 g"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Multiply the given moles times the molar mass of neon, <mathjax>#"20.180 g/mol"#</mathjax> (periodic table).</p>
<p><mathjax>#2.00cancel"mol Ne"xx(20.180"g Ne")/(1cancel"mol Ne")="40.4 g Ne"#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many grams are contained in 2.00 mol of neon?</h1>
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<div class="markdown"><p><mathjax>#"2.00 mol Ne"#</mathjax> has a mass of <mathjax>#"40.4 g"#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Multiply the given moles times the molar mass of neon, <mathjax>#"20.180 g/mol"#</mathjax> (periodic table).</p>
<p><mathjax>#2.00cancel"mol Ne"xx(20.180"g Ne")/(1cancel"mol Ne")="40.4 g Ne"#</mathjax></p></div>
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<div class="markdown"><p>2 moles <mathjax>#xx#</mathjax> 20.2 g/mol = ??</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Assume that:<br/>
n - number of moles<br/>
m - mass of substance<br/>
M - molar mass<br/>
<mathjax>#n = m -: M#</mathjax></p>
<p>2.00 moles (n) of Ne has been provided for you.</p>
<p>Now you have to find the molar mass (M) of Ne. If you look into your periodic table, the molar mass of Neon is 20.2 g/mol.</p>
<p>Final step:<br/>
flip-over the formula and it becomes <mathjax>#m = M xx n#</mathjax>.<br/>
The mass of Neon is: [m = 20.2 g/mol <mathjax>#xx#</mathjax> 2 moles]<br/>
= 40.4 grams.</p>
<p>Therefore, 40.4 grams are contained in 2 moles of Neon.</p></div>
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</article> | How many grams are contained in 2.00 mol of neon? | null |
3,322 | ab9af89a-6ddd-11ea-9140-ccda262736ce | https://socratic.org/questions/what-are-the-oxidation-and-reduction-equations-for-when-zinc-is-placed-in-hcl | Zn(s) + 2 HCl(aq) -> ZnCl2(aq) + H2(g) | start chemical_equation qc_end chemical_equation 9 9 qc_end chemical_equation 13 13 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the oxidation and reduction equations"}] | [{"type":"chemical equation","value":"Zn(s) + 2 HCl(aq) -> ZnCl2(aq) + H2(g)"}] | [{"type":"chemical equation","value":"Zinc"},{"type":"chemical equation","value":"HCl"}] | <h1 class="questionTitle" itemprop="name">What are the oxidation and reduction equations for when zinc is placed in HCl?</h1> | null | Zn(s) + 2 HCl(aq) -> ZnCl2(aq) + H2(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Zn(s) rarr Zn^(2+) + 2e^-#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>And hydronium ion is REDUCED to dihydrogen gas.........</p>
<p><mathjax>#H_3O^+ + e^(-) rarr 1/2H_2(g) +H_2O(l)#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>And we add the equations together to eliminate the electrons.....</p>
<p><mathjax>#Zn(s) + 2H_3O^+ rarrZn^(2+) + H_2(g)uarr +2H_2O(l)#</mathjax></p>
<p>And since chloride anion is conceived to be a spectator in this reaction.....</p>
<p><mathjax>#Zn(s) + 2HCl(aq) rarrZnCl_2(aq) + H_2(g)uarr#</mathjax></p>
<p>Is mass balanced? Is charge balanced? If the answer is NO, then the reaction cannot be considered to represent reality. </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>Zinc is oxidized to <mathjax>#Zn^(2+)#</mathjax>.........</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Zn(s) rarr Zn^(2+) + 2e^-#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>And hydronium ion is REDUCED to dihydrogen gas.........</p>
<p><mathjax>#H_3O^+ + e^(-) rarr 1/2H_2(g) +H_2O(l)#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>And we add the equations together to eliminate the electrons.....</p>
<p><mathjax>#Zn(s) + 2H_3O^+ rarrZn^(2+) + H_2(g)uarr +2H_2O(l)#</mathjax></p>
<p>And since chloride anion is conceived to be a spectator in this reaction.....</p>
<p><mathjax>#Zn(s) + 2HCl(aq) rarrZnCl_2(aq) + H_2(g)uarr#</mathjax></p>
<p>Is mass balanced? Is charge balanced? If the answer is NO, then the reaction cannot be considered to represent reality. </p></div>
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<h1 class="questionTitle" itemprop="name">What are the oxidation and reduction equations for when zinc is placed in HCl?</h1>
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<div class="markdown"><p>Zinc is oxidized to <mathjax>#Zn^(2+)#</mathjax>.........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#Zn(s) rarr Zn^(2+) + 2e^-#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>And hydronium ion is REDUCED to dihydrogen gas.........</p>
<p><mathjax>#H_3O^+ + e^(-) rarr 1/2H_2(g) +H_2O(l)#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>And we add the equations together to eliminate the electrons.....</p>
<p><mathjax>#Zn(s) + 2H_3O^+ rarrZn^(2+) + H_2(g)uarr +2H_2O(l)#</mathjax></p>
<p>And since chloride anion is conceived to be a spectator in this reaction.....</p>
<p><mathjax>#Zn(s) + 2HCl(aq) rarrZnCl_2(aq) + H_2(g)uarr#</mathjax></p>
<p>Is mass balanced? Is charge balanced? If the answer is NO, then the reaction cannot be considered to represent reality. </p></div>
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</article> | What are the oxidation and reduction equations for when zinc is placed in HCl? | null |
3,323 | a8650434-6ddd-11ea-a5d0-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-equation-for-the-ionization-of-benzoic-acid | C6H5COOH(aq) + H2O(l) -> C6H5COO- + H3O+ | start chemical_equation qc_end substance 9 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the ionization"}] | [{"type":"chemical equation","value":"C6H5COOH(aq) + H2O(l) -> C6H5COO- + H3O+"}] | [{"type":"substance name","value":"Benzoic acid"}] | <h1 class="questionTitle" itemprop="name">What is the chemical equation for the ionization of benzoic acid?</h1> | null | C6H5COOH(aq) + H2O(l) -> C6H5COO- + H3O+ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>See this <a href="https://en.wikipedia.org/wiki/Benzoic_acid" rel="nofollow">site.</a> A <mathjax>#pK_a=4.202#</mathjax> is reported. For acetic acid, <mathjax>#H_3C-CO_2H#</mathjax>, <mathjax>#pK_a=4.76#</mathjax>. Can you explain why benzoic acid is a stronger acid than acetic acid?</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#C_6H_5CO_2H(aq) + H_2O(l) rightleftharpoons C_6H_5CO_2^(-) + H_3O^+#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>See this <a href="https://en.wikipedia.org/wiki/Benzoic_acid" rel="nofollow">site.</a> A <mathjax>#pK_a=4.202#</mathjax> is reported. For acetic acid, <mathjax>#H_3C-CO_2H#</mathjax>, <mathjax>#pK_a=4.76#</mathjax>. Can you explain why benzoic acid is a stronger acid than acetic acid?</p></div>
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<h1 class="questionTitle" itemprop="name">What is the chemical equation for the ionization of benzoic acid?</h1>
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anor277
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<div class="markdown"><p><mathjax>#C_6H_5CO_2H(aq) + H_2O(l) rightleftharpoons C_6H_5CO_2^(-) + H_3O^+#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>See this <a href="https://en.wikipedia.org/wiki/Benzoic_acid" rel="nofollow">site.</a> A <mathjax>#pK_a=4.202#</mathjax> is reported. For acetic acid, <mathjax>#H_3C-CO_2H#</mathjax>, <mathjax>#pK_a=4.76#</mathjax>. Can you explain why benzoic acid is a stronger acid than acetic acid?</p></div>
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</article> | What is the chemical equation for the ionization of benzoic acid? | null |
3,324 | ac7bd63a-6ddd-11ea-bbc6-ccda262736ce | https://socratic.org/questions/how-do-you-determine-the-formula-for-aluminium-hydroxide | Al(OH)3 | start chemical_formula qc_end substance 7 8 qc_end end | [{"type":"other","value":"Chemical Formula [OF] aluminium hydroxide [IN] default"}] | [{"type":"chemical equation","value":"Al(OH)3"}] | [{"type":"substance name","value":"Aluminium hydroxide"}] | <h1 class="questionTitle" itemprop="name">How do you determine the formula for aluminium hydroxide? </h1> | null | Al(OH)3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Aluminum has a chemical symbol of <mathjax>#Al#</mathjax>, with a charge of 3+. <br/>
Hydroxide is a molecule consisting of two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> - hydrogen and oxygen. The chemical formula for hydroxide is <mathjax>#OH^(1-)#</mathjax>.</p>
<p>Aluminum is the cation, while hydroxide is the anion, thus we get <mathjax>#Al(OH)_"3"#</mathjax>. This is because we still "criss-cross" the charges as it is an ionic compound - electrons are transferred - so aluminum's charge of 3+ goes to hydroxide while its charge of 1- goes to aluminum. </p>
<p>The 1 is redundant so we don't write it in.</p>
<p>Hope this helps :)</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>"Criss-cross" the charges of each atom because it is an ionic compound. This will get you <mathjax>#Al(OH)_"3"#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Aluminum has a chemical symbol of <mathjax>#Al#</mathjax>, with a charge of 3+. <br/>
Hydroxide is a molecule consisting of two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> - hydrogen and oxygen. The chemical formula for hydroxide is <mathjax>#OH^(1-)#</mathjax>.</p>
<p>Aluminum is the cation, while hydroxide is the anion, thus we get <mathjax>#Al(OH)_"3"#</mathjax>. This is because we still "criss-cross" the charges as it is an ionic compound - electrons are transferred - so aluminum's charge of 3+ goes to hydroxide while its charge of 1- goes to aluminum. </p>
<p>The 1 is redundant so we don't write it in.</p>
<p>Hope this helps :)</p></div>
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<h1 class="questionTitle" itemprop="name">How do you determine the formula for aluminium hydroxide? </h1>
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<div class="markdown"><p>"Criss-cross" the charges of each atom because it is an ionic compound. This will get you <mathjax>#Al(OH)_"3"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Aluminum has a chemical symbol of <mathjax>#Al#</mathjax>, with a charge of 3+. <br/>
Hydroxide is a molecule consisting of two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> - hydrogen and oxygen. The chemical formula for hydroxide is <mathjax>#OH^(1-)#</mathjax>.</p>
<p>Aluminum is the cation, while hydroxide is the anion, thus we get <mathjax>#Al(OH)_"3"#</mathjax>. This is because we still "criss-cross" the charges as it is an ionic compound - electrons are transferred - so aluminum's charge of 3+ goes to hydroxide while its charge of 1- goes to aluminum. </p>
<p>The 1 is redundant so we don't write it in.</p>
<p>Hope this helps :)</p></div>
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</article> | How do you determine the formula for aluminium hydroxide? | null |
3,325 | aadf95af-6ddd-11ea-b069-ccda262736ce | https://socratic.org/questions/what-is-the-decomposition-of-hydrogen-peroxide-to-generate-oxygen | 2 H2O2(aq) ->[MnO2] 2 H2O(l) + O2(g) | start chemical_equation qc_end substance 5 6 qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the decomposition"}] | [{"type":"chemical equation","value":"2 H2O2(aq) ->[MnO2] 2 H2O(l) + O2(g)"}] | [{"type":"substance name","value":"hydrogen peroxide"},{"type":"substance name","value":"hydrogen peroxide"}] | <h1 class="questionTitle" itemprop="name">What is the decomposition of hydrogen peroxide to generate oxygen? </h1> | null | 2 H2O2(aq) ->[MnO2] 2 H2O(l) + O2(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Two moles of hydrogen peroxide <mathjax>#"H"_2 "O"_2#</mathjax> decomposes to produce two moles of water <mathjax>#"H"_2 "O"#</mathjax> and one mole of oxygen gas <mathjax>#"O"_2 (g)#</mathjax>, which then bubbles off. </p>
<p>Both heating and the presence of a heterolytic catalyst <mathjax>#"MnO"_2#</mathjax>- which provides an activation site for the reaction to proceed- can speed up this process.</p></div>
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<div class="markdown"><p><mathjax>#2 color(white)(l) "H"_2 "O"_2 (aq) stackrel("MnO"_2)(to) 2color(white)(l) "H"_2 "O" (l) + "O"_2 (g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Two moles of hydrogen peroxide <mathjax>#"H"_2 "O"_2#</mathjax> decomposes to produce two moles of water <mathjax>#"H"_2 "O"#</mathjax> and one mole of oxygen gas <mathjax>#"O"_2 (g)#</mathjax>, which then bubbles off. </p>
<p>Both heating and the presence of a heterolytic catalyst <mathjax>#"MnO"_2#</mathjax>- which provides an activation site for the reaction to proceed- can speed up this process.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the decomposition of hydrogen peroxide to generate oxygen? </h1>
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<div class="markdown"><p><mathjax>#2 color(white)(l) "H"_2 "O"_2 (aq) stackrel("MnO"_2)(to) 2color(white)(l) "H"_2 "O" (l) + "O"_2 (g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Two moles of hydrogen peroxide <mathjax>#"H"_2 "O"_2#</mathjax> decomposes to produce two moles of water <mathjax>#"H"_2 "O"#</mathjax> and one mole of oxygen gas <mathjax>#"O"_2 (g)#</mathjax>, which then bubbles off. </p>
<p>Both heating and the presence of a heterolytic catalyst <mathjax>#"MnO"_2#</mathjax>- which provides an activation site for the reaction to proceed- can speed up this process.</p></div>
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</article> | What is the decomposition of hydrogen peroxide to generate oxygen? | null |
3,326 | ab1b6dc8-6ddd-11ea-8800-ccda262736ce | https://socratic.org/questions/a-beaker-with-1-50-10-2-ml-of-acetic-acid-buffer-with-a-ph-of-5-000-is-sitting-o | 0.30 | start physical_unit 31 32 ph none qc_end physical_unit 9 11 3 6 volume qc_end physical_unit 9 11 16 16 ph qc_end physical_unit 31 32 34 35 molarity qc_end physical_unit 45 46 39 40 volume qc_end physical_unit 45 46 43 44 molarity qc_end physical_unit 9 10 62 62 pka qc_end end | [{"type":"physical unit","value":"Changed pH [OF] this buffer"}] | [{"type":"physical unit","value":"0.30"}] | [{"type":"physical unit","value":"Volume [OF] acetic acid buffer [=] \\pu{1.50 × 10^2 mL}"},{"type":"physical unit","value":"pH [OF] acetic acid buffer [=] \\pu{5.000}"},{"type":"physical unit","value":"Total molarity [OF] this buffer [=] \\pu{0.100 M}"},{"type":"physical unit","value":"Volume [OF] HCl solution [=] \\pu{7.00 mL}"},{"type":"physical unit","value":"Molarity [OF] HCl solution [=] \\pu{0.360 M}"},{"type":"physical unit","value":"pKa [OF] acetic acid [=] \\pu{4.740}"}] | <h1 class="questionTitle" itemprop="name">A beaker with #1.50 xx 10^2"mL"# of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 7.00 mL of a 0.360 M HCl solution to the beaker?</h1> | <div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>How much will the pH change? The pKa of acetic acid is 4.740.</p></div>
</h2>
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</div> | 0.30 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>!! VERY LONG ANSWER !!</strong></p>
<p>The first thing to do here is to figure out the <strong>concentrations</strong> of acetic acid and of acetate anions present in the buffer. </p>
<p>You know that a weak acid/conjugate base buffer has</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))))) ->#</mathjax> <em>the <strong>Henderson - Hasselbalch equation</strong></em></p>
</blockquote>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#5.000 = 4.740 + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#</mathjax></p>
</blockquote>
<p>Even without doing any calculations, you should be able to say that this buffer contains <strong>more conjugate base</strong> than weak acid. This is the case because the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the buffer is <strong>greater</strong> than the <mathjax>#"p"K_a#</mathjax> of the acid. </p>
<p>You can rewrite the equation as</p>
<blockquote>
<p><mathjax>#log((["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 5.000 - 4.740#</mathjax></p>
<p><mathjax>#log((["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 0.26#</mathjax></p>
</blockquote>
<p>This will be equivalent to</p>
<blockquote>
<p><mathjax>#10^log((["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 10^0.26#</mathjax></p>
<p><mathjax>#(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = 1.8197#</mathjax></p>
</blockquote>
<p>This implies that</p>
<blockquote>
<p><mathjax>#["CH"_3"COO"^(-)] = 1.8197 * ["CH"_3"COOH"]#</mathjax></p>
</blockquote>
<p>You also know that the <strong>total <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of weak acid and conjugate base is equal to <mathjax>#"0.100 M"#</mathjax>, so you can say that</p>
<blockquote>
<p><mathjax>#["CH"_3"COOH"] + ["CH"_3"COO"^(-)] = "0.100 M"#</mathjax></p>
</blockquote>
<p>You now have two equations with two unknowns. For simplicity, I'll use</p>
<blockquote>
<p><mathjax>#["CH"_3"COOH"] = x" "#</mathjax> and <mathjax>#" "["CH"_3"COO"^(-)] = y#</mathjax></p>
</blockquote>
<p>You have</p>
<blockquote>
<p><mathjax>#x =0.100 - y#</mathjax></p>
</blockquote>
<p>Plug this into the fist equation to get</p>
<blockquote>
<p><mathjax>#y = 1.8197 * (0.100 - y)#</mathjax></p>
</blockquote>
<p>Rearrange to get</p>
<blockquote>
<p><mathjax>#2.8197 * y = 0.18197#</mathjax></p>
</blockquote>
<p>This will be equivalent to</p>
<blockquote>
<p><mathjax>#y = 0.18197/2.8197 = 0.06454#</mathjax></p>
</blockquote>
<p>This means that </p>
<blockquote>
<p><mathjax>#x = 0.100 - 0.06454 = 0.03546#</mathjax></p>
</blockquote>
<p>Therefore, you can say that the buffer contains</p>
<blockquote>
<p><mathjax>#["CH"_3"COOH"] = "0.03546 M"" "#</mathjax> and <mathjax>#" " ["CH"_3"COO"^(-)] = "0.06454 M"#</mathjax></p>
</blockquote>
<p>As predicted, the buffer contains <strong>more conjugate base</strong> than weak acid. </p>
<p>Now, you are adding </p>
<blockquote>
<p><mathjax>#7.00 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.360 moles H"_3"O"^(+))/(1color(red)(cancel(color(black)("L"))))="0.002520 moles H"_3"O"^(+)#</mathjax></p>
</blockquote>
<p>to the buffer by way of the hydrochloric acid solution. The hydronium cations will react with the acetate anions to produce acetic acid and water</p>
<blockquote>
<p><mathjax>#"CH"_ 3"COO"_ ((aq))^(-) + "H"_ 3"O"_ ((aq))^(+) -> "CH"_ 3"COOH"_ ((aq)) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>Use the molarity of the acetate anions and the volume of the buffer to calculate the <strong>number of moles</strong> of conjugate base and of weak acid present in the buffer</p>
<blockquote>
<p><mathjax>#1.50 * 10^2 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.06454 moles CH"_3"COO"^(-))/(1color(red)(cancel(color(black)("L"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = "0.009681 moles CH"_3"COO"^(-)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p><mathjax>#1.50 * 10^2 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.03546 moles CH"_3"COOH")/(1color(red)(cancel(color(black)("L"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = "0.005319 moles CH"_3"COOH"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>The reaction will consume hydronium cations and acetate anions in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> and produce acetic acid in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong>, so you can say that <em>after</em> the <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> is complete, the solution will contain </p>
<blockquote>
<p><mathjax>#n_ ("H"_ 3"O"^(+)) = "0 moles" ->#</mathjax> <em><strong>completely consumed</strong></em></p>
<p><mathjax>#n_ ("CH"_ 3"COO"^(-)) = "0.009681 moles" - "0.002520 moles"#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = "0.007161 moles CH"_3"COO"^(-)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p><mathjax>#n_( "CH"_3 "COOH") = "0.005319 moles" + "0.002520 moles"#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = "0.007839 moles CH"_3"COOH"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>The <strong>total volume</strong> of the solution will be</p>
<blockquote>
<p><mathjax>#V_"total" = 1.50 * 10^2"mL" + "7.00 mL" = "157 mL"#</mathjax></p>
</blockquote>
<p>The new concentrations of the weak acid and of the conjugate base will be </p>
<blockquote>
<p><mathjax>#["CH"_3"COOH"] = "0.007839 moles"/(157 * 10^(-3)"L") = "0.04993 M"#</mathjax></p>
<p><mathjax>#["CH"_3"COO"^(-)] = "0.007161 moles"/(157 * 10^(-3)"L") = "0.04561M"#</mathjax></p>
</blockquote>
<p>Finally, use the <strong>Henderson - Hasselbalch equation</strong> to calculate the new pH of the solution </p>
<blockquote>
<p><mathjax>#"pH" = 4.740 + log( (0.04561 color(red)(cancel(color(black)("M"))))/(0.04993color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#"pH" = 4.701#</mathjax></p>
</blockquote>
<p>Therefore, the pH of the solution changed by </p>
<blockquote>
<p><mathjax>#Delta_"pH" = |4.701 - 5.000| = color(darkgreen)(ul(color(black)(0.299)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <em>decimal places</em>, the number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> you have for your volumes and molarities. </p>
<p><em>So, does the result make sense?</em></p>
<p>You added hydrochloric acid, a <strong>strong acid</strong>, to the buffer, so you should expect its pH to <strong>decrease</strong> as a result. </p>
<p>Notice that the final concentration of the weak acid is <strong>greater</strong> than that of the weak base, which is why you have </p>
<blockquote>
<p><mathjax>#4.701 < 4.740" "#</mathjax> or <mathjax>#" ""pH" < "p"K_a#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#Delta_"pH" = 0.299#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>!! VERY LONG ANSWER !!</strong></p>
<p>The first thing to do here is to figure out the <strong>concentrations</strong> of acetic acid and of acetate anions present in the buffer. </p>
<p>You know that a weak acid/conjugate base buffer has</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))))) ->#</mathjax> <em>the <strong>Henderson - Hasselbalch equation</strong></em></p>
</blockquote>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#5.000 = 4.740 + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#</mathjax></p>
</blockquote>
<p>Even without doing any calculations, you should be able to say that this buffer contains <strong>more conjugate base</strong> than weak acid. This is the case because the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the buffer is <strong>greater</strong> than the <mathjax>#"p"K_a#</mathjax> of the acid. </p>
<p>You can rewrite the equation as</p>
<blockquote>
<p><mathjax>#log((["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 5.000 - 4.740#</mathjax></p>
<p><mathjax>#log((["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 0.26#</mathjax></p>
</blockquote>
<p>This will be equivalent to</p>
<blockquote>
<p><mathjax>#10^log((["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 10^0.26#</mathjax></p>
<p><mathjax>#(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = 1.8197#</mathjax></p>
</blockquote>
<p>This implies that</p>
<blockquote>
<p><mathjax>#["CH"_3"COO"^(-)] = 1.8197 * ["CH"_3"COOH"]#</mathjax></p>
</blockquote>
<p>You also know that the <strong>total <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of weak acid and conjugate base is equal to <mathjax>#"0.100 M"#</mathjax>, so you can say that</p>
<blockquote>
<p><mathjax>#["CH"_3"COOH"] + ["CH"_3"COO"^(-)] = "0.100 M"#</mathjax></p>
</blockquote>
<p>You now have two equations with two unknowns. For simplicity, I'll use</p>
<blockquote>
<p><mathjax>#["CH"_3"COOH"] = x" "#</mathjax> and <mathjax>#" "["CH"_3"COO"^(-)] = y#</mathjax></p>
</blockquote>
<p>You have</p>
<blockquote>
<p><mathjax>#x =0.100 - y#</mathjax></p>
</blockquote>
<p>Plug this into the fist equation to get</p>
<blockquote>
<p><mathjax>#y = 1.8197 * (0.100 - y)#</mathjax></p>
</blockquote>
<p>Rearrange to get</p>
<blockquote>
<p><mathjax>#2.8197 * y = 0.18197#</mathjax></p>
</blockquote>
<p>This will be equivalent to</p>
<blockquote>
<p><mathjax>#y = 0.18197/2.8197 = 0.06454#</mathjax></p>
</blockquote>
<p>This means that </p>
<blockquote>
<p><mathjax>#x = 0.100 - 0.06454 = 0.03546#</mathjax></p>
</blockquote>
<p>Therefore, you can say that the buffer contains</p>
<blockquote>
<p><mathjax>#["CH"_3"COOH"] = "0.03546 M"" "#</mathjax> and <mathjax>#" " ["CH"_3"COO"^(-)] = "0.06454 M"#</mathjax></p>
</blockquote>
<p>As predicted, the buffer contains <strong>more conjugate base</strong> than weak acid. </p>
<p>Now, you are adding </p>
<blockquote>
<p><mathjax>#7.00 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.360 moles H"_3"O"^(+))/(1color(red)(cancel(color(black)("L"))))="0.002520 moles H"_3"O"^(+)#</mathjax></p>
</blockquote>
<p>to the buffer by way of the hydrochloric acid solution. The hydronium cations will react with the acetate anions to produce acetic acid and water</p>
<blockquote>
<p><mathjax>#"CH"_ 3"COO"_ ((aq))^(-) + "H"_ 3"O"_ ((aq))^(+) -> "CH"_ 3"COOH"_ ((aq)) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>Use the molarity of the acetate anions and the volume of the buffer to calculate the <strong>number of moles</strong> of conjugate base and of weak acid present in the buffer</p>
<blockquote>
<p><mathjax>#1.50 * 10^2 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.06454 moles CH"_3"COO"^(-))/(1color(red)(cancel(color(black)("L"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = "0.009681 moles CH"_3"COO"^(-)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p><mathjax>#1.50 * 10^2 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.03546 moles CH"_3"COOH")/(1color(red)(cancel(color(black)("L"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = "0.005319 moles CH"_3"COOH"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>The reaction will consume hydronium cations and acetate anions in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> and produce acetic acid in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong>, so you can say that <em>after</em> the <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> is complete, the solution will contain </p>
<blockquote>
<p><mathjax>#n_ ("H"_ 3"O"^(+)) = "0 moles" ->#</mathjax> <em><strong>completely consumed</strong></em></p>
<p><mathjax>#n_ ("CH"_ 3"COO"^(-)) = "0.009681 moles" - "0.002520 moles"#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = "0.007161 moles CH"_3"COO"^(-)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p><mathjax>#n_( "CH"_3 "COOH") = "0.005319 moles" + "0.002520 moles"#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = "0.007839 moles CH"_3"COOH"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>The <strong>total volume</strong> of the solution will be</p>
<blockquote>
<p><mathjax>#V_"total" = 1.50 * 10^2"mL" + "7.00 mL" = "157 mL"#</mathjax></p>
</blockquote>
<p>The new concentrations of the weak acid and of the conjugate base will be </p>
<blockquote>
<p><mathjax>#["CH"_3"COOH"] = "0.007839 moles"/(157 * 10^(-3)"L") = "0.04993 M"#</mathjax></p>
<p><mathjax>#["CH"_3"COO"^(-)] = "0.007161 moles"/(157 * 10^(-3)"L") = "0.04561M"#</mathjax></p>
</blockquote>
<p>Finally, use the <strong>Henderson - Hasselbalch equation</strong> to calculate the new pH of the solution </p>
<blockquote>
<p><mathjax>#"pH" = 4.740 + log( (0.04561 color(red)(cancel(color(black)("M"))))/(0.04993color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#"pH" = 4.701#</mathjax></p>
</blockquote>
<p>Therefore, the pH of the solution changed by </p>
<blockquote>
<p><mathjax>#Delta_"pH" = |4.701 - 5.000| = color(darkgreen)(ul(color(black)(0.299)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <em>decimal places</em>, the number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> you have for your volumes and molarities. </p>
<p><em>So, does the result make sense?</em></p>
<p>You added hydrochloric acid, a <strong>strong acid</strong>, to the buffer, so you should expect its pH to <strong>decrease</strong> as a result. </p>
<p>Notice that the final concentration of the weak acid is <strong>greater</strong> than that of the weak base, which is why you have </p>
<blockquote>
<p><mathjax>#4.701 < 4.740" "#</mathjax> or <mathjax>#" ""pH" < "p"K_a#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A beaker with #1.50 xx 10^2"mL"# of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 7.00 mL of a 0.360 M HCl solution to the beaker?</h1>
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<div class="markdown"><p>How much will the pH change? The pKa of acetic acid is 4.740.</p></div>
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Stefan V.
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<span class="dateCreated" datetime="2017-03-22T00:28:51" itemprop="dateCreated">
Mar 22, 2017
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<div>
<div class="markdown"><p><mathjax>#Delta_"pH" = 0.299#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>!! VERY LONG ANSWER !!</strong></p>
<p>The first thing to do here is to figure out the <strong>concentrations</strong> of acetic acid and of acetate anions present in the buffer. </p>
<p>You know that a weak acid/conjugate base buffer has</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))))) ->#</mathjax> <em>the <strong>Henderson - Hasselbalch equation</strong></em></p>
</blockquote>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#5.000 = 4.740 + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#</mathjax></p>
</blockquote>
<p>Even without doing any calculations, you should be able to say that this buffer contains <strong>more conjugate base</strong> than weak acid. This is the case because the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the buffer is <strong>greater</strong> than the <mathjax>#"p"K_a#</mathjax> of the acid. </p>
<p>You can rewrite the equation as</p>
<blockquote>
<p><mathjax>#log((["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 5.000 - 4.740#</mathjax></p>
<p><mathjax>#log((["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 0.26#</mathjax></p>
</blockquote>
<p>This will be equivalent to</p>
<blockquote>
<p><mathjax>#10^log((["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 10^0.26#</mathjax></p>
<p><mathjax>#(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = 1.8197#</mathjax></p>
</blockquote>
<p>This implies that</p>
<blockquote>
<p><mathjax>#["CH"_3"COO"^(-)] = 1.8197 * ["CH"_3"COOH"]#</mathjax></p>
</blockquote>
<p>You also know that the <strong>total <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of weak acid and conjugate base is equal to <mathjax>#"0.100 M"#</mathjax>, so you can say that</p>
<blockquote>
<p><mathjax>#["CH"_3"COOH"] + ["CH"_3"COO"^(-)] = "0.100 M"#</mathjax></p>
</blockquote>
<p>You now have two equations with two unknowns. For simplicity, I'll use</p>
<blockquote>
<p><mathjax>#["CH"_3"COOH"] = x" "#</mathjax> and <mathjax>#" "["CH"_3"COO"^(-)] = y#</mathjax></p>
</blockquote>
<p>You have</p>
<blockquote>
<p><mathjax>#x =0.100 - y#</mathjax></p>
</blockquote>
<p>Plug this into the fist equation to get</p>
<blockquote>
<p><mathjax>#y = 1.8197 * (0.100 - y)#</mathjax></p>
</blockquote>
<p>Rearrange to get</p>
<blockquote>
<p><mathjax>#2.8197 * y = 0.18197#</mathjax></p>
</blockquote>
<p>This will be equivalent to</p>
<blockquote>
<p><mathjax>#y = 0.18197/2.8197 = 0.06454#</mathjax></p>
</blockquote>
<p>This means that </p>
<blockquote>
<p><mathjax>#x = 0.100 - 0.06454 = 0.03546#</mathjax></p>
</blockquote>
<p>Therefore, you can say that the buffer contains</p>
<blockquote>
<p><mathjax>#["CH"_3"COOH"] = "0.03546 M"" "#</mathjax> and <mathjax>#" " ["CH"_3"COO"^(-)] = "0.06454 M"#</mathjax></p>
</blockquote>
<p>As predicted, the buffer contains <strong>more conjugate base</strong> than weak acid. </p>
<p>Now, you are adding </p>
<blockquote>
<p><mathjax>#7.00 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.360 moles H"_3"O"^(+))/(1color(red)(cancel(color(black)("L"))))="0.002520 moles H"_3"O"^(+)#</mathjax></p>
</blockquote>
<p>to the buffer by way of the hydrochloric acid solution. The hydronium cations will react with the acetate anions to produce acetic acid and water</p>
<blockquote>
<p><mathjax>#"CH"_ 3"COO"_ ((aq))^(-) + "H"_ 3"O"_ ((aq))^(+) -> "CH"_ 3"COOH"_ ((aq)) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>Use the molarity of the acetate anions and the volume of the buffer to calculate the <strong>number of moles</strong> of conjugate base and of weak acid present in the buffer</p>
<blockquote>
<p><mathjax>#1.50 * 10^2 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.06454 moles CH"_3"COO"^(-))/(1color(red)(cancel(color(black)("L"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = "0.009681 moles CH"_3"COO"^(-)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p><mathjax>#1.50 * 10^2 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.03546 moles CH"_3"COOH")/(1color(red)(cancel(color(black)("L"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = "0.005319 moles CH"_3"COOH"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>The reaction will consume hydronium cations and acetate anions in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> and produce acetic acid in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong>, so you can say that <em>after</em> the <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> is complete, the solution will contain </p>
<blockquote>
<p><mathjax>#n_ ("H"_ 3"O"^(+)) = "0 moles" ->#</mathjax> <em><strong>completely consumed</strong></em></p>
<p><mathjax>#n_ ("CH"_ 3"COO"^(-)) = "0.009681 moles" - "0.002520 moles"#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = "0.007161 moles CH"_3"COO"^(-)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p><mathjax>#n_( "CH"_3 "COOH") = "0.005319 moles" + "0.002520 moles"#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = "0.007839 moles CH"_3"COOH"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>The <strong>total volume</strong> of the solution will be</p>
<blockquote>
<p><mathjax>#V_"total" = 1.50 * 10^2"mL" + "7.00 mL" = "157 mL"#</mathjax></p>
</blockquote>
<p>The new concentrations of the weak acid and of the conjugate base will be </p>
<blockquote>
<p><mathjax>#["CH"_3"COOH"] = "0.007839 moles"/(157 * 10^(-3)"L") = "0.04993 M"#</mathjax></p>
<p><mathjax>#["CH"_3"COO"^(-)] = "0.007161 moles"/(157 * 10^(-3)"L") = "0.04561M"#</mathjax></p>
</blockquote>
<p>Finally, use the <strong>Henderson - Hasselbalch equation</strong> to calculate the new pH of the solution </p>
<blockquote>
<p><mathjax>#"pH" = 4.740 + log( (0.04561 color(red)(cancel(color(black)("M"))))/(0.04993color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#"pH" = 4.701#</mathjax></p>
</blockquote>
<p>Therefore, the pH of the solution changed by </p>
<blockquote>
<p><mathjax>#Delta_"pH" = |4.701 - 5.000| = color(darkgreen)(ul(color(black)(0.299)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <em>decimal places</em>, the number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> you have for your volumes and molarities. </p>
<p><em>So, does the result make sense?</em></p>
<p>You added hydrochloric acid, a <strong>strong acid</strong>, to the buffer, so you should expect its pH to <strong>decrease</strong> as a result. </p>
<p>Notice that the final concentration of the weak acid is <strong>greater</strong> than that of the weak base, which is why you have </p>
<blockquote>
<p><mathjax>#4.701 < 4.740" "#</mathjax> or <mathjax>#" ""pH" < "p"K_a#</mathjax></p>
</blockquote></div>
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</article> | A beaker with #1.50 xx 10^2"mL"# of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 7.00 mL of a 0.360 M HCl solution to the beaker? |
How much will the pH change? The pKa of acetic acid is 4.740.
|
3,327 | a83de969-6ddd-11ea-a9b6-ccda262736ce | https://socratic.org/questions/if-the-volume-of-the-gas-in-a-gas-chamber-is-500-ml-at-277-k-what-will-be-the-vo | 541.5 mL | start physical_unit 4 5 volume ml qc_end physical_unit 4 5 11 12 volume qc_end physical_unit 4 5 14 15 temperature qc_end physical_unit 4 5 32 33 temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] mL"}] | [{"type":"physical unit","value":"541.5 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{500 mL}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{277 K}"},{"type":"physical unit","value":"Temperature2 [OF] the gas [=] \\pu{300 K}"}] | <h1 class="questionTitle" itemprop="name">If the volume of the gas in a gas chamber is 500 mL at 277 K, what will be the volume of the gas in the chamber if the temperature is at 300 K?</h1> | null | 541.5 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Apply <a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a></p>
<p><mathjax>#"Volume"/"Temperature" = "Constant"#</mathjax> at <mathjax>#"constant pressure"#</mathjax></p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p>The initial volume is <mathjax>#V_1=500mL#</mathjax></p>
<p>The initial temperature is <mathjax>#T_1=277K#</mathjax></p>
<p>The final temperature is <mathjax>#T_2=300K#</mathjax></p>
<p>The final volume is</p>
<p><mathjax>#V_2=T_2/T_1*V_1=300/277*500=541.5mL#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The final volume is <mathjax>#=541.5mL#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Apply <a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a></p>
<p><mathjax>#"Volume"/"Temperature" = "Constant"#</mathjax> at <mathjax>#"constant pressure"#</mathjax></p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p>The initial volume is <mathjax>#V_1=500mL#</mathjax></p>
<p>The initial temperature is <mathjax>#T_1=277K#</mathjax></p>
<p>The final temperature is <mathjax>#T_2=300K#</mathjax></p>
<p>The final volume is</p>
<p><mathjax>#V_2=T_2/T_1*V_1=300/277*500=541.5mL#</mathjax></p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">If the volume of the gas in a gas chamber is 500 mL at 277 K, what will be the volume of the gas in the chamber if the temperature is at 300 K?</h1>
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<div class="markdown"><p>The final volume is <mathjax>#=541.5mL#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Apply <a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a></p>
<p><mathjax>#"Volume"/"Temperature" = "Constant"#</mathjax> at <mathjax>#"constant pressure"#</mathjax></p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p>The initial volume is <mathjax>#V_1=500mL#</mathjax></p>
<p>The initial temperature is <mathjax>#T_1=277K#</mathjax></p>
<p>The final temperature is <mathjax>#T_2=300K#</mathjax></p>
<p>The final volume is</p>
<p><mathjax>#V_2=T_2/T_1*V_1=300/277*500=541.5mL#</mathjax></p></div>
</div>
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</article> | If the volume of the gas in a gas chamber is 500 mL at 277 K, what will be the volume of the gas in the chamber if the temperature is at 300 K? | null |
3,328 | a9b17b46-6ddd-11ea-9722-ccda262736ce | https://socratic.org/questions/a-82-7-g-sample-of-dinitrogen-monoxide-is-confined-in-a-2-0-l-vessel-what-is-the | 29.93 atm | start physical_unit 3 6 pressure atm qc_end physical_unit 3 6 1 2 mass qc_end physical_unit 13 13 11 12 volume qc_end physical_unit 3 6 21 22 temperature qc_end end | [{"type":"physical unit","value":"Pressure [OF] dinitrogen monoxide sample [IN] atm"}] | [{"type":"physical unit","value":"29.93 atm"}] | [{"type":"physical unit","value":"Mass [OF] dinitrogen monoxide sample [=] \\pu{82.7 g}"},{"type":"physical unit","value":"Volume [OF] vessel [=] \\pu{2.0 L}"},{"type":"physical unit","value":"Temperature [OF] dinitrogen monoxide sample [=] \\pu{115 ℃}"}] | <h1 class="questionTitle" itemprop="name">A 82.7 g sample of dinitrogen monoxide is confined in a 2.0 L vessel, what is the pressure (in atm) at 115°C?</h1> | null | 29.93 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The only challenge to this problem is selecting the appropriate gas constant; here, we choose <mathjax>#R = 0.0821*L*atm*K^(-1)*mol^(-1)#</mathjax>. This gas constant (and others) would be supplied. Of course, we've got to choose the right one for the job!</p>
<p>So, <mathjax>#P = ((82.7*g)/(44.01*g*mol^-1)xx0.0821*L*atm*K^(-1)*mol^-1xx388K)/(2.0*L) = ?? atm#</mathjax></p>
<p>You will note that we have converted the temperature to absolute.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#P = (nRT)/V#</mathjax>; we assume (reasonably?) ideal gas behaviour for <mathjax>#N_2O#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The only challenge to this problem is selecting the appropriate gas constant; here, we choose <mathjax>#R = 0.0821*L*atm*K^(-1)*mol^(-1)#</mathjax>. This gas constant (and others) would be supplied. Of course, we've got to choose the right one for the job!</p>
<p>So, <mathjax>#P = ((82.7*g)/(44.01*g*mol^-1)xx0.0821*L*atm*K^(-1)*mol^-1xx388K)/(2.0*L) = ?? atm#</mathjax></p>
<p>You will note that we have converted the temperature to absolute.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A 82.7 g sample of dinitrogen monoxide is confined in a 2.0 L vessel, what is the pressure (in atm) at 115°C?</h1>
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anor277
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Nov 2, 2015
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<div class="markdown"><p><mathjax>#P = (nRT)/V#</mathjax>; we assume (reasonably?) ideal gas behaviour for <mathjax>#N_2O#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The only challenge to this problem is selecting the appropriate gas constant; here, we choose <mathjax>#R = 0.0821*L*atm*K^(-1)*mol^(-1)#</mathjax>. This gas constant (and others) would be supplied. Of course, we've got to choose the right one for the job!</p>
<p>So, <mathjax>#P = ((82.7*g)/(44.01*g*mol^-1)xx0.0821*L*atm*K^(-1)*mol^-1xx388K)/(2.0*L) = ?? atm#</mathjax></p>
<p>You will note that we have converted the temperature to absolute.</p></div>
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</article> | A 82.7 g sample of dinitrogen monoxide is confined in a 2.0 L vessel, what is the pressure (in atm) at 115°C? | null |
3,329 | aa391e7d-6ddd-11ea-9cee-ccda262736ce | https://socratic.org/questions/if-a-gas-pressure-gauge-reads-15-mm-hg-what-is-the-pressure-in-atmospheres | 0.02 atmospheres | start physical_unit 2 2 pressure atm qc_end physical_unit 2 2 6 7 pressure qc_end end | [{"type":"physical unit","value":"Pressure [OF] the gas [IN] atmospheres"}] | [{"type":"physical unit","value":"0.02 atmospheres"}] | [{"type":"physical unit","value":"Pressure [OF] the gas [=] \\pu{15 mmHg}"}] | <h1 class="questionTitle" itemprop="name">If a gas pressure gauge reads 15 mm Hg, what is the pressure in atmospheres? </h1> | null | 0.02 atmospheres | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>760 mm Hg = 1 atmosphere. </p>
<p><mathjax># 15/760 = atmosphere#</mathjax></p>
<p># 15/760 = 0.0197368 atmospheres rounding off to 2 significant digits gives </p>
<p>0.020 atmospheres </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>0,020 atmospheres </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>760 mm Hg = 1 atmosphere. </p>
<p><mathjax># 15/760 = atmosphere#</mathjax></p>
<p># 15/760 = 0.0197368 atmospheres rounding off to 2 significant digits gives </p>
<p>0.020 atmospheres </p></div>
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<h1 class="questionTitle" itemprop="name">If a gas pressure gauge reads 15 mm Hg, what is the pressure in atmospheres? </h1>
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<div class="markdown"><p>0,020 atmospheres </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>760 mm Hg = 1 atmosphere. </p>
<p><mathjax># 15/760 = atmosphere#</mathjax></p>
<p># 15/760 = 0.0197368 atmospheres rounding off to 2 significant digits gives </p>
<p>0.020 atmospheres </p></div>
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Nathan L.
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Jun 12, 2017
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<div class="markdown"><p><mathjax>#0.020#</mathjax> <mathjax>#"atm"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To convert between millimeters Mercury (<mathjax>#"mm Hg"#</mathjax>) and atmospheres (<mathjax>#"atm"#</mathjax>), we use the conversion factor</p>
<p><mathjax>#(1"atm")/(760"mm Hg")#</mathjax></p>
<p>That is, <mathjax>#1#</mathjax> atmosphere equals <mathjax>#760#</mathjax> millimeters Mercury.</p>
<p>Using <em>dimensional analysis</em>, we can convert the given pressure:</p>
<p><mathjax>#15#</mathjax> <mathjax>#cancel("mm Hg")((1"atm")/(760cancel("mm Hg"))) = color(red)(0.020#</mathjax> <mathjax>#color(red)("atm"#</mathjax></p>
<p><mathjax>#15#</mathjax> <mathjax>#"mm Hg"#</mathjax> is equal to <mathjax>#0.020#</mathjax> <mathjax>#"atm"#</mathjax>, rounded to <mathjax>#color(blue)(2#</mathjax> <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>, the amount given in the problem.</p></div>
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</article> | If a gas pressure gauge reads 15 mm Hg, what is the pressure in atmospheres? | null |
3,330 | a951d1ae-6ddd-11ea-adc2-ccda262736ce | https://socratic.org/questions/what-is-the-net-ionic-equation-for-the-reaction-of-aqueous-lead-ii-nitrate-with- | Pb^2+(aq) + 2 Br-(aq) -> PbBr2(s) | start chemical_equation qc_end end | [{"type":"other","value":"Chemical Equation [OF] the net ionic equation"}] | [{"type":"chemical equation","value":"Pb^2+(aq) + 2 Br-(aq) -> PbBr2(s)"}] | [{"type":"substance name","value":"lead(II) nitrate aqueous"},{"type":"substance name","value":"sodium bromide aqueous"}] | <h1 class="questionTitle" itemprop="name">What is the net ionic equation for the reaction of aqueous lead(II) nitrate with aqueous sodium bromide?</h1> | null | Pb^2+(aq) + 2 Br-(aq) -> PbBr2(s) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Lead(II) nitrate, <mathjax>#"Pb"("NO"_3)_2#</mathjax>, and sodium bromide, <mathjax>#"NaBr"#</mathjax>, are <strong>soluble</strong> in aqueous solution, which means that they dissociate completely to form cations and anions when dissolved in water.</p>
<blockquote>
<p><mathjax>#"Pb"("NO" _ 3)_ (2(aq)) -> "Pb"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-)#</mathjax></p>
<p><mathjax>#"NaBr"_ ((aq)) -> "Na"_ ((aq))^(+) +"Br"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>When these two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are mixed, lead(II) bromide, <mathjax>#"PbBr"_2#</mathjax>, an <strong>insoluble</strong> ionic compound, and sodium nitrate, <mathjax>#"NaNO"_3#</mathjax>, another soluble ionic compound, will be formed.</p>
<p>The <strong>complete ionic equation</strong> will look like this</p>
<blockquote>
<p><mathjax>#"Pb"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-) + 2 xx ["Na"_ ((aq))^(+) +"Br"_ ((aq))^(-)] -> "PbBr"_ (2(s)) darr + 2xx ["Na" _ ((aq))^(+) + "NO"_ (3(aq))^(-)]#</mathjax></p>
</blockquote>
<p>Notice that the nitrate anions, <mathjax>#"NO"_3^(-)#</mathjax>, and the sodium cations, <mathjax>#"Na"^(+)#</mathjax>, exist as <em>ions</em> on both sides of the equation. </p>
<p>This tells you that these two chemical species act as <em>spectator ions</em>. In order to write the <strong>net ionic equation</strong>, you remove the spectator ions and focus on the insoluble solid and the ions that form it</p>
<blockquote>
<p><mathjax>#"Pb"_ ((aq))^(2+) + color(blue)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + color(red)(cancel(color(Black)(2"NO"_ (3(aq))^(-)))) + 2"Br"_ ((aq))^(-) -> "PbBr"_ (2(s)) darr + color(blue)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + color(red)(cancel(color(Black)(2"NO"_ (3(aq))^(-))))#</mathjax></p>
</blockquote>
<p>The net ionic equation fo this reaction will thus be</p>
<blockquote>
<p><mathjax>#"Pb"_ ((aq))^(2+) + 2"Br"_ ((aq))^(-) -> "PbBr"_ (2(s)) darr#</mathjax></p>
</blockquote>
<p>Lead(II) bromide is a <em>white</em> insoluble ionic compound that precipitates out of solution. </p>
<p>
<iframe src="https://www.youtube.com/embed/zk1wWWzObJo?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#"Pb"_ ((aq))^(2+) + 2"Br"_ ((aq))^(-) -> "PbBr"_ (2(s)) darr#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Lead(II) nitrate, <mathjax>#"Pb"("NO"_3)_2#</mathjax>, and sodium bromide, <mathjax>#"NaBr"#</mathjax>, are <strong>soluble</strong> in aqueous solution, which means that they dissociate completely to form cations and anions when dissolved in water.</p>
<blockquote>
<p><mathjax>#"Pb"("NO" _ 3)_ (2(aq)) -> "Pb"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-)#</mathjax></p>
<p><mathjax>#"NaBr"_ ((aq)) -> "Na"_ ((aq))^(+) +"Br"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>When these two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are mixed, lead(II) bromide, <mathjax>#"PbBr"_2#</mathjax>, an <strong>insoluble</strong> ionic compound, and sodium nitrate, <mathjax>#"NaNO"_3#</mathjax>, another soluble ionic compound, will be formed.</p>
<p>The <strong>complete ionic equation</strong> will look like this</p>
<blockquote>
<p><mathjax>#"Pb"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-) + 2 xx ["Na"_ ((aq))^(+) +"Br"_ ((aq))^(-)] -> "PbBr"_ (2(s)) darr + 2xx ["Na" _ ((aq))^(+) + "NO"_ (3(aq))^(-)]#</mathjax></p>
</blockquote>
<p>Notice that the nitrate anions, <mathjax>#"NO"_3^(-)#</mathjax>, and the sodium cations, <mathjax>#"Na"^(+)#</mathjax>, exist as <em>ions</em> on both sides of the equation. </p>
<p>This tells you that these two chemical species act as <em>spectator ions</em>. In order to write the <strong>net ionic equation</strong>, you remove the spectator ions and focus on the insoluble solid and the ions that form it</p>
<blockquote>
<p><mathjax>#"Pb"_ ((aq))^(2+) + color(blue)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + color(red)(cancel(color(Black)(2"NO"_ (3(aq))^(-)))) + 2"Br"_ ((aq))^(-) -> "PbBr"_ (2(s)) darr + color(blue)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + color(red)(cancel(color(Black)(2"NO"_ (3(aq))^(-))))#</mathjax></p>
</blockquote>
<p>The net ionic equation fo this reaction will thus be</p>
<blockquote>
<p><mathjax>#"Pb"_ ((aq))^(2+) + 2"Br"_ ((aq))^(-) -> "PbBr"_ (2(s)) darr#</mathjax></p>
</blockquote>
<p>Lead(II) bromide is a <em>white</em> insoluble ionic compound that precipitates out of solution. </p>
<p>
<iframe src="https://www.youtube.com/embed/zk1wWWzObJo?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<h1 class="questionTitle" itemprop="name">What is the net ionic equation for the reaction of aqueous lead(II) nitrate with aqueous sodium bromide?</h1>
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<div class="markdown"><p><mathjax>#"Pb"_ ((aq))^(2+) + 2"Br"_ ((aq))^(-) -> "PbBr"_ (2(s)) darr#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Lead(II) nitrate, <mathjax>#"Pb"("NO"_3)_2#</mathjax>, and sodium bromide, <mathjax>#"NaBr"#</mathjax>, are <strong>soluble</strong> in aqueous solution, which means that they dissociate completely to form cations and anions when dissolved in water.</p>
<blockquote>
<p><mathjax>#"Pb"("NO" _ 3)_ (2(aq)) -> "Pb"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-)#</mathjax></p>
<p><mathjax>#"NaBr"_ ((aq)) -> "Na"_ ((aq))^(+) +"Br"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>When these two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are mixed, lead(II) bromide, <mathjax>#"PbBr"_2#</mathjax>, an <strong>insoluble</strong> ionic compound, and sodium nitrate, <mathjax>#"NaNO"_3#</mathjax>, another soluble ionic compound, will be formed.</p>
<p>The <strong>complete ionic equation</strong> will look like this</p>
<blockquote>
<p><mathjax>#"Pb"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-) + 2 xx ["Na"_ ((aq))^(+) +"Br"_ ((aq))^(-)] -> "PbBr"_ (2(s)) darr + 2xx ["Na" _ ((aq))^(+) + "NO"_ (3(aq))^(-)]#</mathjax></p>
</blockquote>
<p>Notice that the nitrate anions, <mathjax>#"NO"_3^(-)#</mathjax>, and the sodium cations, <mathjax>#"Na"^(+)#</mathjax>, exist as <em>ions</em> on both sides of the equation. </p>
<p>This tells you that these two chemical species act as <em>spectator ions</em>. In order to write the <strong>net ionic equation</strong>, you remove the spectator ions and focus on the insoluble solid and the ions that form it</p>
<blockquote>
<p><mathjax>#"Pb"_ ((aq))^(2+) + color(blue)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + color(red)(cancel(color(Black)(2"NO"_ (3(aq))^(-)))) + 2"Br"_ ((aq))^(-) -> "PbBr"_ (2(s)) darr + color(blue)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + color(red)(cancel(color(Black)(2"NO"_ (3(aq))^(-))))#</mathjax></p>
</blockquote>
<p>The net ionic equation fo this reaction will thus be</p>
<blockquote>
<p><mathjax>#"Pb"_ ((aq))^(2+) + 2"Br"_ ((aq))^(-) -> "PbBr"_ (2(s)) darr#</mathjax></p>
</blockquote>
<p>Lead(II) bromide is a <em>white</em> insoluble ionic compound that precipitates out of solution. </p>
<p>
<iframe src="https://www.youtube.com/embed/zk1wWWzObJo?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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</article> | What is the net ionic equation for the reaction of aqueous lead(II) nitrate with aqueous sodium bromide? | null |
3,331 | ab06bf73-6ddd-11ea-9fa4-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-the-ionic-compound-composed-of-the-calcium-ion-and-the-p | Ca3(PO4)2 | start chemical_formula qc_end substance 11 12 qc_end substance 15 16 qc_end end | [{"type":"other","value":"Chemical Formula [OF] ionic compound [IN] default"}] | [{"type":"chemical equation","value":"Ca3(PO4)2"}] | [{"type":"substance name","value":"Calcium ion"},{"type":"substance name","value":"Phosphate ion"}] | <h1 class="questionTitle" itemprop="name">What is the formula for the ionic compound composed of the calcium ion and the phosphate ion?</h1> | null | Ca3(PO4)2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Well, the name will be simple, as by <a href="https://socratic.org/chemistry/bonding-basics/ionic-bonding">ionic bonding</a> rules, we first write down the name of the cation, in this case calcium, and then the name of the anion, which in this case is phosphate. </p>
<p>The name is thus: calcium phosphate</p>
<p>Now, we need to criss-cross to get the chemical formula. Calcium usually exists in its <mathjax>#+2#</mathjax> state, while a phosphate ion has a <mathjax>#-3#</mathjax> charge. So, we get:</p>
<p><mathjax>#Ca^(2+)PO_4^(3-)#</mathjax></p>
<p>To criss-cross, we bring down the <mathjax>#3-#</mathjax> oxidation state to calcium, while phosphate will receive the <mathjax>#2+#</mathjax> state. We then remove any signs (positive or negative). The formula is thus:</p>
<p><mathjax>#Ca_3(PO_4)_2#</mathjax> </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#Ca_3(PO_4)_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Well, the name will be simple, as by <a href="https://socratic.org/chemistry/bonding-basics/ionic-bonding">ionic bonding</a> rules, we first write down the name of the cation, in this case calcium, and then the name of the anion, which in this case is phosphate. </p>
<p>The name is thus: calcium phosphate</p>
<p>Now, we need to criss-cross to get the chemical formula. Calcium usually exists in its <mathjax>#+2#</mathjax> state, while a phosphate ion has a <mathjax>#-3#</mathjax> charge. So, we get:</p>
<p><mathjax>#Ca^(2+)PO_4^(3-)#</mathjax></p>
<p>To criss-cross, we bring down the <mathjax>#3-#</mathjax> oxidation state to calcium, while phosphate will receive the <mathjax>#2+#</mathjax> state. We then remove any signs (positive or negative). The formula is thus:</p>
<p><mathjax>#Ca_3(PO_4)_2#</mathjax> </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the formula for the ionic compound composed of the calcium ion and the phosphate ion?</h1>
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Nam D.
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<span class="dateCreated" datetime="2018-06-05T00:23:27" itemprop="dateCreated">
Jun 5, 2018
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<div class="markdown"><p><mathjax>#Ca_3(PO_4)_2#</mathjax></p></div>
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</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Well, the name will be simple, as by <a href="https://socratic.org/chemistry/bonding-basics/ionic-bonding">ionic bonding</a> rules, we first write down the name of the cation, in this case calcium, and then the name of the anion, which in this case is phosphate. </p>
<p>The name is thus: calcium phosphate</p>
<p>Now, we need to criss-cross to get the chemical formula. Calcium usually exists in its <mathjax>#+2#</mathjax> state, while a phosphate ion has a <mathjax>#-3#</mathjax> charge. So, we get:</p>
<p><mathjax>#Ca^(2+)PO_4^(3-)#</mathjax></p>
<p>To criss-cross, we bring down the <mathjax>#3-#</mathjax> oxidation state to calcium, while phosphate will receive the <mathjax>#2+#</mathjax> state. We then remove any signs (positive or negative). The formula is thus:</p>
<p><mathjax>#Ca_3(PO_4)_2#</mathjax> </p></div>
</div>
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</article> | What is the formula for the ionic compound composed of the calcium ion and the phosphate ion? | null |
3,332 | a8d6e69a-6ddd-11ea-a9ad-ccda262736ce | https://socratic.org/questions/how-would-you-calculate-the-enthalpy-of-formation-of-the-reaction-4c3h5n3o9-l-12 | −1784 kJ/mol | start physical_unit 9 10 enthalpy_of_formation kj/mol qc_end chemical_equation 11 23 qc_end physical_unit 9 10 28 29 deltah qc_end end | [{"type":"physical unit","value":"Enthalpy of formation [OF] the reaction [IN] kJ/mol "}] | [{"type":"physical unit","value":"−1784 kJ/mol"}] | [{"type":"chemical equation","value":"4 C3H5N3O9 (l) -> 12 CO2 (g) +10 H2O (g) + 6 N"},{"type":"physical unit","value":"deltaH [OF] the reaction [=] \\pu{-5678 J}"}] | <h1 class="questionTitle" itemprop="name">How would you calculate the enthalpy of formation of the reaction: 4C3H5N3O9 (l) --> 12CO2 (g) +10 H2O (g) + 6N,
where delta H = -5678 J?
</h1> | null | −1784 kJ/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since your question is a little vague to begin with, I'll assume that you must determine the standard <a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of formation of <em>nitroglicerine</em>, <mathjax>#"C"_3"H"_5"N"_3"O"_9#</mathjax>.</p>
<p>The idea here is that you need to use the standard enthalpy change of reaction, <mathjax>#DeltaH^@#</mathjax>, and the standard enthalpy changes of formation of the products to find the standard enthalpy change of formation of interest.</p>
<p>You can find the standard enthalpy changes of formation of carbon dioxide, water, and nitrogen gas here</p>
<p><a href="https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29</a></p>
<p>To find the standard enthalpy change of formation of nitroglicerine, use <strong><a href="http://socratic.org/chemistry/thermochemistry/hess-law">Hess' Law</a></strong>, which tells you that the enthalpy change of a reaction is <em>independent</em> of the path taken and the number of steps needed for that reaction to take place. </p>
<p>This means that you can express the standard enthalpy change of reaction by using the standard enthalpy changes of formation of the reactant and of the products</p>
<blockquote>
<p><mathjax>#DeltaH_"rxn"^@ = sum(n xx DeltaH_"f prod"^@) - sum(m xx DeltaH_"f react"^@)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#n#</mathjax>, <mathjax>#m#</mathjax> - the stoichiometric coefficients of the species that take part in the reaction.</p>
<p>So, the standard enthalpy changes of formation for <strong>one mole</strong> of carbon dioxide, water, and nitrogen gas are</p>
<blockquote>
<p><mathjax>#"CO"_2: -"393.51 kJ/mol"#</mathjax></p>
<p><mathjax>#"H"_2"O": -"241.82 kJ/mol"#</mathjax></p>
<p><mathjax>#"N"_2: " 0 kJ/mol"#</mathjax></p>
</blockquote>
<p>So, your reaction produces</p>
<blockquote>
<ul>
<li><em>12 moles of carbon dioxide</em></li>
<li><em>10 moles of water</em></li>
<li><em>6 moles of nitrogen gas</em></li>
</ul>
</blockquote>
<p>and requires </p>
<blockquote>
<ul>
<li><em>4 moles of nitroglicerine</em></li>
</ul>
</blockquote>
<p>Notice that the enthalpies of formation are given in <em>kilojoules</em> per mole, so convert the enthalpy change of reaction to kilojoules</p>
<blockquote>
<p><mathjax>#-5678color(red)(cancel(color(black)("J"))) * "1 kJ"/(10^3color(red)(cancel(color(black)("J")))) = -"5.678 kJ"#</mathjax></p>
</blockquote>
<p>Plug in your values and solve for <mathjax>#DeltaH_"f nitro"^@#</mathjax></p>
<blockquote>
<p><mathjax>#-"5.678 kJ" = [12color(red)(cancel(color(black)("moles"))) * (-393.51"kJ"/color(red)(cancel(color(black)("mol")))) + 10color(red)(cancel(color(black)("moles"))) * (-241.82"kJ"/color(red)(cancel(color(black)("mole")))) + 6color(red)(cancel(color(black)("moles"))) * 0"kJ"/color(red)(cancel(color(black)("mole")))] - (4 * DeltaH_"f nitro"^@)#</mathjax></p>
</blockquote>
<p>Rearrange to get</p>
<blockquote>
<p><mathjax>#4DeltaH_"f nitro"^@ = -"7140.32 kJ" + "5.678 kJ"#</mathjax></p>
<p><mathjax>#DeltaH_"f nitro"^@ = (-"7134.642 kJ")/4 = color(green)(-"1784 kJ/mol")#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#DeltaH_"f nitro"^@ = -"1784 kJ/mol"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since your question is a little vague to begin with, I'll assume that you must determine the standard <a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of formation of <em>nitroglicerine</em>, <mathjax>#"C"_3"H"_5"N"_3"O"_9#</mathjax>.</p>
<p>The idea here is that you need to use the standard enthalpy change of reaction, <mathjax>#DeltaH^@#</mathjax>, and the standard enthalpy changes of formation of the products to find the standard enthalpy change of formation of interest.</p>
<p>You can find the standard enthalpy changes of formation of carbon dioxide, water, and nitrogen gas here</p>
<p><a href="https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29</a></p>
<p>To find the standard enthalpy change of formation of nitroglicerine, use <strong><a href="http://socratic.org/chemistry/thermochemistry/hess-law">Hess' Law</a></strong>, which tells you that the enthalpy change of a reaction is <em>independent</em> of the path taken and the number of steps needed for that reaction to take place. </p>
<p>This means that you can express the standard enthalpy change of reaction by using the standard enthalpy changes of formation of the reactant and of the products</p>
<blockquote>
<p><mathjax>#DeltaH_"rxn"^@ = sum(n xx DeltaH_"f prod"^@) - sum(m xx DeltaH_"f react"^@)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#n#</mathjax>, <mathjax>#m#</mathjax> - the stoichiometric coefficients of the species that take part in the reaction.</p>
<p>So, the standard enthalpy changes of formation for <strong>one mole</strong> of carbon dioxide, water, and nitrogen gas are</p>
<blockquote>
<p><mathjax>#"CO"_2: -"393.51 kJ/mol"#</mathjax></p>
<p><mathjax>#"H"_2"O": -"241.82 kJ/mol"#</mathjax></p>
<p><mathjax>#"N"_2: " 0 kJ/mol"#</mathjax></p>
</blockquote>
<p>So, your reaction produces</p>
<blockquote>
<ul>
<li><em>12 moles of carbon dioxide</em></li>
<li><em>10 moles of water</em></li>
<li><em>6 moles of nitrogen gas</em></li>
</ul>
</blockquote>
<p>and requires </p>
<blockquote>
<ul>
<li><em>4 moles of nitroglicerine</em></li>
</ul>
</blockquote>
<p>Notice that the enthalpies of formation are given in <em>kilojoules</em> per mole, so convert the enthalpy change of reaction to kilojoules</p>
<blockquote>
<p><mathjax>#-5678color(red)(cancel(color(black)("J"))) * "1 kJ"/(10^3color(red)(cancel(color(black)("J")))) = -"5.678 kJ"#</mathjax></p>
</blockquote>
<p>Plug in your values and solve for <mathjax>#DeltaH_"f nitro"^@#</mathjax></p>
<blockquote>
<p><mathjax>#-"5.678 kJ" = [12color(red)(cancel(color(black)("moles"))) * (-393.51"kJ"/color(red)(cancel(color(black)("mol")))) + 10color(red)(cancel(color(black)("moles"))) * (-241.82"kJ"/color(red)(cancel(color(black)("mole")))) + 6color(red)(cancel(color(black)("moles"))) * 0"kJ"/color(red)(cancel(color(black)("mole")))] - (4 * DeltaH_"f nitro"^@)#</mathjax></p>
</blockquote>
<p>Rearrange to get</p>
<blockquote>
<p><mathjax>#4DeltaH_"f nitro"^@ = -"7140.32 kJ" + "5.678 kJ"#</mathjax></p>
<p><mathjax>#DeltaH_"f nitro"^@ = (-"7134.642 kJ")/4 = color(green)(-"1784 kJ/mol")#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How would you calculate the enthalpy of formation of the reaction: 4C3H5N3O9 (l) --> 12CO2 (g) +10 H2O (g) + 6N,
where delta H = -5678 J?
</h1>
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Stefan V.
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<span class="dateCreated" datetime="2015-10-17T23:22:01" itemprop="dateCreated">
Oct 17, 2015
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<div class="markdown"><p><mathjax>#DeltaH_"f nitro"^@ = -"1784 kJ/mol"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since your question is a little vague to begin with, I'll assume that you must determine the standard <a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of formation of <em>nitroglicerine</em>, <mathjax>#"C"_3"H"_5"N"_3"O"_9#</mathjax>.</p>
<p>The idea here is that you need to use the standard enthalpy change of reaction, <mathjax>#DeltaH^@#</mathjax>, and the standard enthalpy changes of formation of the products to find the standard enthalpy change of formation of interest.</p>
<p>You can find the standard enthalpy changes of formation of carbon dioxide, water, and nitrogen gas here</p>
<p><a href="https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29</a></p>
<p>To find the standard enthalpy change of formation of nitroglicerine, use <strong><a href="http://socratic.org/chemistry/thermochemistry/hess-law">Hess' Law</a></strong>, which tells you that the enthalpy change of a reaction is <em>independent</em> of the path taken and the number of steps needed for that reaction to take place. </p>
<p>This means that you can express the standard enthalpy change of reaction by using the standard enthalpy changes of formation of the reactant and of the products</p>
<blockquote>
<p><mathjax>#DeltaH_"rxn"^@ = sum(n xx DeltaH_"f prod"^@) - sum(m xx DeltaH_"f react"^@)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#n#</mathjax>, <mathjax>#m#</mathjax> - the stoichiometric coefficients of the species that take part in the reaction.</p>
<p>So, the standard enthalpy changes of formation for <strong>one mole</strong> of carbon dioxide, water, and nitrogen gas are</p>
<blockquote>
<p><mathjax>#"CO"_2: -"393.51 kJ/mol"#</mathjax></p>
<p><mathjax>#"H"_2"O": -"241.82 kJ/mol"#</mathjax></p>
<p><mathjax>#"N"_2: " 0 kJ/mol"#</mathjax></p>
</blockquote>
<p>So, your reaction produces</p>
<blockquote>
<ul>
<li><em>12 moles of carbon dioxide</em></li>
<li><em>10 moles of water</em></li>
<li><em>6 moles of nitrogen gas</em></li>
</ul>
</blockquote>
<p>and requires </p>
<blockquote>
<ul>
<li><em>4 moles of nitroglicerine</em></li>
</ul>
</blockquote>
<p>Notice that the enthalpies of formation are given in <em>kilojoules</em> per mole, so convert the enthalpy change of reaction to kilojoules</p>
<blockquote>
<p><mathjax>#-5678color(red)(cancel(color(black)("J"))) * "1 kJ"/(10^3color(red)(cancel(color(black)("J")))) = -"5.678 kJ"#</mathjax></p>
</blockquote>
<p>Plug in your values and solve for <mathjax>#DeltaH_"f nitro"^@#</mathjax></p>
<blockquote>
<p><mathjax>#-"5.678 kJ" = [12color(red)(cancel(color(black)("moles"))) * (-393.51"kJ"/color(red)(cancel(color(black)("mol")))) + 10color(red)(cancel(color(black)("moles"))) * (-241.82"kJ"/color(red)(cancel(color(black)("mole")))) + 6color(red)(cancel(color(black)("moles"))) * 0"kJ"/color(red)(cancel(color(black)("mole")))] - (4 * DeltaH_"f nitro"^@)#</mathjax></p>
</blockquote>
<p>Rearrange to get</p>
<blockquote>
<p><mathjax>#4DeltaH_"f nitro"^@ = -"7140.32 kJ" + "5.678 kJ"#</mathjax></p>
<p><mathjax>#DeltaH_"f nitro"^@ = (-"7134.642 kJ")/4 = color(green)(-"1784 kJ/mol")#</mathjax></p>
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</article> | How would you calculate the enthalpy of formation of the reaction: 4C3H5N3O9 (l) --> 12CO2 (g) +10 H2O (g) + 6N,
where delta H = -5678 J?
| null |
3,333 | a9890968-6ddd-11ea-a5a8-ccda262736ce | https://socratic.org/questions/548658f1581e2a287b982115 | 0.05 kg | start physical_unit 7 7 mass kg qc_end physical_unit 7 7 17 18 pressure qc_end physical_unit 7 7 23 24 temperature qc_end physical_unit 12 12 10 11 volume qc_end end | [{"type":"physical unit","value":"Mass [OF] butane [IN] kg"}] | [{"type":"physical unit","value":"0.05 kg"}] | [{"type":"physical unit","value":"Pressure [OF] butane [=] \\pu{0.963 atm}"},{"type":"physical unit","value":"Temperature [OF] butane [=] \\pu{27 ℃}"},{"type":"physical unit","value":"Volume [OF] cylinder [=] \\pu{20 L}"}] | <h1 class="questionTitle" itemprop="name">What is the mass in kg of butane in a #"20 L"# cylinder with a pressure of #"0.963 atm"# and a temperature of #27^@"C"#?</h1> | null | 0.05 kg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to solve this problem, you will need to use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: </p>
<p><mathjax>#PV=nRT#</mathjax>,</p>
<p>where:</p>
<p><mathjax>#P#</mathjax> is pressure, <mathjax>#V#</mathjax> is volume, <mathjax>#n#</mathjax> is moles, <mathjax>#R#</mathjax> is the gas constant, and <mathjax>#T#</mathjax> is the temperature in Kelvins.</p>
<p>The <a href="http://socratic.org/chemistry/the-behavior-of-gases/gas-laws">gas laws</a> use the Kelvin temperature scale, so the temperature in Celsius must be converted to Kelvins by adding 273.15..</p>
<p>Determine the molar mass of butane, <mathjax>#"C"_4"H"_10"#</mathjax> by multiplying the atomic weight for each element by its subscript. Add the molar masses of each element, and record the answer in g/mol. (You can also look up the molar mass.)</p>
<p><mathjax>#MM_"CH4H10"=(4xx"12.011 g/mol C") + (10xx"1.008 g/mol H")="58.124 g/mol"#</mathjax></p>
<p><strong>Known:</strong></p>
<p><mathjax>#P = "0.963 atm"#</mathjax></p>
<p><mathjax>#V = "20 L"#</mathjax></p>
<p><mathjax>#R ="0.08206 L atm /K mol"#</mathjax></p>
<p><mathjax>#T = "27"^@"C + 273.15 = 300 K"#</mathjax></p>
<p><mathjax>#MM_"butane" = 58.124 "g/mol"#</mathjax></p>
<p><strong>Unknown:</strong></p>
<p>mass of butane</p>
<p><strong>Moles Butane</strong></p>
<p>Rearrange the equation to isolate <mathjax>#n#</mathjax>. Plug in the known values and solve.</p>
<p><mathjax>#n = (PV)/(RT)#</mathjax></p>
<p><mathjax>#n =(0.963 color(red)cancel(color(black)("atm"))xx20 color(red)cancel(color(black)("L")))/(0.08206 color(red)cancel(color(black)("L")) color(red)cancel(color(black)("atm")) color(red)cancel(color(black)("K"))^(-1) "mol"^(-1)xx300color(red)cancel(color(black)("K")))= "0.8 mol C"_4"H"_10"#</mathjax> (rounded to one significant figure because 20 L has only one significant figure)</p>
<p><strong>Determine the mass of butane in the cylinder.</strong></p>
<p><mathjax>#0.8 color(red)cancel(color(black)("mol C"_4"H"_10))xx(58.124 "g C"_4"H"_10)/(1color(red)cancel(color(black)("mol C"_4"H"_10)))="50 g C"_4"H"_10"#</mathjax> (rounded to one significant figure) </p>
<p><strong>Convert mass in grams to kilograms.</strong></p>
<p><mathjax>#50color(red)cancel(color(black)("g C"_4"H"_10))xx(1"kg")/(1000color(red)cancel(color(black)("g")))="0.05 kg C"_4"H"_10"#</mathjax></p>
<p>The mass of butane in the cylinder is <mathjax>#"50 g"#</mathjax>, or <mathjax>#"0.05 kg"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The mass of butane in the cylinder is <mathjax>#"50 g"#</mathjax>, or <mathjax>#"0.05 kg"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to solve this problem, you will need to use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: </p>
<p><mathjax>#PV=nRT#</mathjax>,</p>
<p>where:</p>
<p><mathjax>#P#</mathjax> is pressure, <mathjax>#V#</mathjax> is volume, <mathjax>#n#</mathjax> is moles, <mathjax>#R#</mathjax> is the gas constant, and <mathjax>#T#</mathjax> is the temperature in Kelvins.</p>
<p>The <a href="http://socratic.org/chemistry/the-behavior-of-gases/gas-laws">gas laws</a> use the Kelvin temperature scale, so the temperature in Celsius must be converted to Kelvins by adding 273.15..</p>
<p>Determine the molar mass of butane, <mathjax>#"C"_4"H"_10"#</mathjax> by multiplying the atomic weight for each element by its subscript. Add the molar masses of each element, and record the answer in g/mol. (You can also look up the molar mass.)</p>
<p><mathjax>#MM_"CH4H10"=(4xx"12.011 g/mol C") + (10xx"1.008 g/mol H")="58.124 g/mol"#</mathjax></p>
<p><strong>Known:</strong></p>
<p><mathjax>#P = "0.963 atm"#</mathjax></p>
<p><mathjax>#V = "20 L"#</mathjax></p>
<p><mathjax>#R ="0.08206 L atm /K mol"#</mathjax></p>
<p><mathjax>#T = "27"^@"C + 273.15 = 300 K"#</mathjax></p>
<p><mathjax>#MM_"butane" = 58.124 "g/mol"#</mathjax></p>
<p><strong>Unknown:</strong></p>
<p>mass of butane</p>
<p><strong>Moles Butane</strong></p>
<p>Rearrange the equation to isolate <mathjax>#n#</mathjax>. Plug in the known values and solve.</p>
<p><mathjax>#n = (PV)/(RT)#</mathjax></p>
<p><mathjax>#n =(0.963 color(red)cancel(color(black)("atm"))xx20 color(red)cancel(color(black)("L")))/(0.08206 color(red)cancel(color(black)("L")) color(red)cancel(color(black)("atm")) color(red)cancel(color(black)("K"))^(-1) "mol"^(-1)xx300color(red)cancel(color(black)("K")))= "0.8 mol C"_4"H"_10"#</mathjax> (rounded to one significant figure because 20 L has only one significant figure)</p>
<p><strong>Determine the mass of butane in the cylinder.</strong></p>
<p><mathjax>#0.8 color(red)cancel(color(black)("mol C"_4"H"_10))xx(58.124 "g C"_4"H"_10)/(1color(red)cancel(color(black)("mol C"_4"H"_10)))="50 g C"_4"H"_10"#</mathjax> (rounded to one significant figure) </p>
<p><strong>Convert mass in grams to kilograms.</strong></p>
<p><mathjax>#50color(red)cancel(color(black)("g C"_4"H"_10))xx(1"kg")/(1000color(red)cancel(color(black)("g")))="0.05 kg C"_4"H"_10"#</mathjax></p>
<p>The mass of butane in the cylinder is <mathjax>#"50 g"#</mathjax>, or <mathjax>#"0.05 kg"#</mathjax>.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the mass in kg of butane in a #"20 L"# cylinder with a pressure of #"0.963 atm"# and a temperature of #27^@"C"#?</h1>
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<div class="markdown"><p>The mass of butane in the cylinder is <mathjax>#"50 g"#</mathjax>, or <mathjax>#"0.05 kg"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to solve this problem, you will need to use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: </p>
<p><mathjax>#PV=nRT#</mathjax>,</p>
<p>where:</p>
<p><mathjax>#P#</mathjax> is pressure, <mathjax>#V#</mathjax> is volume, <mathjax>#n#</mathjax> is moles, <mathjax>#R#</mathjax> is the gas constant, and <mathjax>#T#</mathjax> is the temperature in Kelvins.</p>
<p>The <a href="http://socratic.org/chemistry/the-behavior-of-gases/gas-laws">gas laws</a> use the Kelvin temperature scale, so the temperature in Celsius must be converted to Kelvins by adding 273.15..</p>
<p>Determine the molar mass of butane, <mathjax>#"C"_4"H"_10"#</mathjax> by multiplying the atomic weight for each element by its subscript. Add the molar masses of each element, and record the answer in g/mol. (You can also look up the molar mass.)</p>
<p><mathjax>#MM_"CH4H10"=(4xx"12.011 g/mol C") + (10xx"1.008 g/mol H")="58.124 g/mol"#</mathjax></p>
<p><strong>Known:</strong></p>
<p><mathjax>#P = "0.963 atm"#</mathjax></p>
<p><mathjax>#V = "20 L"#</mathjax></p>
<p><mathjax>#R ="0.08206 L atm /K mol"#</mathjax></p>
<p><mathjax>#T = "27"^@"C + 273.15 = 300 K"#</mathjax></p>
<p><mathjax>#MM_"butane" = 58.124 "g/mol"#</mathjax></p>
<p><strong>Unknown:</strong></p>
<p>mass of butane</p>
<p><strong>Moles Butane</strong></p>
<p>Rearrange the equation to isolate <mathjax>#n#</mathjax>. Plug in the known values and solve.</p>
<p><mathjax>#n = (PV)/(RT)#</mathjax></p>
<p><mathjax>#n =(0.963 color(red)cancel(color(black)("atm"))xx20 color(red)cancel(color(black)("L")))/(0.08206 color(red)cancel(color(black)("L")) color(red)cancel(color(black)("atm")) color(red)cancel(color(black)("K"))^(-1) "mol"^(-1)xx300color(red)cancel(color(black)("K")))= "0.8 mol C"_4"H"_10"#</mathjax> (rounded to one significant figure because 20 L has only one significant figure)</p>
<p><strong>Determine the mass of butane in the cylinder.</strong></p>
<p><mathjax>#0.8 color(red)cancel(color(black)("mol C"_4"H"_10))xx(58.124 "g C"_4"H"_10)/(1color(red)cancel(color(black)("mol C"_4"H"_10)))="50 g C"_4"H"_10"#</mathjax> (rounded to one significant figure) </p>
<p><strong>Convert mass in grams to kilograms.</strong></p>
<p><mathjax>#50color(red)cancel(color(black)("g C"_4"H"_10))xx(1"kg")/(1000color(red)cancel(color(black)("g")))="0.05 kg C"_4"H"_10"#</mathjax></p>
<p>The mass of butane in the cylinder is <mathjax>#"50 g"#</mathjax>, or <mathjax>#"0.05 kg"#</mathjax>.</p></div>
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</article> | What is the mass in kg of butane in a #"20 L"# cylinder with a pressure of #"0.963 atm"# and a temperature of #27^@"C"#? | null |
3,334 | a9d8a7db-6ddd-11ea-a8a5-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-normal-rain | 5.70 | start physical_unit 5 6 ph none qc_end substance 5 6 qc_end end | [{"type":"physical unit","value":"pH [OF] normal rain"}] | [{"type":"physical unit","value":"5.70"}] | [{"type":"substance name","value":"Normal rain"}] | <h1 class="questionTitle" itemprop="name">What is the pH of normal rain? </h1> | null | 5.70 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The pH of pure water is 7.0.</p>
<p>However, the pH of normal rain is about 5.7.</p>
<p>Rainwater contains dissolved carbon dioxide from the atmosphere.</p>
<p>Carbon dioxide reacts with water to form carbonic acid.</p>
<p><mathjax>#"CO"_2"(g)" + "H"_2"O(l)" ⇌ underbrace("H"_2"CO"_3"(aq)")_color(red)("carbonic acid")#</mathjax></p>
<p>Carbonic acid is an acid because it reacts with water to form hydronium ions, <mathjax>#"H"_3"O"^"+"#</mathjax>.</p>
<p><mathjax>#"H"_2"CO"_3"(aq)" + "H"_2 "O(l)" ⇌ "H"_3 "O"^"+" + "HCO"_3^"-"#</mathjax></p>
<p>Thus, normal rain is slightly acidic.</p>
<p>In contrast, the pH of typical acid rain is around 4.0 (about 50 times as acidic as normal rain).</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>The pH of normal rain is around 5.7.</p></div>
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</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The pH of pure water is 7.0.</p>
<p>However, the pH of normal rain is about 5.7.</p>
<p>Rainwater contains dissolved carbon dioxide from the atmosphere.</p>
<p>Carbon dioxide reacts with water to form carbonic acid.</p>
<p><mathjax>#"CO"_2"(g)" + "H"_2"O(l)" ⇌ underbrace("H"_2"CO"_3"(aq)")_color(red)("carbonic acid")#</mathjax></p>
<p>Carbonic acid is an acid because it reacts with water to form hydronium ions, <mathjax>#"H"_3"O"^"+"#</mathjax>.</p>
<p><mathjax>#"H"_2"CO"_3"(aq)" + "H"_2 "O(l)" ⇌ "H"_3 "O"^"+" + "HCO"_3^"-"#</mathjax></p>
<p>Thus, normal rain is slightly acidic.</p>
<p>In contrast, the pH of typical acid rain is around 4.0 (about 50 times as acidic as normal rain).</p></div>
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<h1 class="questionTitle" itemprop="name">What is the pH of normal rain? </h1>
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<div class="markdown"><p>The pH of normal rain is around 5.7.</p></div>
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<div class="markdown"><blockquote></blockquote>
<p>The pH of pure water is 7.0.</p>
<p>However, the pH of normal rain is about 5.7.</p>
<p>Rainwater contains dissolved carbon dioxide from the atmosphere.</p>
<p>Carbon dioxide reacts with water to form carbonic acid.</p>
<p><mathjax>#"CO"_2"(g)" + "H"_2"O(l)" ⇌ underbrace("H"_2"CO"_3"(aq)")_color(red)("carbonic acid")#</mathjax></p>
<p>Carbonic acid is an acid because it reacts with water to form hydronium ions, <mathjax>#"H"_3"O"^"+"#</mathjax>.</p>
<p><mathjax>#"H"_2"CO"_3"(aq)" + "H"_2 "O(l)" ⇌ "H"_3 "O"^"+" + "HCO"_3^"-"#</mathjax></p>
<p>Thus, normal rain is slightly acidic.</p>
<p>In contrast, the pH of typical acid rain is around 4.0 (about 50 times as acidic as normal rain).</p></div>
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</article> | What is the pH of normal rain? | null |
3,335 | ac0fd629-6ddd-11ea-87a7-ccda262736ce | https://socratic.org/questions/the-specific-heat-of-ice-is-0-492-cal-g-c-how-many-calories-of-heat-are-required | 959.4 calories | start physical_unit 4 4 heat_energy cal qc_end physical_unit 4 4 6 9 specific_heat qc_end physical_unit 4 4 19 20 mass qc_end physical_unit 4 4 24 25 temperature qc_end physical_unit 4 4 27 28 temperature qc_end end | [{"type":"physical unit","value":"Required heat [OF] ice [IN] calories"}] | [{"type":"physical unit","value":"959.4 calories"}] | [{"type":"physical unit","value":"specific heat [OF] ice [=] \\pu{0.492 cal/(g * ℃)}"},{"type":"physical unit","value":"Mass [OF] ice [=] \\pu{100.0 g}"},{"type":"physical unit","value":"Temperature1 [OF] ice [=] \\pu{-20.0 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] ice [=] \\pu{-0.5 ℃}"}] | <h1 class="questionTitle" itemprop="name">The specific heat of ice is 0.492 cal/(g × °C). How many calories of heat are required to raise 100.0 g of ice from -20.0 °C to -0.5 °C? </h1> | null | 959.4 calories | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And so we simply take the product...</p>
<p><mathjax>#"mass"xx"C"_"specific heat"xxDeltaT=100.0*gxxunderbrace(0.492*cal*g^-1*""^@C^-1)_"quoted specific heat"xx19.5*""^@C=+959.4*cal#</mathjax>....the positive sign indicates heat is ADDED to the system. </p></div>
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<div class="markdown"><p>Well, your ice DOES NOT undergo a phase change....</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And so we simply take the product...</p>
<p><mathjax>#"mass"xx"C"_"specific heat"xxDeltaT=100.0*gxxunderbrace(0.492*cal*g^-1*""^@C^-1)_"quoted specific heat"xx19.5*""^@C=+959.4*cal#</mathjax>....the positive sign indicates heat is ADDED to the system. </p></div>
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<h1 class="questionTitle" itemprop="name">The specific heat of ice is 0.492 cal/(g × °C). How many calories of heat are required to raise 100.0 g of ice from -20.0 °C to -0.5 °C? </h1>
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anor277
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<div class="markdown"><p>Well, your ice DOES NOT undergo a phase change....</p></div>
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<div class="markdown"><p>And so we simply take the product...</p>
<p><mathjax>#"mass"xx"C"_"specific heat"xxDeltaT=100.0*gxxunderbrace(0.492*cal*g^-1*""^@C^-1)_"quoted specific heat"xx19.5*""^@C=+959.4*cal#</mathjax>....the positive sign indicates heat is ADDED to the system. </p></div>
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<div class="markdown"><p><mathjax>#959.4#</mathjax> calories</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We use the <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> equation, which states that,</p>
<p><mathjax>#q=mcDeltaT#</mathjax></p>
<p>where:</p>
<blockquote>
<ul>
<li>
<p><mathjax>#q#</mathjax> is the heat energy supplied in joules</p>
</li>
<li>
<p><mathjax>#m#</mathjax> is the mass of the substance in kilograms</p>
</li>
<li>
<p><mathjax>#c#</mathjax> is the specific heat capacity of the object in joules</p>
</li>
<li>
<p><mathjax>#DeltaT#</mathjax> is the change in temperature</p>
</li>
</ul>
</blockquote>
<p>Here, <mathjax>#DeltaT=-0.5^@"C"-(-20^@"C")=19.5^@"C"#</mathjax>.</p>
<p>So, we get:</p>
<p><mathjax>#q=100color(red)cancelcolor(black)"g"*(0.492 \ "cal")/(color(red)cancelcolor(black)"g"color(red)cancelcolor(black)(""^@"C"))*19.5^@"C"#</mathjax></p>
<p><mathjax>#=959.4 \ "cal"#</mathjax></p></div>
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</article> | The specific heat of ice is 0.492 cal/(g × °C). How many calories of heat are required to raise 100.0 g of ice from -20.0 °C to -0.5 °C? | null |
3,336 | ac885b28-6ddd-11ea-b9c8-ccda262736ce | https://socratic.org/questions/how-many-grams-of-glucose-must-be-added-to-275-g-of-water-in-order-to-prepare-pe | 3.62 grams | start physical_unit 4 4 mass g qc_end physical_unit 12 12 9 10 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] glucose [IN] grams"}] | [{"type":"physical unit","value":"3.62 grams"}] | [{"type":"physical unit","value":"Mass [OF] water [=] \\pu{275 g}"},{"type":"physical unit","value":"Percent by mass concentration [OF] glucose in aqueous glucose solution [=] \\pu{1.30%}"}] | <h1 class="questionTitle" itemprop="name">How many grams of glucose must be added to 275 g of water in order to prepare percent by mass concentration of aqueous glucose solution 1.30% ? </h1> | null | 3.62 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass</strong> tells you the number of grams of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is glucose, present <strong>for every</strong> <mathjax>#"100. g"#</mathjax> of the solution. </p>
<p>A solution that is <mathjax>#"1.30%#</mathjax> glucose <strong>by mass</strong> will contain <mathjax>#"1.30 g"#</mathjax> of glucose for every <mathjax>#"100. g"#</mathjax> of the solution. </p>
<p>So the question is actually asking about the number of grams of glucose that must be added to <mathjax>#"275 g"#</mathjax> of water in order to get <mathjax>#"1.30 g"#</mathjax> of glucose <strong>for every</strong> <mathjax>#"100. g"#</mathjax> of this solution. </p>
<p>If you take <mathjax>#x#</mathjax> <mathjax>#"g"#</mathjax> to be the mass of glucose needed to make this solution, you can say that <strong>after</strong> you add the glucose to the water, the <strong>total mass</strong> of the solution will be </p>
<blockquote>
<p><mathjax>#x quad "g" + "275 g" = (x + 275) quad "g"#</mathjax></p>
</blockquote>
<p>So you know that <mathjax>#x#</mathjax> <mathjax>#"g"#</mathjax> of glucose in <mathjax>#(x + 275)#</mathjax> <mathjax>#"g"#</mathjax> of the solution must be equivalent to <mathjax>#"1.30 g"#</mathjax> of glucose in <mathjax>#"100. g"#</mathjax> of the solution. </p>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#(x color(red)(cancel(color(black)("g glucose"))))/((x + 275) color(red)(cancel(color(black)("g solution")))) = (1.30 color(red)(cancel(color(black)("g glucose"))))/(100. color(red)(cancel(color(black)("g solution"))))#</mathjax></p>
</blockquote>
<p>Rearrange and solve for <mathjax>#x#</mathjax> to get</p>
<blockquote>
<p><mathjax>#100. * x = 1.30 * x + 1.30 * 275#</mathjax></p>
<p><mathjax>#98.7 * x = 357.5 implies x = 357.5/98.7 = 3.6221#</mathjax></p>
</blockquote>
<p>Rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for your values, the answer will be </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("mass of glucose = 3.62 g")))#</mathjax></p>
</blockquote></div>
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<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"3.62 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass</strong> tells you the number of grams of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is glucose, present <strong>for every</strong> <mathjax>#"100. g"#</mathjax> of the solution. </p>
<p>A solution that is <mathjax>#"1.30%#</mathjax> glucose <strong>by mass</strong> will contain <mathjax>#"1.30 g"#</mathjax> of glucose for every <mathjax>#"100. g"#</mathjax> of the solution. </p>
<p>So the question is actually asking about the number of grams of glucose that must be added to <mathjax>#"275 g"#</mathjax> of water in order to get <mathjax>#"1.30 g"#</mathjax> of glucose <strong>for every</strong> <mathjax>#"100. g"#</mathjax> of this solution. </p>
<p>If you take <mathjax>#x#</mathjax> <mathjax>#"g"#</mathjax> to be the mass of glucose needed to make this solution, you can say that <strong>after</strong> you add the glucose to the water, the <strong>total mass</strong> of the solution will be </p>
<blockquote>
<p><mathjax>#x quad "g" + "275 g" = (x + 275) quad "g"#</mathjax></p>
</blockquote>
<p>So you know that <mathjax>#x#</mathjax> <mathjax>#"g"#</mathjax> of glucose in <mathjax>#(x + 275)#</mathjax> <mathjax>#"g"#</mathjax> of the solution must be equivalent to <mathjax>#"1.30 g"#</mathjax> of glucose in <mathjax>#"100. g"#</mathjax> of the solution. </p>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#(x color(red)(cancel(color(black)("g glucose"))))/((x + 275) color(red)(cancel(color(black)("g solution")))) = (1.30 color(red)(cancel(color(black)("g glucose"))))/(100. color(red)(cancel(color(black)("g solution"))))#</mathjax></p>
</blockquote>
<p>Rearrange and solve for <mathjax>#x#</mathjax> to get</p>
<blockquote>
<p><mathjax>#100. * x = 1.30 * x + 1.30 * 275#</mathjax></p>
<p><mathjax>#98.7 * x = 357.5 implies x = 357.5/98.7 = 3.6221#</mathjax></p>
</blockquote>
<p>Rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for your values, the answer will be </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("mass of glucose = 3.62 g")))#</mathjax></p>
</blockquote></div>
</div>
</div>
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Stefan V.
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<span class="dateCreated" datetime="2018-04-18T00:04:04" itemprop="dateCreated">
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<div class="markdown"><p><mathjax>#"3.62 g"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The idea here is that the solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass</strong> tells you the number of grams of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is glucose, present <strong>for every</strong> <mathjax>#"100. g"#</mathjax> of the solution. </p>
<p>A solution that is <mathjax>#"1.30%#</mathjax> glucose <strong>by mass</strong> will contain <mathjax>#"1.30 g"#</mathjax> of glucose for every <mathjax>#"100. g"#</mathjax> of the solution. </p>
<p>So the question is actually asking about the number of grams of glucose that must be added to <mathjax>#"275 g"#</mathjax> of water in order to get <mathjax>#"1.30 g"#</mathjax> of glucose <strong>for every</strong> <mathjax>#"100. g"#</mathjax> of this solution. </p>
<p>If you take <mathjax>#x#</mathjax> <mathjax>#"g"#</mathjax> to be the mass of glucose needed to make this solution, you can say that <strong>after</strong> you add the glucose to the water, the <strong>total mass</strong> of the solution will be </p>
<blockquote>
<p><mathjax>#x quad "g" + "275 g" = (x + 275) quad "g"#</mathjax></p>
</blockquote>
<p>So you know that <mathjax>#x#</mathjax> <mathjax>#"g"#</mathjax> of glucose in <mathjax>#(x + 275)#</mathjax> <mathjax>#"g"#</mathjax> of the solution must be equivalent to <mathjax>#"1.30 g"#</mathjax> of glucose in <mathjax>#"100. g"#</mathjax> of the solution. </p>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#(x color(red)(cancel(color(black)("g glucose"))))/((x + 275) color(red)(cancel(color(black)("g solution")))) = (1.30 color(red)(cancel(color(black)("g glucose"))))/(100. color(red)(cancel(color(black)("g solution"))))#</mathjax></p>
</blockquote>
<p>Rearrange and solve for <mathjax>#x#</mathjax> to get</p>
<blockquote>
<p><mathjax>#100. * x = 1.30 * x + 1.30 * 275#</mathjax></p>
<p><mathjax>#98.7 * x = 357.5 implies x = 357.5/98.7 = 3.6221#</mathjax></p>
</blockquote>
<p>Rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for your values, the answer will be </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("mass of glucose = 3.62 g")))#</mathjax></p>
</blockquote></div>
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</article> | How many grams of glucose must be added to 275 g of water in order to prepare percent by mass concentration of aqueous glucose solution 1.30% ? | null |
3,337 | a846ef70-6ddd-11ea-aadb-ccda262736ce | https://socratic.org/questions/1-caleulate-the-molar-concentration-of-the-solution-in-2-liters-of-which-contain | 0.09 M | start physical_unit 5 6 molarity mol/l qc_end physical_unit 16 17 13 14 mass qc_end physical_unit 5 6 8 9 volume qc_end end | [{"type":"physical unit","value":"Molar concentration [OF] the solution [IN] M"}] | [{"type":"physical unit","value":"0.09 M"}] | [{"type":"physical unit","value":"Mass [OF] lead(II) nitrate [=] \\pu{58 g}"},{"type":"physical unit","value":"Volume [OF] the solution [=] \\pu{2 liters}"}] | <h1 class="questionTitle" itemprop="name">Calculate the molar concentration of the solution in 2 liters of which contained
58 g of lead(II) nitrate?</h1> | null | 0.09 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molar mass of <mathjax>#"Pb"("NO"_3)_2: " 332.2 g/mol"#</mathjax></p>
<p>So</p>
<p><mathjax>#"58 g Pb"("NO"_3)_2 * "1 mol"/("331.2 g Pb"("NO"_3)_2) = "0.0175 mol of Pb"("NO"_3)_2#</mathjax></p>
<p>The formula for the molar concentration is </p>
<p><mathjax>#"M"="mol"/"L"#</mathjax> </p>
<p>Since <mathjax>#"L"#</mathjax> is given and the mol is calculated, </p>
<p><mathjax>#"M" = "0.0175 mol"/"2 L"#</mathjax> </p>
<p>The answer will be around <mathjax>#"0.0875 M"#</mathjax>. However, you need only one sigfig, so you round it, which gives you <mathjax>#"0.09 M"#</mathjax>.</p></div>
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<div class="markdown"><p>The concentration of the solution will be <mathjax>#"0.09 M"#</mathjax>. </p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molar mass of <mathjax>#"Pb"("NO"_3)_2: " 332.2 g/mol"#</mathjax></p>
<p>So</p>
<p><mathjax>#"58 g Pb"("NO"_3)_2 * "1 mol"/("331.2 g Pb"("NO"_3)_2) = "0.0175 mol of Pb"("NO"_3)_2#</mathjax></p>
<p>The formula for the molar concentration is </p>
<p><mathjax>#"M"="mol"/"L"#</mathjax> </p>
<p>Since <mathjax>#"L"#</mathjax> is given and the mol is calculated, </p>
<p><mathjax>#"M" = "0.0175 mol"/"2 L"#</mathjax> </p>
<p>The answer will be around <mathjax>#"0.0875 M"#</mathjax>. However, you need only one sigfig, so you round it, which gives you <mathjax>#"0.09 M"#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">Calculate the molar concentration of the solution in 2 liters of which contained
58 g of lead(II) nitrate?</h1>
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<div class="markdown"><p>The concentration of the solution will be <mathjax>#"0.09 M"#</mathjax>. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The molar mass of <mathjax>#"Pb"("NO"_3)_2: " 332.2 g/mol"#</mathjax></p>
<p>So</p>
<p><mathjax>#"58 g Pb"("NO"_3)_2 * "1 mol"/("331.2 g Pb"("NO"_3)_2) = "0.0175 mol of Pb"("NO"_3)_2#</mathjax></p>
<p>The formula for the molar concentration is </p>
<p><mathjax>#"M"="mol"/"L"#</mathjax> </p>
<p>Since <mathjax>#"L"#</mathjax> is given and the mol is calculated, </p>
<p><mathjax>#"M" = "0.0175 mol"/"2 L"#</mathjax> </p>
<p>The answer will be around <mathjax>#"0.0875 M"#</mathjax>. However, you need only one sigfig, so you round it, which gives you <mathjax>#"0.09 M"#</mathjax>.</p></div>
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</article> | Calculate the molar concentration of the solution in 2 liters of which contained
58 g of lead(II) nitrate? | null |
3,338 | aa718ef0-6ddd-11ea-b556-ccda262736ce | https://socratic.org/questions/57e935457c01497cceefb405 | 100.00 mL | start physical_unit 6 7 volume ml qc_end physical_unit 6 9 3 3 percent qc_end physical_unit 9 9 18 19 volume qc_end end | [{"type":"physical unit","value":"Volume [OF] carbon tetrachloride [IN] mL"}] | [{"type":"physical unit","value":"100.00 mL"}] | [{"type":"physical unit","value":"Percentage [OF] carbon tetrachloride in benzene [=] \\pu{50%}"},{"type":"physical unit","value":"Volume [OF] benzene [=] \\pu{100 mL}"}] | <h1 class="questionTitle" itemprop="name">To make a 50% solution of carbon tetrachloride in benzene, how much carbon tetrachloride should be added to #100*mL# of benzene?</h1> | null | 100.00 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You want a <mathjax>#50%#</mathjax> <mathjax>#v/v#</mathjax> solution of <mathjax>#"carbon tet"#</mathjax> in <mathjax>#"benzene"#</mathjax>. Clearly, we should take EQUAL volumes of each <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> to prepare such a concentration. You started with a <mathjax>#100*mL#</mathjax> of benzene; you thus need a <mathjax>#100*mL#</mathjax> volume of carbon tetrachloride. The volumes should certainly be additive to a first approximation. </p>
<p>Could I prepare a 50% solution of <mathjax>#"carbon tet"#</mathjax> in water? Why or why not?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>Surely you should add <mathjax>#100*mL#</mathjax> of <mathjax>#"carbon tetrachloride"#</mathjax>?</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You want a <mathjax>#50%#</mathjax> <mathjax>#v/v#</mathjax> solution of <mathjax>#"carbon tet"#</mathjax> in <mathjax>#"benzene"#</mathjax>. Clearly, we should take EQUAL volumes of each <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> to prepare such a concentration. You started with a <mathjax>#100*mL#</mathjax> of benzene; you thus need a <mathjax>#100*mL#</mathjax> volume of carbon tetrachloride. The volumes should certainly be additive to a first approximation. </p>
<p>Could I prepare a 50% solution of <mathjax>#"carbon tet"#</mathjax> in water? Why or why not?</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">To make a 50% solution of carbon tetrachloride in benzene, how much carbon tetrachloride should be added to #100*mL# of benzene?</h1>
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<div class="markdown"><p>Surely you should add <mathjax>#100*mL#</mathjax> of <mathjax>#"carbon tetrachloride"#</mathjax>?</p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You want a <mathjax>#50%#</mathjax> <mathjax>#v/v#</mathjax> solution of <mathjax>#"carbon tet"#</mathjax> in <mathjax>#"benzene"#</mathjax>. Clearly, we should take EQUAL volumes of each <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> to prepare such a concentration. You started with a <mathjax>#100*mL#</mathjax> of benzene; you thus need a <mathjax>#100*mL#</mathjax> volume of carbon tetrachloride. The volumes should certainly be additive to a first approximation. </p>
<p>Could I prepare a 50% solution of <mathjax>#"carbon tet"#</mathjax> in water? Why or why not?</p></div>
</div>
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</article> | To make a 50% solution of carbon tetrachloride in benzene, how much carbon tetrachloride should be added to #100*mL# of benzene? | null |
3,339 | ac38f752-6ddd-11ea-ac93-ccda262736ce | https://socratic.org/questions/590cb9d211ef6b19838d0574 | 60 mL | start physical_unit 3 3 volume ml qc_end physical_unit 14 14 10 11 volume qc_end c_other OTHER qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Volume [OF] water [IN] mL"}] | [{"type":"physical unit","value":"60 mL"}] | [{"type":"physical unit","value":"Volume [OF] methane [=] \\pu{40 L}"},{"type":"other","value":"Complete combustion."},{"type":"other","value":"Under standard conditions."}] | <h1 class="questionTitle" itemprop="name">What volume of water results from complete combustion of a #40*L# volume of methane under standard conditions?</h1> | null | 60 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Methane is a room temperature gas....,<mathjax>#"boiling point "=#</mathjax> <mathjax>#-164#</mathjax> <mathjax>#""^@C#</mathjax>. And of course water is a room temperature liquid.....</p>
<p>We assume that the methane gas is under standard conditions, i.e. <mathjax>#298*K#</mathjax>, and <mathjax>#1*atm#</mathjax>. </p>
<p>And this represents <mathjax>#(40*L)/(24.5*L*mol^-1)=1.63*mol#</mathjax> of the hydrocarbon. </p>
<p>Given the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, we get................</p>
<p><mathjax>#(2xx1.63*molxx18.01*g*mol^-1)/(1.0*g*mL^-1)#</mathjax></p>
<p><mathjax>#~=60*mL#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The problem with this question is that we must ASSUME standard conditions........</p>
<p><mathjax>#CH_4(g) + 2O_2(g) rarr CO_2(g) +2H_2O(l)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Methane is a room temperature gas....,<mathjax>#"boiling point "=#</mathjax> <mathjax>#-164#</mathjax> <mathjax>#""^@C#</mathjax>. And of course water is a room temperature liquid.....</p>
<p>We assume that the methane gas is under standard conditions, i.e. <mathjax>#298*K#</mathjax>, and <mathjax>#1*atm#</mathjax>. </p>
<p>And this represents <mathjax>#(40*L)/(24.5*L*mol^-1)=1.63*mol#</mathjax> of the hydrocarbon. </p>
<p>Given the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, we get................</p>
<p><mathjax>#(2xx1.63*molxx18.01*g*mol^-1)/(1.0*g*mL^-1)#</mathjax></p>
<p><mathjax>#~=60*mL#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What volume of water results from complete combustion of a #40*L# volume of methane under standard conditions?</h1>
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<div class="markdown"><p>The problem with this question is that we must ASSUME standard conditions........</p>
<p><mathjax>#CH_4(g) + 2O_2(g) rarr CO_2(g) +2H_2O(l)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Methane is a room temperature gas....,<mathjax>#"boiling point "=#</mathjax> <mathjax>#-164#</mathjax> <mathjax>#""^@C#</mathjax>. And of course water is a room temperature liquid.....</p>
<p>We assume that the methane gas is under standard conditions, i.e. <mathjax>#298*K#</mathjax>, and <mathjax>#1*atm#</mathjax>. </p>
<p>And this represents <mathjax>#(40*L)/(24.5*L*mol^-1)=1.63*mol#</mathjax> of the hydrocarbon. </p>
<p>Given the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, we get................</p>
<p><mathjax>#(2xx1.63*molxx18.01*g*mol^-1)/(1.0*g*mL^-1)#</mathjax></p>
<p><mathjax>#~=60*mL#</mathjax></p></div>
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</article> | What volume of water results from complete combustion of a #40*L# volume of methane under standard conditions? | null |
3,340 | a992060c-6ddd-11ea-a039-ccda262736ce | https://socratic.org/questions/a-sample-of-compressed-methane-has-a-volume-of-648-ml-at-a-pressure-of-503-kpa-t | 1509.00 kPa | start physical_unit 21 22 pressure kpa qc_end physical_unit 21 22 34 35 volume qc_end physical_unit 4 4 15 16 pressure qc_end physical_unit 4 4 9 10 volume qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the methane [IN] kPa"}] | [{"type":"physical unit","value":"1509.00 kPa"}] | [{"type":"physical unit","value":"Volume2 [OF] the methane [=] \\pu{216 mL}"},{"type":"physical unit","value":"Pressure1 [OF] methane sample [=] \\pu{503 kPa}"},{"type":"physical unit","value":"Volume1 [OF] methane sample [=] \\pu{648 mL}"}] | <h1 class="questionTitle" itemprop="name">A sample of compressed methane has a volume of 648 mL at a pressure of 503 kPa. To what pressure would the methane have to be compressed in order to have a volume of 216 mL?</h1> | null | 1509.00 kPa | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We apply <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p><mathjax>#P_1=503kPa#</mathjax></p>
<p><mathjax>#V_1=648mL#</mathjax></p>
<p><mathjax>#V_2=216mL#</mathjax></p>
<p><mathjax>#P_2=V_1/V_2*P_1#</mathjax></p>
<p><mathjax>#=648/216*503=1509kPa#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>The pressure is <mathjax>#=1509kPa#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We apply <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p><mathjax>#P_1=503kPa#</mathjax></p>
<p><mathjax>#V_1=648mL#</mathjax></p>
<p><mathjax>#V_2=216mL#</mathjax></p>
<p><mathjax>#P_2=V_1/V_2*P_1#</mathjax></p>
<p><mathjax>#=648/216*503=1509kPa#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A sample of compressed methane has a volume of 648 mL at a pressure of 503 kPa. To what pressure would the methane have to be compressed in order to have a volume of 216 mL?</h1>
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Narad T.
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Mar 14, 2017
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<div class="markdown"><p>The pressure is <mathjax>#=1509kPa#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We apply <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p><mathjax>#P_1=503kPa#</mathjax></p>
<p><mathjax>#V_1=648mL#</mathjax></p>
<p><mathjax>#V_2=216mL#</mathjax></p>
<p><mathjax>#P_2=V_1/V_2*P_1#</mathjax></p>
<p><mathjax>#=648/216*503=1509kPa#</mathjax></p></div>
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anor277
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<div class="markdown"><p>Well, <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a> holds that <mathjax>#P_1V_1=P_2V_2#</mathjax>............</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So we solve for <mathjax>#P_2=(P_1V_1)/V_2=(503*kPaxx648*mL)/(216*mL)=??*kPa#</mathjax>.</p>
<p>Clearly, the pressure increases, here almost threefold.</p></div>
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</article> | A sample of compressed methane has a volume of 648 mL at a pressure of 503 kPa. To what pressure would the methane have to be compressed in order to have a volume of 216 mL? | null |
3,341 | a96d5c18-6ddd-11ea-a7cb-ccda262736ce | https://socratic.org/questions/59ed788511ef6b285ef05eb5 | 4.30 moles | start physical_unit 4 4 mole mol qc_end physical_unit 11 12 8 9 mole qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mole [OF] water [IN] moles"}] | [{"type":"physical unit","value":"4.30 moles"}] | [{"type":"physical unit","value":"Mole [OF] oxygen gas [=] \\pu{2.15 moles}"},{"type":"other","value":"Excess hydrogen gas."}] | <h1 class="questionTitle" itemprop="name">How many moles of water are produced when #"2.15 moles"# of oxygen gas react with excess hydrogen gas?</h1> | null | 4.30 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"I mol"#</mathjax> of <mathjax>#"O"_2#</mathjax> produces <mathjax>#"2 moles"#</mathjax> of <mathjax>#"H"_2"O"#</mathjax>, so <mathjax>#"2.15 moles"#</mathjax> of <mathjax>#"O"_2#</mathjax> produce </p>
<p><mathjax>#2/1×2.15 = "4.30 moles H"_2"O"#</mathjax> </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#"4.30 moles"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"I mol"#</mathjax> of <mathjax>#"O"_2#</mathjax> produces <mathjax>#"2 moles"#</mathjax> of <mathjax>#"H"_2"O"#</mathjax>, so <mathjax>#"2.15 moles"#</mathjax> of <mathjax>#"O"_2#</mathjax> produce </p>
<p><mathjax>#2/1×2.15 = "4.30 moles H"_2"O"#</mathjax> </p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How many moles of water are produced when #"2.15 moles"# of oxygen gas react with excess hydrogen gas?</h1>
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<div class="markdown"><p><mathjax>#"4.30 moles"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"I mol"#</mathjax> of <mathjax>#"O"_2#</mathjax> produces <mathjax>#"2 moles"#</mathjax> of <mathjax>#"H"_2"O"#</mathjax>, so <mathjax>#"2.15 moles"#</mathjax> of <mathjax>#"O"_2#</mathjax> produce </p>
<p><mathjax>#2/1×2.15 = "4.30 moles H"_2"O"#</mathjax> </p></div>
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</article> | How many moles of water are produced when #"2.15 moles"# of oxygen gas react with excess hydrogen gas? | null |
3,342 | a9440ff4-6ddd-11ea-a9e3-ccda262736ce | https://socratic.org/questions/how-do-you-solve-for-k-using-the-arrhenius-equation-a-first-order-reaction-has-a | 2.34 s^(−1) | start physical_unit 14 14 equilibrium_constant_k s^(-1) qc_end physical_unit 11 14 22 23 activation_barrier qc_end physical_unit 11 14 32 35 pre-exponential_factor qc_end physical_unit 11 14 41 42 temperature qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"k [OF] the reaction [IN] s^(−1)"}] | [{"type":"physical unit","value":"2.34 s^(−1)"}] | [{"type":"physical unit","value":"Ea [OF] a first order reaction [=] \\pu{65.7 kJ/mol}"},{"type":"physical unit","value":"pre-exponential factor A [OF] a first order reaction [=] \\pu{1.31 × 10^12 s^(−1)}"},{"type":"physical unit","value":"Temperature [OF] a first order reaction [=] \\pu{19 ℃}"},{"type":"other","value":"using the Arrhenius Equation"}] | <h1 class="questionTitle" itemprop="name">How do you solve for k using the Arrhenius Equation? A first order reaction has an activation energy of #"E"_a = "65.7 kJ/mol"# and a frequency factor (pre-exponential factor, #"A"#) of #1.31 xx 10^12 "s"^(-1)#. Calculate the rate constant at #19^@ "C"#.</h1> | null | 2.34 s^(−1) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://en.wikipedia.org/wiki/Arrhenius_equation#Equation" rel="nofollow">Arrhenius Equation</a></p>
<blockquote>
<p><mathjax>#k = "A" * e^(-"E"_a//(R * T))#</mathjax></p>
</blockquote>
<p>gives the relationship between the following quantities</p>
<ul>
<li>The rate constant <mathjax>#k#</mathjax> as seen in the <a href="https://socratic.org/chemistry/chemical-kinetics/rate-law">rate law</a> of the reaction;</li>
<li>The pre-exponential factor <mathjax>#"A"#</mathjax> for this particular reaction;</li>
<li>The activation energy <mathjax>#"E"_a#</mathjax> of this reaction;</li>
<li>The absolute temperature <mathjax>#T#</mathjax> under which the reaction takes place.</li>
</ul>
<p>Whereas the ideal gas constant is also involved in the calculation. The ideal gas constant takes various units, with a different numerical value for each. It would thus be necessary to dimensional analysis while calculating the exponent part of the reaction. The exponent shall end up without a unit. This example takes <mathjax>#R = 8.314 color(white)(l) color(navy)("J") //("mol" * "K")#</mathjax>.</p>
<p>Note that the Arrhenius equation requires an absolute temperature <mathjax>#T#</mathjax> in degree Kelvins <mathjax>#"K"#</mathjax> whereas the question supplied the temperature in degrees Celsius. Numerically add <mathjax>#273(.15)#</mathjax> to the temperature in degrees Celsius to get the absolute temperature in Kelvins. </p>
<p><mathjax>#T = 19 + 273 = 292 color(white)(l) "K"#</mathjax></p>
<p>The question gives the activation energy in kilojoules; however, the ideal gas constant demands the unit joule.</p>
<p><mathjax>#"E"_a = 65.7 color(white)(l) "kJ" = 65.7 xx color(navy)(10^3 color(white)(l) "J"#</mathjax></p>
<p>Substitute the real values and calculate the exponent part of the expression:</p>
<blockquote>
<p><mathjax>#-"E"_a / (R * T) = (65.7 xx color(navy)(10^3 color(white)(l) "J") * "mol"^(-1))/(8.314 color(white)(l) color(navy)("J") * "mol"^(-1) * "K"^(-1) * 292 color(white)(l) "K") #</mathjax><br/>
<mathjax>#color(white)(-"E"_a / (R * T)) ~~-27.0 color(white)(l) color(lightgreen)("(dimensionless)")#</mathjax></p>
</blockquote>
<p>Make sure that all units cancel out such that the exponent part is dimensionless. Evaluate the rest of the equation to find the rate constant <mathjax>#k#</mathjax></p>
<blockquote>
<p><mathjax>#k = "A" * e^(-"E"_a//(R * T))#</mathjax><br/>
<mathjax>#color(white)(k) = 1.31 xx 10^(12) color(white)(l) color(navy)(s^(-1)) xx e^(-27.0)#</mathjax><br/>
<mathjax>#color(white)(k) ~~ 2.34 color(white)(l) color(navy)(s^(-1))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#k ~~ 2.34 color(white)(l) "s"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://en.wikipedia.org/wiki/Arrhenius_equation#Equation" rel="nofollow">Arrhenius Equation</a></p>
<blockquote>
<p><mathjax>#k = "A" * e^(-"E"_a//(R * T))#</mathjax></p>
</blockquote>
<p>gives the relationship between the following quantities</p>
<ul>
<li>The rate constant <mathjax>#k#</mathjax> as seen in the <a href="https://socratic.org/chemistry/chemical-kinetics/rate-law">rate law</a> of the reaction;</li>
<li>The pre-exponential factor <mathjax>#"A"#</mathjax> for this particular reaction;</li>
<li>The activation energy <mathjax>#"E"_a#</mathjax> of this reaction;</li>
<li>The absolute temperature <mathjax>#T#</mathjax> under which the reaction takes place.</li>
</ul>
<p>Whereas the ideal gas constant is also involved in the calculation. The ideal gas constant takes various units, with a different numerical value for each. It would thus be necessary to dimensional analysis while calculating the exponent part of the reaction. The exponent shall end up without a unit. This example takes <mathjax>#R = 8.314 color(white)(l) color(navy)("J") //("mol" * "K")#</mathjax>.</p>
<p>Note that the Arrhenius equation requires an absolute temperature <mathjax>#T#</mathjax> in degree Kelvins <mathjax>#"K"#</mathjax> whereas the question supplied the temperature in degrees Celsius. Numerically add <mathjax>#273(.15)#</mathjax> to the temperature in degrees Celsius to get the absolute temperature in Kelvins. </p>
<p><mathjax>#T = 19 + 273 = 292 color(white)(l) "K"#</mathjax></p>
<p>The question gives the activation energy in kilojoules; however, the ideal gas constant demands the unit joule.</p>
<p><mathjax>#"E"_a = 65.7 color(white)(l) "kJ" = 65.7 xx color(navy)(10^3 color(white)(l) "J"#</mathjax></p>
<p>Substitute the real values and calculate the exponent part of the expression:</p>
<blockquote>
<p><mathjax>#-"E"_a / (R * T) = (65.7 xx color(navy)(10^3 color(white)(l) "J") * "mol"^(-1))/(8.314 color(white)(l) color(navy)("J") * "mol"^(-1) * "K"^(-1) * 292 color(white)(l) "K") #</mathjax><br/>
<mathjax>#color(white)(-"E"_a / (R * T)) ~~-27.0 color(white)(l) color(lightgreen)("(dimensionless)")#</mathjax></p>
</blockquote>
<p>Make sure that all units cancel out such that the exponent part is dimensionless. Evaluate the rest of the equation to find the rate constant <mathjax>#k#</mathjax></p>
<blockquote>
<p><mathjax>#k = "A" * e^(-"E"_a//(R * T))#</mathjax><br/>
<mathjax>#color(white)(k) = 1.31 xx 10^(12) color(white)(l) color(navy)(s^(-1)) xx e^(-27.0)#</mathjax><br/>
<mathjax>#color(white)(k) ~~ 2.34 color(white)(l) color(navy)(s^(-1))#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How do you solve for k using the Arrhenius Equation? A first order reaction has an activation energy of #"E"_a = "65.7 kJ/mol"# and a frequency factor (pre-exponential factor, #"A"#) of #1.31 xx 10^12 "s"^(-1)#. Calculate the rate constant at #19^@ "C"#.</h1>
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<div class="markdown"><p><mathjax>#k ~~ 2.34 color(white)(l) "s"^(-1)#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://en.wikipedia.org/wiki/Arrhenius_equation#Equation" rel="nofollow">Arrhenius Equation</a></p>
<blockquote>
<p><mathjax>#k = "A" * e^(-"E"_a//(R * T))#</mathjax></p>
</blockquote>
<p>gives the relationship between the following quantities</p>
<ul>
<li>The rate constant <mathjax>#k#</mathjax> as seen in the <a href="https://socratic.org/chemistry/chemical-kinetics/rate-law">rate law</a> of the reaction;</li>
<li>The pre-exponential factor <mathjax>#"A"#</mathjax> for this particular reaction;</li>
<li>The activation energy <mathjax>#"E"_a#</mathjax> of this reaction;</li>
<li>The absolute temperature <mathjax>#T#</mathjax> under which the reaction takes place.</li>
</ul>
<p>Whereas the ideal gas constant is also involved in the calculation. The ideal gas constant takes various units, with a different numerical value for each. It would thus be necessary to dimensional analysis while calculating the exponent part of the reaction. The exponent shall end up without a unit. This example takes <mathjax>#R = 8.314 color(white)(l) color(navy)("J") //("mol" * "K")#</mathjax>.</p>
<p>Note that the Arrhenius equation requires an absolute temperature <mathjax>#T#</mathjax> in degree Kelvins <mathjax>#"K"#</mathjax> whereas the question supplied the temperature in degrees Celsius. Numerically add <mathjax>#273(.15)#</mathjax> to the temperature in degrees Celsius to get the absolute temperature in Kelvins. </p>
<p><mathjax>#T = 19 + 273 = 292 color(white)(l) "K"#</mathjax></p>
<p>The question gives the activation energy in kilojoules; however, the ideal gas constant demands the unit joule.</p>
<p><mathjax>#"E"_a = 65.7 color(white)(l) "kJ" = 65.7 xx color(navy)(10^3 color(white)(l) "J"#</mathjax></p>
<p>Substitute the real values and calculate the exponent part of the expression:</p>
<blockquote>
<p><mathjax>#-"E"_a / (R * T) = (65.7 xx color(navy)(10^3 color(white)(l) "J") * "mol"^(-1))/(8.314 color(white)(l) color(navy)("J") * "mol"^(-1) * "K"^(-1) * 292 color(white)(l) "K") #</mathjax><br/>
<mathjax>#color(white)(-"E"_a / (R * T)) ~~-27.0 color(white)(l) color(lightgreen)("(dimensionless)")#</mathjax></p>
</blockquote>
<p>Make sure that all units cancel out such that the exponent part is dimensionless. Evaluate the rest of the equation to find the rate constant <mathjax>#k#</mathjax></p>
<blockquote>
<p><mathjax>#k = "A" * e^(-"E"_a//(R * T))#</mathjax><br/>
<mathjax>#color(white)(k) = 1.31 xx 10^(12) color(white)(l) color(navy)(s^(-1)) xx e^(-27.0)#</mathjax><br/>
<mathjax>#color(white)(k) ~~ 2.34 color(white)(l) color(navy)(s^(-1))#</mathjax></p>
</blockquote></div>
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</article> | How do you solve for k using the Arrhenius Equation? A first order reaction has an activation energy of #"E"_a = "65.7 kJ/mol"# and a frequency factor (pre-exponential factor, #"A"#) of #1.31 xx 10^12 "s"^(-1)#. Calculate the rate constant at #19^@ "C"#. | null |
3,343 | aa694bae-6ddd-11ea-a7bf-ccda262736ce | https://socratic.org/questions/burning-then-sample-of-an-organic-compound-with-the-weight-of-0-00480-g-it-is-af | CH2O | start chemical_formula qc_end physical_unit 20 20 16 17 mass qc_end physical_unit 26 26 22 23 mass qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"CH2O"}] | [{"type":"physical unit","value":"Weight [OF] the organic compound sample [=] \\pu{0.00480 g}"},{"type":"physical unit","value":"Weight [OF] CO2 [=] \\pu{0.00703 g}"},{"type":"physical unit","value":"Weight [OF] H2O [=] \\pu{0.00283 g}"},{"type":"other","value":"The compound consists of hydrogen, carbon and oxygen."}] | <h1 class="questionTitle" itemprop="name">Burning then sample of an organic compound with the weight of 0.00480 g it is afforded: 0.00703 g of the CO2 and 0.00283 g of the H2O, Determine the empirical formula of the compound which consists of hydrogen, carbon and oxygen?</h1> | null | CH2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We can calculate the masses of <mathjax>#"C"#</mathjax> and <mathjax>#"H"#</mathjax> from the masses of their oxides (<mathjax>#"CO"_2#</mathjax> and <mathjax>#"H"_2"O"#</mathjax>).</p>
<p><mathjax>#"Mass of C" = 7.03 color(red)(cancel(color(black)("mg CO"_2))) × "12.01 mg C"/(44.01 color(red)(cancel(color(black)("mg CO"_2)))) = "1.918 g C"#</mathjax></p>
<p><mathjax>#"Mass of H" = 0.283 color(red)(cancel(color(black)("mg H"_2"O"))) × "2.016 mg H"/(18.02 color(red)(cancel(color(black)("mg H"_2"O")))) = "0.3166 mg H"#</mathjax></p>
<p><mathjax>#"Mass of O" = "Mass of compound - mass of C - mass of O" = "4.80 mg - 1.918 mg - 0.3166 mg" = "2.565 mg"#</mathjax></p>
<blockquote></blockquote>
<p>Now, we must convert these masses to moles and find their ratios.</p>
<p>From here on, I like to summarize the calculations in a table.</p>
<p><mathjax>#"Element"color(white)(X) "Mass/mg"color(white)(X) "Millimoles"color(white)(m) "Ratio"color(white)(m)"Integers"#</mathjax><br/>
<mathjax>#stackrel(—————————————————-———)(color(white)(ll)"C" color(white)(XXXmm)1.918 color(white)(mmm)0.1597
color(white)(Xlll)1color(white)(Xmmml)1#</mathjax><br/>
<mathjax>#color(white)(ll)"H" color(white)(XXXXm)0.3166 color(white)(mml)0.3145 color(white)(mlll)1.969 color(white)(XXl)2#</mathjax><br/>
<mathjax>#color(white)(ll)"O" color(white)(mmmml)color(white)(ll)2.565color(white)(mmll)0.1603color(white)(mm)1.004color(white)(mml)1#</mathjax></p>
<p>The empirical formula is <mathjax>#"CH"_2"O"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The empirical formula is <mathjax>#"CH"_2"O"#</mathjax>. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We can calculate the masses of <mathjax>#"C"#</mathjax> and <mathjax>#"H"#</mathjax> from the masses of their oxides (<mathjax>#"CO"_2#</mathjax> and <mathjax>#"H"_2"O"#</mathjax>).</p>
<p><mathjax>#"Mass of C" = 7.03 color(red)(cancel(color(black)("mg CO"_2))) × "12.01 mg C"/(44.01 color(red)(cancel(color(black)("mg CO"_2)))) = "1.918 g C"#</mathjax></p>
<p><mathjax>#"Mass of H" = 0.283 color(red)(cancel(color(black)("mg H"_2"O"))) × "2.016 mg H"/(18.02 color(red)(cancel(color(black)("mg H"_2"O")))) = "0.3166 mg H"#</mathjax></p>
<p><mathjax>#"Mass of O" = "Mass of compound - mass of C - mass of O" = "4.80 mg - 1.918 mg - 0.3166 mg" = "2.565 mg"#</mathjax></p>
<blockquote></blockquote>
<p>Now, we must convert these masses to moles and find their ratios.</p>
<p>From here on, I like to summarize the calculations in a table.</p>
<p><mathjax>#"Element"color(white)(X) "Mass/mg"color(white)(X) "Millimoles"color(white)(m) "Ratio"color(white)(m)"Integers"#</mathjax><br/>
<mathjax>#stackrel(—————————————————-———)(color(white)(ll)"C" color(white)(XXXmm)1.918 color(white)(mmm)0.1597
color(white)(Xlll)1color(white)(Xmmml)1#</mathjax><br/>
<mathjax>#color(white)(ll)"H" color(white)(XXXXm)0.3166 color(white)(mml)0.3145 color(white)(mlll)1.969 color(white)(XXl)2#</mathjax><br/>
<mathjax>#color(white)(ll)"O" color(white)(mmmml)color(white)(ll)2.565color(white)(mmll)0.1603color(white)(mm)1.004color(white)(mml)1#</mathjax></p>
<p>The empirical formula is <mathjax>#"CH"_2"O"#</mathjax>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Burning then sample of an organic compound with the weight of 0.00480 g it is afforded: 0.00703 g of the CO2 and 0.00283 g of the H2O, Determine the empirical formula of the compound which consists of hydrogen, carbon and oxygen?</h1>
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Ernest Z.
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<div class="markdown"><p>The empirical formula is <mathjax>#"CH"_2"O"#</mathjax>. </p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We can calculate the masses of <mathjax>#"C"#</mathjax> and <mathjax>#"H"#</mathjax> from the masses of their oxides (<mathjax>#"CO"_2#</mathjax> and <mathjax>#"H"_2"O"#</mathjax>).</p>
<p><mathjax>#"Mass of C" = 7.03 color(red)(cancel(color(black)("mg CO"_2))) × "12.01 mg C"/(44.01 color(red)(cancel(color(black)("mg CO"_2)))) = "1.918 g C"#</mathjax></p>
<p><mathjax>#"Mass of H" = 0.283 color(red)(cancel(color(black)("mg H"_2"O"))) × "2.016 mg H"/(18.02 color(red)(cancel(color(black)("mg H"_2"O")))) = "0.3166 mg H"#</mathjax></p>
<p><mathjax>#"Mass of O" = "Mass of compound - mass of C - mass of O" = "4.80 mg - 1.918 mg - 0.3166 mg" = "2.565 mg"#</mathjax></p>
<blockquote></blockquote>
<p>Now, we must convert these masses to moles and find their ratios.</p>
<p>From here on, I like to summarize the calculations in a table.</p>
<p><mathjax>#"Element"color(white)(X) "Mass/mg"color(white)(X) "Millimoles"color(white)(m) "Ratio"color(white)(m)"Integers"#</mathjax><br/>
<mathjax>#stackrel(—————————————————-———)(color(white)(ll)"C" color(white)(XXXmm)1.918 color(white)(mmm)0.1597
color(white)(Xlll)1color(white)(Xmmml)1#</mathjax><br/>
<mathjax>#color(white)(ll)"H" color(white)(XXXXm)0.3166 color(white)(mml)0.3145 color(white)(mlll)1.969 color(white)(XXl)2#</mathjax><br/>
<mathjax>#color(white)(ll)"O" color(white)(mmmml)color(white)(ll)2.565color(white)(mmll)0.1603color(white)(mm)1.004color(white)(mml)1#</mathjax></p>
<p>The empirical formula is <mathjax>#"CH"_2"O"#</mathjax>.</p></div>
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</article> | Burning then sample of an organic compound with the weight of 0.00480 g it is afforded: 0.00703 g of the CO2 and 0.00283 g of the H2O, Determine the empirical formula of the compound which consists of hydrogen, carbon and oxygen? | null |
3,344 | a8caff40-6ddd-11ea-be22-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-hydrogen-in-one-mole-of-aluminum-hydroxide | 3.03 g | start physical_unit 5 5 mass g qc_end physical_unit 10 11 7 8 mole qc_end physical_unit 5 5 18 19 molar_mass qc_end end | [{"type":"physical unit","value":"Mass [OF] hydrogen [IN] g"}] | [{"type":"physical unit","value":"3.03 g"}] | [{"type":"physical unit","value":"Mole [OF] aluminum hydroxide [=] \\pu{1 mole}"},{"type":"physical unit","value":"Molar mass [OF] hydrogen [=] \\pu{1.01 g/mol}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of hydrogen in one mole of aluminum hydroxide?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The molar mass of hydrogen is 1.01 g/mol </p></div>
</h2>
</div>
</div> | 3.03 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is figure out the <em>chemical formula</em> for aluminium hydroxide. </p>
<p>Aluminium is located in group <mathjax>#13#</mathjax> of <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, and forms <mathjax>#3+#</mathjax> cations, <mathjax>#"Al"^(3+)#</mathjax>. The hydroxide anion, <mathjax>#"OH"^(-)#</mathjax>, carries a <mathjax>#1-#</mathjax> charge, which means that a <strong>formula unit</strong> of aluminium hydroxide will look like this </p>
<blockquote>
<p><mathjax>#["Al"^(3+)] + color(blue)(3)["OH"^(-)] -> "Al"("OH")_color(blue)(3)#</mathjax></p>
</blockquote>
<p>Now, you can figure out the <strong>mass</strong> of hydrogen present in <mathjax>#1#</mathjax> <strong>mole</strong> of aluminium hydroxide by first determining how many <em>moles</em> if hydrogen you get in <mathjax>#1#</mathjax> mole of aluminium hydroxide. </p>
<p>Since <mathjax>#1#</mathjax> <strong>mole</strong> of aluminium hydroxide contains <mathjax>#color(blue)(3)#</mathjax> <strong>moles</strong> of hydroxide anions, which in turn contain <mathjax>#1#</mathjax> <strong>mole</strong> of hydrogen <em>each</em>, you can say that you will have</p>
<blockquote>
<p><mathjax>#"1 mole Al"("OH")_ color(blue)(3) -> color(blue)(3)color(white)(a)"moles OH"^(-) -> color(blue)(3)color(white)(a)"moles H"#</mathjax></p>
</blockquote>
<p>The problem tells you that the <strong>molar mass</strong> of hydrogen is equal to <mathjax>#"1.01 g mol"^(-1)#</mathjax>. This means that <mathjax>#1#</mathjax> <strong>mole</strong> of hydrogen has a mass of <mathjax>#"1.01 g"#</mathjax>. </p>
<p>You can thus say that one mole of aluminium hydroxide contains </p>
<blockquote>
<p><mathjax>#color(blue)(3)color(red)(cancel(color(black)("moles H"))) * "1.01 g"/(1color(red)(cancel(color(black)("mole H")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("3.03 g H")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"3.03 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is figure out the <em>chemical formula</em> for aluminium hydroxide. </p>
<p>Aluminium is located in group <mathjax>#13#</mathjax> of <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, and forms <mathjax>#3+#</mathjax> cations, <mathjax>#"Al"^(3+)#</mathjax>. The hydroxide anion, <mathjax>#"OH"^(-)#</mathjax>, carries a <mathjax>#1-#</mathjax> charge, which means that a <strong>formula unit</strong> of aluminium hydroxide will look like this </p>
<blockquote>
<p><mathjax>#["Al"^(3+)] + color(blue)(3)["OH"^(-)] -> "Al"("OH")_color(blue)(3)#</mathjax></p>
</blockquote>
<p>Now, you can figure out the <strong>mass</strong> of hydrogen present in <mathjax>#1#</mathjax> <strong>mole</strong> of aluminium hydroxide by first determining how many <em>moles</em> if hydrogen you get in <mathjax>#1#</mathjax> mole of aluminium hydroxide. </p>
<p>Since <mathjax>#1#</mathjax> <strong>mole</strong> of aluminium hydroxide contains <mathjax>#color(blue)(3)#</mathjax> <strong>moles</strong> of hydroxide anions, which in turn contain <mathjax>#1#</mathjax> <strong>mole</strong> of hydrogen <em>each</em>, you can say that you will have</p>
<blockquote>
<p><mathjax>#"1 mole Al"("OH")_ color(blue)(3) -> color(blue)(3)color(white)(a)"moles OH"^(-) -> color(blue)(3)color(white)(a)"moles H"#</mathjax></p>
</blockquote>
<p>The problem tells you that the <strong>molar mass</strong> of hydrogen is equal to <mathjax>#"1.01 g mol"^(-1)#</mathjax>. This means that <mathjax>#1#</mathjax> <strong>mole</strong> of hydrogen has a mass of <mathjax>#"1.01 g"#</mathjax>. </p>
<p>You can thus say that one mole of aluminium hydroxide contains </p>
<blockquote>
<p><mathjax>#color(blue)(3)color(red)(cancel(color(black)("moles H"))) * "1.01 g"/(1color(red)(cancel(color(black)("mole H")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("3.03 g H")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the mass of hydrogen in one mole of aluminum hydroxide?</h1>
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<div class="markdown"><p>The molar mass of hydrogen is 1.01 g/mol </p></div>
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Stefan V.
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<div class="markdown"><p><mathjax>#"3.03 g"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is figure out the <em>chemical formula</em> for aluminium hydroxide. </p>
<p>Aluminium is located in group <mathjax>#13#</mathjax> of <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, and forms <mathjax>#3+#</mathjax> cations, <mathjax>#"Al"^(3+)#</mathjax>. The hydroxide anion, <mathjax>#"OH"^(-)#</mathjax>, carries a <mathjax>#1-#</mathjax> charge, which means that a <strong>formula unit</strong> of aluminium hydroxide will look like this </p>
<blockquote>
<p><mathjax>#["Al"^(3+)] + color(blue)(3)["OH"^(-)] -> "Al"("OH")_color(blue)(3)#</mathjax></p>
</blockquote>
<p>Now, you can figure out the <strong>mass</strong> of hydrogen present in <mathjax>#1#</mathjax> <strong>mole</strong> of aluminium hydroxide by first determining how many <em>moles</em> if hydrogen you get in <mathjax>#1#</mathjax> mole of aluminium hydroxide. </p>
<p>Since <mathjax>#1#</mathjax> <strong>mole</strong> of aluminium hydroxide contains <mathjax>#color(blue)(3)#</mathjax> <strong>moles</strong> of hydroxide anions, which in turn contain <mathjax>#1#</mathjax> <strong>mole</strong> of hydrogen <em>each</em>, you can say that you will have</p>
<blockquote>
<p><mathjax>#"1 mole Al"("OH")_ color(blue)(3) -> color(blue)(3)color(white)(a)"moles OH"^(-) -> color(blue)(3)color(white)(a)"moles H"#</mathjax></p>
</blockquote>
<p>The problem tells you that the <strong>molar mass</strong> of hydrogen is equal to <mathjax>#"1.01 g mol"^(-1)#</mathjax>. This means that <mathjax>#1#</mathjax> <strong>mole</strong> of hydrogen has a mass of <mathjax>#"1.01 g"#</mathjax>. </p>
<p>You can thus say that one mole of aluminium hydroxide contains </p>
<blockquote>
<p><mathjax>#color(blue)(3)color(red)(cancel(color(black)("moles H"))) * "1.01 g"/(1color(red)(cancel(color(black)("mole H")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("3.03 g H")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | What is the mass of hydrogen in one mole of aluminum hydroxide? |
The molar mass of hydrogen is 1.01 g/mol
|
3,345 | abe2d02e-6ddd-11ea-a02e-ccda262736ce | https://socratic.org/questions/5890346f11ef6b566c597f24 | C6H5OH + 7 O2 -> 6 CO2 + 3 H2O | start chemical_equation qc_end substance 7 7 qc_end c_other Stoichiometrically qc_end end | [{"type":"other","value":"Chemical Equation [OF] the combustion"}] | [{"type":"chemical equation","value":"C6H5OH + 7 O2 -> 6 CO2 + 3 H2O"}] | [{"type":"substance name","value":"Ethanol"},{"type":"other","value":"Stoichiometrically."}] | <h1 class="questionTitle" itemprop="name">How do we represent the combustion of ethanol stoichiometrically?</h1> | null | C6H5OH + 7 O2 -> 6 CO2 + 3 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All hydrocarbons are known to combust to give carbon dioxide and water. So we write the equation accordingly:</p>
<p><mathjax>#C_6H_5OH + 7O_2 rarr 6CO_2 + 3H_2O#</mathjax></p>
<p>Are charge and mass balanced? For every reactant particle, is there a corresponding reactant particle? If there is not, there should be, and you know that you have furrher work to do. The typical order of operations is: balance the carbons; then balance the hydrogens; and lastly balance the oxygens......</p>
<p><mathjax>#C_6H_5OH + O_2 rarr 6CO_2 + H_2O#</mathjax>; <mathjax>#"carbons balanced."#</mathjax></p>
<p><mathjax>#C_6H_5OH +O_2 rarr 6CO_2 + 3H_2O#</mathjax>; <mathjax>#"hydrogens balanced."#</mathjax></p>
<p><mathjax>#C_6H_5OH + 7O_2 rarr 6CO_2 + 3H_2O#</mathjax>; <mathjax>#"oxygens balanced."#</mathjax> </p>
<p>The reaction is now stoichiometrically balanced. </p>
<p>Will this reaction be exothermic? How do you represent the combustion of ethanol, <mathjax>#C_2H_5OH#</mathjax>, of hexane, <mathjax>#C_6H_14#</mathjax>? The hexane combustion is a trickier proposition on the basis of arithmetic. Try it out. </p>
<p>Especially with aromatics, sometimes incomplete combustion occurs to give <mathjax>#C#</mathjax>, as soot, and <mathjax>#CO#</mathjax> as combustion products. </p>
<p>For more of the same see <a href="https://socratic.org/questions/can-you-explain-balancing-chemical-equations-in-detail#365584">here.</a> </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p>Balance mass; balance charge.</p>
<p><mathjax>#C_6H_5OH + 7O_2 rarr 6CO_2 + 3H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All hydrocarbons are known to combust to give carbon dioxide and water. So we write the equation accordingly:</p>
<p><mathjax>#C_6H_5OH + 7O_2 rarr 6CO_2 + 3H_2O#</mathjax></p>
<p>Are charge and mass balanced? For every reactant particle, is there a corresponding reactant particle? If there is not, there should be, and you know that you have furrher work to do. The typical order of operations is: balance the carbons; then balance the hydrogens; and lastly balance the oxygens......</p>
<p><mathjax>#C_6H_5OH + O_2 rarr 6CO_2 + H_2O#</mathjax>; <mathjax>#"carbons balanced."#</mathjax></p>
<p><mathjax>#C_6H_5OH +O_2 rarr 6CO_2 + 3H_2O#</mathjax>; <mathjax>#"hydrogens balanced."#</mathjax></p>
<p><mathjax>#C_6H_5OH + 7O_2 rarr 6CO_2 + 3H_2O#</mathjax>; <mathjax>#"oxygens balanced."#</mathjax> </p>
<p>The reaction is now stoichiometrically balanced. </p>
<p>Will this reaction be exothermic? How do you represent the combustion of ethanol, <mathjax>#C_2H_5OH#</mathjax>, of hexane, <mathjax>#C_6H_14#</mathjax>? The hexane combustion is a trickier proposition on the basis of arithmetic. Try it out. </p>
<p>Especially with aromatics, sometimes incomplete combustion occurs to give <mathjax>#C#</mathjax>, as soot, and <mathjax>#CO#</mathjax> as combustion products. </p>
<p>For more of the same see <a href="https://socratic.org/questions/can-you-explain-balancing-chemical-equations-in-detail#365584">here.</a> </p></div>
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<h1 class="questionTitle" itemprop="name">How do we represent the combustion of ethanol stoichiometrically?</h1>
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Stephen G.
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<div class="markdown"><p>Balance mass; balance charge.</p>
<p><mathjax>#C_6H_5OH + 7O_2 rarr 6CO_2 + 3H_2O#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All hydrocarbons are known to combust to give carbon dioxide and water. So we write the equation accordingly:</p>
<p><mathjax>#C_6H_5OH + 7O_2 rarr 6CO_2 + 3H_2O#</mathjax></p>
<p>Are charge and mass balanced? For every reactant particle, is there a corresponding reactant particle? If there is not, there should be, and you know that you have furrher work to do. The typical order of operations is: balance the carbons; then balance the hydrogens; and lastly balance the oxygens......</p>
<p><mathjax>#C_6H_5OH + O_2 rarr 6CO_2 + H_2O#</mathjax>; <mathjax>#"carbons balanced."#</mathjax></p>
<p><mathjax>#C_6H_5OH +O_2 rarr 6CO_2 + 3H_2O#</mathjax>; <mathjax>#"hydrogens balanced."#</mathjax></p>
<p><mathjax>#C_6H_5OH + 7O_2 rarr 6CO_2 + 3H_2O#</mathjax>; <mathjax>#"oxygens balanced."#</mathjax> </p>
<p>The reaction is now stoichiometrically balanced. </p>
<p>Will this reaction be exothermic? How do you represent the combustion of ethanol, <mathjax>#C_2H_5OH#</mathjax>, of hexane, <mathjax>#C_6H_14#</mathjax>? The hexane combustion is a trickier proposition on the basis of arithmetic. Try it out. </p>
<p>Especially with aromatics, sometimes incomplete combustion occurs to give <mathjax>#C#</mathjax>, as soot, and <mathjax>#CO#</mathjax> as combustion products. </p>
<p>For more of the same see <a href="https://socratic.org/questions/can-you-explain-balancing-chemical-equations-in-detail#365584">here.</a> </p></div>
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</article> | How do we represent the combustion of ethanol stoichiometrically? | null |
3,346 | ab2386d5-6ddd-11ea-a782-ccda262736ce | https://socratic.org/questions/how-do-you-balance-nh-4-2co-3-nh-3-co-2-h-2o | (NH4)2CO3 -> 2 NH3 + CO2 + H2O | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"(NH4)2CO3 -> 2 NH3 + CO2 + H2O"}] | [{"type":"chemical equation","value":"(NH4)2CO3 -> NH3 + CO2 + H2O"}] | <h1 class="questionTitle" itemprop="name">How do you balance #(NH_4)_2CO_3 -> NH_3 + CO_2 + H_2O#?</h1> | null | (NH4)2CO3 -> 2 NH3 + CO2 + H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><ol>
<li>First is to account the initial number of atoms as reference, both the reactant and the products using the T balance; </li>
<li>Present the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> according to its appearance in the reactant side and copy the same in the product side; though it's no rule on this; <br/>
<mathjax>#Reactant TTProduct#</mathjax><br/>
<mathjax>#N=2#</mathjax>...........l<mathjax>#N=cancel1 2#</mathjax><br/>
<mathjax>#H=8#</mathjax>...........l<mathjax>#H=cancel5 8#</mathjax><br/>
<mathjax>#C=1#</mathjax>............l<mathjax>#C=1#</mathjax><br/>
<mathjax>#O=3#</mathjax>............l<mathjax>#O=3#</mathjax></li>
<li>Generally, complex <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> are always considered first and work your way to the least one;</li>
<li>Every change you make with coefficients, make sure to recount the number of atoms of the involve elements and record using your T balance and of course cancelling the previous counts;</li>
<li>In this case, putting 2 as coefficient in the first product of the reaction will suffice the equation.</li>
<li>The balanced equation is<br/>
<mathjax>#(NH_4)_2CO_3->2NH_3+CO_2+H_2O#</mathjax> </li>
</ol></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>This equation can be balanced using trial and error method. The most common method of <a href="http://socratic.org/chemistry/chemical-reactions/balancing-chemical-equations">balancing chemical equations</a>.</p>
<p><mathjax>#(NH_4)_2CO_3 -> 2NH_3 + CO_2 + H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><ol>
<li>First is to account the initial number of atoms as reference, both the reactant and the products using the T balance; </li>
<li>Present the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> according to its appearance in the reactant side and copy the same in the product side; though it's no rule on this; <br/>
<mathjax>#Reactant TTProduct#</mathjax><br/>
<mathjax>#N=2#</mathjax>...........l<mathjax>#N=cancel1 2#</mathjax><br/>
<mathjax>#H=8#</mathjax>...........l<mathjax>#H=cancel5 8#</mathjax><br/>
<mathjax>#C=1#</mathjax>............l<mathjax>#C=1#</mathjax><br/>
<mathjax>#O=3#</mathjax>............l<mathjax>#O=3#</mathjax></li>
<li>Generally, complex <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> are always considered first and work your way to the least one;</li>
<li>Every change you make with coefficients, make sure to recount the number of atoms of the involve elements and record using your T balance and of course cancelling the previous counts;</li>
<li>In this case, putting 2 as coefficient in the first product of the reaction will suffice the equation.</li>
<li>The balanced equation is<br/>
<mathjax>#(NH_4)_2CO_3->2NH_3+CO_2+H_2O#</mathjax> </li>
</ol></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How do you balance #(NH_4)_2CO_3 -> NH_3 + CO_2 + H_2O#?</h1>
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Mar 15, 2016
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<div class="markdown"><p>This equation can be balanced using trial and error method. The most common method of <a href="http://socratic.org/chemistry/chemical-reactions/balancing-chemical-equations">balancing chemical equations</a>.</p>
<p><mathjax>#(NH_4)_2CO_3 -> 2NH_3 + CO_2 + H_2O#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><ol>
<li>First is to account the initial number of atoms as reference, both the reactant and the products using the T balance; </li>
<li>Present the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> according to its appearance in the reactant side and copy the same in the product side; though it's no rule on this; <br/>
<mathjax>#Reactant TTProduct#</mathjax><br/>
<mathjax>#N=2#</mathjax>...........l<mathjax>#N=cancel1 2#</mathjax><br/>
<mathjax>#H=8#</mathjax>...........l<mathjax>#H=cancel5 8#</mathjax><br/>
<mathjax>#C=1#</mathjax>............l<mathjax>#C=1#</mathjax><br/>
<mathjax>#O=3#</mathjax>............l<mathjax>#O=3#</mathjax></li>
<li>Generally, complex <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> are always considered first and work your way to the least one;</li>
<li>Every change you make with coefficients, make sure to recount the number of atoms of the involve elements and record using your T balance and of course cancelling the previous counts;</li>
<li>In this case, putting 2 as coefficient in the first product of the reaction will suffice the equation.</li>
<li>The balanced equation is<br/>
<mathjax>#(NH_4)_2CO_3->2NH_3+CO_2+H_2O#</mathjax> </li>
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</article> | How do you balance #(NH_4)_2CO_3 -> NH_3 + CO_2 + H_2O#? | null |
3,347 | a8411cb7-6ddd-11ea-9378-ccda262736ce | https://socratic.org/questions/58f1f5f37c014946026fb9fa | 3/2 | start physical_unit 14 14 coefficient none qc_end substance 4 4 qc_end end | [{"type":"physical unit","value":"Coefficient [OF] hydrogen"}] | [{"type":"physical unit","value":"3/2"}] | [{"type":"substance name","value":"Ammonia"}] | <h1 class="questionTitle" itemprop="name">In the synthesis of ammonia from its constituent elements, what is the coefficient of hydrogen in the stoichiometric equation?</h1> | null | 3/2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>i.e. <mathjax>#1/2N_2(g)+3/2H_2(g) rightleftharpoons NH_3(g)#</mathjax></p>
<p>This is a high pressure process, which could also be represented as,</p>
<p><mathjax>#N_2 + 3H_2 rightleftharpoons 2NH_3#</mathjax></p>
<p>The reaction is normally carried out at 200 atm and high temperatures, and catalyzed on metal or metal oxide surfaces. It is one of the most important industrial reactions, in that without such a reaction we would not be able to grow nitrogenous fertilizer with which to grow food.</p>
<p>Given the bond energy of the <mathjax>#N-=N#</mathjax> bond, about the only thing this reaction has got going for it is the relative involatility/solubility of ammonia compared to the parent gases. Even a small turnover at equilibrium could be removed from the process, and the equilibrium reshunted to the right to make more product. </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p>Well, I would write <mathjax>#3/2#</mathjax>........</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>i.e. <mathjax>#1/2N_2(g)+3/2H_2(g) rightleftharpoons NH_3(g)#</mathjax></p>
<p>This is a high pressure process, which could also be represented as,</p>
<p><mathjax>#N_2 + 3H_2 rightleftharpoons 2NH_3#</mathjax></p>
<p>The reaction is normally carried out at 200 atm and high temperatures, and catalyzed on metal or metal oxide surfaces. It is one of the most important industrial reactions, in that without such a reaction we would not be able to grow nitrogenous fertilizer with which to grow food.</p>
<p>Given the bond energy of the <mathjax>#N-=N#</mathjax> bond, about the only thing this reaction has got going for it is the relative involatility/solubility of ammonia compared to the parent gases. Even a small turnover at equilibrium could be removed from the process, and the equilibrium reshunted to the right to make more product. </p></div>
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<h1 class="questionTitle" itemprop="name">In the synthesis of ammonia from its constituent elements, what is the coefficient of hydrogen in the stoichiometric equation?</h1>
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<div class="markdown"><p>Well, I would write <mathjax>#3/2#</mathjax>........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>i.e. <mathjax>#1/2N_2(g)+3/2H_2(g) rightleftharpoons NH_3(g)#</mathjax></p>
<p>This is a high pressure process, which could also be represented as,</p>
<p><mathjax>#N_2 + 3H_2 rightleftharpoons 2NH_3#</mathjax></p>
<p>The reaction is normally carried out at 200 atm and high temperatures, and catalyzed on metal or metal oxide surfaces. It is one of the most important industrial reactions, in that without such a reaction we would not be able to grow nitrogenous fertilizer with which to grow food.</p>
<p>Given the bond energy of the <mathjax>#N-=N#</mathjax> bond, about the only thing this reaction has got going for it is the relative involatility/solubility of ammonia compared to the parent gases. Even a small turnover at equilibrium could be removed from the process, and the equilibrium reshunted to the right to make more product. </p></div>
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</article> | In the synthesis of ammonia from its constituent elements, what is the coefficient of hydrogen in the stoichiometric equation? | null |
3,348 | a9af7824-6ddd-11ea-9946-ccda262736ce | https://socratic.org/questions/how-many-liters-of-oxygen-are-needed-to-exactly-react-with-25-8-g-of-methane-at- | 70 liters | start physical_unit 4 4 volume l qc_end chemical_equation 17 25 qc_end c_other OTHER qc_end physical_unit 14 14 11 12 mass qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] oxygen [IN] liters"}] | [{"type":"physical unit","value":"70 liters"}] | [{"type":"chemical equation","value":"CH4(g) + 2 O2(g) -> CO2(g) + 2 H2O(l)"},{"type":"other","value":"Exactly react."},{"type":"physical unit","value":"Mass [OF] methane [=] \\pu{25.8 g}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">How many liters of oxygen are needed to exactly react with 25.8 g of methane at STP? #CH_4(g) + 2O_2(g) -> CO_2(g) + 2H_2O(l)#?</h1> | null | 70 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And so we gots the combustion equation...</p>
<p><mathjax>#CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(l) + Delta#</mathjax></p>
<p>And so <mathjax>#"moles of methane"=(25.8*g)/(16.04*g*mol^-1)=1.536*mol#</mathjax>..</p>
<p>And so we require TWICE this molar quantity of dioxygen gas...</p>
<p>And this constitutes a VOLUME of....</p>
<p><mathjax>#2xx1.536*molxx22.4*L*mol^-1~=70*L#</mathjax></p>
<p>And why should we study this reaction? Well, it heats our homes, and cooks our dinner, and sometimes drives our cabs. In short it allows us to live beyond 30...</p></div>
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<div class="markdown"><p>Well at <mathjax>#"STP"#</mathjax> one mole of gas occupies <mathjax>#22.4*L#</mathjax> (this depends on syllabus; luckily, the definition is included as supplementary material on most exam papers....)</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And so we gots the combustion equation...</p>
<p><mathjax>#CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(l) + Delta#</mathjax></p>
<p>And so <mathjax>#"moles of methane"=(25.8*g)/(16.04*g*mol^-1)=1.536*mol#</mathjax>..</p>
<p>And so we require TWICE this molar quantity of dioxygen gas...</p>
<p>And this constitutes a VOLUME of....</p>
<p><mathjax>#2xx1.536*molxx22.4*L*mol^-1~=70*L#</mathjax></p>
<p>And why should we study this reaction? Well, it heats our homes, and cooks our dinner, and sometimes drives our cabs. In short it allows us to live beyond 30...</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many liters of oxygen are needed to exactly react with 25.8 g of methane at STP? #CH_4(g) + 2O_2(g) -> CO_2(g) + 2H_2O(l)#?</h1>
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<div class="markdown"><p>Well at <mathjax>#"STP"#</mathjax> one mole of gas occupies <mathjax>#22.4*L#</mathjax> (this depends on syllabus; luckily, the definition is included as supplementary material on most exam papers....)</p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And so we gots the combustion equation...</p>
<p><mathjax>#CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(l) + Delta#</mathjax></p>
<p>And so <mathjax>#"moles of methane"=(25.8*g)/(16.04*g*mol^-1)=1.536*mol#</mathjax>..</p>
<p>And so we require TWICE this molar quantity of dioxygen gas...</p>
<p>And this constitutes a VOLUME of....</p>
<p><mathjax>#2xx1.536*molxx22.4*L*mol^-1~=70*L#</mathjax></p>
<p>And why should we study this reaction? Well, it heats our homes, and cooks our dinner, and sometimes drives our cabs. In short it allows us to live beyond 30...</p></div>
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</article> | How many liters of oxygen are needed to exactly react with 25.8 g of methane at STP? #CH_4(g) + 2O_2(g) -> CO_2(g) + 2H_2O(l)#? | null |
3,349 | aced6d46-6ddd-11ea-b4ae-ccda262736ce | https://socratic.org/questions/if-sample-1-contains-2-98-moles-of-hydrogen-at-35-1-degrees-c-and-2-3-atm-in-a-3 | 4.12 moles | start physical_unit 7 7 mole mol qc_end physical_unit 7 7 4 5 mole qc_end physical_unit 7 7 9 11 temperature qc_end physical_unit 7 7 13 14 pressure qc_end physical_unit 19 19 17 18 volume qc_end physical_unit 19 19 28 29 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mole2 [OF] hydrogen [IN] moles"}] | [{"type":"physical unit","value":"4.12 moles"}] | [{"type":"physical unit","value":"Mole1 [OF] hydrogen [=] \\pu{2.98 moles}"},{"type":"physical unit","value":"Temperature1 [OF] hydrogen [=] \\pu{35.1 degrees C}"},{"type":"physical unit","value":"Pressure1 [OF] hydrogen [=] \\pu{2.3 atm}"},{"type":"physical unit","value":"Volume1 [OF] container [=] \\pu{32.8 L}"},{"type":"physical unit","value":"Volume2 [OF] container [=] \\pu{45.3 liters}"},{"type":"other","value":"Under the same conditions."}] | <h1 class="questionTitle" itemprop="name">If Sample #1 contains 2.98 moles of hydrogen at 35.1 degrees C and 2.3 atm in a 32.8 L container. How many moles of hydrogen are in a 45.3 liter container under the same conditions? Thank you for helping.</h1> | null | 4.12 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that because the temperature and the pressure of the gas <strong>remain constant</strong>, you can use the fact that the volume of the container is <em>directly proportional</em> to the number of moles of gas as given by <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/avogadro-s-law">Avogadro's Law</a></strong>. </p>
<p>Mathematically, this is can be written as</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(V_1/n_1 = V_2/n_2)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#V_1#</mathjax> and <mathjax>#n_1#</mathjax> represent the volume and number of moles of gas at an initial state</li>
<li><mathjax>#V_2#</mathjax> and <mathjax>#n_2#</mathjax> represent the volume and the number of moles of gas at a final state</li>
</ul>
</blockquote>
<p>This means that when temperature and pressure are kept constant, <strong>increasing</strong> the number of moles of gas present in the container will cause its volume to <strong>increase</strong>.</p>
<p>Similarly, <strong>decreasing</strong> the number of moles of gas present in the container will cause its volume to <strong>decrease</strong>. </p>
<p>In your case, the volume of the container increased</p>
<blockquote>
<p><mathjax>#"32.8 L " -> " 45.3 L"#</mathjax></p>
</blockquote>
<p>you can say that the number of moles of gas present in the container must have <strong>increased</strong>. </p>
<p>Rearrange the equation to solve for <mathjax>#n_2#</mathjax></p>
<blockquote>
<p><mathjax>#V_1/n_1 = V_2/n_2 implies n_2 = V_2/V_1 * n_1#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#n_2 = (45.3 color(red)(cancel(color(black)("L"))))/(32.8color(red)(cancel(color(black)("L")))) * "2.98 moles" = color(darkgreen)(ul(color(black)("4.12 moles")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"4.12 moles"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that because the temperature and the pressure of the gas <strong>remain constant</strong>, you can use the fact that the volume of the container is <em>directly proportional</em> to the number of moles of gas as given by <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/avogadro-s-law">Avogadro's Law</a></strong>. </p>
<p>Mathematically, this is can be written as</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(V_1/n_1 = V_2/n_2)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#V_1#</mathjax> and <mathjax>#n_1#</mathjax> represent the volume and number of moles of gas at an initial state</li>
<li><mathjax>#V_2#</mathjax> and <mathjax>#n_2#</mathjax> represent the volume and the number of moles of gas at a final state</li>
</ul>
</blockquote>
<p>This means that when temperature and pressure are kept constant, <strong>increasing</strong> the number of moles of gas present in the container will cause its volume to <strong>increase</strong>.</p>
<p>Similarly, <strong>decreasing</strong> the number of moles of gas present in the container will cause its volume to <strong>decrease</strong>. </p>
<p>In your case, the volume of the container increased</p>
<blockquote>
<p><mathjax>#"32.8 L " -> " 45.3 L"#</mathjax></p>
</blockquote>
<p>you can say that the number of moles of gas present in the container must have <strong>increased</strong>. </p>
<p>Rearrange the equation to solve for <mathjax>#n_2#</mathjax></p>
<blockquote>
<p><mathjax>#V_1/n_1 = V_2/n_2 implies n_2 = V_2/V_1 * n_1#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#n_2 = (45.3 color(red)(cancel(color(black)("L"))))/(32.8color(red)(cancel(color(black)("L")))) * "2.98 moles" = color(darkgreen)(ul(color(black)("4.12 moles")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If Sample #1 contains 2.98 moles of hydrogen at 35.1 degrees C and 2.3 atm in a 32.8 L container. How many moles of hydrogen are in a 45.3 liter container under the same conditions? Thank you for helping.</h1>
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Stefan V.
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May 14, 2017
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<div class="markdown"><p><mathjax>#"4.12 moles"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that because the temperature and the pressure of the gas <strong>remain constant</strong>, you can use the fact that the volume of the container is <em>directly proportional</em> to the number of moles of gas as given by <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/avogadro-s-law">Avogadro's Law</a></strong>. </p>
<p>Mathematically, this is can be written as</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(V_1/n_1 = V_2/n_2)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#V_1#</mathjax> and <mathjax>#n_1#</mathjax> represent the volume and number of moles of gas at an initial state</li>
<li><mathjax>#V_2#</mathjax> and <mathjax>#n_2#</mathjax> represent the volume and the number of moles of gas at a final state</li>
</ul>
</blockquote>
<p>This means that when temperature and pressure are kept constant, <strong>increasing</strong> the number of moles of gas present in the container will cause its volume to <strong>increase</strong>.</p>
<p>Similarly, <strong>decreasing</strong> the number of moles of gas present in the container will cause its volume to <strong>decrease</strong>. </p>
<p>In your case, the volume of the container increased</p>
<blockquote>
<p><mathjax>#"32.8 L " -> " 45.3 L"#</mathjax></p>
</blockquote>
<p>you can say that the number of moles of gas present in the container must have <strong>increased</strong>. </p>
<p>Rearrange the equation to solve for <mathjax>#n_2#</mathjax></p>
<blockquote>
<p><mathjax>#V_1/n_1 = V_2/n_2 implies n_2 = V_2/V_1 * n_1#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#n_2 = (45.3 color(red)(cancel(color(black)("L"))))/(32.8color(red)(cancel(color(black)("L")))) * "2.98 moles" = color(darkgreen)(ul(color(black)("4.12 moles")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | If Sample #1 contains 2.98 moles of hydrogen at 35.1 degrees C and 2.3 atm in a 32.8 L container. How many moles of hydrogen are in a 45.3 liter container under the same conditions? Thank you for helping. | null |
3,350 | ab6823b5-6ddd-11ea-9c4f-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-solution-contains-250-moles-of-kcl-in-8-00-l-of-soluti | 0.03 M | start physical_unit 11 11 molarity mol/l qc_end physical_unit 11 11 8 9 mole qc_end physical_unit 6 6 13 14 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] KCl solution [IN] M"}] | [{"type":"physical unit","value":"0.03 M"}] | [{"type":"physical unit","value":"Mole [OF] KCl [=] \\pu{0.250 moles}"},{"type":"physical unit","value":"Volume [OF] KCl solution [=] \\pu{8.00 L}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of a solution contains .250 moles of #KCl# in 8.00 L of solution?</h1> | null | 0.03 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus <mathjax>#"Molarity"=(0.250*mol)/(8.00*L)~=0.03*mol*L^-1#</mathjax> with respect to <mathjax>#KCl#</mathjax>.</p>
<p>How many grams of salt are dissolved in a litre volume of solution?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#"Molarity"="Moles of solute"/"Volume of solution"~=0.03*mol*L^-1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus <mathjax>#"Molarity"=(0.250*mol)/(8.00*L)~=0.03*mol*L^-1#</mathjax> with respect to <mathjax>#KCl#</mathjax>.</p>
<p>How many grams of salt are dissolved in a litre volume of solution?</p></div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of a solution contains .250 moles of #KCl# in 8.00 L of solution?</h1>
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<div class="markdown"><p><mathjax>#"Molarity"="Moles of solute"/"Volume of solution"~=0.03*mol*L^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And thus <mathjax>#"Molarity"=(0.250*mol)/(8.00*L)~=0.03*mol*L^-1#</mathjax> with respect to <mathjax>#KCl#</mathjax>.</p>
<p>How many grams of salt are dissolved in a litre volume of solution?</p></div>
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</article> | What is the molarity of a solution contains .250 moles of #KCl# in 8.00 L of solution? | null |
3,351 | ab50f5c6-6ddd-11ea-9e79-ccda262736ce | https://socratic.org/questions/how-do-you-balance-c-3h-6-o-2-co-h-2o | C3H6 + 3 O2 -> 3 CO + 3 H2O | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"C3H6 + 3 O2 -> 3 CO + 3 H2O"}] | [{"type":"chemical equation","value":"C3H6 + O2 -> CO + H2O"}] | <h1 class="questionTitle" itemprop="name">How do you balance #C_3H_6 + O_2 -> CO + H_2O#?</h1> | null | C3H6 + 3 O2 -> 3 CO + 3 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To balance, we add molecules such as to ensure that the number of atoms of each element is the same on each side of the equation.</p>
<p>This equation represents the incomplete combustion of propane as insufficient oxygen must have been present since carbon monoxide was form instead of carbon dioxide which would form if there was sufficient oxygen to ensure complete combustion.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#C_3H_6+3O_2->3CO+3H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To balance, we add molecules such as to ensure that the number of atoms of each element is the same on each side of the equation.</p>
<p>This equation represents the incomplete combustion of propane as insufficient oxygen must have been present since carbon monoxide was form instead of carbon dioxide which would form if there was sufficient oxygen to ensure complete combustion.</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you balance #C_3H_6 + O_2 -> CO + H_2O#?</h1>
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Trevor Ryan.
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Stefan V.
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<span class="dateCreated" datetime="2016-01-29T19:20:20" itemprop="dateCreated">
Jan 29, 2016
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<div class="markdown"><p><mathjax>#C_3H_6+3O_2->3CO+3H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To balance, we add molecules such as to ensure that the number of atoms of each element is the same on each side of the equation.</p>
<p>This equation represents the incomplete combustion of propane as insufficient oxygen must have been present since carbon monoxide was form instead of carbon dioxide which would form if there was sufficient oxygen to ensure complete combustion.</p></div>
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<span class="dateCreated" datetime="2016-01-29T19:25:21" itemprop="dateCreated">
Jan 29, 2016
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<div class="markdown"><p><mathjax>#C_3H_6 + 3O_2 rarr 3CO + 3H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We have three carbons on the reagents' side and only one on the products' side, so we need to balance that by making that equal.</p>
<p><mathjax>#C_3H_6 + O_2 rarr 3CO + H_2O#</mathjax> </p>
<p>We have six hydrogens on the reagents' side and only 2 on the products' side so we need to balance that by making them equal.</p>
<p><mathjax>#C_3H_6 + O_2 rarr 3CO + 3H_2O#</mathjax></p>
<p>We have 2 oxygens on the reagents' side and 6 on the products' side, so we need to balance that by making them equal, now, since the reagent's side has only one substance with oxygen it'll be easier to make that equal, so,</p>
<p><mathjax>#C_3H_6 + 3O_2 rarr 3CO + 3H_2O#</mathjax></p>
<p>Do note that if you had equalized the first time around by</p>
<p><mathjax>#1/3C_3H_6 + O_2 rarr CO + H_2O#</mathjax></p>
<p>It'd be already balanced, but it's considered good practice to balance equations by having integer coefficients, which we could have solved by multiplying everything by 3</p>
<p><mathjax>#C_3H_6 + 3O_2 rarr 3CO + 3H_2O#</mathjax></p></div>
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</article> | How do you balance #C_3H_6 + O_2 -> CO + H_2O#? | null |
3,352 | acaa0bd4-6ddd-11ea-befa-ccda262736ce | https://socratic.org/questions/the-volume-of-400-ml-of-chlorine-gas-at-400-mm-hg-is-decreased-to-200-ml-at-cons | 800.00 mmHg | start physical_unit 6 7 pressure mmhg qc_end physical_unit 6 7 3 4 volume qc_end physical_unit 6 7 9 10 pressure qc_end physical_unit 6 7 14 15 volume qc_end c_other constant_temperature qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] chlorine gas [IN] mmHg"}] | [{"type":"physical unit","value":"800.00 mmHg"}] | [{"type":"physical unit","value":"Volume1 [OF] chlorine gas [=] \\pu{400 mL}"},{"type":"physical unit","value":"Pressure1 [OF] chlorine gas [=] \\pu{400 mmHg}"},{"type":"physical unit","value":"Volume2 [OF] chlorine gas [=] \\pu{200 mL}"},{"type":"other","value":"ConstantTemperature"}] | <h1 class="questionTitle" itemprop="name">The volume of 400 mL of chlorine gas at 400 mm Hg is decreased to 200 mL at constant temperature. What is the new gas pressure? </h1> | null | 800.00 mmHg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>, which states that at constant mass and temperature, the volume of a gas is inversely proportional to the pressure.</p>
<p>The equation for Boyle's law is <mathjax>#P_1V_1=P_2V_2"#</mathjax>.</p>
<p><strong>Given</strong><br/>
<mathjax>#V_1="400 mL"#</mathjax><br/>
<mathjax>#P_1="400 mmHg"#</mathjax><br/>
<mathjax>#V_2="200 mL"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#P_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#P_2#</mathjax>. Substitute the given values into the equation and solve.</p>
<p><mathjax>#P_1V_1=P_2V_2"#</mathjax></p>
<p><mathjax>#P_2=(P_1V_1)/V_2#</mathjax></p>
<p><mathjax>#P_2=((400"mmHg") * (400"mL"))/(200"mL")="800 mmHg"#</mathjax></p></div>
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<div class="answerSummary">
<div>
<div class="markdown"><p>The new <a href="http://socratic.org/chemistry/the-behavior-of-gases/gas-pressure">gas pressure</a> is 800 mmHg.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>, which states that at constant mass and temperature, the volume of a gas is inversely proportional to the pressure.</p>
<p>The equation for Boyle's law is <mathjax>#P_1V_1=P_2V_2"#</mathjax>.</p>
<p><strong>Given</strong><br/>
<mathjax>#V_1="400 mL"#</mathjax><br/>
<mathjax>#P_1="400 mmHg"#</mathjax><br/>
<mathjax>#V_2="200 mL"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#P_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#P_2#</mathjax>. Substitute the given values into the equation and solve.</p>
<p><mathjax>#P_1V_1=P_2V_2"#</mathjax></p>
<p><mathjax>#P_2=(P_1V_1)/V_2#</mathjax></p>
<p><mathjax>#P_2=((400"mmHg") * (400"mL"))/(200"mL")="800 mmHg"#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">The volume of 400 mL of chlorine gas at 400 mm Hg is decreased to 200 mL at constant temperature. What is the new gas pressure? </h1>
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<div class="markdown"><p>The new <a href="http://socratic.org/chemistry/the-behavior-of-gases/gas-pressure">gas pressure</a> is 800 mmHg.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>, which states that at constant mass and temperature, the volume of a gas is inversely proportional to the pressure.</p>
<p>The equation for Boyle's law is <mathjax>#P_1V_1=P_2V_2"#</mathjax>.</p>
<p><strong>Given</strong><br/>
<mathjax>#V_1="400 mL"#</mathjax><br/>
<mathjax>#P_1="400 mmHg"#</mathjax><br/>
<mathjax>#V_2="200 mL"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#P_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#P_2#</mathjax>. Substitute the given values into the equation and solve.</p>
<p><mathjax>#P_1V_1=P_2V_2"#</mathjax></p>
<p><mathjax>#P_2=(P_1V_1)/V_2#</mathjax></p>
<p><mathjax>#P_2=((400"mmHg") * (400"mL"))/(200"mL")="800 mmHg"#</mathjax></p></div>
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</article> | The volume of 400 mL of chlorine gas at 400 mm Hg is decreased to 200 mL at constant temperature. What is the new gas pressure? | null |
3,353 | ab0a6915-6ddd-11ea-8f59-ccda262736ce | https://socratic.org/questions/how-many-ml-of-o-2-gas-at-25-c-and-755-mm-hg-pressure-can-be-produced-from-the-t | 82.9 mL | start physical_unit 4 5 volume ml qc_end physical_unit 4 5 7 8 temperature qc_end physical_unit 4 5 10 11 pressure qc_end physical_unit 24 24 21 22 mass qc_end end | [{"type":"physical unit","value":"Volume [OF] O2 gas [IN] mL"}] | [{"type":"physical unit","value":"82.9 mL"}] | [{"type":"physical unit","value":"Temperature [OF] O2 gas [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Pressure [OF] O2 gas [=] \\pu{755 mmHg}"},{"type":"physical unit","value":"Mass [OF] KClO3 [=] \\pu{0.300 grams}"}] | <h1 class="questionTitle" itemprop="name">How many mL of #O_2# gas at 25°C and 755 mm Hg pressure can be produced from the thermal decomposition of 0.300 grams of #KClO_3#?</h1> | null | 82.9 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Write the balanced chemical equation.</strong></p>
<p>The balanced equation is</p>
<p><mathjax>#"2KClO"_3 → "2KCl" + "3O"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Strategy</strong></p>
<p>The problem is to convert grams of <mathjax>#"KClO"_3#</mathjax> to millilitres of <mathjax>#"O"_2#</mathjax>.</p>
<p>We can use the flow chart below to help us.</p>
<p><img alt="Flow Chart" src="https://useruploads.socratic.org/CMNjQO19TnKQdvFfQouh_Flow%20Chart.jpg"/><br/>
(Adapted from www.lsua.us)</p>
<blockquote></blockquote>
<p>The process is:</p>
<p><strong>(a)</strong> Use the <strong>molar mass</strong> to convert mass of <mathjax>#"KClO"_3#</mathjax> to moles of <mathjax>#"KClO"_3#</mathjax>.<br/>
<strong>(b)</strong> Use the <strong>molar ratio</strong> (from the balanced equation) to convert moles of <mathjax>#"KClO"_3#</mathjax> to moles of <mathjax>#"O"_2#</mathjax>.<br/>
<strong>(e)</strong> Use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a></strong> to convert moles of <mathjax>#"O"_2#</mathjax> to volume of <mathjax>#"O"_2#</mathjax>.</p>
<blockquote></blockquote>
<p>In equation form,</p>
<p><mathjax>#"grams of KClO"_3 stackrelcolor(blue)("molar mass"color(white)(ml)) (→) "moles of KClO"_3 stackrelcolor(blue)("molar ratio"color(white)(ml))→ "moles of O"_2 stackrelcolor(blue)("Ideal Gas Law"color(white)(ml))(→) "volume of O"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>The Calculations</strong></p>
<p><strong>(a) Moles of <mathjax>#"KClO"_3#</mathjax></strong></p>
<p><mathjax>#0.300 color(red)(cancel(color(black)("g KClO"_3))) × ("1 mol KClO"_3)/( 122.55 color(red)(cancel(color(black)("g KClO"_3)))) = "0.002 448 mol KClO"_3 #</mathjax></p>
<blockquote></blockquote>
<p><strong>(b) Moles of <mathjax>#"O"_2#</mathjax></strong></p>
<p><mathjax>#0.002 448color(red)(cancel(color(black)("mol KClO"_3))) × ("3 mol O"_2)/(2 color(red)(cancel(color(black)("mol KClO"_3)))) = "0.003 672 mol O"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>(c) Volume of <mathjax>#"O"_2#</mathjax></strong></p>
<p>The Ideal Gas Law is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this to give</p>
<p><mathjax>#V = (nRT)/P#</mathjax></p>
<p><mathjax>#n = "0.003 672 mol"#</mathjax><br/>
<mathjax>#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "(25 + 273.15) K" = "298.15 K"#</mathjax></p>
<p><mathjax>#P = 755 color(red)(cancel(color(black)("mmHg"))) × "1 atm"/(760 color(red)(cancel(color(black)("mmHg")))) = "0.9934 atm"#</mathjax></p>
<p>∴ <mathjax>#V = ("0.003 672" color(red)(cancel(color(black)("mol"))) × "0.082 06"color(white)(l) "L"·color(red)(cancel(color(black)("atm·K"^"-1""mol"^"-1"))) × 273.15color(red)(cancel(color(black)( "K"))))/(0.9934 color(red)(cancel(color(black)("atm")))) = "0.0829 L" = " 82.9 mL"#</mathjax></p>
<p>The volume of <mathjax>#"O"_2#</mathjax> produced is <mathjax>#"82.9 mL"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The reaction will produce 82.9 mL of <mathjax>#"O"_2#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Write the balanced chemical equation.</strong></p>
<p>The balanced equation is</p>
<p><mathjax>#"2KClO"_3 → "2KCl" + "3O"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Strategy</strong></p>
<p>The problem is to convert grams of <mathjax>#"KClO"_3#</mathjax> to millilitres of <mathjax>#"O"_2#</mathjax>.</p>
<p>We can use the flow chart below to help us.</p>
<p><img alt="Flow Chart" src="https://useruploads.socratic.org/CMNjQO19TnKQdvFfQouh_Flow%20Chart.jpg"/><br/>
(Adapted from www.lsua.us)</p>
<blockquote></blockquote>
<p>The process is:</p>
<p><strong>(a)</strong> Use the <strong>molar mass</strong> to convert mass of <mathjax>#"KClO"_3#</mathjax> to moles of <mathjax>#"KClO"_3#</mathjax>.<br/>
<strong>(b)</strong> Use the <strong>molar ratio</strong> (from the balanced equation) to convert moles of <mathjax>#"KClO"_3#</mathjax> to moles of <mathjax>#"O"_2#</mathjax>.<br/>
<strong>(e)</strong> Use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a></strong> to convert moles of <mathjax>#"O"_2#</mathjax> to volume of <mathjax>#"O"_2#</mathjax>.</p>
<blockquote></blockquote>
<p>In equation form,</p>
<p><mathjax>#"grams of KClO"_3 stackrelcolor(blue)("molar mass"color(white)(ml)) (→) "moles of KClO"_3 stackrelcolor(blue)("molar ratio"color(white)(ml))→ "moles of O"_2 stackrelcolor(blue)("Ideal Gas Law"color(white)(ml))(→) "volume of O"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>The Calculations</strong></p>
<p><strong>(a) Moles of <mathjax>#"KClO"_3#</mathjax></strong></p>
<p><mathjax>#0.300 color(red)(cancel(color(black)("g KClO"_3))) × ("1 mol KClO"_3)/( 122.55 color(red)(cancel(color(black)("g KClO"_3)))) = "0.002 448 mol KClO"_3 #</mathjax></p>
<blockquote></blockquote>
<p><strong>(b) Moles of <mathjax>#"O"_2#</mathjax></strong></p>
<p><mathjax>#0.002 448color(red)(cancel(color(black)("mol KClO"_3))) × ("3 mol O"_2)/(2 color(red)(cancel(color(black)("mol KClO"_3)))) = "0.003 672 mol O"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>(c) Volume of <mathjax>#"O"_2#</mathjax></strong></p>
<p>The Ideal Gas Law is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this to give</p>
<p><mathjax>#V = (nRT)/P#</mathjax></p>
<p><mathjax>#n = "0.003 672 mol"#</mathjax><br/>
<mathjax>#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "(25 + 273.15) K" = "298.15 K"#</mathjax></p>
<p><mathjax>#P = 755 color(red)(cancel(color(black)("mmHg"))) × "1 atm"/(760 color(red)(cancel(color(black)("mmHg")))) = "0.9934 atm"#</mathjax></p>
<p>∴ <mathjax>#V = ("0.003 672" color(red)(cancel(color(black)("mol"))) × "0.082 06"color(white)(l) "L"·color(red)(cancel(color(black)("atm·K"^"-1""mol"^"-1"))) × 273.15color(red)(cancel(color(black)( "K"))))/(0.9934 color(red)(cancel(color(black)("atm")))) = "0.0829 L" = " 82.9 mL"#</mathjax></p>
<p>The volume of <mathjax>#"O"_2#</mathjax> produced is <mathjax>#"82.9 mL"#</mathjax>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many mL of #O_2# gas at 25°C and 755 mm Hg pressure can be produced from the thermal decomposition of 0.300 grams of #KClO_3#?</h1>
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Ernest Z.
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<span class="dateCreated" datetime="2016-06-22T00:20:34" itemprop="dateCreated">
Jun 22, 2016
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<div class="markdown"><p>The reaction will produce 82.9 mL of <mathjax>#"O"_2#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Write the balanced chemical equation.</strong></p>
<p>The balanced equation is</p>
<p><mathjax>#"2KClO"_3 → "2KCl" + "3O"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Strategy</strong></p>
<p>The problem is to convert grams of <mathjax>#"KClO"_3#</mathjax> to millilitres of <mathjax>#"O"_2#</mathjax>.</p>
<p>We can use the flow chart below to help us.</p>
<p><img alt="Flow Chart" src="https://useruploads.socratic.org/CMNjQO19TnKQdvFfQouh_Flow%20Chart.jpg"/><br/>
(Adapted from www.lsua.us)</p>
<blockquote></blockquote>
<p>The process is:</p>
<p><strong>(a)</strong> Use the <strong>molar mass</strong> to convert mass of <mathjax>#"KClO"_3#</mathjax> to moles of <mathjax>#"KClO"_3#</mathjax>.<br/>
<strong>(b)</strong> Use the <strong>molar ratio</strong> (from the balanced equation) to convert moles of <mathjax>#"KClO"_3#</mathjax> to moles of <mathjax>#"O"_2#</mathjax>.<br/>
<strong>(e)</strong> Use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a></strong> to convert moles of <mathjax>#"O"_2#</mathjax> to volume of <mathjax>#"O"_2#</mathjax>.</p>
<blockquote></blockquote>
<p>In equation form,</p>
<p><mathjax>#"grams of KClO"_3 stackrelcolor(blue)("molar mass"color(white)(ml)) (→) "moles of KClO"_3 stackrelcolor(blue)("molar ratio"color(white)(ml))→ "moles of O"_2 stackrelcolor(blue)("Ideal Gas Law"color(white)(ml))(→) "volume of O"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>The Calculations</strong></p>
<p><strong>(a) Moles of <mathjax>#"KClO"_3#</mathjax></strong></p>
<p><mathjax>#0.300 color(red)(cancel(color(black)("g KClO"_3))) × ("1 mol KClO"_3)/( 122.55 color(red)(cancel(color(black)("g KClO"_3)))) = "0.002 448 mol KClO"_3 #</mathjax></p>
<blockquote></blockquote>
<p><strong>(b) Moles of <mathjax>#"O"_2#</mathjax></strong></p>
<p><mathjax>#0.002 448color(red)(cancel(color(black)("mol KClO"_3))) × ("3 mol O"_2)/(2 color(red)(cancel(color(black)("mol KClO"_3)))) = "0.003 672 mol O"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>(c) Volume of <mathjax>#"O"_2#</mathjax></strong></p>
<p>The Ideal Gas Law is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this to give</p>
<p><mathjax>#V = (nRT)/P#</mathjax></p>
<p><mathjax>#n = "0.003 672 mol"#</mathjax><br/>
<mathjax>#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "(25 + 273.15) K" = "298.15 K"#</mathjax></p>
<p><mathjax>#P = 755 color(red)(cancel(color(black)("mmHg"))) × "1 atm"/(760 color(red)(cancel(color(black)("mmHg")))) = "0.9934 atm"#</mathjax></p>
<p>∴ <mathjax>#V = ("0.003 672" color(red)(cancel(color(black)("mol"))) × "0.082 06"color(white)(l) "L"·color(red)(cancel(color(black)("atm·K"^"-1""mol"^"-1"))) × 273.15color(red)(cancel(color(black)( "K"))))/(0.9934 color(red)(cancel(color(black)("atm")))) = "0.0829 L" = " 82.9 mL"#</mathjax></p>
<p>The volume of <mathjax>#"O"_2#</mathjax> produced is <mathjax>#"82.9 mL"#</mathjax>.</p></div>
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</article> | How many mL of #O_2# gas at 25°C and 755 mm Hg pressure can be produced from the thermal decomposition of 0.300 grams of #KClO_3#? | null |
3,354 | aa4391ee-6ddd-11ea-b67b-ccda262736ce | https://socratic.org/questions/how-many-grams-of-solute-are-present-in-50-ml-of-0-360-m-sodium-chloride | 1.05 grams | start physical_unit 4 4 mass g qc_end physical_unit 13 14 11 12 molarity qc_end end | [{"type":"physical unit","value":"Mass [OF] solute [IN] grams"}] | [{"type":"physical unit","value":"1.05 grams"}] | [{"type":"physical unit","value":"Volume [OF] sodium chloride solution [=] \\pu{50 mL}"},{"type":"physical unit","value":"Molarity [OF] sodium chloride solution [=] \\pu{0.360 M}"}] | <h1 class="questionTitle" itemprop="name">How many grams of solute are present in 50 mL of 0.360 M sodium chloride? </h1> | null | 1.05 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's start off with the equation for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a>:</p>
<p><img alt="www.chemtech.org" src="https://useruploads.socratic.org/WODdMpkjTZSr4s8dgEU5_12-molarity.gif"/> </p>
<p>We are given the molarity and the volume of solution. The only issue is that the volume is given in mL instead of L. This issue can be fixed by using the following conversion factor:</p>
<p><mathjax>#color(white)(aaaaaaaaaaaaaaaaaaaaa)1000mL = 1L#</mathjax></p>
<p>Therefore, if we divide 50mL by 1000mL we will obtain a value of <strong>0.05L.</strong></p>
<p>Next, the equation has to be rearranged to solve for the moles of solute:</p>
<p><strong>Moles of solute = Molarity <mathjax>#xx#</mathjax> Liters of solution</strong></p>
<p>Now, multiply 0.360 M by 0.05:</p>
<p><mathjax>#("0.360 mol")/("1 L") xx "0.05 L" = "0.018 mol"#</mathjax></p>
<p>To obtain the mass of solute, we will need to the molar mass of NaCl, which is 58.44 g/mol:</p>
<p>Finally, multiply the number of moles by 58.44 g/mol</p>
<p><mathjax>#0.018 cancel"mol" xx (58.44g)/(1cancel"mol")#</mathjax></p>
<p>Boom, here it is:<br/>
<mathjax>#1.05g#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><strong>There are 1.05g of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present.</strong></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's start off with the equation for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a>:</p>
<p><img alt="www.chemtech.org" src="https://useruploads.socratic.org/WODdMpkjTZSr4s8dgEU5_12-molarity.gif"/> </p>
<p>We are given the molarity and the volume of solution. The only issue is that the volume is given in mL instead of L. This issue can be fixed by using the following conversion factor:</p>
<p><mathjax>#color(white)(aaaaaaaaaaaaaaaaaaaaa)1000mL = 1L#</mathjax></p>
<p>Therefore, if we divide 50mL by 1000mL we will obtain a value of <strong>0.05L.</strong></p>
<p>Next, the equation has to be rearranged to solve for the moles of solute:</p>
<p><strong>Moles of solute = Molarity <mathjax>#xx#</mathjax> Liters of solution</strong></p>
<p>Now, multiply 0.360 M by 0.05:</p>
<p><mathjax>#("0.360 mol")/("1 L") xx "0.05 L" = "0.018 mol"#</mathjax></p>
<p>To obtain the mass of solute, we will need to the molar mass of NaCl, which is 58.44 g/mol:</p>
<p>Finally, multiply the number of moles by 58.44 g/mol</p>
<p><mathjax>#0.018 cancel"mol" xx (58.44g)/(1cancel"mol")#</mathjax></p>
<p>Boom, here it is:<br/>
<mathjax>#1.05g#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of solute are present in 50 mL of 0.360 M sodium chloride? </h1>
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Stefan V.
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<div class="markdown"><p><strong>There are 1.05g of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present.</strong></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's start off with the equation for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a>:</p>
<p><img alt="www.chemtech.org" src="https://useruploads.socratic.org/WODdMpkjTZSr4s8dgEU5_12-molarity.gif"/> </p>
<p>We are given the molarity and the volume of solution. The only issue is that the volume is given in mL instead of L. This issue can be fixed by using the following conversion factor:</p>
<p><mathjax>#color(white)(aaaaaaaaaaaaaaaaaaaaa)1000mL = 1L#</mathjax></p>
<p>Therefore, if we divide 50mL by 1000mL we will obtain a value of <strong>0.05L.</strong></p>
<p>Next, the equation has to be rearranged to solve for the moles of solute:</p>
<p><strong>Moles of solute = Molarity <mathjax>#xx#</mathjax> Liters of solution</strong></p>
<p>Now, multiply 0.360 M by 0.05:</p>
<p><mathjax>#("0.360 mol")/("1 L") xx "0.05 L" = "0.018 mol"#</mathjax></p>
<p>To obtain the mass of solute, we will need to the molar mass of NaCl, which is 58.44 g/mol:</p>
<p>Finally, multiply the number of moles by 58.44 g/mol</p>
<p><mathjax>#0.018 cancel"mol" xx (58.44g)/(1cancel"mol")#</mathjax></p>
<p>Boom, here it is:<br/>
<mathjax>#1.05g#</mathjax></p></div>
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</article> | How many grams of solute are present in 50 mL of 0.360 M sodium chloride? | null |
3,355 | aacae744-6ddd-11ea-8ecc-ccda262736ce | https://socratic.org/questions/if-1-000g-of-tin-metal-reacts-with-0-640g-of-fluorine-gas-what-is-the-empirical- | SnF4 | start chemical_formula qc_end physical_unit 4 5 1 2 mass qc_end physical_unit 11 12 8 9 mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] the product [IN] empirical"}] | [{"type":"chemical equation","value":"SnF4"}] | [{"type":"physical unit","value":"Mass [OF] tin metal [=] \\pu{1.000 g}"},{"type":"physical unit","value":"Mass [OF] fluorine gas [=] \\pu{0.640 g}"}] | <h1 class="questionTitle" itemprop="name">If 1.000g of tin metal reacts with 0.640g of fluorine gas, what is the empirical formula of the product?</h1> | null | SnF4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <strong>empirical formula</strong> of a given compound tells you the <strong>smallest whole number ratio</strong> in which its constituent <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> combine to form said compound. </p>
<p>In your case, an unknown compound is said to contain <em>tin</em>, <mathjax>#"Sn"#</mathjax>, and <em>fluorine</em>, <mathjax>#"F"#</mathjax>. This compound is synthesized by reacting tin metal with <strong>fluorine gas</strong>, <mathjax>#"F"_2#</mathjax>.</p>
<p>It's important to realize that fluorine gas exists as <em>diatomic molecules</em>, and so its <strong>molar mass</strong> will be <em>twice as big</em> as the molar mass of fluorine, <mathjax>#"F"#</mathjax>. </p>
<p>Grab a periodic table and look for tin and fluorine. Their molar masses are </p>
<blockquote>
<p><mathjax>#M_("M Sn") = "118.72 g mol"^(-1)#</mathjax></p>
<p><mathjax>#M_("M F") = "18.998 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>The molar mass of fluorine gas will thus be </p>
<blockquote>
<p><mathjax>#M_("M F"_ 2) = 2 xx M_("M F")#</mathjax></p>
<p><mathjax>#M_("M F"_ 2) = 2 xx "18.998 g mol"^(-1) = "37.996 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>This means that your sample of fluorine gas contained </p>
<blockquote>
<p><mathjax>#0.640 color(red)(cancel(color(black)("g"))) * "1 mole F"_2/(37.996color(red)(cancel(color(black)("g")))) = "0.016844 moles F"_2#</mathjax></p>
</blockquote>
<p>Since every <em>molecule</em> of <mathjax>#"F"_2#</mathjax> contains <mathjax>#2#</mathjax> <strong>atoms</strong> of <mathjax>#"F"#</mathjax></p>
<p><img alt="http://www.alamy.com/stock-photo-fluormolekl-f2-fluorine-molecule-f2-29433539.html" src="https://useruploads.socratic.org/twLlskvsT3yYgeFNiEgG_fluormolekl.f2.fluorine.molecule.f2.BKTPNR.jpg"/> </p>
<p>you know that your unknown compound will contain </p>
<blockquote>
<p><mathjax>#0.016844 color(red)(cancel(color(black)("moles F"_2))) * "2 moles F"/(1color(red)(cancel(color(black)("mole F"_2)))) = "0.033688 moles F"#</mathjax></p>
</blockquote>
<p>The sample will also contain </p>
<blockquote>
<p><mathjax>#1.000 color(red)(cancel(color(black)("g"))) * "1 mole Sn"/(118.72 color(red)(cancel(color(black)("g")))) = "0.0084232 moles Sn"#</mathjax></p>
</blockquote>
<p>To find the <strong>mole ratio</strong> that exists between the two elements in the unknown compound, divide both values by the <em>smallest one</em> </p>
<blockquote>
<p><mathjax>#"For Sn: " (0.0084232 color(red)(cancel(color(black)("moles"))))/(0.0084232color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For F: " (0.033688color(red)(cancel(color(black)("moles"))))/(0.0084232color(red)(cancel(color(black)("moles")))) = 3.999 ~~ 4#</mathjax></p>
</blockquote>
<p>Since <mathjax>#1:4#</mathjax> is already the <strong>smallest whole number ratio</strong> that can exist here, you can say that this compound has the <strong>empirical formula</strong></p>
<blockquote>
<p><mathjax>#"Sn"_1"F"_4 implies color(green)(|bar(ul(color(white)(a/a)color(black)("SnF"_4)color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"SnF"_4#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <strong>empirical formula</strong> of a given compound tells you the <strong>smallest whole number ratio</strong> in which its constituent <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> combine to form said compound. </p>
<p>In your case, an unknown compound is said to contain <em>tin</em>, <mathjax>#"Sn"#</mathjax>, and <em>fluorine</em>, <mathjax>#"F"#</mathjax>. This compound is synthesized by reacting tin metal with <strong>fluorine gas</strong>, <mathjax>#"F"_2#</mathjax>.</p>
<p>It's important to realize that fluorine gas exists as <em>diatomic molecules</em>, and so its <strong>molar mass</strong> will be <em>twice as big</em> as the molar mass of fluorine, <mathjax>#"F"#</mathjax>. </p>
<p>Grab a periodic table and look for tin and fluorine. Their molar masses are </p>
<blockquote>
<p><mathjax>#M_("M Sn") = "118.72 g mol"^(-1)#</mathjax></p>
<p><mathjax>#M_("M F") = "18.998 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>The molar mass of fluorine gas will thus be </p>
<blockquote>
<p><mathjax>#M_("M F"_ 2) = 2 xx M_("M F")#</mathjax></p>
<p><mathjax>#M_("M F"_ 2) = 2 xx "18.998 g mol"^(-1) = "37.996 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>This means that your sample of fluorine gas contained </p>
<blockquote>
<p><mathjax>#0.640 color(red)(cancel(color(black)("g"))) * "1 mole F"_2/(37.996color(red)(cancel(color(black)("g")))) = "0.016844 moles F"_2#</mathjax></p>
</blockquote>
<p>Since every <em>molecule</em> of <mathjax>#"F"_2#</mathjax> contains <mathjax>#2#</mathjax> <strong>atoms</strong> of <mathjax>#"F"#</mathjax></p>
<p><img alt="http://www.alamy.com/stock-photo-fluormolekl-f2-fluorine-molecule-f2-29433539.html" src="https://useruploads.socratic.org/twLlskvsT3yYgeFNiEgG_fluormolekl.f2.fluorine.molecule.f2.BKTPNR.jpg"/> </p>
<p>you know that your unknown compound will contain </p>
<blockquote>
<p><mathjax>#0.016844 color(red)(cancel(color(black)("moles F"_2))) * "2 moles F"/(1color(red)(cancel(color(black)("mole F"_2)))) = "0.033688 moles F"#</mathjax></p>
</blockquote>
<p>The sample will also contain </p>
<blockquote>
<p><mathjax>#1.000 color(red)(cancel(color(black)("g"))) * "1 mole Sn"/(118.72 color(red)(cancel(color(black)("g")))) = "0.0084232 moles Sn"#</mathjax></p>
</blockquote>
<p>To find the <strong>mole ratio</strong> that exists between the two elements in the unknown compound, divide both values by the <em>smallest one</em> </p>
<blockquote>
<p><mathjax>#"For Sn: " (0.0084232 color(red)(cancel(color(black)("moles"))))/(0.0084232color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For F: " (0.033688color(red)(cancel(color(black)("moles"))))/(0.0084232color(red)(cancel(color(black)("moles")))) = 3.999 ~~ 4#</mathjax></p>
</blockquote>
<p>Since <mathjax>#1:4#</mathjax> is already the <strong>smallest whole number ratio</strong> that can exist here, you can say that this compound has the <strong>empirical formula</strong></p>
<blockquote>
<p><mathjax>#"Sn"_1"F"_4 implies color(green)(|bar(ul(color(white)(a/a)color(black)("SnF"_4)color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">If 1.000g of tin metal reacts with 0.640g of fluorine gas, what is the empirical formula of the product?</h1>
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Stefan V.
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Aug 7, 2016
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<div class="markdown"><p><mathjax>#"SnF"_4#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <strong>empirical formula</strong> of a given compound tells you the <strong>smallest whole number ratio</strong> in which its constituent <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> combine to form said compound. </p>
<p>In your case, an unknown compound is said to contain <em>tin</em>, <mathjax>#"Sn"#</mathjax>, and <em>fluorine</em>, <mathjax>#"F"#</mathjax>. This compound is synthesized by reacting tin metal with <strong>fluorine gas</strong>, <mathjax>#"F"_2#</mathjax>.</p>
<p>It's important to realize that fluorine gas exists as <em>diatomic molecules</em>, and so its <strong>molar mass</strong> will be <em>twice as big</em> as the molar mass of fluorine, <mathjax>#"F"#</mathjax>. </p>
<p>Grab a periodic table and look for tin and fluorine. Their molar masses are </p>
<blockquote>
<p><mathjax>#M_("M Sn") = "118.72 g mol"^(-1)#</mathjax></p>
<p><mathjax>#M_("M F") = "18.998 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>The molar mass of fluorine gas will thus be </p>
<blockquote>
<p><mathjax>#M_("M F"_ 2) = 2 xx M_("M F")#</mathjax></p>
<p><mathjax>#M_("M F"_ 2) = 2 xx "18.998 g mol"^(-1) = "37.996 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>This means that your sample of fluorine gas contained </p>
<blockquote>
<p><mathjax>#0.640 color(red)(cancel(color(black)("g"))) * "1 mole F"_2/(37.996color(red)(cancel(color(black)("g")))) = "0.016844 moles F"_2#</mathjax></p>
</blockquote>
<p>Since every <em>molecule</em> of <mathjax>#"F"_2#</mathjax> contains <mathjax>#2#</mathjax> <strong>atoms</strong> of <mathjax>#"F"#</mathjax></p>
<p><img alt="http://www.alamy.com/stock-photo-fluormolekl-f2-fluorine-molecule-f2-29433539.html" src="https://useruploads.socratic.org/twLlskvsT3yYgeFNiEgG_fluormolekl.f2.fluorine.molecule.f2.BKTPNR.jpg"/> </p>
<p>you know that your unknown compound will contain </p>
<blockquote>
<p><mathjax>#0.016844 color(red)(cancel(color(black)("moles F"_2))) * "2 moles F"/(1color(red)(cancel(color(black)("mole F"_2)))) = "0.033688 moles F"#</mathjax></p>
</blockquote>
<p>The sample will also contain </p>
<blockquote>
<p><mathjax>#1.000 color(red)(cancel(color(black)("g"))) * "1 mole Sn"/(118.72 color(red)(cancel(color(black)("g")))) = "0.0084232 moles Sn"#</mathjax></p>
</blockquote>
<p>To find the <strong>mole ratio</strong> that exists between the two elements in the unknown compound, divide both values by the <em>smallest one</em> </p>
<blockquote>
<p><mathjax>#"For Sn: " (0.0084232 color(red)(cancel(color(black)("moles"))))/(0.0084232color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For F: " (0.033688color(red)(cancel(color(black)("moles"))))/(0.0084232color(red)(cancel(color(black)("moles")))) = 3.999 ~~ 4#</mathjax></p>
</blockquote>
<p>Since <mathjax>#1:4#</mathjax> is already the <strong>smallest whole number ratio</strong> that can exist here, you can say that this compound has the <strong>empirical formula</strong></p>
<blockquote>
<p><mathjax>#"Sn"_1"F"_4 implies color(green)(|bar(ul(color(white)(a/a)color(black)("SnF"_4)color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</article> | If 1.000g of tin metal reacts with 0.640g of fluorine gas, what is the empirical formula of the product? | null |
3,356 | ad0b51c0-6ddd-11ea-93ec-ccda262736ce | https://socratic.org/questions/how-do-you-calculate-the-number-of-moles-in-0-135-grams-of-hno-3 | 2.14 × 10^(-3) moles | start physical_unit 12 12 mole mol qc_end physical_unit 12 12 9 10 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] HNO3 [IN] moles"}] | [{"type":"physical unit","value":"2.14 × 10^(-3) moles"}] | [{"type":"physical unit","value":"Mass [OF] HNO3 [=] \\pu{0.135 grams}"}] | <h1 class="questionTitle" itemprop="name">How do you calculate the number of moles in 0.135 grams of #HNO_3#?</h1> | null | 2.14 × 10^(-3) moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to calculate the moles in 0.135 grams of <mathjax>#HNO_3#</mathjax></p>
<p>We begin by finding the molar mass of <mathjax>#HNO_3#</mathjax> using the atomic masses from <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>. </p>
<p><mathjax>#H = 1 x 1.01 = 1.01#</mathjax><br/>
<mathjax>#N = 1 x 14.01 = 14.01#</mathjax><br/>
<mathjax>#O= 3 x 15.99 = 47.97#</mathjax></p>
<p><mathjax>#1.01 + 14.01 + 47.97 = 62.99 g/(mol)#</mathjax></p>
<p>Now we can use a conversion factor to convert<br/>
grams to moles </p>
<p><mathjax>#0.135 cancel(g HNO_3) x (1 mol)/(62.99cancel(g HNO_3 )#</mathjax></p>
<p><mathjax>#=0.00214 mol HNO_3#</mathjax></p>
<p>or </p>
<p><mathjax>#2.14x10^-3 mol HNO_3#</mathjax> </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#=0.00214 mol HNO_3#</mathjax></p>
<p>or </p>
<p><mathjax>#2.14x10^-3 mol HNO_3#</mathjax> </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to calculate the moles in 0.135 grams of <mathjax>#HNO_3#</mathjax></p>
<p>We begin by finding the molar mass of <mathjax>#HNO_3#</mathjax> using the atomic masses from <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>. </p>
<p><mathjax>#H = 1 x 1.01 = 1.01#</mathjax><br/>
<mathjax>#N = 1 x 14.01 = 14.01#</mathjax><br/>
<mathjax>#O= 3 x 15.99 = 47.97#</mathjax></p>
<p><mathjax>#1.01 + 14.01 + 47.97 = 62.99 g/(mol)#</mathjax></p>
<p>Now we can use a conversion factor to convert<br/>
grams to moles </p>
<p><mathjax>#0.135 cancel(g HNO_3) x (1 mol)/(62.99cancel(g HNO_3 )#</mathjax></p>
<p><mathjax>#=0.00214 mol HNO_3#</mathjax></p>
<p>or </p>
<p><mathjax>#2.14x10^-3 mol HNO_3#</mathjax> </p></div>
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<h1 class="questionTitle" itemprop="name">How do you calculate the number of moles in 0.135 grams of #HNO_3#?</h1>
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<div class="markdown"><p><mathjax>#=0.00214 mol HNO_3#</mathjax></p>
<p>or </p>
<p><mathjax>#2.14x10^-3 mol HNO_3#</mathjax> </p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to calculate the moles in 0.135 grams of <mathjax>#HNO_3#</mathjax></p>
<p>We begin by finding the molar mass of <mathjax>#HNO_3#</mathjax> using the atomic masses from <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>. </p>
<p><mathjax>#H = 1 x 1.01 = 1.01#</mathjax><br/>
<mathjax>#N = 1 x 14.01 = 14.01#</mathjax><br/>
<mathjax>#O= 3 x 15.99 = 47.97#</mathjax></p>
<p><mathjax>#1.01 + 14.01 + 47.97 = 62.99 g/(mol)#</mathjax></p>
<p>Now we can use a conversion factor to convert<br/>
grams to moles </p>
<p><mathjax>#0.135 cancel(g HNO_3) x (1 mol)/(62.99cancel(g HNO_3 )#</mathjax></p>
<p><mathjax>#=0.00214 mol HNO_3#</mathjax></p>
<p>or </p>
<p><mathjax>#2.14x10^-3 mol HNO_3#</mathjax> </p></div>
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Edmund Liew
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<div class="markdown"><p>Assume that:<br/>
n = number of moles <br/>
m = mass of substance<br/>
M = molar mass</p>
<p><mathjax>#n = m -: M#</mathjax></p>
<p>0.135 grams (m) of <mathjax>#HNO_3#</mathjax> has been provided for you.</p>
<p>Now you need to find the molar mass (M). If you look closely at the compound formula, you can see that Hydrogen, Nitrogen and Oxygen are present. The molar mass of: Hydrogen is 1.0 g/mol, Nitrogen is 14.0 g/mol and Oxygen 16.0 g/mol.<br/>
Therefore, the molar mass of <mathjax>#HNO_3#</mathjax> is:<br/>
<mathjax>#[1 xx 1.0 + 1 xx 14.0 + 3 xx 16.0]#</mathjax> = 63.0 g/mol.</p>
<p>Looking back at the <mathjax>#n = m -: M#</mathjax> formula,<br/>
the number of moles (n) present is: <br/>
<mathjax>#[n = 0.135 -: 63.0]#</mathjax> = 0.00214 moles.<br/>
Note: Answer rounded to 3 <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>.</p></div>
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</article> | How do you calculate the number of moles in 0.135 grams of #HNO_3#? | null |
3,357 | a99c0c9c-6ddd-11ea-b136-ccda262736ce | https://socratic.org/questions/how-do-you-balance-ni-c-4h-8n-2o-2-ni-c-4h-8n-2o-2-2 | Ni + 2 C4H8N2O2 -> Ni(C4H8N2O2)2 | start chemical_equation qc_end chemical_equation 4 8 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"Ni + 2 C4H8N2O2 -> Ni(C4H8N2O2)2"}] | [{"type":"chemical equation","value":"Ni + C4H8N2O2 -> Ni(C4H8N2O2)2"}] | <h1 class="questionTitle" itemprop="name">How do you balance #Ni + C_4H_8N_2O_2 -> Ni(C_4H_8N_2O_2)_2#?</h1> | null | Ni + 2 C4H8N2O2 -> Ni(C4H8N2O2)2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Ni + C_4H_8N_2O_2 -> Ni(C_4H_8N_2O_2)_2#</mathjax><br/>
as there are 1 molecule nickel, 4 molecules of carbon, 8 molecules of hydrogen, 2 molecules of nitrogen, and 2 molecules of oxygen in left hand side<br/>
but 1 molecule of nickel, 8 molecules of carbon, 16 molecules of hydrogen, 4 molecules of nitrogen, and 4 molecules of oxygen<br/>
so, by multiplying <mathjax>#C_4H_8N_2O_2#</mathjax> by <mathjax>#2#</mathjax> the equation will be balanced. </p></div>
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<div class="markdown"><p><mathjax>#Ni + 2C_4H_8N_2O_2 -> Ni(C_4H_8N_2O_2)_2#</mathjax> </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Ni + C_4H_8N_2O_2 -> Ni(C_4H_8N_2O_2)_2#</mathjax><br/>
as there are 1 molecule nickel, 4 molecules of carbon, 8 molecules of hydrogen, 2 molecules of nitrogen, and 2 molecules of oxygen in left hand side<br/>
but 1 molecule of nickel, 8 molecules of carbon, 16 molecules of hydrogen, 4 molecules of nitrogen, and 4 molecules of oxygen<br/>
so, by multiplying <mathjax>#C_4H_8N_2O_2#</mathjax> by <mathjax>#2#</mathjax> the equation will be balanced. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you balance #Ni + C_4H_8N_2O_2 -> Ni(C_4H_8N_2O_2)_2#?</h1>
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Sihan Tawsik
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<div class="markdown"><p><mathjax>#Ni + 2C_4H_8N_2O_2 -> Ni(C_4H_8N_2O_2)_2#</mathjax> </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#Ni + C_4H_8N_2O_2 -> Ni(C_4H_8N_2O_2)_2#</mathjax><br/>
as there are 1 molecule nickel, 4 molecules of carbon, 8 molecules of hydrogen, 2 molecules of nitrogen, and 2 molecules of oxygen in left hand side<br/>
but 1 molecule of nickel, 8 molecules of carbon, 16 molecules of hydrogen, 4 molecules of nitrogen, and 4 molecules of oxygen<br/>
so, by multiplying <mathjax>#C_4H_8N_2O_2#</mathjax> by <mathjax>#2#</mathjax> the equation will be balanced. </p></div>
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Stefan V.
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<div class="markdown"><p><mathjax>#"Ni"_text((aq])^(2+) + 2"C"_4"H"_8"N"_2"O"_text(2(aq]) -> "Ni"("C"_4"H"_7"N"_2"O"_2)_text(2(s]) darr + 2"H"_text((aq])^(+)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The starting chemical equation given to you is actually <strong>incorrect</strong>. </p>
<p>This reaction takes place in <em>aqueous solution</em> and it involves the nickel(II) cation, <mathjax>#"Ni"^(2+)#</mathjax>, <strong>not</strong> nickel metal, <mathjax>#"Ni"#</mathjax>. </p>
<p>Reacting nickel(II) cations with <em>dimethylglyoxime</em>, <mathjax>#"C"_4"H"_8"N"_2"O"_2#</mathjax>, will produce an <strong>insolube solid</strong> called <em>nickel dimethylglyoxime</em>, <mathjax>#"Ni"("C"_4"H"_8"N"_2"O"_2)_2#</mathjax>, which will precipitate out of solution. </p>
<p>The nickel(II) cations can be delivered to the solution by a soluble salt like <em>nickel(II) nitrate</em>, <mathjax>#"Ni"("NO"_3)_2#</mathjax>. </p>
<p>The actual <strong>balanced</strong> chemical equation for this reaction can be written like this </p>
<blockquote>
<p><mathjax>#"Ni"_text((aq])^(2+) + 2"C"_4"H"_8"N"_2"O"_text(2(aq]) -> "Ni"("C"_4"H"_7"N"_2"O"_2)_text(2(s]) darr + 2"H"_text((aq])^(+)#</mathjax></p>
</blockquote>
<p>The reaction involves two dimethylglyoxime molecules acting as <strong>chelating agents</strong> to form the nickel dymethylglyoxime complex. </p>
<p><img alt="http://www.che.tohoku.ac.jp/~analchem/precipitate/index_e.html" src="https://useruploads.socratic.org/mnCgsY3TEeWURODXa8JS_react.gif"/> </p>
<p>When the nickel(II) cations are delivered via nickel(II) nitrate, the reaction will produce </p>
<blockquote>
<p><mathjax>#"Ni"("NO"_3)_text(2(aq]) + 2"C"_4"H"_8"N"_2"O"_text(2(aq]) -> "Ni"("C"_4"H"_7"N"_2"O"_2)_text(2(s]) darr + 2"HNO"_text(3(aq])#</mathjax></p>
</blockquote>
<p>The nitrate anions, <mathjax>#"NO"_3^(-)#</mathjax>, are <em>spectator ions</em> in the reaction, meaning that they can be found on both sides of the chemical equation. </p>
<p>This reaction is used as a <em>confirmation test</em> for the presence of nickel(II) cations. Nickel dymethylglyoxime is a red precipitate. </p>
<p><img alt="https://commons.wikimedia.org" src="https://useruploads.socratic.org/V3ZKd0eVR0WmWRjTxRyE_1024px-Nickel_Dimethylglyoxime_2014.05.08.jpg"/> </p></div>
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</article> | How do you balance #Ni + C_4H_8N_2O_2 -> Ni(C_4H_8N_2O_2)_2#? | null |
3,358 | a8a7bca8-6ddd-11ea-b130-ccda262736ce | https://socratic.org/questions/58f95c4d11ef6b2bf8975543 | 30.00 moles | start physical_unit 13 13 mole mol qc_end physical_unit 8 8 4 5 mole qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mole [OF] dioxygen [IN] moles"}] | [{"type":"physical unit","value":"30.00 moles"}] | [{"type":"physical unit","value":"Mole [OF] glucose [=] \\pu{5 mole}"},{"type":"other","value":"Completely combust."}] | <h1 class="questionTitle" itemprop="name">To completely combust a 5 mole quantity of glucose, how many moles of dioxygen are required?</h1> | null | 30.00 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is this stoichiometrically balanced?</p>
<p>And thus, clearly, we require a 30 mole quantity of dioxygen gas for complete combustion of a 5 mole quantity of glucose. What masses do these quantities represent?</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>Well first we write a stoichiometrically balanced equation...</p>
<p><mathjax>#C_6H_12O_6(s) + 6O_2(g) rarr 6CO_2(g) + 6H_2O(l)#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is this stoichiometrically balanced?</p>
<p>And thus, clearly, we require a 30 mole quantity of dioxygen gas for complete combustion of a 5 mole quantity of glucose. What masses do these quantities represent?</p></div>
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<div class="markdown"><p>Well first we write a stoichiometrically balanced equation...</p>
<p><mathjax>#C_6H_12O_6(s) + 6O_2(g) rarr 6CO_2(g) + 6H_2O(l)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is this stoichiometrically balanced?</p>
<p>And thus, clearly, we require a 30 mole quantity of dioxygen gas for complete combustion of a 5 mole quantity of glucose. What masses do these quantities represent?</p></div>
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</article> | To completely combust a 5 mole quantity of glucose, how many moles of dioxygen are required? | null |
3,359 | a9cac47b-6ddd-11ea-8aeb-ccda262736ce | https://socratic.org/questions/565f20dc11ef6b029208fbc7 | PO4^3- + HCl(aq) -> HPO4^2- + Cl-; HPO4^2- + HCl(aq) -> H2PO4- + Cl-; H2PO4- + HCl(aq) -> H3PO4(aq) + Cl- | start chemical_equation qc_end substance 2 3 qc_end substance 6 7 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"PO4^3- + HCl(aq) -> HPO4^2- + Cl-; HPO4^2- + HCl(aq) -> H2PO4- + Cl-; H2PO4- + HCl(aq) -> H3PO4(aq) + Cl-"}] | [{"type":"substance name","value":"Phosphate salts"},{"type":"substance name","value":"Hydrochloric acid"}] | <h1 class="questionTitle" itemprop="name">How do phosphate salts react with hydrochloric acid?</h1> | null | PO4^3- + HCl(aq) -> HPO4^2- + Cl-; HPO4^2- + HCl(aq) -> H2PO4- + Cl-; H2PO4- + HCl(aq) -> H3PO4(aq) + Cl- | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So how do you do it?</p>
<p>Any chemical reaction CONSERVES mass AND charge. What do I mean? With respect to conservation of mass, If I start with 10 g of reactants (from all sources), AT MOST I can get 10 g products (and in practice I am not even going to get that). To reiterate, mass is conserved in every chemical reaction. A quantity of sodium phosphate represents a specific mass, and is conserved throughout the reaction.</p>
<p>Not only mass is conserved, but CHARGE is conserved. The 3 reactions reflect this. On the LHS (left hand side) of (i) there was a charge of -3; on the RHS, there is also a charge of -3, as required. </p>
<p>Now it may seem that chemical reactions are quite complicated, but if you look at them, they will always be balanced; that is they conserve mass and charge. It is your job to make chemical sense, and to observe conservation. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#i.#</mathjax> <mathjax>#PO_4^(3-) + HCl(aq) rarr HPO_4^(2-) + Cl^-#</mathjax><br/>
<mathjax>#ii.#</mathjax> <mathjax>#HPO_4^(2-) + HCl(aq) rarr H_2PO_4^(-) + Cl^-#</mathjax><br/>
<mathjax>#iii.#</mathjax> <mathjax>#H_2PO_4^(-) + HCl(aq) rarr H_3PO_4(aq) + Cl^-#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So how do you do it?</p>
<p>Any chemical reaction CONSERVES mass AND charge. What do I mean? With respect to conservation of mass, If I start with 10 g of reactants (from all sources), AT MOST I can get 10 g products (and in practice I am not even going to get that). To reiterate, mass is conserved in every chemical reaction. A quantity of sodium phosphate represents a specific mass, and is conserved throughout the reaction.</p>
<p>Not only mass is conserved, but CHARGE is conserved. The 3 reactions reflect this. On the LHS (left hand side) of (i) there was a charge of -3; on the RHS, there is also a charge of -3, as required. </p>
<p>Now it may seem that chemical reactions are quite complicated, but if you look at them, they will always be balanced; that is they conserve mass and charge. It is your job to make chemical sense, and to observe conservation. </p></div>
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<h1 class="questionTitle" itemprop="name">How do phosphate salts react with hydrochloric acid?</h1>
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<div class="markdown"><p><mathjax>#i.#</mathjax> <mathjax>#PO_4^(3-) + HCl(aq) rarr HPO_4^(2-) + Cl^-#</mathjax><br/>
<mathjax>#ii.#</mathjax> <mathjax>#HPO_4^(2-) + HCl(aq) rarr H_2PO_4^(-) + Cl^-#</mathjax><br/>
<mathjax>#iii.#</mathjax> <mathjax>#H_2PO_4^(-) + HCl(aq) rarr H_3PO_4(aq) + Cl^-#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So how do you do it?</p>
<p>Any chemical reaction CONSERVES mass AND charge. What do I mean? With respect to conservation of mass, If I start with 10 g of reactants (from all sources), AT MOST I can get 10 g products (and in practice I am not even going to get that). To reiterate, mass is conserved in every chemical reaction. A quantity of sodium phosphate represents a specific mass, and is conserved throughout the reaction.</p>
<p>Not only mass is conserved, but CHARGE is conserved. The 3 reactions reflect this. On the LHS (left hand side) of (i) there was a charge of -3; on the RHS, there is also a charge of -3, as required. </p>
<p>Now it may seem that chemical reactions are quite complicated, but if you look at them, they will always be balanced; that is they conserve mass and charge. It is your job to make chemical sense, and to observe conservation. </p></div>
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</article> | How do phosphate salts react with hydrochloric acid? | null |
3,360 | a95e1615-6ddd-11ea-97e3-ccda262736ce | https://socratic.org/questions/5943fc4fb72cff72ff2454de | 11.90 | start physical_unit 5 5 ph none qc_end physical_unit 11 11 15 17 pkb qc_end physical_unit 11 11 9 10 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] ammonia solution"}] | [{"type":"physical unit","value":"11.90"}] | [{"type":"physical unit","value":"pKb [OF] ammonia [=] \\pu{4.3 × 10^(-5)}"},{"type":"physical unit","value":"Molarity [OF] ammonia solution [=] \\pu{1.5 mol/L}"}] | <h1 class="questionTitle" itemprop="name">What is #pH# of a solution nominally composed of #1.5*mol*L^-1# ammonia? #pK_b# #"ammonia"=4.3xx10^-5#</h1> | null | 11.90 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We interrogate the equilibrium.......</p>
<p><mathjax>#NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^(+) + HO^(-)#</mathjax>,</p>
<p>for which <mathjax>#K_b=([NH_4^+][HO^-])/([NH_3(aq)])=4.3xx10^-5#</mathjax></p>
<p>ANd if we propose that <mathjax>#x*mol*L^-1#</mathjax>of <mathjax>#"ammonia"#</mathjax> associates, then we can rewrite the <mathjax>#K_b#</mathjax> expression to give.......</p>
<p><mathjax>#x^2/(1.5-x)=4.3xx10^-5#</mathjax>, and if we make the approx., that <mathjax>#1.5">>"x#</mathjax>, then.......</p>
<p><mathjax>#x~=sqrt(1.5xx4.3xx10^-5)#</mathjax></p>
<p><mathjax>#x_1=0.00803*mol*L^-1#</mathjax></p>
<p><mathjax>#x_2=0.00801*mol*L^-1#</mathjax> (we plug the first approx. back into the expression...)</p>
<p><mathjax>#x_3=0.00801*mol*L^-1#</mathjax>. Since the approximations have converged </p>
<p>And thus <mathjax>#[NH_4^+]=[HO^-]=0.00801*mol*L^-1#</mathjax>, and <mathjax>#[NH_3(aq)]=1.492*mol*L^-1#</mathjax>.</p>
<p>Now <mathjax>#pOH=2.10#</mathjax>, and thus <mathjax>#pH=14-2.10=11.90#</mathjax> </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#pH=11.90#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We interrogate the equilibrium.......</p>
<p><mathjax>#NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^(+) + HO^(-)#</mathjax>,</p>
<p>for which <mathjax>#K_b=([NH_4^+][HO^-])/([NH_3(aq)])=4.3xx10^-5#</mathjax></p>
<p>ANd if we propose that <mathjax>#x*mol*L^-1#</mathjax>of <mathjax>#"ammonia"#</mathjax> associates, then we can rewrite the <mathjax>#K_b#</mathjax> expression to give.......</p>
<p><mathjax>#x^2/(1.5-x)=4.3xx10^-5#</mathjax>, and if we make the approx., that <mathjax>#1.5">>"x#</mathjax>, then.......</p>
<p><mathjax>#x~=sqrt(1.5xx4.3xx10^-5)#</mathjax></p>
<p><mathjax>#x_1=0.00803*mol*L^-1#</mathjax></p>
<p><mathjax>#x_2=0.00801*mol*L^-1#</mathjax> (we plug the first approx. back into the expression...)</p>
<p><mathjax>#x_3=0.00801*mol*L^-1#</mathjax>. Since the approximations have converged </p>
<p>And thus <mathjax>#[NH_4^+]=[HO^-]=0.00801*mol*L^-1#</mathjax>, and <mathjax>#[NH_3(aq)]=1.492*mol*L^-1#</mathjax>.</p>
<p>Now <mathjax>#pOH=2.10#</mathjax>, and thus <mathjax>#pH=14-2.10=11.90#</mathjax> </p></div>
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<h1 class="questionTitle" itemprop="name">What is #pH# of a solution nominally composed of #1.5*mol*L^-1# ammonia? #pK_b# #"ammonia"=4.3xx10^-5#</h1>
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<div class="markdown"><p><mathjax>#pH=11.90#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We interrogate the equilibrium.......</p>
<p><mathjax>#NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^(+) + HO^(-)#</mathjax>,</p>
<p>for which <mathjax>#K_b=([NH_4^+][HO^-])/([NH_3(aq)])=4.3xx10^-5#</mathjax></p>
<p>ANd if we propose that <mathjax>#x*mol*L^-1#</mathjax>of <mathjax>#"ammonia"#</mathjax> associates, then we can rewrite the <mathjax>#K_b#</mathjax> expression to give.......</p>
<p><mathjax>#x^2/(1.5-x)=4.3xx10^-5#</mathjax>, and if we make the approx., that <mathjax>#1.5">>"x#</mathjax>, then.......</p>
<p><mathjax>#x~=sqrt(1.5xx4.3xx10^-5)#</mathjax></p>
<p><mathjax>#x_1=0.00803*mol*L^-1#</mathjax></p>
<p><mathjax>#x_2=0.00801*mol*L^-1#</mathjax> (we plug the first approx. back into the expression...)</p>
<p><mathjax>#x_3=0.00801*mol*L^-1#</mathjax>. Since the approximations have converged </p>
<p>And thus <mathjax>#[NH_4^+]=[HO^-]=0.00801*mol*L^-1#</mathjax>, and <mathjax>#[NH_3(aq)]=1.492*mol*L^-1#</mathjax>.</p>
<p>Now <mathjax>#pOH=2.10#</mathjax>, and thus <mathjax>#pH=14-2.10=11.90#</mathjax> </p></div>
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</article> | What is #pH# of a solution nominally composed of #1.5*mol*L^-1# ammonia? #pK_b# #"ammonia"=4.3xx10^-5# | null |
3,361 | ad06239e-6ddd-11ea-b7a9-ccda262736ce | https://socratic.org/questions/when-lead-l-nitrate-reacts-with-sodium-iodide-what-is-the-formula-for-the-produc | PbI2 | start chemical_formula qc_end substance 1 2 qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] lead (lI) iodide [IN] default"}] | [{"type":"chemical equation","value":"PbI2"}] | [{"type":"substance name","value":"Lead(l) nitrate"},{"type":"substance name","value":"Sodium iodide"}] | <h1 class="questionTitle" itemprop="name">When lead(l) nitrate reacts with sodium iodide, what is the formula for the product lead (lI) iodide?</h1> | null | PbI2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Pb^(2+) + 2I^(-) rarr PbI_2(s)darr#</mathjax></p>
<p><mathjax>#"Lead(II) iodide"#</mathjax> or (more rarely) <mathjax>#"plumbous iodide"#</mathjax> is a water insoluble, fine yellow powder.</p>
<p>In the given reaction, both mass and charge have been conserved. What does this mean?</p>
<p>The reaction of halides with lead nitrate is often a useful way to identify an unknown halide. <mathjax>#PbCl_2#</mathjax> forms an insoluble white precipitate; <mathjax>#PbBr_2#</mathjax> forms an insoluble cream-coloured precipitate; and <mathjax>#PbI_2#</mathjax> forms an insoluble bright yellow precipitate.</p></div>
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<div class="markdown"><p><mathjax>#"Lead(II) iodide"#</mathjax> <mathjax>#-=#</mathjax> <mathjax>#PbI_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Pb^(2+) + 2I^(-) rarr PbI_2(s)darr#</mathjax></p>
<p><mathjax>#"Lead(II) iodide"#</mathjax> or (more rarely) <mathjax>#"plumbous iodide"#</mathjax> is a water insoluble, fine yellow powder.</p>
<p>In the given reaction, both mass and charge have been conserved. What does this mean?</p>
<p>The reaction of halides with lead nitrate is often a useful way to identify an unknown halide. <mathjax>#PbCl_2#</mathjax> forms an insoluble white precipitate; <mathjax>#PbBr_2#</mathjax> forms an insoluble cream-coloured precipitate; and <mathjax>#PbI_2#</mathjax> forms an insoluble bright yellow precipitate.</p></div>
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<h1 class="questionTitle" itemprop="name">When lead(l) nitrate reacts with sodium iodide, what is the formula for the product lead (lI) iodide?</h1>
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<div class="markdown"><p><mathjax>#"Lead(II) iodide"#</mathjax> <mathjax>#-=#</mathjax> <mathjax>#PbI_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Pb^(2+) + 2I^(-) rarr PbI_2(s)darr#</mathjax></p>
<p><mathjax>#"Lead(II) iodide"#</mathjax> or (more rarely) <mathjax>#"plumbous iodide"#</mathjax> is a water insoluble, fine yellow powder.</p>
<p>In the given reaction, both mass and charge have been conserved. What does this mean?</p>
<p>The reaction of halides with lead nitrate is often a useful way to identify an unknown halide. <mathjax>#PbCl_2#</mathjax> forms an insoluble white precipitate; <mathjax>#PbBr_2#</mathjax> forms an insoluble cream-coloured precipitate; and <mathjax>#PbI_2#</mathjax> forms an insoluble bright yellow precipitate.</p></div>
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</article> | When lead(l) nitrate reacts with sodium iodide, what is the formula for the product lead (lI) iodide? | null |
3,362 | aa0eaaf4-6ddd-11ea-a1ea-ccda262736ce | https://socratic.org/questions/a-sample-of-argon-gas-at-520-mm-hg-expands-from-0-150-l-to-0-300-l-if-the-temper | 260.00 mmHg | start physical_unit 1 4 pressure mmhg qc_end physical_unit 1 4 6 7 pressure qc_end physical_unit 1 4 10 11 volume qc_end physical_unit 1 4 13 14 volume qc_end c_other constant_temperature qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] argon gas sample [IN] mmHg"}] | [{"type":"physical unit","value":"260.00 mmHg"}] | [{"type":"physical unit","value":"Pressure1 [OF] argon gas sample [=] \\pu{520 mmHg}"},{"type":"physical unit","value":"Volume1 [OF] argon gas sample [=] \\pu{0.150 L}"},{"type":"physical unit","value":"Volume2 [OF] argon gas sample [=] \\pu{0.300 L}"},{"type":"other","value":"ConstantTemperature"}] | <h1 class="questionTitle" itemprop="name">A sample of argon gas at 520 mm Hg expands from 0.150 L to 0.300 L. If the temperature remains constant, what is the final pressure in mm Hg?</h1> | null | 260.00 mmHg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To solve this problem, we can use the pressure-volume relationship of gases, illustrated by <em><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a></em>:</p>
<p><mathjax>#P_1V_1 = P_2V_2#</mathjax></p>
<p>Since we're trying to find the final pressure (<mathjax>#P_2#</mathjax>), let's rearrange the equation to solve for <mathjax>#P_2#</mathjax>:</p>
<p><mathjax>#P_2 = (P_1V_1)/(V_2)#</mathjax></p>
<p>Plugging in known values, we have</p>
<p><mathjax>#P_2 = ((520"mm Hg")(0.150cancel("L")))/(0.300cancel("L")) = color(red)(260#</mathjax> <mathjax>#color(red)("mm Hg"#</mathjax></p>
<p>The final pressure is thus <mathjax>#260#</mathjax> millimeters mercury.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#260#</mathjax> <mathjax>#"mm Hg"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To solve this problem, we can use the pressure-volume relationship of gases, illustrated by <em><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a></em>:</p>
<p><mathjax>#P_1V_1 = P_2V_2#</mathjax></p>
<p>Since we're trying to find the final pressure (<mathjax>#P_2#</mathjax>), let's rearrange the equation to solve for <mathjax>#P_2#</mathjax>:</p>
<p><mathjax>#P_2 = (P_1V_1)/(V_2)#</mathjax></p>
<p>Plugging in known values, we have</p>
<p><mathjax>#P_2 = ((520"mm Hg")(0.150cancel("L")))/(0.300cancel("L")) = color(red)(260#</mathjax> <mathjax>#color(red)("mm Hg"#</mathjax></p>
<p>The final pressure is thus <mathjax>#260#</mathjax> millimeters mercury.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A sample of argon gas at 520 mm Hg expands from 0.150 L to 0.300 L. If the temperature remains constant, what is the final pressure in mm Hg?</h1>
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Nathan L.
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Jun 11, 2017
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<div class="markdown"><p><mathjax>#260#</mathjax> <mathjax>#"mm Hg"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To solve this problem, we can use the pressure-volume relationship of gases, illustrated by <em><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a></em>:</p>
<p><mathjax>#P_1V_1 = P_2V_2#</mathjax></p>
<p>Since we're trying to find the final pressure (<mathjax>#P_2#</mathjax>), let's rearrange the equation to solve for <mathjax>#P_2#</mathjax>:</p>
<p><mathjax>#P_2 = (P_1V_1)/(V_2)#</mathjax></p>
<p>Plugging in known values, we have</p>
<p><mathjax>#P_2 = ((520"mm Hg")(0.150cancel("L")))/(0.300cancel("L")) = color(red)(260#</mathjax> <mathjax>#color(red)("mm Hg"#</mathjax></p>
<p>The final pressure is thus <mathjax>#260#</mathjax> millimeters mercury.</p></div>
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</article> | A sample of argon gas at 520 mm Hg expands from 0.150 L to 0.300 L. If the temperature remains constant, what is the final pressure in mm Hg? | null |
3,363 | a8cab110-6ddd-11ea-a698-ccda262736ce | https://socratic.org/questions/how-many-moles-of-zinc-oxide-is-produced-if-4-44-moles-of-zinc-reacts-with-exces | 4.44 moles | start physical_unit 4 5 mole mol qc_end physical_unit 4 4 9 10 mole qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mole [OF] zinc oxide [IN] moles"}] | [{"type":"physical unit","value":"4.44 moles"}] | [{"type":"physical unit","value":"Mole [OF] zinc [=] \\pu{4.44 moles}"},{"type":"other","value":"Excess oxygen."}] | <h1 class="questionTitle" itemprop="name">How many moles of zinc oxide is produced if 4.44 moles of zinc reacts with excess oxygen?</h1> | null | 4.44 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As for all these problems, we first need a stoichiometrically balanced equation,</p>
<p><mathjax>#Zn(s) + 1/2O_2(g) rarr ZnO(s)#</mathjax></p>
<p>And thus between the oxide and the starting metal, there is clearly <mathjax>#1:1#</mathjax> equivalence. So if there are 4.44 moles of zinc, clearly there are 4.44 moles of oxide. If 10 moles of zinc oxide were produced, then how moles of zinc were used?</p></div>
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<div class="markdown"><p>Why <mathjax>#4.44#</mathjax> <mathjax>#mol#</mathjax> precisely. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As for all these problems, we first need a stoichiometrically balanced equation,</p>
<p><mathjax>#Zn(s) + 1/2O_2(g) rarr ZnO(s)#</mathjax></p>
<p>And thus between the oxide and the starting metal, there is clearly <mathjax>#1:1#</mathjax> equivalence. So if there are 4.44 moles of zinc, clearly there are 4.44 moles of oxide. If 10 moles of zinc oxide were produced, then how moles of zinc were used?</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of zinc oxide is produced if 4.44 moles of zinc reacts with excess oxygen?</h1>
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anor277
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Dec 19, 2016
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<div class="markdown"><p>Why <mathjax>#4.44#</mathjax> <mathjax>#mol#</mathjax> precisely. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As for all these problems, we first need a stoichiometrically balanced equation,</p>
<p><mathjax>#Zn(s) + 1/2O_2(g) rarr ZnO(s)#</mathjax></p>
<p>And thus between the oxide and the starting metal, there is clearly <mathjax>#1:1#</mathjax> equivalence. So if there are 4.44 moles of zinc, clearly there are 4.44 moles of oxide. If 10 moles of zinc oxide were produced, then how moles of zinc were used?</p></div>
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</article> | How many moles of zinc oxide is produced if 4.44 moles of zinc reacts with excess oxygen? | null |
3,364 | aadf95b0-6ddd-11ea-88f3-ccda262736ce | https://socratic.org/questions/what-is-the-partial-pressure-of-oxygen-collected-over-water-if-the-temperature-i | 717.53 mmHg | start physical_unit 6 6 partial_pressure mmhg qc_end physical_unit 6 6 14 15 temperature qc_end physical_unit 19 19 22 23 total_pressure qc_end substance 9 9 qc_end end | [{"type":"physical unit","value":"Partial pressure [OF] oxygen [IN] mmHg"}] | [{"type":"physical unit","value":"717.53 mmHg"}] | [{"type":"physical unit","value":"Temperature [OF] oxygen [=] \\pu{20.0 ℃}"},{"type":"physical unit","value":"Total pressure [OF] the gas [=] \\pu{735 mmHg}"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">What is the partial pressure of oxygen collected over water if the temperature is 20.0°C and the total gas pressure is 735 mmHg?</h1> | null | 717.53 mmHg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to realize that the oxygen, which is collected <em>over water</em>, will be mixed with <strong>water vapor</strong>. </p>
<p>This means that the *<em>total pressure8</em> of the mixture will have two components </p>
<blockquote>
<ul>
<li><em>the <a href="http://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of oxygen</em></li>
<li><em>the vapor pressure of water at</em> <mathjax>#20.0^@"C"#</mathjax></li>
</ul>
</blockquote>
<p>The vapor pressure of water at that temperature is approximately <mathjax>#"14.47 mmHg"#</mathjax> </p>
<p><a href="http://www.endmemo.com/chem/vaporpressurewater.php" rel="nofollow" target="_blank">http://www.endmemo.com/chem/vaporpressurewater.php</a></p>
<p>This means that you will have</p>
<blockquote>
<p><mathjax>#P_"total" = P_"oxygen" + P_"water"#</mathjax></p>
<p><mathjax>#P_"oxygen" = "735 mmHg" - "17.47 mmHg" = color(green)("718 mmHg")#</mathjax></p>
</blockquote></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"718 mmHg"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to realize that the oxygen, which is collected <em>over water</em>, will be mixed with <strong>water vapor</strong>. </p>
<p>This means that the *<em>total pressure8</em> of the mixture will have two components </p>
<blockquote>
<ul>
<li><em>the <a href="http://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of oxygen</em></li>
<li><em>the vapor pressure of water at</em> <mathjax>#20.0^@"C"#</mathjax></li>
</ul>
</blockquote>
<p>The vapor pressure of water at that temperature is approximately <mathjax>#"14.47 mmHg"#</mathjax> </p>
<p><a href="http://www.endmemo.com/chem/vaporpressurewater.php" rel="nofollow" target="_blank">http://www.endmemo.com/chem/vaporpressurewater.php</a></p>
<p>This means that you will have</p>
<blockquote>
<p><mathjax>#P_"total" = P_"oxygen" + P_"water"#</mathjax></p>
<p><mathjax>#P_"oxygen" = "735 mmHg" - "17.47 mmHg" = color(green)("718 mmHg")#</mathjax></p>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">What is the partial pressure of oxygen collected over water if the temperature is 20.0°C and the total gas pressure is 735 mmHg?</h1>
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Stefan V.
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Nov 22, 2015
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<div class="markdown"><p><mathjax>#"718 mmHg"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to realize that the oxygen, which is collected <em>over water</em>, will be mixed with <strong>water vapor</strong>. </p>
<p>This means that the *<em>total pressure8</em> of the mixture will have two components </p>
<blockquote>
<ul>
<li><em>the <a href="http://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of oxygen</em></li>
<li><em>the vapor pressure of water at</em> <mathjax>#20.0^@"C"#</mathjax></li>
</ul>
</blockquote>
<p>The vapor pressure of water at that temperature is approximately <mathjax>#"14.47 mmHg"#</mathjax> </p>
<p><a href="http://www.endmemo.com/chem/vaporpressurewater.php" rel="nofollow" target="_blank">http://www.endmemo.com/chem/vaporpressurewater.php</a></p>
<p>This means that you will have</p>
<blockquote>
<p><mathjax>#P_"total" = P_"oxygen" + P_"water"#</mathjax></p>
<p><mathjax>#P_"oxygen" = "735 mmHg" - "17.47 mmHg" = color(green)("718 mmHg")#</mathjax></p>
</blockquote></div>
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</article> | What is the partial pressure of oxygen collected over water if the temperature is 20.0°C and the total gas pressure is 735 mmHg? | null |
3,365 | ab0188d2-6ddd-11ea-a1d6-ccda262736ce | https://socratic.org/questions/what-is-the-volume-occupied-by-3-0-10-23-molecules-of-bromine-gas-at-stp | 11.16 L | start physical_unit 11 12 volume l qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] bromine gas [IN] L"}] | [{"type":"physical unit","value":"11.16 L"}] | [{"type":"physical unit","value":"Number [OF] bromine gas molecules [=] \\pu{3.0 × 10^23}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">What is the volume occupied by #3.0*10^23# molecules of bromine gas at STP?</h1> | null | 11.16 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#color(orange)("Because we are at STP, we will have to use the ideal gas law:")#</mathjax><br/>
<img alt="slideplayer.com" src="https://useruploads.socratic.org/OZDt7aeS7WCBHM21KnaQ_slide_2.jpg"/> </p>
<p><strong>At standard temperature and pressure, the temperature is 273K and the pressure is 1 atm.</strong> </p>
<p>Next, list your known and unknown variables. Our only unknown is the volume of <mathjax>#Br_2(g)#</mathjax>. Our known variables are P,R, and T. </p>
<p>We don't necessarily have <mathjax>#n#</mathjax> yet, but we're given a value that will lead us to the number of moles of bromine gas. All we have to do is convert molecules to moles using the conversion factor below:<img alt="www.slideshare.net" src="https://useruploads.socratic.org/mYBbnllEQ56NBDgaY0dM_moles-5-728.jpg"/> </p>
<p>We'll use the second conversion factor because we can cancel out molecules and end up with moles:</p>
<p><mathjax>#3.0xx10^(23)cancel"molecules"xx(1mol)/(6.02x10^(23)cancel"molecules")#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.498 mol#</mathjax></p>
<p>Now all of the variables have good units! All that's left to do is rearrange the equation and solve for V like so: </p>
<p><mathjax>#V = (nxxRxxT)/P#</mathjax><br/>
<mathjax>#V = (0.498cancel"mol"xx0.0821(Lxxcancel"atm")/(cancel"mol"xxcancel"K")xx(273cancelK))/(1cancel"atm")#</mathjax><br/>
<mathjax>#V = 11 L#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#color(magenta)("11 L")#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#color(orange)("Because we are at STP, we will have to use the ideal gas law:")#</mathjax><br/>
<img alt="slideplayer.com" src="https://useruploads.socratic.org/OZDt7aeS7WCBHM21KnaQ_slide_2.jpg"/> </p>
<p><strong>At standard temperature and pressure, the temperature is 273K and the pressure is 1 atm.</strong> </p>
<p>Next, list your known and unknown variables. Our only unknown is the volume of <mathjax>#Br_2(g)#</mathjax>. Our known variables are P,R, and T. </p>
<p>We don't necessarily have <mathjax>#n#</mathjax> yet, but we're given a value that will lead us to the number of moles of bromine gas. All we have to do is convert molecules to moles using the conversion factor below:<img alt="www.slideshare.net" src="https://useruploads.socratic.org/mYBbnllEQ56NBDgaY0dM_moles-5-728.jpg"/> </p>
<p>We'll use the second conversion factor because we can cancel out molecules and end up with moles:</p>
<p><mathjax>#3.0xx10^(23)cancel"molecules"xx(1mol)/(6.02x10^(23)cancel"molecules")#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.498 mol#</mathjax></p>
<p>Now all of the variables have good units! All that's left to do is rearrange the equation and solve for V like so: </p>
<p><mathjax>#V = (nxxRxxT)/P#</mathjax><br/>
<mathjax>#V = (0.498cancel"mol"xx0.0821(Lxxcancel"atm")/(cancel"mol"xxcancel"K")xx(273cancelK))/(1cancel"atm")#</mathjax><br/>
<mathjax>#V = 11 L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the volume occupied by #3.0*10^23# molecules of bromine gas at STP?</h1>
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<div class="markdown"><p><mathjax>#color(magenta)("11 L")#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#color(orange)("Because we are at STP, we will have to use the ideal gas law:")#</mathjax><br/>
<img alt="slideplayer.com" src="https://useruploads.socratic.org/OZDt7aeS7WCBHM21KnaQ_slide_2.jpg"/> </p>
<p><strong>At standard temperature and pressure, the temperature is 273K and the pressure is 1 atm.</strong> </p>
<p>Next, list your known and unknown variables. Our only unknown is the volume of <mathjax>#Br_2(g)#</mathjax>. Our known variables are P,R, and T. </p>
<p>We don't necessarily have <mathjax>#n#</mathjax> yet, but we're given a value that will lead us to the number of moles of bromine gas. All we have to do is convert molecules to moles using the conversion factor below:<img alt="www.slideshare.net" src="https://useruploads.socratic.org/mYBbnllEQ56NBDgaY0dM_moles-5-728.jpg"/> </p>
<p>We'll use the second conversion factor because we can cancel out molecules and end up with moles:</p>
<p><mathjax>#3.0xx10^(23)cancel"molecules"xx(1mol)/(6.02x10^(23)cancel"molecules")#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.498 mol#</mathjax></p>
<p>Now all of the variables have good units! All that's left to do is rearrange the equation and solve for V like so: </p>
<p><mathjax>#V = (nxxRxxT)/P#</mathjax><br/>
<mathjax>#V = (0.498cancel"mol"xx0.0821(Lxxcancel"atm")/(cancel"mol"xxcancel"K")xx(273cancelK))/(1cancel"atm")#</mathjax><br/>
<mathjax>#V = 11 L#</mathjax></p></div>
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</article> | What is the volume occupied by #3.0*10^23# molecules of bromine gas at STP? | null |
3,366 | aa0e3658-6ddd-11ea-89f8-ccda262736ce | https://socratic.org/questions/how-do-you-balance-hbr-o-2-br-2-h-2o | 4 HBr + O2 -> 2 Br2 + 2 H2O | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"4 HBr + O2 -> 2 Br2 + 2 H2O"}] | [{"type":"chemical equation","value":"HBr + O2 -> Br2 + H2O"}] | <h1 class="questionTitle" itemprop="name">How do you balance #HBr + O_2 -> Br_2 + H_2O#?</h1> | null | 4 HBr + O2 -> 2 Br2 + 2 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Add molecules (balancing numbers in front) to ensure that the number of atoms of each element is the same on the left and right hand sides of the equation.</p></div>
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<div class="markdown"><p><mathjax>#4HBr+O_2->2Br_2+2H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Add molecules (balancing numbers in front) to ensure that the number of atoms of each element is the same on the left and right hand sides of the equation.</p></div>
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<div class="markdown"><p><mathjax>#4HBr+O_2->2Br_2+2H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Add molecules (balancing numbers in front) to ensure that the number of atoms of each element is the same on the left and right hand sides of the equation.</p></div>
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<div class="markdown"><p><mathjax>#4"HBr"+"O"_2→2"Br"_2+2"H"_2"O"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This equation is one of those where you need to balance O at the very beginning. </p>
<p>Otherwise as a thumb-rule one leaves H and O to taken up at the last.</p>
<p>Observe that in the unbalanced equation, on the reactants side there are 2 atoms of oxygen, whereas there is single oxygen atom on the products side.</p>
<p>Therefore to balance O on both sides, we need to put 2 in front of water molecule. This fixes 4 hydrogen atoms on the right side of the equation.</p>
<p>To balance 4H atoms in the products, we need 4 molecules HBr on the reactants side.</p>
<p>Now we are left with balancing of bromine atoms. Clearly to balance 4 Br atoms on the reactants side, we need to put 2 in front of <mathjax>#" Br"_2 ,#</mathjax> to balance the complete equation.</p></div>
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</article> | How do you balance #HBr + O_2 -> Br_2 + H_2O#? | null |
3,367 | a85d9b90-6ddd-11ea-8c3f-ccda262736ce | https://socratic.org/questions/a-serving-of-cheez-its-releases-130-kcal-1-kcal-4-18-kj-when-digested-by-your-bo | 78.93 ℃ | start physical_unit 29 29 temperature °c qc_end physical_unit 15 16 5 6 heat_energy qc_end c_other OTHER qc_end physical_unit 29 29 26 27 mass qc_end physical_unit 29 29 31 32 temperature qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] water [IN] ℃"}] | [{"type":"physical unit","value":"78.93 ℃"}] | [{"type":"physical unit","value":"Released energy [OF] your body [=] \\pu{130 kcal}"},{"type":"other","value":"1 kcal = 4.18 kJ"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{2.5 kg}"},{"type":"physical unit","value":"Temperature1 [OF] water [=] \\pu{27 ℃}"}] | <h1 class="questionTitle" itemprop="name">A serving of Cheez-Its releases 130 kcal (1 kcal = 4.18 kJ) when digested by your body. If this same amount of energy were transferred to 2.5 kg of water at 27°C ,what would the final temperature be?</h1> | null | 78.93 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Energy transferred (W) is given by </p>
<p><mathjax>#W=mcDeltaT#</mathjax>,</p>
<p>where <mathjax>#m#</mathjax> is the mass, <mathjax>#c#</mathjax> is the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity, and <mathjax>#Delta T#</mathjax> is the change in temperature.</p>
<p>So in this case,</p>
<p><mathjax>#DeltaT=W/(mc)=(4180Jxx130)/((2,5kg)*(4186J//kg.K))=51,925^@C#</mathjax>.</p>
<p>Therefore the final temperature of the water will be </p>
<p><mathjax>#T_f=27+51,925=78,925^@C#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#78,925^@C#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Energy transferred (W) is given by </p>
<p><mathjax>#W=mcDeltaT#</mathjax>,</p>
<p>where <mathjax>#m#</mathjax> is the mass, <mathjax>#c#</mathjax> is the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity, and <mathjax>#Delta T#</mathjax> is the change in temperature.</p>
<p>So in this case,</p>
<p><mathjax>#DeltaT=W/(mc)=(4180Jxx130)/((2,5kg)*(4186J//kg.K))=51,925^@C#</mathjax>.</p>
<p>Therefore the final temperature of the water will be </p>
<p><mathjax>#T_f=27+51,925=78,925^@C#</mathjax>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A serving of Cheez-Its releases 130 kcal (1 kcal = 4.18 kJ) when digested by your body. If this same amount of energy were transferred to 2.5 kg of water at 27°C ,what would the final temperature be?</h1>
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<span class="dateCreated" datetime="2015-12-25T15:51:45" itemprop="dateCreated">
Dec 25, 2015
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<div class="markdown"><p><mathjax>#78,925^@C#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Energy transferred (W) is given by </p>
<p><mathjax>#W=mcDeltaT#</mathjax>,</p>
<p>where <mathjax>#m#</mathjax> is the mass, <mathjax>#c#</mathjax> is the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity, and <mathjax>#Delta T#</mathjax> is the change in temperature.</p>
<p>So in this case,</p>
<p><mathjax>#DeltaT=W/(mc)=(4180Jxx130)/((2,5kg)*(4186J//kg.K))=51,925^@C#</mathjax>.</p>
<p>Therefore the final temperature of the water will be </p>
<p><mathjax>#T_f=27+51,925=78,925^@C#</mathjax>.</p></div>
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</article> | A serving of Cheez-Its releases 130 kcal (1 kcal = 4.18 kJ) when digested by your body. If this same amount of energy were transferred to 2.5 kg of water at 27°C ,what would the final temperature be? | null |
3,368 | aba5d626-6ddd-11ea-8215-ccda262736ce | https://socratic.org/questions/a-70-hci-solution-has-the-strength-of-35-meq-ml-how-many-ml-are-needed-to-prepar | 1.45 mL | start physical_unit 2 3 volume ml qc_end physical_unit 2 2 1 1 percent qc_end physical_unit 2 3 8 9 strength qc_end physical_unit 2 3 17 18 mass qc_end end | [{"type":"physical unit","value":"Volume [OF] HCI solution [IN] mL"}] | [{"type":"physical unit","value":"1.45 mL"}] | [{"type":"physical unit","value":"Percent [OF] HCI in solution [=] \\pu{70%}"},{"type":"physical unit","value":"Strength [OF] HCI solution [=] \\pu{35 mEq/mL}"},{"type":"physical unit","value":"Mass [OF] HCI solution [=] \\pu{50 mEq}"}] | <h1 class="questionTitle" itemprop="name">A 70% HCI solution has the strength of 35 mEq/mL. How many mL are needed to prepare 50 mEq?</h1> | null | 1.45 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We take the quotient........</p>
<p><mathjax>#(50*cancel"mEq")/(35*cancel"mEq"*"mL"^-1)=1.43*1/(1/("mL"))=1.43*"mL"...#</mathjax></p>
<p>I think this is a chemically unreasonable concentration. Conc. <mathjax>#HCl#</mathjax> as used in the lab is approx. <mathjax>#10*mol*L^-1#</mathjax> and is <mathjax>#38-40%#</mathjax> <mathjax>#w/w#</mathjax>.</p>
<p>I would not want to handle hydrochloric acid of this concentration. </p></div>
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<div class="markdown"><p>Well.....<mathjax>#1.45*mL#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We take the quotient........</p>
<p><mathjax>#(50*cancel"mEq")/(35*cancel"mEq"*"mL"^-1)=1.43*1/(1/("mL"))=1.43*"mL"...#</mathjax></p>
<p>I think this is a chemically unreasonable concentration. Conc. <mathjax>#HCl#</mathjax> as used in the lab is approx. <mathjax>#10*mol*L^-1#</mathjax> and is <mathjax>#38-40%#</mathjax> <mathjax>#w/w#</mathjax>.</p>
<p>I would not want to handle hydrochloric acid of this concentration. </p></div>
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<h1 class="questionTitle" itemprop="name">A 70% HCI solution has the strength of 35 mEq/mL. How many mL are needed to prepare 50 mEq?</h1>
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<div class="markdown"><p>Well.....<mathjax>#1.45*mL#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We take the quotient........</p>
<p><mathjax>#(50*cancel"mEq")/(35*cancel"mEq"*"mL"^-1)=1.43*1/(1/("mL"))=1.43*"mL"...#</mathjax></p>
<p>I think this is a chemically unreasonable concentration. Conc. <mathjax>#HCl#</mathjax> as used in the lab is approx. <mathjax>#10*mol*L^-1#</mathjax> and is <mathjax>#38-40%#</mathjax> <mathjax>#w/w#</mathjax>.</p>
<p>I would not want to handle hydrochloric acid of this concentration. </p></div>
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</article> | A 70% HCI solution has the strength of 35 mEq/mL. How many mL are needed to prepare 50 mEq? | null |
3,369 | ac26fc5b-6ddd-11ea-aea6-ccda262736ce | https://socratic.org/questions/how-can-i-balance-this-chemical-equations-aluminum-and-hydrochloric-acid-react-t | 2 Al(s) + 6 HCl(aq) -> 3 H2(g) + 2 AlCl3(aq) | start chemical_equation qc_end substance 7 7 qc_end substance 9 10 qc_end substance 14 15 qc_end substance 17 18 qc_end end | [{"type":"other","value":"Chemical Equation [OF] this chemical equations"}] | [{"type":"chemical equation","value":"2 Al(s) + 6 HCl(aq) -> 3 H2(g) + 2 AlCl3(aq)"}] | [{"type":"substance name","value":"Aluminum"},{"type":"substance name","value":"Hydrochloric acid"},{"type":"substance name","value":"Aluminum chloride"},{"type":"substance name","value":"Hydrogen gas"}] | <h1 class="questionTitle" itemprop="name">How can I balance this chemical equations? Aluminum and hydrochloric acid react to form aluminum chloride and hydrogen gas. </h1> | null | 2 Al(s) + 6 HCl(aq) -> 3 H2(g) + 2 AlCl3(aq) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This reaction is between a metal and an acid which typically results in a salt and the release of hydrogen gas. The unbalanced reaction is </p>
<blockquote>
<p><mathjax>#Al + HCl -> H_2 + AlCl_3#</mathjax>.</p>
</blockquote>
<p>This is a <strong>redox reaction</strong>, whose half-reactions are and become:</p>
<blockquote>
<p><mathjax>#2("Al"(s) -> "Al"^(3+)(aq) + cancel(3e^(-)))#</mathjax><br/>
<mathjax>#3(2"H"^(+)(aq) + cancel(2e^(-)) -> "H"_2(g))#</mathjax><br/>
<mathjax>#"-----------------------------------------------"#</mathjax><br/>
<mathjax>#2"Al"(s) + 6"H"^(+)(aq) -> 3"H"_2(g) + 2"Al"^(3+)(aq)#</mathjax></p>
</blockquote>
<p>Aluminum <strong>oxidizes</strong> as <mathjax>#"Al" -> "Al"^(3+)#</mathjax>, while hydrogen <strong>reduces</strong> as <mathjax>#2"H"^(+) -> "H"_2^0#</mathjax>. </p>
<p>If we add back the spectator <mathjax>#"Cl"^(-)#</mathjax>, we get:</p>
<blockquote>
<p><mathjax>#color(blue)(2"Al"(s) + 6"HCl"(aq) -> 3"H"_2(g) + 2"AlCl"_3(aq))#</mathjax></p>
</blockquote>
<p>I hope this was helpful.</p>
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<div class="markdown"><p><mathjax>#color(blue)(2"Al"(s) + 6"HCl"(aq) -> 3"H"_2(g) + 2"AlCl"_3(aq))#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This reaction is between a metal and an acid which typically results in a salt and the release of hydrogen gas. The unbalanced reaction is </p>
<blockquote>
<p><mathjax>#Al + HCl -> H_2 + AlCl_3#</mathjax>.</p>
</blockquote>
<p>This is a <strong>redox reaction</strong>, whose half-reactions are and become:</p>
<blockquote>
<p><mathjax>#2("Al"(s) -> "Al"^(3+)(aq) + cancel(3e^(-)))#</mathjax><br/>
<mathjax>#3(2"H"^(+)(aq) + cancel(2e^(-)) -> "H"_2(g))#</mathjax><br/>
<mathjax>#"-----------------------------------------------"#</mathjax><br/>
<mathjax>#2"Al"(s) + 6"H"^(+)(aq) -> 3"H"_2(g) + 2"Al"^(3+)(aq)#</mathjax></p>
</blockquote>
<p>Aluminum <strong>oxidizes</strong> as <mathjax>#"Al" -> "Al"^(3+)#</mathjax>, while hydrogen <strong>reduces</strong> as <mathjax>#2"H"^(+) -> "H"_2^0#</mathjax>. </p>
<p>If we add back the spectator <mathjax>#"Cl"^(-)#</mathjax>, we get:</p>
<blockquote>
<p><mathjax>#color(blue)(2"Al"(s) + 6"HCl"(aq) -> 3"H"_2(g) + 2"AlCl"_3(aq))#</mathjax></p>
</blockquote>
<p>I hope this was helpful.</p>
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<h1 class="questionTitle" itemprop="name">How can I balance this chemical equations? Aluminum and hydrochloric acid react to form aluminum chloride and hydrogen gas. </h1>
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<span class="dateCreated" datetime="2014-05-13T17:15:41" itemprop="dateCreated">
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<div class="markdown"><p><mathjax>#color(blue)(2"Al"(s) + 6"HCl"(aq) -> 3"H"_2(g) + 2"AlCl"_3(aq))#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This reaction is between a metal and an acid which typically results in a salt and the release of hydrogen gas. The unbalanced reaction is </p>
<blockquote>
<p><mathjax>#Al + HCl -> H_2 + AlCl_3#</mathjax>.</p>
</blockquote>
<p>This is a <strong>redox reaction</strong>, whose half-reactions are and become:</p>
<blockquote>
<p><mathjax>#2("Al"(s) -> "Al"^(3+)(aq) + cancel(3e^(-)))#</mathjax><br/>
<mathjax>#3(2"H"^(+)(aq) + cancel(2e^(-)) -> "H"_2(g))#</mathjax><br/>
<mathjax>#"-----------------------------------------------"#</mathjax><br/>
<mathjax>#2"Al"(s) + 6"H"^(+)(aq) -> 3"H"_2(g) + 2"Al"^(3+)(aq)#</mathjax></p>
</blockquote>
<p>Aluminum <strong>oxidizes</strong> as <mathjax>#"Al" -> "Al"^(3+)#</mathjax>, while hydrogen <strong>reduces</strong> as <mathjax>#2"H"^(+) -> "H"_2^0#</mathjax>. </p>
<p>If we add back the spectator <mathjax>#"Cl"^(-)#</mathjax>, we get:</p>
<blockquote>
<p><mathjax>#color(blue)(2"Al"(s) + 6"HCl"(aq) -> 3"H"_2(g) + 2"AlCl"_3(aq))#</mathjax></p>
</blockquote>
<p>I hope this was helpful.</p>
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</article> | How can I balance this chemical equations? Aluminum and hydrochloric acid react to form aluminum chloride and hydrogen gas. | null |
3,370 | abf07c80-6ddd-11ea-815d-ccda262736ce | https://socratic.org/questions/the-pressure-of-a-gas-is-1-34-atm-at-a-temperature-of-298-k-what-will-the-pressu | 2.30 atm | start physical_unit 4 4 pressure atm qc_end physical_unit 4 4 6 7 pressure qc_end physical_unit 4 4 12 13 temperature qc_end physical_unit 4 4 25 26 temperature qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the gas [IN] atm"}] | [{"type":"physical unit","value":"2.30 atm"}] | [{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{1.34 atm}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{298 K}"},{"type":"physical unit","value":"Temperature2 [OF] the gas [=] \\pu{512 K}"}] | <h1 class="questionTitle" itemprop="name">The pressure of a gas is 1.34 atm at a temperature of 298 K. What will the pressure be if the temperature is increased to 512 K? </h1> | null | 2.30 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As Temperature is increased, Pressure is increased. Pressure is directly proportional to Temperature.</p>
<p><mathjax>#P_1#</mathjax> / <mathjax>#T_1#</mathjax> = <mathjax>#P_2#</mathjax> / <mathjax>#T_2#</mathjax> </p>
<p><mathjax>#P_1#</mathjax> = 1.34 atm , <mathjax>#T_1#</mathjax> = 298 K</p>
<p><mathjax>#P_2#</mathjax> = X atm , <mathjax>#T_2#</mathjax> = 512 K</p>
<p>1.34 atm / 298 K = X atm / 512 K</p>
<p>rearranging we get,</p>
<p>1.34 atm x 512 K = 298 K . X atm</p>
<p>686.08 K atm = 298 K . X atm</p>
<p>X atm = 686.08 K atm / 298 K </p>
<p>X atm = 2.30 atm</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>2.30 atm</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As Temperature is increased, Pressure is increased. Pressure is directly proportional to Temperature.</p>
<p><mathjax>#P_1#</mathjax> / <mathjax>#T_1#</mathjax> = <mathjax>#P_2#</mathjax> / <mathjax>#T_2#</mathjax> </p>
<p><mathjax>#P_1#</mathjax> = 1.34 atm , <mathjax>#T_1#</mathjax> = 298 K</p>
<p><mathjax>#P_2#</mathjax> = X atm , <mathjax>#T_2#</mathjax> = 512 K</p>
<p>1.34 atm / 298 K = X atm / 512 K</p>
<p>rearranging we get,</p>
<p>1.34 atm x 512 K = 298 K . X atm</p>
<p>686.08 K atm = 298 K . X atm</p>
<p>X atm = 686.08 K atm / 298 K </p>
<p>X atm = 2.30 atm</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">The pressure of a gas is 1.34 atm at a temperature of 298 K. What will the pressure be if the temperature is increased to 512 K? </h1>
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<div class="markdown"><p>2.30 atm</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As Temperature is increased, Pressure is increased. Pressure is directly proportional to Temperature.</p>
<p><mathjax>#P_1#</mathjax> / <mathjax>#T_1#</mathjax> = <mathjax>#P_2#</mathjax> / <mathjax>#T_2#</mathjax> </p>
<p><mathjax>#P_1#</mathjax> = 1.34 atm , <mathjax>#T_1#</mathjax> = 298 K</p>
<p><mathjax>#P_2#</mathjax> = X atm , <mathjax>#T_2#</mathjax> = 512 K</p>
<p>1.34 atm / 298 K = X atm / 512 K</p>
<p>rearranging we get,</p>
<p>1.34 atm x 512 K = 298 K . X atm</p>
<p>686.08 K atm = 298 K . X atm</p>
<p>X atm = 686.08 K atm / 298 K </p>
<p>X atm = 2.30 atm</p></div>
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</article> | The pressure of a gas is 1.34 atm at a temperature of 298 K. What will the pressure be if the temperature is increased to 512 K? | null |
3,371 | ab0dec35-6ddd-11ea-a7b7-ccda262736ce | https://socratic.org/questions/how-would-you-write-the-balanced-chemical-equation-for-the-reaction-of-khp-with- | 1,2-{HO(O=C)}{K+−O(O=C)}C6H4 + NaOH(aq) -> 1,2−{Na+−O(O=C)}{K+−O(O=C)}C6H4 + H2O(aq) | start chemical_equation qc_end chemical_equation 12 12 qc_end chemical_equation 14 14 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reaction"}] | [{"type":"chemical equation","value":"1,2-{HO(O=C)}{K+−O(O=C)}C6H4 + NaOH(aq) -> 1,2−{Na+−O(O=C)}{K+−O(O=C)}C6H4 + H2O(aq)"}] | [{"type":"chemical equation","value":"KHP"},{"type":"chemical equation","value":"NaOH"}] | <h1 class="questionTitle" itemprop="name">How would you write the balanced chemical equation for the reaction of KHP with NaOH?</h1> | null | 1,2-{HO(O=C)}{K+−O(O=C)}C6H4 + NaOH(aq) -> 1,2−{Na+−O(O=C)}{K+−O(O=C)}C6H4 + H2O(aq) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"KHP=potassium hydrogenpthalate"#</mathjax></p>
<p><mathjax>#"KHP"#</mathjax> is the mono-potassium salt of phthalic acid <mathjax>#1,2-{HO(O=C)}_2C_6H_4#</mathjax>, i.e. benzenedicarboxlic acid.</p></div>
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<div class="markdown"><p><mathjax>#1,2-{HO(O=C)}{K^(+)""^(-)O(O=C)}C_6H_4 + NaOH(aq) rarr 1,2-{Na^(+)""^(-)O(O=C)}{K^(+)""^(-)O(O=C)}C_6H_4 +H_2O(aq)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"KHP=potassium hydrogenpthalate"#</mathjax></p>
<p><mathjax>#"KHP"#</mathjax> is the mono-potassium salt of phthalic acid <mathjax>#1,2-{HO(O=C)}_2C_6H_4#</mathjax>, i.e. benzenedicarboxlic acid.</p></div>
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<h1 class="questionTitle" itemprop="name">How would you write the balanced chemical equation for the reaction of KHP with NaOH?</h1>
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anor277
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<div class="markdown"><p><mathjax>#1,2-{HO(O=C)}{K^(+)""^(-)O(O=C)}C_6H_4 + NaOH(aq) rarr 1,2-{Na^(+)""^(-)O(O=C)}{K^(+)""^(-)O(O=C)}C_6H_4 +H_2O(aq)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"KHP=potassium hydrogenpthalate"#</mathjax></p>
<p><mathjax>#"KHP"#</mathjax> is the mono-potassium salt of phthalic acid <mathjax>#1,2-{HO(O=C)}_2C_6H_4#</mathjax>, i.e. benzenedicarboxlic acid.</p></div>
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</article> | How would you write the balanced chemical equation for the reaction of KHP with NaOH? | null |
3,372 | a83a8d10-6ddd-11ea-9e28-ccda262736ce | https://socratic.org/questions/5905fe4ab72cff21362a7a84 | 7.60 L | start physical_unit 8 8 volume l qc_end c_other constant_temperature qc_end physical_unit 8 8 4 5 volume qc_end physical_unit 8 8 10 11 pressure qc_end physical_unit 8 8 19 20 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] gas [IN] L"}] | [{"type":"physical unit","value":"7.60 L"}] | [{"type":"other","value":"ConstantTemperature"},{"type":"physical unit","value":"Volume1 [OF] gas [=] \\pu{3.8 L}"},{"type":"physical unit","value":"Pressure1 [OF] gas [=] \\pu{765 mmHg}"},{"type":"physical unit","value":"Pressure2 [OF] gas [=] \\pu{0.500 atm}"}] | <h1 class="questionTitle" itemprop="name">At constant temperature, a #3.8*L# volume of gas at #765*mm*Hg#, was expanded to give a pressure of #0.500*atm#. What is the new volume?</h1> | null | 7.60 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need to know that <mathjax>#1*atm#</mathjax> will support a column of mercury that is <mathjax>#760*mm#</mathjax> high, and thus a mercury column may used as a highly visual measurement of pressure:</p>
<p><mathjax>#1*atm-=760*mm*Hg#</mathjax> (note that given the schemozzle that occurs when you spill mercury in the lab, you really should not put mercury under a pressure of MORE than 1 atmosphere). Here.......</p>
<p><mathjax>#P_1=(765*mm*Hg)/(760*mm*Hg*atm^-1)=1.01*atm.#</mathjax></p>
<p>We solve for <mathjax>#V_2=(P_1V_1)/P_2=(1.01*atmxx3.8*L)/(0.500*atm)=7.60*L#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#P_1V_1=P_2V_2#</mathjax> under conditions of constant temperature. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need to know that <mathjax>#1*atm#</mathjax> will support a column of mercury that is <mathjax>#760*mm#</mathjax> high, and thus a mercury column may used as a highly visual measurement of pressure:</p>
<p><mathjax>#1*atm-=760*mm*Hg#</mathjax> (note that given the schemozzle that occurs when you spill mercury in the lab, you really should not put mercury under a pressure of MORE than 1 atmosphere). Here.......</p>
<p><mathjax>#P_1=(765*mm*Hg)/(760*mm*Hg*atm^-1)=1.01*atm.#</mathjax></p>
<p>We solve for <mathjax>#V_2=(P_1V_1)/P_2=(1.01*atmxx3.8*L)/(0.500*atm)=7.60*L#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">At constant temperature, a #3.8*L# volume of gas at #765*mm*Hg#, was expanded to give a pressure of #0.500*atm#. What is the new volume?</h1>
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anor277
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<div class="markdown"><p><mathjax>#P_1V_1=P_2V_2#</mathjax> under conditions of constant temperature. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need to know that <mathjax>#1*atm#</mathjax> will support a column of mercury that is <mathjax>#760*mm#</mathjax> high, and thus a mercury column may used as a highly visual measurement of pressure:</p>
<p><mathjax>#1*atm-=760*mm*Hg#</mathjax> (note that given the schemozzle that occurs when you spill mercury in the lab, you really should not put mercury under a pressure of MORE than 1 atmosphere). Here.......</p>
<p><mathjax>#P_1=(765*mm*Hg)/(760*mm*Hg*atm^-1)=1.01*atm.#</mathjax></p>
<p>We solve for <mathjax>#V_2=(P_1V_1)/P_2=(1.01*atmxx3.8*L)/(0.500*atm)=7.60*L#</mathjax></p></div>
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</article> | At constant temperature, a #3.8*L# volume of gas at #765*mm*Hg#, was expanded to give a pressure of #0.500*atm#. What is the new volume? | null |
3,373 | a8cc85c6-6ddd-11ea-aca8-ccda262736ce | https://socratic.org/questions/what-volume-will-12-0-g-of-oxygen-gas-o-2-occupy-at-25-c-and-a-pressure-of-52-7- | 17.6 L | start physical_unit 6 7 volume l qc_end physical_unit 6 7 3 4 mass qc_end physical_unit 6 7 11 12 temperature qc_end physical_unit 6 7 17 18 pressure qc_end end | [{"type":"physical unit","value":"Volume [OF] O2 gas [IN] L"}] | [{"type":"physical unit","value":"17.6 L"}] | [{"type":"physical unit","value":"Mass [OF] O2 gas [=] \\pu{12.0 g}"},{"type":"physical unit","value":"Temperature [OF] O2 gas [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Pressure [OF] O2 gas [=] \\pu{52.7 kPa}"}] | <h1 class="questionTitle" itemprop="name">What volume will 12.0 g of oxygen gas (#O_2#) occupy at 25°C and a pressure of 52.7 kPa?</h1> | null | 17.6 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a classic example of how an <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation practice problem looks like. </p>
<p>The <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong> equation looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(| bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)))|)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume occupied by the gas<br/>
<mathjax>#n#</mathjax> - the number of moles of gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given as <mathjax>#0.0821("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the <strong>absolute temperature</strong> of the gas</p>
<p>Now, notice that the problem does not provide you with the <em>number of moles</em> of gas present in the sample. However, it does provide with the mass of the sample and the <em>identity of the gas</em>. </p>
<p>This means that you can use its <strong>molar mass</strong> as a conversion factor to go from <em>grams</em> to <em>moles</em></p>
<blockquote>
<p><mathjax>#12.0 color(red)(cancel(color(black)("g"))) * overbrace("1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))))^(color(purple)("molar mass of molecular oxygen")) = "0.375 moles O"_2#</mathjax></p>
</blockquote>
<p>Your biggest challenge when it comes to this type of problems will often be to make sure that the <strong>units</strong> given to you match those used in the expression of the <em>ideal gas constant</em> ,<mathjax>#R#</mathjax>. </p>
<p>In this case, you have to make sure that you convert the temperature from <em>degrees Celsius</em> to <em>Kelvin</em> and the pressure from <em>kilopascals</em> to <em>atm</em> by using the conversion factors</p>
<blockquote>
<p><mathjax>#T["K"] = t[""^@"C"] + 273.15#</mathjax></p>
<p><mathjax>#"1 atm " = " 101.325 kPa"#</mathjax></p>
</blockquote>
<p>Rearrange the ideal gas law equation to solve for <mathjax>#V#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies V = (nRT)/P#</mathjax></p>
</blockquote>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#V = (0.375color(red)(cancel(color(black)("moles"))) * 0.0821 (color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 25.0)color(red)(cancel(color(black)("K"))))/(52.7/101.325color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#V = color(green)(|bar(ul(color(green)("17.6 L"))|) ->#</mathjax> <em>rounded to three <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong></em></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"17.6 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a classic example of how an <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation practice problem looks like. </p>
<p>The <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong> equation looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(| bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)))|)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume occupied by the gas<br/>
<mathjax>#n#</mathjax> - the number of moles of gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given as <mathjax>#0.0821("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the <strong>absolute temperature</strong> of the gas</p>
<p>Now, notice that the problem does not provide you with the <em>number of moles</em> of gas present in the sample. However, it does provide with the mass of the sample and the <em>identity of the gas</em>. </p>
<p>This means that you can use its <strong>molar mass</strong> as a conversion factor to go from <em>grams</em> to <em>moles</em></p>
<blockquote>
<p><mathjax>#12.0 color(red)(cancel(color(black)("g"))) * overbrace("1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))))^(color(purple)("molar mass of molecular oxygen")) = "0.375 moles O"_2#</mathjax></p>
</blockquote>
<p>Your biggest challenge when it comes to this type of problems will often be to make sure that the <strong>units</strong> given to you match those used in the expression of the <em>ideal gas constant</em> ,<mathjax>#R#</mathjax>. </p>
<p>In this case, you have to make sure that you convert the temperature from <em>degrees Celsius</em> to <em>Kelvin</em> and the pressure from <em>kilopascals</em> to <em>atm</em> by using the conversion factors</p>
<blockquote>
<p><mathjax>#T["K"] = t[""^@"C"] + 273.15#</mathjax></p>
<p><mathjax>#"1 atm " = " 101.325 kPa"#</mathjax></p>
</blockquote>
<p>Rearrange the ideal gas law equation to solve for <mathjax>#V#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies V = (nRT)/P#</mathjax></p>
</blockquote>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#V = (0.375color(red)(cancel(color(black)("moles"))) * 0.0821 (color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 25.0)color(red)(cancel(color(black)("K"))))/(52.7/101.325color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#V = color(green)(|bar(ul(color(green)("17.6 L"))|) ->#</mathjax> <em>rounded to three <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong></em></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What volume will 12.0 g of oxygen gas (#O_2#) occupy at 25°C and a pressure of 52.7 kPa?</h1>
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Stefan V.
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Mar 5, 2016
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<div class="markdown"><p><mathjax>#"17.6 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a classic example of how an <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation practice problem looks like. </p>
<p>The <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong> equation looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(| bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)))|)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume occupied by the gas<br/>
<mathjax>#n#</mathjax> - the number of moles of gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given as <mathjax>#0.0821("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the <strong>absolute temperature</strong> of the gas</p>
<p>Now, notice that the problem does not provide you with the <em>number of moles</em> of gas present in the sample. However, it does provide with the mass of the sample and the <em>identity of the gas</em>. </p>
<p>This means that you can use its <strong>molar mass</strong> as a conversion factor to go from <em>grams</em> to <em>moles</em></p>
<blockquote>
<p><mathjax>#12.0 color(red)(cancel(color(black)("g"))) * overbrace("1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))))^(color(purple)("molar mass of molecular oxygen")) = "0.375 moles O"_2#</mathjax></p>
</blockquote>
<p>Your biggest challenge when it comes to this type of problems will often be to make sure that the <strong>units</strong> given to you match those used in the expression of the <em>ideal gas constant</em> ,<mathjax>#R#</mathjax>. </p>
<p>In this case, you have to make sure that you convert the temperature from <em>degrees Celsius</em> to <em>Kelvin</em> and the pressure from <em>kilopascals</em> to <em>atm</em> by using the conversion factors</p>
<blockquote>
<p><mathjax>#T["K"] = t[""^@"C"] + 273.15#</mathjax></p>
<p><mathjax>#"1 atm " = " 101.325 kPa"#</mathjax></p>
</blockquote>
<p>Rearrange the ideal gas law equation to solve for <mathjax>#V#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies V = (nRT)/P#</mathjax></p>
</blockquote>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#V = (0.375color(red)(cancel(color(black)("moles"))) * 0.0821 (color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 25.0)color(red)(cancel(color(black)("K"))))/(52.7/101.325color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#V = color(green)(|bar(ul(color(green)("17.6 L"))|) ->#</mathjax> <em>rounded to three <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong></em></p>
</blockquote></div>
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</article> | What volume will 12.0 g of oxygen gas (#O_2#) occupy at 25°C and a pressure of 52.7 kPa? | null |
3,374 | a8b30be6-6ddd-11ea-949d-ccda262736ce | https://socratic.org/questions/using-the-equation-2al-3h2so4-al2-so4-3-3h2-starting-with-15-0-mol-of-al-determi | 22.5 moles | start physical_unit 12 12 theoretical_yield mol qc_end physical_unit 4 4 15 16 mole qc_end chemical_equation 3 12 qc_end end | [{"type":"physical unit","value":"Theoretical yield [OF] H2 [IN] moles"}] | [{"type":"physical unit","value":"22.5 moles"}] | [{"type":"physical unit","value":"Mole [OF] Al [=] \\pu{15.0 moles}"},{"type":"chemical equation","value":"2 Al(s) + 3 H2SO4(aq) -> Al2(SO4)3(aq) + 3 H2(g)"}] | <h1 class="questionTitle" itemprop="name">Using the equation #2"Al"(s) + 3"H"_2"SO"_4(aq) -> "Al"_2("SO"_4)_3(aq) + 3"H"_2(g)#. Starting with #15.0# moles of #"Al"#, determine the theoretical yield of #"H"_2# ?</h1> | null | 22.5 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You first find the ratio between your unknown and your known, unknown on top and known on the bottom. </p>
<p>So with this question, your unknown is <mathjax>#"H"_2#</mathjax> and your known is <mathjax>#"Al"#</mathjax>. Your ratio is <mathjax>#3/2#</mathjax>. </p>
<p>Then, you take the number of moles of your known, <mathjax>#15#</mathjax> moles of <mathjax>#"Al"#</mathjax>, and multiply it by that ratio and you get the number of moles of your unknown, which in this case is <mathjax>#22.5#</mathjax> moles of <mathjax>#"H"_2#</mathjax></p>
<p><mathjax>#"15 mol Al" * "3 mol H"_2/"2 mol Al" = "22.5 mol H"_2#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"22.5 moles H"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You first find the ratio between your unknown and your known, unknown on top and known on the bottom. </p>
<p>So with this question, your unknown is <mathjax>#"H"_2#</mathjax> and your known is <mathjax>#"Al"#</mathjax>. Your ratio is <mathjax>#3/2#</mathjax>. </p>
<p>Then, you take the number of moles of your known, <mathjax>#15#</mathjax> moles of <mathjax>#"Al"#</mathjax>, and multiply it by that ratio and you get the number of moles of your unknown, which in this case is <mathjax>#22.5#</mathjax> moles of <mathjax>#"H"_2#</mathjax></p>
<p><mathjax>#"15 mol Al" * "3 mol H"_2/"2 mol Al" = "22.5 mol H"_2#</mathjax></p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">Using the equation #2"Al"(s) + 3"H"_2"SO"_4(aq) -> "Al"_2("SO"_4)_3(aq) + 3"H"_2(g)#. Starting with #15.0# moles of #"Al"#, determine the theoretical yield of #"H"_2# ?</h1>
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GOOSE313
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Stefan V.
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Feb 26, 2018
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<div class="markdown"><p><mathjax>#"22.5 moles H"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You first find the ratio between your unknown and your known, unknown on top and known on the bottom. </p>
<p>So with this question, your unknown is <mathjax>#"H"_2#</mathjax> and your known is <mathjax>#"Al"#</mathjax>. Your ratio is <mathjax>#3/2#</mathjax>. </p>
<p>Then, you take the number of moles of your known, <mathjax>#15#</mathjax> moles of <mathjax>#"Al"#</mathjax>, and multiply it by that ratio and you get the number of moles of your unknown, which in this case is <mathjax>#22.5#</mathjax> moles of <mathjax>#"H"_2#</mathjax></p>
<p><mathjax>#"15 mol Al" * "3 mol H"_2/"2 mol Al" = "22.5 mol H"_2#</mathjax></p></div>
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</article> | Using the equation #2"Al"(s) + 3"H"_2"SO"_4(aq) -> "Al"_2("SO"_4)_3(aq) + 3"H"_2(g)#. Starting with #15.0# moles of #"Al"#, determine the theoretical yield of #"H"_2# ? | null |
3,375 | ac8e9a00-6ddd-11ea-b20d-ccda262736ce | https://socratic.org/questions/if-the-pressure-is-8-314-10-4-pa-r-is-8-314-the-temperature-is-100-k-and-the-num | 3.00 m^3 | start physical_unit 22 22 volume m^3 qc_end physical_unit 22 22 4 7 pressure qc_end physical_unit 22 22 4 4 constant_r qc_end physical_unit 22 22 14 15 temperature qc_end end | [{"type":"physical unit","value":"Volume [OF] the gas [IN] m^3"}] | [{"type":"physical unit","value":"3.00 m^3"}] | [{"type":"physical unit","value":"Pressure [OF] the gas [=] \\pu{8.314 × 10^4 Pa}"},{"type":"physical unit","value":"R [OF] the gas [=] \\pu{8.314}"},{"type":"physical unit","value":"Temperature [OF] the gas [=] \\pu{100 K}"},{"type":"physical unit","value":"Mole [OF] the gas [=] \\pu{300 moles }"}] | <h1 class="questionTitle" itemprop="name">If the pressure is #8.314 * 10^4# #Pa#, #R# is #8.314#, the temperature is #100# #K#, and the number of moles of gas (#n#) is #300#, what is the volume of a gas in #m^3#?</h1> | null | 3.00 m^3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong> equation, which looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
<p>Here you have</p>
<blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the number of moles of gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em><br/>
<mathjax>#T#</mathjax> - the <strong>absolute temperature</strong> of the gas</p>
</blockquote>
<p>Now, it's important to notice that the problem doesn't provide you with the <strong>units</strong> used for the universal gas constant. This means that you're going to have to do some research and see what units of <mathjax>#R#</mathjax>, if any, match the <em>value</em> of <mathjax>#R#</mathjax> and the units for <strong>pressure</strong> and <strong>volume</strong> given to you. </p>
<p><a href="http://www.cpp.edu/~lllee/gasconstant.pdf" rel="nofollow" target="_blank">http://www.cpp.edu/~lllee/gasconstant.pdf</a></p>
<p>As you can see, you have</p>
<blockquote>
<p><mathjax>#R = 8.314color(white)(.) ("Pa m"^3)/("mol K")#</mathjax></p>
</blockquote>
<p>In this case, the units and the value match those given to you, so this is what you'll use in your calculations. </p>
<p>Your goal here is to figure out the <strong>volume</strong> of the gas, so rearrange the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation to solve for <mathjax>#V#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies V = (nRT)/P#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#V = (300 color(red)(cancel(color(black)("moles"))) * 8.314(color(red)(cancel(color(black)("Pa"))) * "m"^3)/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 100color(red)(cancel(color(black)("K"))))/(8.314 * 10^4 color(red)(cancel(color(black)("Pa")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("3 m"^3)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer must be rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a></strong>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"3 m"^3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong> equation, which looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
<p>Here you have</p>
<blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the number of moles of gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em><br/>
<mathjax>#T#</mathjax> - the <strong>absolute temperature</strong> of the gas</p>
</blockquote>
<p>Now, it's important to notice that the problem doesn't provide you with the <strong>units</strong> used for the universal gas constant. This means that you're going to have to do some research and see what units of <mathjax>#R#</mathjax>, if any, match the <em>value</em> of <mathjax>#R#</mathjax> and the units for <strong>pressure</strong> and <strong>volume</strong> given to you. </p>
<p><a href="http://www.cpp.edu/~lllee/gasconstant.pdf" rel="nofollow" target="_blank">http://www.cpp.edu/~lllee/gasconstant.pdf</a></p>
<p>As you can see, you have</p>
<blockquote>
<p><mathjax>#R = 8.314color(white)(.) ("Pa m"^3)/("mol K")#</mathjax></p>
</blockquote>
<p>In this case, the units and the value match those given to you, so this is what you'll use in your calculations. </p>
<p>Your goal here is to figure out the <strong>volume</strong> of the gas, so rearrange the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation to solve for <mathjax>#V#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies V = (nRT)/P#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#V = (300 color(red)(cancel(color(black)("moles"))) * 8.314(color(red)(cancel(color(black)("Pa"))) * "m"^3)/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 100color(red)(cancel(color(black)("K"))))/(8.314 * 10^4 color(red)(cancel(color(black)("Pa")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("3 m"^3)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer must be rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a></strong>. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If the pressure is #8.314 * 10^4# #Pa#, #R# is #8.314#, the temperature is #100# #K#, and the number of moles of gas (#n#) is #300#, what is the volume of a gas in #m^3#?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-07-28T12:18:54" itemprop="dateCreated">
Jul 28, 2016
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<div class="markdown"><p><mathjax>#"3 m"^3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong> equation, which looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
<p>Here you have</p>
<blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the number of moles of gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em><br/>
<mathjax>#T#</mathjax> - the <strong>absolute temperature</strong> of the gas</p>
</blockquote>
<p>Now, it's important to notice that the problem doesn't provide you with the <strong>units</strong> used for the universal gas constant. This means that you're going to have to do some research and see what units of <mathjax>#R#</mathjax>, if any, match the <em>value</em> of <mathjax>#R#</mathjax> and the units for <strong>pressure</strong> and <strong>volume</strong> given to you. </p>
<p><a href="http://www.cpp.edu/~lllee/gasconstant.pdf" rel="nofollow" target="_blank">http://www.cpp.edu/~lllee/gasconstant.pdf</a></p>
<p>As you can see, you have</p>
<blockquote>
<p><mathjax>#R = 8.314color(white)(.) ("Pa m"^3)/("mol K")#</mathjax></p>
</blockquote>
<p>In this case, the units and the value match those given to you, so this is what you'll use in your calculations. </p>
<p>Your goal here is to figure out the <strong>volume</strong> of the gas, so rearrange the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation to solve for <mathjax>#V#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies V = (nRT)/P#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#V = (300 color(red)(cancel(color(black)("moles"))) * 8.314(color(red)(cancel(color(black)("Pa"))) * "m"^3)/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 100color(red)(cancel(color(black)("K"))))/(8.314 * 10^4 color(red)(cancel(color(black)("Pa")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("3 m"^3)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer must be rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a></strong>. </p></div>
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</article> | If the pressure is #8.314 * 10^4# #Pa#, #R# is #8.314#, the temperature is #100# #K#, and the number of moles of gas (#n#) is #300#, what is the volume of a gas in #m^3#? | null |
3,376 | a9ace050-6ddd-11ea-81f9-ccda262736ce | https://socratic.org/questions/how-would-you-use-the-henderson-hasselbalch-equation-to-calculate-the-ratio-of-h | 10^1.05 | start physical_unit 12 14 ratio none qc_end c_other OTHER qc_end physical_unit 16 16 21 21 ph qc_end end | [{"type":"physical unit","value":"Ratio [OF] H2CO3 to HCO3- "}] | [{"type":"physical unit","value":"10^1.05"}] | [{"type":"other","value":"Henderson-Hasselbalch equation."},{"type":"physical unit","value":"pH [OF] blood [=] \\pu{7.40}"}] | <h1 class="questionTitle" itemprop="name">How would you use the Henderson-Hasselbalch equation to calculate the ratio of H2CO3 to HCO3- in blood having a pH of 7.40?
</h1> | null | 10^1.05 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If <mathjax>#pH= 7.40#</mathjax>, then <mathjax>#log_(10){[HC(=O)O^-]/[HC(=O)OH]}#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.05#</mathjax>. This reasonable because the <mathjax>#pH#</mathjax> is higher than the <mathjax>#pK_a#</mathjax> and there should be more carbonate than carbonic acid.</p>
<p>Thus <mathjax>#{[HC(=O)O^-]/[HC(=O)OH]}#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^(1.05)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#pH = pK_a + log_(10){[A^-]/[HA]}#</mathjax>. Clearly, we need <mathjax>#pK_a#</mathjax> for <mathjax>#H_2CO_3#</mathjax>, i.e. <mathjax>#pK_a = 6.35.#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If <mathjax>#pH= 7.40#</mathjax>, then <mathjax>#log_(10){[HC(=O)O^-]/[HC(=O)OH]}#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.05#</mathjax>. This reasonable because the <mathjax>#pH#</mathjax> is higher than the <mathjax>#pK_a#</mathjax> and there should be more carbonate than carbonic acid.</p>
<p>Thus <mathjax>#{[HC(=O)O^-]/[HC(=O)OH]}#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^(1.05)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How would you use the Henderson-Hasselbalch equation to calculate the ratio of H2CO3 to HCO3- in blood having a pH of 7.40?
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<div class="markdown"><p><mathjax>#pH = pK_a + log_(10){[A^-]/[HA]}#</mathjax>. Clearly, we need <mathjax>#pK_a#</mathjax> for <mathjax>#H_2CO_3#</mathjax>, i.e. <mathjax>#pK_a = 6.35.#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>If <mathjax>#pH= 7.40#</mathjax>, then <mathjax>#log_(10){[HC(=O)O^-]/[HC(=O)OH]}#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.05#</mathjax>. This reasonable because the <mathjax>#pH#</mathjax> is higher than the <mathjax>#pK_a#</mathjax> and there should be more carbonate than carbonic acid.</p>
<p>Thus <mathjax>#{[HC(=O)O^-]/[HC(=O)OH]}#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^(1.05)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p></div>
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</article> | How would you use the Henderson-Hasselbalch equation to calculate the ratio of H2CO3 to HCO3- in blood having a pH of 7.40?
| null |
3,377 | ab0e129a-6ddd-11ea-8a5f-ccda262736ce | https://socratic.org/questions/how-many-liters-of-a-0-2-m-naoh-solution-are-needed-in-order-to-have-1-0-moles-o | 5.00 liters | start physical_unit 7 8 volume l qc_end physical_unit 7 7 15 16 mole qc_end physical_unit 7 8 5 6 molarity qc_end end | [{"type":"physical unit","value":"Volume [OF] NaOH solution [IN] liters"}] | [{"type":"physical unit","value":"5.00 liters"}] | [{"type":"physical unit","value":"Mole [OF] NaOH [=] \\pu{1.0 moles}"},{"type":"physical unit","value":"Molarity [OF] NaOH solution [=] \\pu{0.2 M}"}] | <h1 class="questionTitle" itemprop="name">How many liters of a 0.2 M NaOH solution are needed in order to have 1.0 moles of NaOH?</h1> | null | 5.00 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax>.</p>
<p>Given that the typical units of <mathjax>#"concentration"#</mathjax> are <mathjax>#mol*L^-1#</mathjax>, to get <mathjax>#"concentration"#</mathjax> or <mathjax>#"amount of moles"#</mathjax> or <mathjax>#"volume of solution"#</mathjax> we just have to take the appropriate product or quotient.</p>
<p>We want <mathjax>#1.0*mol#</mathjax> <mathjax>#NaOH#</mathjax>: </p>
<p>this the product <mathjax>#"Volume of solution "xx" Concentration"#</mathjax>.</p>
<p>So we need the quotient: <mathjax>#"Moles of solute"/"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.0*cancel(mol))/(0.2*cancel(mol)*L^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#5*L#</mathjax>. </p>
<p>(i.e. <mathjax>#1/(1*L^-1)=1*L)#</mathjax></p>
<p>Note that we go to such trouble in including the units in these calculations as an extra check on our arithmetic. Sometimes you ask yourself should I divide or should I multiply. Dimensional analysis answers our question. We wanted an answer in <mathjax>#"litres"#</mathjax>, and we got one. This persuades us that we did the calculation right. </p></div>
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<div class="markdown"><p><mathjax>#"5 litres"#</mathjax> of solution are required.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax>.</p>
<p>Given that the typical units of <mathjax>#"concentration"#</mathjax> are <mathjax>#mol*L^-1#</mathjax>, to get <mathjax>#"concentration"#</mathjax> or <mathjax>#"amount of moles"#</mathjax> or <mathjax>#"volume of solution"#</mathjax> we just have to take the appropriate product or quotient.</p>
<p>We want <mathjax>#1.0*mol#</mathjax> <mathjax>#NaOH#</mathjax>: </p>
<p>this the product <mathjax>#"Volume of solution "xx" Concentration"#</mathjax>.</p>
<p>So we need the quotient: <mathjax>#"Moles of solute"/"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.0*cancel(mol))/(0.2*cancel(mol)*L^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#5*L#</mathjax>. </p>
<p>(i.e. <mathjax>#1/(1*L^-1)=1*L)#</mathjax></p>
<p>Note that we go to such trouble in including the units in these calculations as an extra check on our arithmetic. Sometimes you ask yourself should I divide or should I multiply. Dimensional analysis answers our question. We wanted an answer in <mathjax>#"litres"#</mathjax>, and we got one. This persuades us that we did the calculation right. </p></div>
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<h1 class="questionTitle" itemprop="name">How many liters of a 0.2 M NaOH solution are needed in order to have 1.0 moles of NaOH?</h1>
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<div class="markdown"><p><mathjax>#"5 litres"#</mathjax> of solution are required.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax>.</p>
<p>Given that the typical units of <mathjax>#"concentration"#</mathjax> are <mathjax>#mol*L^-1#</mathjax>, to get <mathjax>#"concentration"#</mathjax> or <mathjax>#"amount of moles"#</mathjax> or <mathjax>#"volume of solution"#</mathjax> we just have to take the appropriate product or quotient.</p>
<p>We want <mathjax>#1.0*mol#</mathjax> <mathjax>#NaOH#</mathjax>: </p>
<p>this the product <mathjax>#"Volume of solution "xx" Concentration"#</mathjax>.</p>
<p>So we need the quotient: <mathjax>#"Moles of solute"/"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.0*cancel(mol))/(0.2*cancel(mol)*L^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#5*L#</mathjax>. </p>
<p>(i.e. <mathjax>#1/(1*L^-1)=1*L)#</mathjax></p>
<p>Note that we go to such trouble in including the units in these calculations as an extra check on our arithmetic. Sometimes you ask yourself should I divide or should I multiply. Dimensional analysis answers our question. We wanted an answer in <mathjax>#"litres"#</mathjax>, and we got one. This persuades us that we did the calculation right. </p></div>
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</article> | How many liters of a 0.2 M NaOH solution are needed in order to have 1.0 moles of NaOH? | null |
3,378 | a842ca98-6ddd-11ea-9466-ccda262736ce | https://socratic.org/questions/hydrofluoric-acid-hf-reacts-with-silicon-dioxide-to-produce-silicon-tetraflourid | 4 HF + SiO2 -> SiF4 + 2 H2O | start chemical_equation qc_end chemical_equation 2 2 qc_end substance 5 6 qc_end substance 9 10 qc_end substance 12 12 qc_end end | [{"type":"other","value":"Chemical Equation [OF] this reaction"}] | [{"type":"chemical equation","value":"4 HF + SiO2 -> SiF4 + 2 H2O"}] | [{"type":"chemical equation","value":"HF"},{"type":"substance name","value":"Silicon dioxide"},{"type":"substance name","value":"Silicon tetraflouride"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">Hydrofluoric acid (HF) reacts with silicon dioxide to produce silicon tetraflouride and water. How do you write the balanced equation for this reaction? </h1> | null | 4 HF + SiO2 -> SiF4 + 2 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing the chemical formulas of the <strong>reactants</strong>, which are </p>
<blockquote>
<ul>
<li><em>hydrofluoric acid</em>, <mathjax>#"HF"#</mathjax></li>
<li><em>silicon dioxide</em>, <mathjax>#"SiO"_2#</mathjax></li>
</ul>
</blockquote>
<p>Do the same for the <strong>products</strong>, which are </p>
<blockquote>
<ul>
<li><em>silicon tetrafluoride</em>, <mathjax>#"SiF"_4#</mathjax></li>
<li><em>water</em>, <mathjax>#"H"_2"O"#</mathjax></li>
</ul>
</blockquote>
<p>Before writing the <em>unbalanced chemical equation</em> for this reaction, it's important to note that this reaction is actually used to <strong>decorate glass</strong> by inducing etching patterns on it surface. </p>
<p>Aqueous hydrofluoric acid will react with the glass, which is represented by silicon dioxide, and produce silicon tetrafluoride, a gas that will be given off, and liquid water. </p>
<p>The unbalanced chemical equation will thus be </p>
<blockquote>
<p><mathjax>#"HF"_text((aq]) + "SiO"_text(2(s]) -> "SiF"_text(4(g]) uarr + "H"_2"O"_text((L])#</mathjax></p>
</blockquote>
<p>To balance this equation, all you have to do is multiply the hydrofluoric acid by <mathjax>#4#</mathjax> and the water by <mathjax>#2#</mathjax>. This will ensure that <strong>all the atoms</strong> present on the reactants' side are accounted for on the products' side. </p>
<blockquote>
<p><mathjax>#4"HF"_text((aq]) + "SiO"_text(2(s]) -> "SiF"_text(4(g]) uarr + 2"H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>I think that this reaction can also produce aqueous <em>hexafluorosilicic acid</em>, <mathjax>#"H"_2"SiF"_6#</mathjax>, and water.</p>
<p>Mind you, and this should go without saying, hydrofluoric acid is <strong>extremely, extremely dangerous</strong>! <em>A lot</em> of horrific injuries have occurred while using hydrofluoric acid for glass etching projects!</p>
<p>Here's a very cool video showing the reaction between a glass light bulb and hydrofluoric acid </p>
<p>
<iframe src="https://www.youtube.com/embed/6ZBwluyR2Tc?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#4"HF"_text((aq]) + "SiO"_text(2(s]) -> "SiF"_text(4(g]) uarr + 2"H"_2"O"_text((l])#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing the chemical formulas of the <strong>reactants</strong>, which are </p>
<blockquote>
<ul>
<li><em>hydrofluoric acid</em>, <mathjax>#"HF"#</mathjax></li>
<li><em>silicon dioxide</em>, <mathjax>#"SiO"_2#</mathjax></li>
</ul>
</blockquote>
<p>Do the same for the <strong>products</strong>, which are </p>
<blockquote>
<ul>
<li><em>silicon tetrafluoride</em>, <mathjax>#"SiF"_4#</mathjax></li>
<li><em>water</em>, <mathjax>#"H"_2"O"#</mathjax></li>
</ul>
</blockquote>
<p>Before writing the <em>unbalanced chemical equation</em> for this reaction, it's important to note that this reaction is actually used to <strong>decorate glass</strong> by inducing etching patterns on it surface. </p>
<p>Aqueous hydrofluoric acid will react with the glass, which is represented by silicon dioxide, and produce silicon tetrafluoride, a gas that will be given off, and liquid water. </p>
<p>The unbalanced chemical equation will thus be </p>
<blockquote>
<p><mathjax>#"HF"_text((aq]) + "SiO"_text(2(s]) -> "SiF"_text(4(g]) uarr + "H"_2"O"_text((L])#</mathjax></p>
</blockquote>
<p>To balance this equation, all you have to do is multiply the hydrofluoric acid by <mathjax>#4#</mathjax> and the water by <mathjax>#2#</mathjax>. This will ensure that <strong>all the atoms</strong> present on the reactants' side are accounted for on the products' side. </p>
<blockquote>
<p><mathjax>#4"HF"_text((aq]) + "SiO"_text(2(s]) -> "SiF"_text(4(g]) uarr + 2"H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>I think that this reaction can also produce aqueous <em>hexafluorosilicic acid</em>, <mathjax>#"H"_2"SiF"_6#</mathjax>, and water.</p>
<p>Mind you, and this should go without saying, hydrofluoric acid is <strong>extremely, extremely dangerous</strong>! <em>A lot</em> of horrific injuries have occurred while using hydrofluoric acid for glass etching projects!</p>
<p>Here's a very cool video showing the reaction between a glass light bulb and hydrofluoric acid </p>
<p>
<iframe src="https://www.youtube.com/embed/6ZBwluyR2Tc?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Hydrofluoric acid (HF) reacts with silicon dioxide to produce silicon tetraflouride and water. How do you write the balanced equation for this reaction? </h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#4"HF"_text((aq]) + "SiO"_text(2(s]) -> "SiF"_text(4(g]) uarr + 2"H"_2"O"_text((l])#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing the chemical formulas of the <strong>reactants</strong>, which are </p>
<blockquote>
<ul>
<li><em>hydrofluoric acid</em>, <mathjax>#"HF"#</mathjax></li>
<li><em>silicon dioxide</em>, <mathjax>#"SiO"_2#</mathjax></li>
</ul>
</blockquote>
<p>Do the same for the <strong>products</strong>, which are </p>
<blockquote>
<ul>
<li><em>silicon tetrafluoride</em>, <mathjax>#"SiF"_4#</mathjax></li>
<li><em>water</em>, <mathjax>#"H"_2"O"#</mathjax></li>
</ul>
</blockquote>
<p>Before writing the <em>unbalanced chemical equation</em> for this reaction, it's important to note that this reaction is actually used to <strong>decorate glass</strong> by inducing etching patterns on it surface. </p>
<p>Aqueous hydrofluoric acid will react with the glass, which is represented by silicon dioxide, and produce silicon tetrafluoride, a gas that will be given off, and liquid water. </p>
<p>The unbalanced chemical equation will thus be </p>
<blockquote>
<p><mathjax>#"HF"_text((aq]) + "SiO"_text(2(s]) -> "SiF"_text(4(g]) uarr + "H"_2"O"_text((L])#</mathjax></p>
</blockquote>
<p>To balance this equation, all you have to do is multiply the hydrofluoric acid by <mathjax>#4#</mathjax> and the water by <mathjax>#2#</mathjax>. This will ensure that <strong>all the atoms</strong> present on the reactants' side are accounted for on the products' side. </p>
<blockquote>
<p><mathjax>#4"HF"_text((aq]) + "SiO"_text(2(s]) -> "SiF"_text(4(g]) uarr + 2"H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>I think that this reaction can also produce aqueous <em>hexafluorosilicic acid</em>, <mathjax>#"H"_2"SiF"_6#</mathjax>, and water.</p>
<p>Mind you, and this should go without saying, hydrofluoric acid is <strong>extremely, extremely dangerous</strong>! <em>A lot</em> of horrific injuries have occurred while using hydrofluoric acid for glass etching projects!</p>
<p>Here's a very cool video showing the reaction between a glass light bulb and hydrofluoric acid </p>
<p>
<iframe src="https://www.youtube.com/embed/6ZBwluyR2Tc?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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</article> | Hydrofluoric acid (HF) reacts with silicon dioxide to produce silicon tetraflouride and water. How do you write the balanced equation for this reaction? | null |
3,379 | abc93bee-6ddd-11ea-bd7c-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-number-of-sodium-na-in-pure-sodium-na | 0 | start physical_unit 7 7 oxidation_number none qc_end chemical_equation 7 7 qc_end end | [{"type":"physical unit","value":"Oxidation number [OF] Na"}] | [{"type":"physical unit","value":"0"}] | [{"type":"chemical equation","value":"Na"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation number of sodium (Na) in pure sodium, Na?</h1> | null | 0 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Redox reaction are conceived to involve the transfer of electrons. If a species gains electrons it is said to have been <mathjax>#"REDUCED"#</mathjax>. If a species loses electrons is is said to have been <mathjax>#"OXIDIZED"#</mathjax>. The oxidation number of an element in its standard state is thus manifestly <mathjax>#"ZERO"#</mathjax>. What are the <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> of <mathjax>#O_2#</mathjax>, and <mathjax>#F_2#</mathjax>, and <mathjax>#Ca#</mathjax>?</p>
<p>The notions of oxidation and reduction have been flying pretty fast at you. Remember that these are conceptual notions, that do not have real significance. Assignment of oxidation numbers allows us to balance complicated redox equations. So it might be an idea to suspend your disbelief when we say that <mathjax>#Mn^(2+)#</mathjax> loses 5 electrons when it is oxidized to permanaganate ion, <mathjax>#Mn(VII+)#</mathjax>, and focus on the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>.</p>
<p><mathjax>#Mn^(2+)+4H_2OrarrMnO_4^(-)+8H^(+)+5e^(-)#</mathjax></p>
<p>Are mass and charge conserved here? If not, they should be. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>In sodium metal, the oxidation number is <mathjax>#0#</mathjax>. Elements are <mathjax>#"zerovalent"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Redox reaction are conceived to involve the transfer of electrons. If a species gains electrons it is said to have been <mathjax>#"REDUCED"#</mathjax>. If a species loses electrons is is said to have been <mathjax>#"OXIDIZED"#</mathjax>. The oxidation number of an element in its standard state is thus manifestly <mathjax>#"ZERO"#</mathjax>. What are the <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> of <mathjax>#O_2#</mathjax>, and <mathjax>#F_2#</mathjax>, and <mathjax>#Ca#</mathjax>?</p>
<p>The notions of oxidation and reduction have been flying pretty fast at you. Remember that these are conceptual notions, that do not have real significance. Assignment of oxidation numbers allows us to balance complicated redox equations. So it might be an idea to suspend your disbelief when we say that <mathjax>#Mn^(2+)#</mathjax> loses 5 electrons when it is oxidized to permanaganate ion, <mathjax>#Mn(VII+)#</mathjax>, and focus on the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>.</p>
<p><mathjax>#Mn^(2+)+4H_2OrarrMnO_4^(-)+8H^(+)+5e^(-)#</mathjax></p>
<p>Are mass and charge conserved here? If not, they should be. </p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation number of sodium (Na) in pure sodium, Na?</h1>
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<div class="markdown"><p>In sodium metal, the oxidation number is <mathjax>#0#</mathjax>. Elements are <mathjax>#"zerovalent"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Redox reaction are conceived to involve the transfer of electrons. If a species gains electrons it is said to have been <mathjax>#"REDUCED"#</mathjax>. If a species loses electrons is is said to have been <mathjax>#"OXIDIZED"#</mathjax>. The oxidation number of an element in its standard state is thus manifestly <mathjax>#"ZERO"#</mathjax>. What are the <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> of <mathjax>#O_2#</mathjax>, and <mathjax>#F_2#</mathjax>, and <mathjax>#Ca#</mathjax>?</p>
<p>The notions of oxidation and reduction have been flying pretty fast at you. Remember that these are conceptual notions, that do not have real significance. Assignment of oxidation numbers allows us to balance complicated redox equations. So it might be an idea to suspend your disbelief when we say that <mathjax>#Mn^(2+)#</mathjax> loses 5 electrons when it is oxidized to permanaganate ion, <mathjax>#Mn(VII+)#</mathjax>, and focus on the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>.</p>
<p><mathjax>#Mn^(2+)+4H_2OrarrMnO_4^(-)+8H^(+)+5e^(-)#</mathjax></p>
<p>Are mass and charge conserved here? If not, they should be. </p></div>
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</article> | What is the oxidation number of sodium (Na) in pure sodium, Na? | null |
3,380 | ab67d668-6ddd-11ea-8990-ccda262736ce | https://socratic.org/questions/a-quantity-of-a-gas-at-a-temperature-of-223-k-has-a-volume-of-100-0-dm-3-to-what | 412.57 K | start physical_unit 21 22 temperature k qc_end physical_unit 21 22 9 10 temperature qc_end physical_unit 21 22 15 16 volume qc_end physical_unit 21 22 36 37 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the gas [IN] K"}] | [{"type":"physical unit","value":"412.57 K"}] | [{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{223 K}"},{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{100.0 dm^3}"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{185 dm^3}"},{"type":"other","value":"The pressure is kept constant."}] | <h1 class="questionTitle" itemprop="name">A quantity of a gas at a temperature of #223# #K# has a volume of #100.0# #dm^3# To what temperature must the gas be raised, while the pressure is kept constant, to give a volume of #185# #dm^3#?</h1> | null | 412.57 K | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Let us recall the gas law"#</mathjax></p>
<p><mathjax>#"PV = nRT"#</mathjax></p>
<p>Our question is that a quantity of a gas at a temperature of <br/>
223K has a volume of <mathjax>#"100.0 dm"^3#</mathjax>. </p>
<p>To what temperature must the gas be raised, while the pressure is kept constant, to give a volume of <mathjax>#"185 dm"^3#</mathjax> ?</p>
<p>So let;s rearrange the formula </p>
<p><mathjax>#P = "(nRT)"/V#</mathjax></p>
<p><mathjax>#Where V = 100.0dm^3#</mathjax></p>
<p><mathjax>#"Where p = is unknown or x"#</mathjax></p>
<p><mathjax>#"Where R = gas constant law which is 0.0821L"#</mathjax></p>
<p><mathjax>#"Where n = n is number of moles of gas"#</mathjax></p>
<p>So let us plug in the variables</p>
<p><mathjax>#"100 dm"^3#</mathjax> should be converted in litres first.</p>
<blockquote>
<p><mathjax>#"1 dm"^3 = "1 L"#</mathjax></p>
</blockquote>
<p>So</p>
<blockquote>
<p><mathjax>#"100 dm"^3 = "100 L"#</mathjax></p>
</blockquote>
<p><mathjax># (n * 0.0821L * 223K) /(100L) = "18.309n"/"100L" = 0.18309n #</mathjax></p>
<p>So if pressure is constant to get a 185dm^3 </p>
<p>The expression would be </p>
<p><mathjax>#185dm^3 = (n * 0.0821L * T)/(0.18309n)#</mathjax></p>
<p>Cut out the both n</p>
<p><mathjax>#185dm^3 = 0.0821 * T/"0.18309"#</mathjax></p>
<p><mathjax>#"so T = Let's solve your equation step-by-step."#</mathjax></p>
<p><mathjax>#185="0.0821t"/0.18309#</mathjax></p>
<p>Step 1: Multiply both sides by 0.18309.</p>
<p><mathjax>#185="0.0821t"/0.18309#</mathjax></p>
<p><mathjax>#(185) * (0.18309) = ("0.0821t" * 0.18309) * (0.18309) = 33.87165=0.0821t#</mathjax></p>
<p>Step 2: Flip the equation.</p>
<p><mathjax>#0.0821t=33.87165#</mathjax></p>
<p>Step 3: Divide both sides by 0.0821.</p>
<p><mathjax>#"0.0821t"/0.0821 = 33.87165/0.0821#</mathjax></p>
<p><mathjax>#t=412.565773#</mathjax></p>
<p>Answer:</p>
<p><mathjax>#T= "412.6 K"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"412.6 K"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Let us recall the gas law"#</mathjax></p>
<p><mathjax>#"PV = nRT"#</mathjax></p>
<p>Our question is that a quantity of a gas at a temperature of <br/>
223K has a volume of <mathjax>#"100.0 dm"^3#</mathjax>. </p>
<p>To what temperature must the gas be raised, while the pressure is kept constant, to give a volume of <mathjax>#"185 dm"^3#</mathjax> ?</p>
<p>So let;s rearrange the formula </p>
<p><mathjax>#P = "(nRT)"/V#</mathjax></p>
<p><mathjax>#Where V = 100.0dm^3#</mathjax></p>
<p><mathjax>#"Where p = is unknown or x"#</mathjax></p>
<p><mathjax>#"Where R = gas constant law which is 0.0821L"#</mathjax></p>
<p><mathjax>#"Where n = n is number of moles of gas"#</mathjax></p>
<p>So let us plug in the variables</p>
<p><mathjax>#"100 dm"^3#</mathjax> should be converted in litres first.</p>
<blockquote>
<p><mathjax>#"1 dm"^3 = "1 L"#</mathjax></p>
</blockquote>
<p>So</p>
<blockquote>
<p><mathjax>#"100 dm"^3 = "100 L"#</mathjax></p>
</blockquote>
<p><mathjax># (n * 0.0821L * 223K) /(100L) = "18.309n"/"100L" = 0.18309n #</mathjax></p>
<p>So if pressure is constant to get a 185dm^3 </p>
<p>The expression would be </p>
<p><mathjax>#185dm^3 = (n * 0.0821L * T)/(0.18309n)#</mathjax></p>
<p>Cut out the both n</p>
<p><mathjax>#185dm^3 = 0.0821 * T/"0.18309"#</mathjax></p>
<p><mathjax>#"so T = Let's solve your equation step-by-step."#</mathjax></p>
<p><mathjax>#185="0.0821t"/0.18309#</mathjax></p>
<p>Step 1: Multiply both sides by 0.18309.</p>
<p><mathjax>#185="0.0821t"/0.18309#</mathjax></p>
<p><mathjax>#(185) * (0.18309) = ("0.0821t" * 0.18309) * (0.18309) = 33.87165=0.0821t#</mathjax></p>
<p>Step 2: Flip the equation.</p>
<p><mathjax>#0.0821t=33.87165#</mathjax></p>
<p>Step 3: Divide both sides by 0.0821.</p>
<p><mathjax>#"0.0821t"/0.0821 = 33.87165/0.0821#</mathjax></p>
<p><mathjax>#t=412.565773#</mathjax></p>
<p>Answer:</p>
<p><mathjax>#T= "412.6 K"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A quantity of a gas at a temperature of #223# #K# has a volume of #100.0# #dm^3# To what temperature must the gas be raised, while the pressure is kept constant, to give a volume of #185# #dm^3#?</h1>
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Professor Sam
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Stefan V.
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Feb 10, 2017
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<div>
<div class="markdown"><p><mathjax>#"412.6 K"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Let us recall the gas law"#</mathjax></p>
<p><mathjax>#"PV = nRT"#</mathjax></p>
<p>Our question is that a quantity of a gas at a temperature of <br/>
223K has a volume of <mathjax>#"100.0 dm"^3#</mathjax>. </p>
<p>To what temperature must the gas be raised, while the pressure is kept constant, to give a volume of <mathjax>#"185 dm"^3#</mathjax> ?</p>
<p>So let;s rearrange the formula </p>
<p><mathjax>#P = "(nRT)"/V#</mathjax></p>
<p><mathjax>#Where V = 100.0dm^3#</mathjax></p>
<p><mathjax>#"Where p = is unknown or x"#</mathjax></p>
<p><mathjax>#"Where R = gas constant law which is 0.0821L"#</mathjax></p>
<p><mathjax>#"Where n = n is number of moles of gas"#</mathjax></p>
<p>So let us plug in the variables</p>
<p><mathjax>#"100 dm"^3#</mathjax> should be converted in litres first.</p>
<blockquote>
<p><mathjax>#"1 dm"^3 = "1 L"#</mathjax></p>
</blockquote>
<p>So</p>
<blockquote>
<p><mathjax>#"100 dm"^3 = "100 L"#</mathjax></p>
</blockquote>
<p><mathjax># (n * 0.0821L * 223K) /(100L) = "18.309n"/"100L" = 0.18309n #</mathjax></p>
<p>So if pressure is constant to get a 185dm^3 </p>
<p>The expression would be </p>
<p><mathjax>#185dm^3 = (n * 0.0821L * T)/(0.18309n)#</mathjax></p>
<p>Cut out the both n</p>
<p><mathjax>#185dm^3 = 0.0821 * T/"0.18309"#</mathjax></p>
<p><mathjax>#"so T = Let's solve your equation step-by-step."#</mathjax></p>
<p><mathjax>#185="0.0821t"/0.18309#</mathjax></p>
<p>Step 1: Multiply both sides by 0.18309.</p>
<p><mathjax>#185="0.0821t"/0.18309#</mathjax></p>
<p><mathjax>#(185) * (0.18309) = ("0.0821t" * 0.18309) * (0.18309) = 33.87165=0.0821t#</mathjax></p>
<p>Step 2: Flip the equation.</p>
<p><mathjax>#0.0821t=33.87165#</mathjax></p>
<p>Step 3: Divide both sides by 0.0821.</p>
<p><mathjax>#"0.0821t"/0.0821 = 33.87165/0.0821#</mathjax></p>
<p><mathjax>#t=412.565773#</mathjax></p>
<p>Answer:</p>
<p><mathjax>#T= "412.6 K"#</mathjax></p></div>
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</article> | A quantity of a gas at a temperature of #223# #K# has a volume of #100.0# #dm^3# To what temperature must the gas be raised, while the pressure is kept constant, to give a volume of #185# #dm^3#? | null |
3,381 | aadfbcfe-6ddd-11ea-9b94-ccda262736ce | https://socratic.org/questions/what-would-be-the-empirical-formula-of-ethane | CH3 | start chemical_formula qc_end substance 7 7 qc_end end | [{"type":"other","value":"Chemical Formula [OF] ethane [IN] empirical"}] | [{"type":"chemical equation","value":"CH3"}] | [{"type":"substance name","value":"Ethane"}] | <h1 class="questionTitle" itemprop="name"> What would be the empirical formula of ethane?
</h1> | null | CH3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Empirical formulas are the simplest ratio of the atoms in a compound. Ethane has a molecular formula of <mathjax>#C_2H_6#</mathjax>. Both of the number of carbons and hydrogens are divisible by 2, so to get the empirical formula we are trying to find their lowest ratio, which in this case is <mathjax>#CH_3#</mathjax>.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#CH_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Empirical formulas are the simplest ratio of the atoms in a compound. Ethane has a molecular formula of <mathjax>#C_2H_6#</mathjax>. Both of the number of carbons and hydrogens are divisible by 2, so to get the empirical formula we are trying to find their lowest ratio, which in this case is <mathjax>#CH_3#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name"> What would be the empirical formula of ethane?
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Oliver S.
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<div class="markdown"><p><mathjax>#CH_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Empirical formulas are the simplest ratio of the atoms in a compound. Ethane has a molecular formula of <mathjax>#C_2H_6#</mathjax>. Both of the number of carbons and hydrogens are divisible by 2, so to get the empirical formula we are trying to find their lowest ratio, which in this case is <mathjax>#CH_3#</mathjax>.</p></div>
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3,382 | ad0eab9e-6ddd-11ea-a861-ccda262736ce | https://socratic.org/questions/a-hot-air-balloon-is-filled-with-1-89-x-10-2-liters-of-air-at-21-c-if-atmospheri | 700.00 K | start physical_unit 25 26 temperature k qc_end physical_unit 25 26 14 15 temperature qc_end physical_unit 25 26 7 10 volume qc_end physical_unit 25 26 35 38 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the air [IN] K"}] | [{"type":"physical unit","value":"700.00 K"}] | [{"type":"physical unit","value":"Temperature1 [OF] the air [=] \\pu{21 ℃}"},{"type":"physical unit","value":"Volume1 [OF] the air [=] \\pu{1.89 × 10^2 liters}"},{"type":"physical unit","value":"Volume2 [OF] the air [=] \\pu{4.5 × 10^2 liters}"},{"type":"other","value":"Atmospheric pressure does not change,"}] | <h1 class="questionTitle" itemprop="name">A hot air balloon is filled with #1.89 x 10^2# liters of air at 21 C. If atmospheric pressure does not change, how hot must the air become in order to increase the volume to #4.5 x 10^2# liters? </h1> | null | 700.00 K | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Charles' Law"#</mathjax> states that a given quantity of gas has a volume directly proportional to the absolute temperature:</p>
<p><mathjax>#VpropT#</mathjax>, else <mathjax>#V=kT#</mathjax>. Solving for <mathjax>#k#</mathjax>, <mathjax>#V_1/T_1=V_2/T_2#</mathjax>, and <mathjax>#T_2=V_2/V_1xxT_1#</mathjax></p>
<p><mathjax>#T_2=(4.5xx10^2*L)/(1.89xx10^2*L)xx294*K#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??K#</mathjax></p></div>
</div>
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<div>
<div class="markdown"><p>We use <mathjax>#"Charles' Law"#</mathjax> <mathjax>#V_2/T_2=V_1/T_1#</mathjax>. <mathjax>#T_2#</mathjax> will at least double <mathjax>#T_1#</mathjax>. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Charles' Law"#</mathjax> states that a given quantity of gas has a volume directly proportional to the absolute temperature:</p>
<p><mathjax>#VpropT#</mathjax>, else <mathjax>#V=kT#</mathjax>. Solving for <mathjax>#k#</mathjax>, <mathjax>#V_1/T_1=V_2/T_2#</mathjax>, and <mathjax>#T_2=V_2/V_1xxT_1#</mathjax></p>
<p><mathjax>#T_2=(4.5xx10^2*L)/(1.89xx10^2*L)xx294*K#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??K#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A hot air balloon is filled with #1.89 x 10^2# liters of air at 21 C. If atmospheric pressure does not change, how hot must the air become in order to increase the volume to #4.5 x 10^2# liters? </h1>
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anor277
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<div class="markdown"><p>We use <mathjax>#"Charles' Law"#</mathjax> <mathjax>#V_2/T_2=V_1/T_1#</mathjax>. <mathjax>#T_2#</mathjax> will at least double <mathjax>#T_1#</mathjax>. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Charles' Law"#</mathjax> states that a given quantity of gas has a volume directly proportional to the absolute temperature:</p>
<p><mathjax>#VpropT#</mathjax>, else <mathjax>#V=kT#</mathjax>. Solving for <mathjax>#k#</mathjax>, <mathjax>#V_1/T_1=V_2/T_2#</mathjax>, and <mathjax>#T_2=V_2/V_1xxT_1#</mathjax></p>
<p><mathjax>#T_2=(4.5xx10^2*L)/(1.89xx10^2*L)xx294*K#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??K#</mathjax></p></div>
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</article> | A hot air balloon is filled with #1.89 x 10^2# liters of air at 21 C. If atmospheric pressure does not change, how hot must the air become in order to increase the volume to #4.5 x 10^2# liters? | null |
3,383 | acb6a16e-6ddd-11ea-a793-ccda262736ce | https://socratic.org/questions/how-i-can-calculate-the-number-of-hydrogen-atoms-in-one-gallon-deionized-water-t | 3.04 × 10^26 | start physical_unit 7 8 number none qc_end physical_unit 12 13 10 11 volume qc_end end | [{"type":"physical unit","value":"Number [OF] hydrogen atoms"}] | [{"type":"physical unit","value":"3.04 × 10^26"}] | [{"type":"physical unit","value":"Volume [OF] deionized water [=] \\pu{1 gallon}"}] | <h1 class="questionTitle" itemprop="name">How I can calculate the number of hydrogen atoms in one gallon deionized water? thank you so much. </h1> | null | 3.04 × 10^26 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Convert gallons of water to litres</strong></p>
<p><mathjax>#"Volume" = 1 color(red)(cancel(color(black)("gal"))) × "4.55 L"/(1 color(red)(cancel(color(black)("gal")))) = "4.55 L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Convert litres of water to grams of water</strong></p>
<p><mathjax>#"Grams of water" = 4.55 color(red)(cancel(color(black)("L"))) × (1000 color(red)(cancel(color(black)("mL"))))/(1 color(red)(cancel(color(black)("L")))) × "0.9982 g"/(1 color(red)(cancel(color(black)("L")))) = "4542 g"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Convert grams of water to moles of water</strong></p>
<p><mathjax>#"Moles of water" = 4542 color(red)(cancel(color(black)("g"))) × "1 mol"/(18.02 color(red)(cancel(color(black)("g")))) = "252.0 mol"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4. Convert moles of water to molecules of water</strong></p>
<p><mathjax>#"Molecules of water" = 252.0 color(red)(cancel(color(black)("mol"))) × (6.022 × 10^23 "molecules")/(1 color(red)(cancel(color(black)("mol"))))#</mathjax></p>
<p><mathjax>#= 1.518 × 10^26color(white)(l) "molecules"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 5. Convert molecules of water to atoms of hydrogen</strong></p>
<p><mathjax>#"Atoms of H" = 1.518 × 10^26color(red)(cancel(color(black)("molecules H"_2"O"))) × "2 atoms H"/(1 color(red)(cancel(color(black)("molecule H"_2"O"))))#</mathjax></p>
<p><mathjax>#= 3.04 × 10^26color(white)(l) "atoms H"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><blockquote></blockquote>
<p>There are <mathjax>#3.04 × 10^26color(white)(l) "atoms of H"#</mathjax> in 1 gal of water.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Convert gallons of water to litres</strong></p>
<p><mathjax>#"Volume" = 1 color(red)(cancel(color(black)("gal"))) × "4.55 L"/(1 color(red)(cancel(color(black)("gal")))) = "4.55 L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Convert litres of water to grams of water</strong></p>
<p><mathjax>#"Grams of water" = 4.55 color(red)(cancel(color(black)("L"))) × (1000 color(red)(cancel(color(black)("mL"))))/(1 color(red)(cancel(color(black)("L")))) × "0.9982 g"/(1 color(red)(cancel(color(black)("L")))) = "4542 g"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Convert grams of water to moles of water</strong></p>
<p><mathjax>#"Moles of water" = 4542 color(red)(cancel(color(black)("g"))) × "1 mol"/(18.02 color(red)(cancel(color(black)("g")))) = "252.0 mol"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4. Convert moles of water to molecules of water</strong></p>
<p><mathjax>#"Molecules of water" = 252.0 color(red)(cancel(color(black)("mol"))) × (6.022 × 10^23 "molecules")/(1 color(red)(cancel(color(black)("mol"))))#</mathjax></p>
<p><mathjax>#= 1.518 × 10^26color(white)(l) "molecules"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 5. Convert molecules of water to atoms of hydrogen</strong></p>
<p><mathjax>#"Atoms of H" = 1.518 × 10^26color(red)(cancel(color(black)("molecules H"_2"O"))) × "2 atoms H"/(1 color(red)(cancel(color(black)("molecule H"_2"O"))))#</mathjax></p>
<p><mathjax>#= 3.04 × 10^26color(white)(l) "atoms H"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How I can calculate the number of hydrogen atoms in one gallon deionized water? thank you so much. </h1>
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<p>There are <mathjax>#3.04 × 10^26color(white)(l) "atoms of H"#</mathjax> in 1 gal of water.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Convert gallons of water to litres</strong></p>
<p><mathjax>#"Volume" = 1 color(red)(cancel(color(black)("gal"))) × "4.55 L"/(1 color(red)(cancel(color(black)("gal")))) = "4.55 L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Convert litres of water to grams of water</strong></p>
<p><mathjax>#"Grams of water" = 4.55 color(red)(cancel(color(black)("L"))) × (1000 color(red)(cancel(color(black)("mL"))))/(1 color(red)(cancel(color(black)("L")))) × "0.9982 g"/(1 color(red)(cancel(color(black)("L")))) = "4542 g"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Convert grams of water to moles of water</strong></p>
<p><mathjax>#"Moles of water" = 4542 color(red)(cancel(color(black)("g"))) × "1 mol"/(18.02 color(red)(cancel(color(black)("g")))) = "252.0 mol"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4. Convert moles of water to molecules of water</strong></p>
<p><mathjax>#"Molecules of water" = 252.0 color(red)(cancel(color(black)("mol"))) × (6.022 × 10^23 "molecules")/(1 color(red)(cancel(color(black)("mol"))))#</mathjax></p>
<p><mathjax>#= 1.518 × 10^26color(white)(l) "molecules"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 5. Convert molecules of water to atoms of hydrogen</strong></p>
<p><mathjax>#"Atoms of H" = 1.518 × 10^26color(red)(cancel(color(black)("molecules H"_2"O"))) × "2 atoms H"/(1 color(red)(cancel(color(black)("molecule H"_2"O"))))#</mathjax></p>
<p><mathjax>#= 3.04 × 10^26color(white)(l) "atoms H"#</mathjax></p></div>
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</article> | How I can calculate the number of hydrogen atoms in one gallon deionized water? thank you so much. | null |
3,384 | acb62ac1-6ddd-11ea-8b2b-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-equation-for-butane-burning-completely-in-air | C4H10(g) + 13/2 O2(g) -> 4 CO2(g) + 5 H2O(l) | start chemical_equation qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Equation [OF] the chemical equation"}] | [{"type":"chemical equation","value":"C4H10(g) + 13/2 O2(g) -> 4 CO2(g) + 5 H2O(l)"}] | [{"type":"other","value":"Butane burns completely in air"}] | <h1 class="questionTitle" itemprop="name">What is the chemical equation for butane burning completely in air?</h1> | null | C4H10(g) + 13/2 O2(g) -> 4 CO2(g) + 5 H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#C_4H_10(g) + 13/2O_2(g) rarr 4CO_2(g) + 5H_2O(l)+Delta#</mathjax></p>
<p>If you like, you can double the equation to remove the half-integral coefficient. What does the <mathjax>#Delta#</mathjax> symbol on the reactant side represent?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Butane is completely combusted to carbon dioxide and water........</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#C_4H_10(g) + 13/2O_2(g) rarr 4CO_2(g) + 5H_2O(l)+Delta#</mathjax></p>
<p>If you like, you can double the equation to remove the half-integral coefficient. What does the <mathjax>#Delta#</mathjax> symbol on the reactant side represent?</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the chemical equation for butane burning completely in air?</h1>
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<div class="markdown"><p>Butane is completely combusted to carbon dioxide and water........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#C_4H_10(g) + 13/2O_2(g) rarr 4CO_2(g) + 5H_2O(l)+Delta#</mathjax></p>
<p>If you like, you can double the equation to remove the half-integral coefficient. What does the <mathjax>#Delta#</mathjax> symbol on the reactant side represent?</p></div>
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<a href="https://socratic.org/answers/460229" itemprop="url">Answer link</a>
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Nathan L.
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<span class="dateCreated" datetime="2017-08-04T18:38:24" itemprop="dateCreated">
Aug 4, 2017
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<div class="markdown"><p><mathjax>#"C"_4"H"_10(g) + 13/2"O"_2(g) rarr 4"CO"_2(g) + 5"H"_2"O"(g)#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Butane (<mathjax>#"C"_4"H"_10#</mathjax>) is undergoing <em>complete combustion</em> here, so we can write our generic hydrocarbon combustion equation:</p>
<blockquote>
<p><mathjax>#ul("hydrocarbon" + "oxygen" rarr "carbon dioxide" + "water (vapor)"#</mathjax> (complete combustion)</p>
</blockquote>
<p>Now we fill in the reagents and balance the equation:</p>
<blockquote>
<p><mathjax>#"C"_4"H"_10(g) + "O"_2(g) rarr "CO"_2(g) + "H"_2"O"(g)#</mathjax> (unbalanced)</p>
<p><mathjax>#color(red)(ulbar(|stackrel(" ")(" ""C"_4"H"_10(g) + 13/2"O"_2(g) rarr 4"CO"_2(g) + 5"H"_2"O"(g)" ")|)#</mathjax></p>
</blockquote></div>
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</article> | What is the chemical equation for butane burning completely in air? | null |
3,385 | a8c077ef-6ddd-11ea-b15d-ccda262736ce | https://socratic.org/questions/587ccb727c01494254dbaf39 | 2 NaN3(s) -> 2 Na(s) + 3 N2(g) | start chemical_equation qc_end chemical_equation 4 4 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 NaN3(s) -> 2 Na(s) + 3 N2(g)"}] | [{"type":"chemical equation","value":"NaN3"}] | <h1 class="questionTitle" itemprop="name">How does sodium azide, #NaN_3#, decompose?</h1> | null | 2 NaN3(s) -> 2 Na(s) + 3 N2(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is mass balanced? Is charge balanced? Is this a redox equation? </p>
<p>This reaction is used pratically to inflate an airbag after an auto collision. Why should the reaction do this?</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#2 NaN_3(s) → 2Na(s) + 3 N_2(g)#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is mass balanced? Is charge balanced? Is this a redox equation? </p>
<p>This reaction is used pratically to inflate an airbag after an auto collision. Why should the reaction do this?</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How does sodium azide, #NaN_3#, decompose?</h1>
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<div class="markdown"><p><mathjax>#2 NaN_3(s) → 2Na(s) + 3 N_2(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is mass balanced? Is charge balanced? Is this a redox equation? </p>
<p>This reaction is used pratically to inflate an airbag after an auto collision. Why should the reaction do this?</p></div>
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</article> | How does sodium azide, #NaN_3#, decompose? | null |
3,386 | ad120168-6ddd-11ea-bb74-ccda262736ce | https://socratic.org/questions/a-gas-occupies-50-l-at-a-pressure-of-2-atm-what-is-the-volume-when-the-pressure- | 10.00 L | start physical_unit 1 1 volume l qc_end physical_unit 1 1 3 4 volume qc_end physical_unit 1 1 9 10 pressure qc_end physical_unit 1 1 21 22 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"10.00 L"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{50 L}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{2 atm}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{10 atm}"}] | <h1 class="questionTitle" itemprop="name">A gas occupies 50 L at a pressure of 2 atm. What is the volume when the pressure is increased to 10 atm?</h1> | null | 10.00 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to calculate the new volume of a gas when its pressure is increased.</p>
<p>To solve this, we can use the pressure-volume relationship of gases, illustrated by <em><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a></em>:</p>
<p><mathjax>#P_1V_1 = P_2V_2#</mathjax></p>
<p>where</p>
<ul>
<li>
<p><mathjax>#P#</mathjax> is the initial (<mathjax>#"1"#</mathjax>) and final (<mathjax>#"2"#</mathjax>) pressure of the gas, and</p>
</li>
<li>
<p><mathjax>#V#</mathjax> is the initial (<mathjax>#"1"#</mathjax>) and final (<mathjax>#"2"#</mathjax>) volume of the gas.</p>
</li>
</ul>
<p>Let's plug in our known variables, and rearrange this equation to solve for the final volume, <mathjax>#V_2#</mathjax>:</p>
<p><mathjax>#V_2 = (P_1V_1)/(P_2) = ((2cancel("atm"))(50"L"))/(10cancel("atm")) = color(red)(10"L"#</mathjax></p>
<p>Thus, when the pressure of the gas was <em>increased</em> by a factor of <mathjax>#5#</mathjax> (<mathjax>#2"atm"#</mathjax> to <mathjax>#10"atm"#</mathjax>), the volume <em>decreased</em> by a factor of <mathjax>#5#</mathjax> (<mathjax>#50"L"#</mathjax> to <mathjax>#10"L"#</mathjax>).</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#10#</mathjax> <mathjax>#"L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to calculate the new volume of a gas when its pressure is increased.</p>
<p>To solve this, we can use the pressure-volume relationship of gases, illustrated by <em><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a></em>:</p>
<p><mathjax>#P_1V_1 = P_2V_2#</mathjax></p>
<p>where</p>
<ul>
<li>
<p><mathjax>#P#</mathjax> is the initial (<mathjax>#"1"#</mathjax>) and final (<mathjax>#"2"#</mathjax>) pressure of the gas, and</p>
</li>
<li>
<p><mathjax>#V#</mathjax> is the initial (<mathjax>#"1"#</mathjax>) and final (<mathjax>#"2"#</mathjax>) volume of the gas.</p>
</li>
</ul>
<p>Let's plug in our known variables, and rearrange this equation to solve for the final volume, <mathjax>#V_2#</mathjax>:</p>
<p><mathjax>#V_2 = (P_1V_1)/(P_2) = ((2cancel("atm"))(50"L"))/(10cancel("atm")) = color(red)(10"L"#</mathjax></p>
<p>Thus, when the pressure of the gas was <em>increased</em> by a factor of <mathjax>#5#</mathjax> (<mathjax>#2"atm"#</mathjax> to <mathjax>#10"atm"#</mathjax>), the volume <em>decreased</em> by a factor of <mathjax>#5#</mathjax> (<mathjax>#50"L"#</mathjax> to <mathjax>#10"L"#</mathjax>).</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">A gas occupies 50 L at a pressure of 2 atm. What is the volume when the pressure is increased to 10 atm?</h1>
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Nathan L.
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May 31, 2017
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<div class="markdown"><p><mathjax>#10#</mathjax> <mathjax>#"L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to calculate the new volume of a gas when its pressure is increased.</p>
<p>To solve this, we can use the pressure-volume relationship of gases, illustrated by <em><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a></em>:</p>
<p><mathjax>#P_1V_1 = P_2V_2#</mathjax></p>
<p>where</p>
<ul>
<li>
<p><mathjax>#P#</mathjax> is the initial (<mathjax>#"1"#</mathjax>) and final (<mathjax>#"2"#</mathjax>) pressure of the gas, and</p>
</li>
<li>
<p><mathjax>#V#</mathjax> is the initial (<mathjax>#"1"#</mathjax>) and final (<mathjax>#"2"#</mathjax>) volume of the gas.</p>
</li>
</ul>
<p>Let's plug in our known variables, and rearrange this equation to solve for the final volume, <mathjax>#V_2#</mathjax>:</p>
<p><mathjax>#V_2 = (P_1V_1)/(P_2) = ((2cancel("atm"))(50"L"))/(10cancel("atm")) = color(red)(10"L"#</mathjax></p>
<p>Thus, when the pressure of the gas was <em>increased</em> by a factor of <mathjax>#5#</mathjax> (<mathjax>#2"atm"#</mathjax> to <mathjax>#10"atm"#</mathjax>), the volume <em>decreased</em> by a factor of <mathjax>#5#</mathjax> (<mathjax>#50"L"#</mathjax> to <mathjax>#10"L"#</mathjax>).</p></div>
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</article> | A gas occupies 50 L at a pressure of 2 atm. What is the volume when the pressure is increased to 10 atm? | null |
3,387 | aa3b414c-6ddd-11ea-be28-ccda262736ce | https://socratic.org/questions/a-gas-is-held-at-3-8-atm-and-500-k-if-the-pressure-is-then-decreased-to-1-2-atm- | 157.89 K | start physical_unit 1 1 temperature k qc_end physical_unit 1 1 8 9 temperature qc_end physical_unit 1 1 5 6 pressure qc_end physical_unit 1 1 17 18 pressure qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] gas [IN] K"}] | [{"type":"physical unit","value":"157.89 K"}] | [{"type":"physical unit","value":"Temperature1 [OF] gas [=] \\pu{500 K}"},{"type":"physical unit","value":"Pressure1 [OF] gas [=] \\pu{3.8 atm}"},{"type":"physical unit","value":"Pressure2 [OF] gas [=] \\pu{1.2 atm}"}] | <h1 class="questionTitle" itemprop="name">A gas is held at 3.8 atm and 500 K. If the pressure is then decreased to 1.2 atm, what will the new temperature be?</h1> | null | 157.89 K | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This question involves Gay-Lussac's law to answer, which states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.</p>
<p>This means that as the pressure goes up, the temperature also goes up, and vice-versa. </p>
<p><strong>The equation to use is:</strong></p>
<p><mathjax>#P_1/(T_1)=P_2/(T_2)#</mathjax>, </p>
<p>where <mathjax>#P#</mathjax> is pressure in atm, and <mathjax>#T#</mathjax> is temperature in Kelvins.</p>
<p><strong>Given/Known Information</strong><br/>
<mathjax>#P_1="3.8 atm"#</mathjax><br/>
<mathjax>#T_2="500 K"#</mathjax><br/>
<mathjax>#P_2="1.2 atm"#</mathjax></p>
<p><strong>Unknown:</strong> <mathjax>#T_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax>.</p>
<p><mathjax>#T_2=(P_2T_1)/(P_1)#</mathjax></p>
<p><mathjax>#T_2=(1.2color(red)cancel(color(black)("atm"))xx500 "K")/(3.8color(red)cancel(color(black)("atm")))~~"200 K"#</mathjax> rounded to 1 significant figure</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The final temperature will be <mathjax>#~~200color(white)(.)"K"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This question involves Gay-Lussac's law to answer, which states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.</p>
<p>This means that as the pressure goes up, the temperature also goes up, and vice-versa. </p>
<p><strong>The equation to use is:</strong></p>
<p><mathjax>#P_1/(T_1)=P_2/(T_2)#</mathjax>, </p>
<p>where <mathjax>#P#</mathjax> is pressure in atm, and <mathjax>#T#</mathjax> is temperature in Kelvins.</p>
<p><strong>Given/Known Information</strong><br/>
<mathjax>#P_1="3.8 atm"#</mathjax><br/>
<mathjax>#T_2="500 K"#</mathjax><br/>
<mathjax>#P_2="1.2 atm"#</mathjax></p>
<p><strong>Unknown:</strong> <mathjax>#T_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax>.</p>
<p><mathjax>#T_2=(P_2T_1)/(P_1)#</mathjax></p>
<p><mathjax>#T_2=(1.2color(red)cancel(color(black)("atm"))xx500 "K")/(3.8color(red)cancel(color(black)("atm")))~~"200 K"#</mathjax> rounded to 1 significant figure</p></div>
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<h1 class="questionTitle" itemprop="name">A gas is held at 3.8 atm and 500 K. If the pressure is then decreased to 1.2 atm, what will the new temperature be?</h1>
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<div class="markdown"><p>The final temperature will be <mathjax>#~~200color(white)(.)"K"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This question involves Gay-Lussac's law to answer, which states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.</p>
<p>This means that as the pressure goes up, the temperature also goes up, and vice-versa. </p>
<p><strong>The equation to use is:</strong></p>
<p><mathjax>#P_1/(T_1)=P_2/(T_2)#</mathjax>, </p>
<p>where <mathjax>#P#</mathjax> is pressure in atm, and <mathjax>#T#</mathjax> is temperature in Kelvins.</p>
<p><strong>Given/Known Information</strong><br/>
<mathjax>#P_1="3.8 atm"#</mathjax><br/>
<mathjax>#T_2="500 K"#</mathjax><br/>
<mathjax>#P_2="1.2 atm"#</mathjax></p>
<p><strong>Unknown:</strong> <mathjax>#T_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax>.</p>
<p><mathjax>#T_2=(P_2T_1)/(P_1)#</mathjax></p>
<p><mathjax>#T_2=(1.2color(red)cancel(color(black)("atm"))xx500 "K")/(3.8color(red)cancel(color(black)("atm")))~~"200 K"#</mathjax> rounded to 1 significant figure</p></div>
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</article> | A gas is held at 3.8 atm and 500 K. If the pressure is then decreased to 1.2 atm, what will the new temperature be? | null |
3,388 | a8e90040-6ddd-11ea-9907-ccda262736ce | https://socratic.org/questions/what-is-the-balanced-net-ionic-equation-for-sodium-hydroxide-and-acetylsalicylic | C6H4OCOCH3COOH(aq) + OH-(aq) -> C6H4OCOCH3COO-(aq) + H2O(l) | start chemical_equation qc_end substance 8 9 qc_end substance 11 12 qc_end end | [{"type":"other","value":"Chemical Equation [OF] net ionic equation"}] | [{"type":"chemical equation","value":"C6H4OCOCH3COOH(aq) + OH-(aq) -> C6H4OCOCH3COO-(aq) + H2O(l)"}] | [{"type":"substance name","value":"Sodium hydroxide"},{"type":"substance name","value":"Acetylsalicylic acid"}] | <h1 class="questionTitle" itemprop="name">What is the balanced net ionic equation for sodium hydroxide and acetylsalicylic acid?</h1> | null | C6H4OCOCH3COOH(aq) + OH-(aq) -> C6H4OCOCH3COO-(aq) + H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium hydroxide and acetylsalicylic acid neutralize each other in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> to produce aqueous sodium acetylsalicylate and water. </p>
<p><img alt="http://ibalchemy.com/d-2/" src="https://useruploads.socratic.org/VL79m2PCQfu8jZNiWMA0_Aspirin-Solubility.jpg"/> </p>
<p>The <strong>molecular equation</strong> looks like this</p>
<blockquote>
<p><mathjax>#"C"_ 6"H"_ 4 "OCOCH"_ 3"COOH" _ ((aq)) + "NaOH"_ ((aq)) -> "C"_ 6"H"_ 4 "OCOCH"_ 3"COONa"_ ((aq)) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>The <strong>complete ionic equation</strong> looks like this</p>
<blockquote>
<p><mathjax>#"C"_ 6"H"_ 4 "OCOCH"_ 3"COOH" _ ((aq)) + "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "C"_ 6"H"_ 4 "OCOCH"_ 3"COO" _ ((aq))^(-) + "Na"_ ((aq))^(+) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>The sodium cations are <em>spectator ions</em> here, i.e. they are present on both sides of the chemical equation, so you can eliminate them</p>
<blockquote>
<p><mathjax>#"C"_ 6"H"_ 4 "OCOCH"_ 3"COOH" _ ((aq)) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + "OH"_ ((aq))^(-)-> "C"_ 6"H"_ 4 "OCOCH"_ 3"COO" _ ((aq))^(-) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>and write the <strong>net ionic equation</strong> that describes this reaction. </p>
<blockquote>
<p><mathjax>#"C"_ 6"H"_ 4 "OCOCH"_ 3"COOH" _ ((aq)) + "OH"_ ((aq))^(-)-> "C"_ 6"H"_ 4 "OCOCH"_ 3"COO" _ ((aq))^(-) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"C"_ 6"H"_ 4 "OCOCH"_ 3"COOH" _ ((aq)) + "OH"_ ((aq))^(-)-> "C"_ 6"H"_ 4 "OCOCH"_ 3"COO" _ ((aq))^(-) + "H"_ 2"O"_ ((l))#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium hydroxide and acetylsalicylic acid neutralize each other in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> to produce aqueous sodium acetylsalicylate and water. </p>
<p><img alt="http://ibalchemy.com/d-2/" src="https://useruploads.socratic.org/VL79m2PCQfu8jZNiWMA0_Aspirin-Solubility.jpg"/> </p>
<p>The <strong>molecular equation</strong> looks like this</p>
<blockquote>
<p><mathjax>#"C"_ 6"H"_ 4 "OCOCH"_ 3"COOH" _ ((aq)) + "NaOH"_ ((aq)) -> "C"_ 6"H"_ 4 "OCOCH"_ 3"COONa"_ ((aq)) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>The <strong>complete ionic equation</strong> looks like this</p>
<blockquote>
<p><mathjax>#"C"_ 6"H"_ 4 "OCOCH"_ 3"COOH" _ ((aq)) + "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "C"_ 6"H"_ 4 "OCOCH"_ 3"COO" _ ((aq))^(-) + "Na"_ ((aq))^(+) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>The sodium cations are <em>spectator ions</em> here, i.e. they are present on both sides of the chemical equation, so you can eliminate them</p>
<blockquote>
<p><mathjax>#"C"_ 6"H"_ 4 "OCOCH"_ 3"COOH" _ ((aq)) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + "OH"_ ((aq))^(-)-> "C"_ 6"H"_ 4 "OCOCH"_ 3"COO" _ ((aq))^(-) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>and write the <strong>net ionic equation</strong> that describes this reaction. </p>
<blockquote>
<p><mathjax>#"C"_ 6"H"_ 4 "OCOCH"_ 3"COOH" _ ((aq)) + "OH"_ ((aq))^(-)-> "C"_ 6"H"_ 4 "OCOCH"_ 3"COO" _ ((aq))^(-) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the balanced net ionic equation for sodium hydroxide and acetylsalicylic acid?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2018-02-11T14:14:17" itemprop="dateCreated">
Feb 11, 2018
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<div class="markdown"><p><mathjax>#"C"_ 6"H"_ 4 "OCOCH"_ 3"COOH" _ ((aq)) + "OH"_ ((aq))^(-)-> "C"_ 6"H"_ 4 "OCOCH"_ 3"COO" _ ((aq))^(-) + "H"_ 2"O"_ ((l))#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium hydroxide and acetylsalicylic acid neutralize each other in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> to produce aqueous sodium acetylsalicylate and water. </p>
<p><img alt="http://ibalchemy.com/d-2/" src="https://useruploads.socratic.org/VL79m2PCQfu8jZNiWMA0_Aspirin-Solubility.jpg"/> </p>
<p>The <strong>molecular equation</strong> looks like this</p>
<blockquote>
<p><mathjax>#"C"_ 6"H"_ 4 "OCOCH"_ 3"COOH" _ ((aq)) + "NaOH"_ ((aq)) -> "C"_ 6"H"_ 4 "OCOCH"_ 3"COONa"_ ((aq)) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>The <strong>complete ionic equation</strong> looks like this</p>
<blockquote>
<p><mathjax>#"C"_ 6"H"_ 4 "OCOCH"_ 3"COOH" _ ((aq)) + "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "C"_ 6"H"_ 4 "OCOCH"_ 3"COO" _ ((aq))^(-) + "Na"_ ((aq))^(+) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>The sodium cations are <em>spectator ions</em> here, i.e. they are present on both sides of the chemical equation, so you can eliminate them</p>
<blockquote>
<p><mathjax>#"C"_ 6"H"_ 4 "OCOCH"_ 3"COOH" _ ((aq)) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + "OH"_ ((aq))^(-)-> "C"_ 6"H"_ 4 "OCOCH"_ 3"COO" _ ((aq))^(-) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>and write the <strong>net ionic equation</strong> that describes this reaction. </p>
<blockquote>
<p><mathjax>#"C"_ 6"H"_ 4 "OCOCH"_ 3"COOH" _ ((aq)) + "OH"_ ((aq))^(-)-> "C"_ 6"H"_ 4 "OCOCH"_ 3"COO" _ ((aq))^(-) + "H"_ 2"O"_ ((l))#</mathjax></p>
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</article> | What is the balanced net ionic equation for sodium hydroxide and acetylsalicylic acid? | null |
3,389 | a8ee0470-6ddd-11ea-912b-ccda262736ce | https://socratic.org/questions/when-the-following-equation-is-balanced-k-s-h-2o-l-koh-aq-h-2-g-what-is-the-coef | 1 | start physical_unit 12 12 coefficient none qc_end chemical_equation 6 12 qc_end end | [{"type":"physical unit","value":"Coefficient [OF] H2"}] | [{"type":"physical unit","value":"1"}] | [{"type":"chemical equation","value":"K(s) + H2O(l) -> KOH(aq) + H2(g)"}] | <h1 class="questionTitle" itemprop="name">When the following equation is balanced, #K(s) + H_2O(l) -> KOH(aq) + H_2(g)#, what is the coefficient of #H_2#?</h1> | null | 1 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#K(s)+color(blue)(H_2)O(l)->KOcolor(blue)(H)(aq)+color(blue)(H_2)(g)#</mathjax></p>
<p>By looking at this equation, we can notice that we have <mathjax>#color(blue)(2H)#</mathjax> in the left side and <mathjax>#color(blue)(3H)#</mathjax> to the right side.</p>
<p>In order to balance this equation, we can multiply the <mathjax>#H_2#</mathjax> by <mathjax>#color(red)(1/2)#</mathjax>. Then we get:</p>
<p><mathjax>#K(s)+color(blue)(H_2)O(l)->KOcolor(blue)(H)(aq)+color(red)(1/2)color(blue)(H_2)(g)#</mathjax></p>
<p>This way, we will have <mathjax>#color(blue)(2H)#</mathjax> in each side.</p>
<p>Since we prefer to have the coefficients as whole numbers and not fractions, we can multiply the entire reaction by <mathjax>#color(green)(2)#</mathjax> so we get:</p>
<p><mathjax>#color(green)(2)K(s)+color(green)(2)color(blue)(H_2)O(l)->color(green)(2)KOcolor(blue)(H)(aq)+color(blue)(H_2)(g)#</mathjax></p></div>
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<div class="markdown"><p>The coefficient go <mathjax>#H_2#</mathjax> is <mathjax>#1#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#K(s)+color(blue)(H_2)O(l)->KOcolor(blue)(H)(aq)+color(blue)(H_2)(g)#</mathjax></p>
<p>By looking at this equation, we can notice that we have <mathjax>#color(blue)(2H)#</mathjax> in the left side and <mathjax>#color(blue)(3H)#</mathjax> to the right side.</p>
<p>In order to balance this equation, we can multiply the <mathjax>#H_2#</mathjax> by <mathjax>#color(red)(1/2)#</mathjax>. Then we get:</p>
<p><mathjax>#K(s)+color(blue)(H_2)O(l)->KOcolor(blue)(H)(aq)+color(red)(1/2)color(blue)(H_2)(g)#</mathjax></p>
<p>This way, we will have <mathjax>#color(blue)(2H)#</mathjax> in each side.</p>
<p>Since we prefer to have the coefficients as whole numbers and not fractions, we can multiply the entire reaction by <mathjax>#color(green)(2)#</mathjax> so we get:</p>
<p><mathjax>#color(green)(2)K(s)+color(green)(2)color(blue)(H_2)O(l)->color(green)(2)KOcolor(blue)(H)(aq)+color(blue)(H_2)(g)#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">When the following equation is balanced, #K(s) + H_2O(l) -> KOH(aq) + H_2(g)#, what is the coefficient of #H_2#?</h1>
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Dr. Hayek
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<div class="markdown"><p>The coefficient go <mathjax>#H_2#</mathjax> is <mathjax>#1#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#K(s)+color(blue)(H_2)O(l)->KOcolor(blue)(H)(aq)+color(blue)(H_2)(g)#</mathjax></p>
<p>By looking at this equation, we can notice that we have <mathjax>#color(blue)(2H)#</mathjax> in the left side and <mathjax>#color(blue)(3H)#</mathjax> to the right side.</p>
<p>In order to balance this equation, we can multiply the <mathjax>#H_2#</mathjax> by <mathjax>#color(red)(1/2)#</mathjax>. Then we get:</p>
<p><mathjax>#K(s)+color(blue)(H_2)O(l)->KOcolor(blue)(H)(aq)+color(red)(1/2)color(blue)(H_2)(g)#</mathjax></p>
<p>This way, we will have <mathjax>#color(blue)(2H)#</mathjax> in each side.</p>
<p>Since we prefer to have the coefficients as whole numbers and not fractions, we can multiply the entire reaction by <mathjax>#color(green)(2)#</mathjax> so we get:</p>
<p><mathjax>#color(green)(2)K(s)+color(green)(2)color(blue)(H_2)O(l)->color(green)(2)KOcolor(blue)(H)(aq)+color(blue)(H_2)(g)#</mathjax></p></div>
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</article> | When the following equation is balanced, #K(s) + H_2O(l) -> KOH(aq) + H_2(g)#, what is the coefficient of #H_2#? | null |
3,390 | a8aa2d5d-6ddd-11ea-9da0-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-barium-hydroxide | Ba(OH)2 | start chemical_formula qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] barium hydroxide [IN] default"}] | [{"type":"chemical equation","value":"Ba(OH)2"}] | [{"type":"substance name","value":"Barium hydroxide"}] | <h1 class="questionTitle" itemprop="name">What is the formula for barium hydroxide?</h1> | null | Ba(OH)2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Barium is an element in Group 2, so it has two <a href="http://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a>, and that it's charge is 2+. </p>
<p>Hydroxide is a polyatomic negative ion with a valency of 1, and that it's charge is 1-. </p>
<p>As barium is a metal while hydroxide is a non-metal, they combine to form an ionic compound. </p>
<p>To form a neutral compound, balance the positive and negative charges by applying LCM,and the number of positive and negative charges must be equal.</p>
<p><mathjax>#Ba^(2+)#</mathjax> and <mathjax>#(OH)^(2-)#</mathjax> </p>
<p><mathjax>#Ba^(2+)#</mathjax> <mathjax>#(OH)^(2-)#</mathjax> <mathjax>#(OH)^(2-)#</mathjax> </p>
<p>As seen from above, one barium ion and two hydroxide ions form barium hydroxide. Thus, the formula of barium hydroxide is <mathjax>#Ba(OH)^2#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#Ba(OH)^2#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Barium is an element in Group 2, so it has two <a href="http://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a>, and that it's charge is 2+. </p>
<p>Hydroxide is a polyatomic negative ion with a valency of 1, and that it's charge is 1-. </p>
<p>As barium is a metal while hydroxide is a non-metal, they combine to form an ionic compound. </p>
<p>To form a neutral compound, balance the positive and negative charges by applying LCM,and the number of positive and negative charges must be equal.</p>
<p><mathjax>#Ba^(2+)#</mathjax> and <mathjax>#(OH)^(2-)#</mathjax> </p>
<p><mathjax>#Ba^(2+)#</mathjax> <mathjax>#(OH)^(2-)#</mathjax> <mathjax>#(OH)^(2-)#</mathjax> </p>
<p>As seen from above, one barium ion and two hydroxide ions form barium hydroxide. Thus, the formula of barium hydroxide is <mathjax>#Ba(OH)^2#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the formula for barium hydroxide?</h1>
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<a class="topContributorPic" href="/users/angel-f"><img alt="" class="" src="https://lh3.googleusercontent.com/-LHG51GU3E4k/AAAAAAAAAAI/AAAAAAAAAAA/AB6qoq0n6VltfbPpExuskzLUsZferyiZgg/mo/photo.jpg?sz=50" title=""/></a>
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<div class="markdown"><p><mathjax>#Ba(OH)^2#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Barium is an element in Group 2, so it has two <a href="http://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a>, and that it's charge is 2+. </p>
<p>Hydroxide is a polyatomic negative ion with a valency of 1, and that it's charge is 1-. </p>
<p>As barium is a metal while hydroxide is a non-metal, they combine to form an ionic compound. </p>
<p>To form a neutral compound, balance the positive and negative charges by applying LCM,and the number of positive and negative charges must be equal.</p>
<p><mathjax>#Ba^(2+)#</mathjax> and <mathjax>#(OH)^(2-)#</mathjax> </p>
<p><mathjax>#Ba^(2+)#</mathjax> <mathjax>#(OH)^(2-)#</mathjax> <mathjax>#(OH)^(2-)#</mathjax> </p>
<p>As seen from above, one barium ion and two hydroxide ions form barium hydroxide. Thus, the formula of barium hydroxide is <mathjax>#Ba(OH)^2#</mathjax></p></div>
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</article> | What is the formula for barium hydroxide? | null |
3,391 | ad2020ac-6ddd-11ea-a7dd-ccda262736ce | https://socratic.org/questions/an-inflated-balloon-has-a-volume-of-5-75-l-at-a-temperature-of-22-c-at-what-temp | 2.85 degrees Celsius | start physical_unit 1 2 temperature °c qc_end physical_unit 1 2 7 8 volume qc_end physical_unit 1 2 13 14 temperature qc_end physical_unit 1 2 29 30 volume qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the inflated balloon [IN] degrees Celsius"}] | [{"type":"physical unit","value":"2.85 degrees Celsius"}] | [{"type":"physical unit","value":"Volume1 [OF] the inflated balloon [=] \\pu{5.75 L}"},{"type":"physical unit","value":"Temperature1 [OF] the inflated balloon [=] \\pu{22 ℃}"},{"type":"physical unit","value":"Volume2 [OF] the inflated balloon [=] \\pu{5.38 L}"}] | <h1 class="questionTitle" itemprop="name">An inflated balloon has a volume of 5.75 L at a temperature of 22°C. At what temperature in degrees Celsius will the volume of the balloon decrease to 5.38 L?</h1> | null | 2.85 degrees Celsius | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a> which states that at constant pressure, the volume is directly proportional to the temperature in Kelvins. The equation to use is <mathjax>#V_1/T_1=V_2/T_2#</mathjax>.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#V_1="5.75 L"#</mathjax><br/>
<mathjax>#T_1="22"^@"C""+273.15=295 K"#</mathjax><br/>
<mathjax>#V_2="5.38 L"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#T_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax>. Substitute the known values into the equation and solve. Convert the temperature from Kelvins to degrees Celsius.</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p><mathjax>#T_2=(T_1V_2)/V_1#</mathjax></p>
<p><mathjax>#T_2=((295"K")*(5.38cancel"L"))/(5.75cancel"L")#</mathjax></p>
<p><mathjax>#T_2="276 K"#</mathjax></p>
<p>Convert Kelvins to <mathjax>#""^@"C""#</mathjax>.</p>
<p><mathjax>#""^@"C""="K-273.15#</mathjax></p>
<p><mathjax>#""^@"C"=276-273.15="2.85"^@"C"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The volume will be at <mathjax>#"5.38 L"#</mathjax> when the temperature is at <mathjax>#color(blue)("2.85"^@"C"")#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a> which states that at constant pressure, the volume is directly proportional to the temperature in Kelvins. The equation to use is <mathjax>#V_1/T_1=V_2/T_2#</mathjax>.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#V_1="5.75 L"#</mathjax><br/>
<mathjax>#T_1="22"^@"C""+273.15=295 K"#</mathjax><br/>
<mathjax>#V_2="5.38 L"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#T_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax>. Substitute the known values into the equation and solve. Convert the temperature from Kelvins to degrees Celsius.</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p><mathjax>#T_2=(T_1V_2)/V_1#</mathjax></p>
<p><mathjax>#T_2=((295"K")*(5.38cancel"L"))/(5.75cancel"L")#</mathjax></p>
<p><mathjax>#T_2="276 K"#</mathjax></p>
<p>Convert Kelvins to <mathjax>#""^@"C""#</mathjax>.</p>
<p><mathjax>#""^@"C""="K-273.15#</mathjax></p>
<p><mathjax>#""^@"C"=276-273.15="2.85"^@"C"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">An inflated balloon has a volume of 5.75 L at a temperature of 22°C. At what temperature in degrees Celsius will the volume of the balloon decrease to 5.38 L?</h1>
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<div class="markdown"><p>The volume will be at <mathjax>#"5.38 L"#</mathjax> when the temperature is at <mathjax>#color(blue)("2.85"^@"C"")#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a> which states that at constant pressure, the volume is directly proportional to the temperature in Kelvins. The equation to use is <mathjax>#V_1/T_1=V_2/T_2#</mathjax>.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#V_1="5.75 L"#</mathjax><br/>
<mathjax>#T_1="22"^@"C""+273.15=295 K"#</mathjax><br/>
<mathjax>#V_2="5.38 L"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#T_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax>. Substitute the known values into the equation and solve. Convert the temperature from Kelvins to degrees Celsius.</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p><mathjax>#T_2=(T_1V_2)/V_1#</mathjax></p>
<p><mathjax>#T_2=((295"K")*(5.38cancel"L"))/(5.75cancel"L")#</mathjax></p>
<p><mathjax>#T_2="276 K"#</mathjax></p>
<p>Convert Kelvins to <mathjax>#""^@"C""#</mathjax>.</p>
<p><mathjax>#""^@"C""="K-273.15#</mathjax></p>
<p><mathjax>#""^@"C"=276-273.15="2.85"^@"C"#</mathjax></p></div>
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</article> | An inflated balloon has a volume of 5.75 L at a temperature of 22°C. At what temperature in degrees Celsius will the volume of the balloon decrease to 5.38 L? | null |
3,392 | a967b6a5-6ddd-11ea-852f-ccda262736ce | https://socratic.org/questions/58914bd111ef6b566c5b6ad2 | 9.98 g | start physical_unit 2 2 mass g qc_end c_other OTHER qc_end physical_unit 13 14 11 12 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] dioxygen [IN] g"}] | [{"type":"physical unit","value":"9.98 g"}] | [{"type":"other","value":"Complete combustion."},{"type":"physical unit","value":"Mass [OF] methane gas [=] \\pu{2.5 g}"}] | <h1 class="questionTitle" itemprop="name">How much dioxygen is required for the complete combustion of a #2.5*g# methane gas? </h1> | null | 9.98 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You need (i) a stoichiometrically balanced equation:</p>
<p><mathjax>#CH_4 + 2O_2 rarr CO_2 +2H_2O#</mathjax></p>
<p>And (ii), the molar quantity of methane that undergoes combustion:</p>
<p><mathjax>#(2.50*g)/(16.01*g*mol^-1)=0.156*mol#</mathjax></p>
<p>Given the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, <mathjax>#2xx0.156*molxx32.00*g*mol^-1=??g#</mathjax> dioxygen gas will combust. </p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Under <mathjax>#12*g#</mathjax> of dioxygen gas will be required for complete combustion. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You need (i) a stoichiometrically balanced equation:</p>
<p><mathjax>#CH_4 + 2O_2 rarr CO_2 +2H_2O#</mathjax></p>
<p>And (ii), the molar quantity of methane that undergoes combustion:</p>
<p><mathjax>#(2.50*g)/(16.01*g*mol^-1)=0.156*mol#</mathjax></p>
<p>Given the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, <mathjax>#2xx0.156*molxx32.00*g*mol^-1=??g#</mathjax> dioxygen gas will combust. </p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">How much dioxygen is required for the complete combustion of a #2.5*g# methane gas? </h1>
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<div class="markdown"><p>Under <mathjax>#12*g#</mathjax> of dioxygen gas will be required for complete combustion. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You need (i) a stoichiometrically balanced equation:</p>
<p><mathjax>#CH_4 + 2O_2 rarr CO_2 +2H_2O#</mathjax></p>
<p>And (ii), the molar quantity of methane that undergoes combustion:</p>
<p><mathjax>#(2.50*g)/(16.01*g*mol^-1)=0.156*mol#</mathjax></p>
<p>Given the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, <mathjax>#2xx0.156*molxx32.00*g*mol^-1=??g#</mathjax> dioxygen gas will combust. </p></div>
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</article> | How much dioxygen is required for the complete combustion of a #2.5*g# methane gas? | null |
3,393 | a9090776-6ddd-11ea-b9ad-ccda262736ce | https://socratic.org/questions/the-temperature-of-24-iiters-of-a-certain-gas-is-300-k-if-we-change-the-temperat | 32 L | start physical_unit 26 27 volume l qc_end physical_unit 6 8 10 11 temperature qc_end physical_unit 6 8 3 4 volume qc_end physical_unit 26 27 18 19 temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"32 L"}] | [{"type":"physical unit","value":"Temperature1 [OF] a certain gas [=] \\pu{300 K}"},{"type":"physical unit","value":"Volume1 [OF] a certain gas [=] \\pu{24 Iiters}"},{"type":"physical unit","value":"Temperature2 [OF] the gas [=] \\pu{400 K}"}] | <h1 class="questionTitle" itemprop="name">The temperature of 24 Iiters of a certain gas is 300 K. If we change the temperature to 400 K, what is the new volume of the gas?</h1> | null | 32 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of Charle's law, which states that the volume of a given amount of gas kept at constant pressure will vary directly with the temperature in Kelvins. The equation to use is <mathjax>#V_1/T_1=V_2/T_2#</mathjax>.</p>
<p><strong>Given</strong><br/>
<mathjax>#V_1="24 L"#</mathjax><br/>
<mathjax>#T_1="300 K"#</mathjax><br/>
<mathjax>#T_2="400 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#V_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax>. Substitute the given values into the equation and solve.</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p><mathjax>#V_2=(V_1T_2)/T_1#</mathjax></p>
<p><mathjax>#V_2=((24"L"xx400cancel"K"))/(300cancel"K")="32 K"#</mathjax></p>
<p><mathjax>#V_2="32 L"#</mathjax></p></div>
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<div class="answerSummary">
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<div class="markdown"><p>The second volume is 32 L.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of Charle's law, which states that the volume of a given amount of gas kept at constant pressure will vary directly with the temperature in Kelvins. The equation to use is <mathjax>#V_1/T_1=V_2/T_2#</mathjax>.</p>
<p><strong>Given</strong><br/>
<mathjax>#V_1="24 L"#</mathjax><br/>
<mathjax>#T_1="300 K"#</mathjax><br/>
<mathjax>#T_2="400 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#V_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax>. Substitute the given values into the equation and solve.</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p><mathjax>#V_2=(V_1T_2)/T_1#</mathjax></p>
<p><mathjax>#V_2=((24"L"xx400cancel"K"))/(300cancel"K")="32 K"#</mathjax></p>
<p><mathjax>#V_2="32 L"#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">The temperature of 24 Iiters of a certain gas is 300 K. If we change the temperature to 400 K, what is the new volume of the gas?</h1>
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<div class="markdown"><p>The second volume is 32 L.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of Charle's law, which states that the volume of a given amount of gas kept at constant pressure will vary directly with the temperature in Kelvins. The equation to use is <mathjax>#V_1/T_1=V_2/T_2#</mathjax>.</p>
<p><strong>Given</strong><br/>
<mathjax>#V_1="24 L"#</mathjax><br/>
<mathjax>#T_1="300 K"#</mathjax><br/>
<mathjax>#T_2="400 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#V_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax>. Substitute the given values into the equation and solve.</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p><mathjax>#V_2=(V_1T_2)/T_1#</mathjax></p>
<p><mathjax>#V_2=((24"L"xx400cancel"K"))/(300cancel"K")="32 K"#</mathjax></p>
<p><mathjax>#V_2="32 L"#</mathjax></p></div>
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</article> | The temperature of 24 Iiters of a certain gas is 300 K. If we change the temperature to 400 K, what is the new volume of the gas? | null |
3,394 | ab9ab214-6ddd-11ea-9672-ccda262736ce | https://socratic.org/questions/how-many-grams-of-diphosphorus-trioxide-p-2o3-are-required-to-produce-10-2-moles | 560.72 grams | start physical_unit 6 6 mass g qc_end chemical_equation 19 25 qc_end physical_unit 15 15 11 12 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] P2O3 [IN] grams"}] | [{"type":"physical unit","value":"560.72 grams"}] | [{"type":"chemical equation","value":"P2O3 + 3 H2O -> 2 H3PO3"},{"type":"physical unit","value":"Mole [OF] H3PO3 [=] \\pu{10.2 moles}"}] | <h1 class="questionTitle" itemprop="name">How many grams of diphosphorus trioxide, #P_2O3#, are required to produce 10.2 moles of phosphorous acid, #H_3PO_3# in the reaction #P_2O_3 + 3H_2O -> 2H_3PO_3#?</h1> | null | 560.72 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong><br/>
<mathjax>#"P"_2"O"_3 + "3H"_2"O"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2H"_3"PO"_3"#</mathjax></p>
<p>First use the balanced equation to determine <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#"H"_3"PO"_3"#</mathjax> and <mathjax>#"P"_2"O"_3"#</mathjax>. This ratio will be used to determine the moles <mathjax>#"P"_2"O"_3"#</mathjax> required to produce <mathjax>#10.2#</mathjax> moles <mathjax>#"H"_3"PO"_3"#</mathjax>.</p>
<p>Mole Ratio Between <mathjax>#"P"_2"O"_3"#</mathjax> and <mathjax>#"H"_3"PO"_3"#</mathjax> from the balanced equation.</p>
<p><mathjax>#(1"mol P"_2"O"_3)/(2"mol H"_3"PO"_3)#</mathjax> and <mathjax>#(2"mol H"_3"PO"_3)/(1"mol P"_2"O"_3)#</mathjax></p>
<p>Multiply the moles <mathjax>#"H"_3"PO"_3"#</mathjax> by the molar mass that cancels <mathjax>#"H"_3"PO"_3"#</mathjax> and leaves <mathjax>#"P"_2"O"_3"#</mathjax></p>
<p><mathjax>#10.2color(red)cancel(color(black)("mol H"_3"PO"_3))xx(1"mol P"_2"O"_3)/(2color(red)cancel(color(black)("mol H"_3"PO"_3)))="5.10 mol P"_2"O"_3#</mathjax></p>
<p>Now that the moles <mathjax>#"P"_2"O"_3"#</mathjax> required to produce <mathjax>#"10.2 mol H"_3"PO"_3"#</mathjax> are known, multiply the number of moles by its molar mass, <mathjax>#"109.945 g/mol"#</mathjax>. <a href="https://www.ncbi.nlm.nih.gov/pccompound?term=P2O3" rel="nofollow" target="_blank">https://www.ncbi.nlm.nih.gov/pccompound?term=P2O3</a> This will give the mass in grams needed for <mathjax>#"P"_2"O"_3"#</mathjax> to produce <mathjax>#"10.2 mol H"_3"PO"_3"#</mathjax>.</p>
<p><mathjax>#5.10color(red)cancel(color(black)("mol P"_2"O"_3))xx(109.945"g P"_2"O"_3)/(1color(red)cancel(color(black)("mol P"_2"O"_3)))="561 g P"_2"O"_3#</mathjax> rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"561 g P"_2"O"_3"#</mathjax> are required to produce <mathjax>#"10.2 mol H"_3"PO"_3"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong><br/>
<mathjax>#"P"_2"O"_3 + "3H"_2"O"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2H"_3"PO"_3"#</mathjax></p>
<p>First use the balanced equation to determine <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#"H"_3"PO"_3"#</mathjax> and <mathjax>#"P"_2"O"_3"#</mathjax>. This ratio will be used to determine the moles <mathjax>#"P"_2"O"_3"#</mathjax> required to produce <mathjax>#10.2#</mathjax> moles <mathjax>#"H"_3"PO"_3"#</mathjax>.</p>
<p>Mole Ratio Between <mathjax>#"P"_2"O"_3"#</mathjax> and <mathjax>#"H"_3"PO"_3"#</mathjax> from the balanced equation.</p>
<p><mathjax>#(1"mol P"_2"O"_3)/(2"mol H"_3"PO"_3)#</mathjax> and <mathjax>#(2"mol H"_3"PO"_3)/(1"mol P"_2"O"_3)#</mathjax></p>
<p>Multiply the moles <mathjax>#"H"_3"PO"_3"#</mathjax> by the molar mass that cancels <mathjax>#"H"_3"PO"_3"#</mathjax> and leaves <mathjax>#"P"_2"O"_3"#</mathjax></p>
<p><mathjax>#10.2color(red)cancel(color(black)("mol H"_3"PO"_3))xx(1"mol P"_2"O"_3)/(2color(red)cancel(color(black)("mol H"_3"PO"_3)))="5.10 mol P"_2"O"_3#</mathjax></p>
<p>Now that the moles <mathjax>#"P"_2"O"_3"#</mathjax> required to produce <mathjax>#"10.2 mol H"_3"PO"_3"#</mathjax> are known, multiply the number of moles by its molar mass, <mathjax>#"109.945 g/mol"#</mathjax>. <a href="https://www.ncbi.nlm.nih.gov/pccompound?term=P2O3" rel="nofollow" target="_blank">https://www.ncbi.nlm.nih.gov/pccompound?term=P2O3</a> This will give the mass in grams needed for <mathjax>#"P"_2"O"_3"#</mathjax> to produce <mathjax>#"10.2 mol H"_3"PO"_3"#</mathjax>.</p>
<p><mathjax>#5.10color(red)cancel(color(black)("mol P"_2"O"_3))xx(109.945"g P"_2"O"_3)/(1color(red)cancel(color(black)("mol P"_2"O"_3)))="561 g P"_2"O"_3#</mathjax> rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">How many grams of diphosphorus trioxide, #P_2O3#, are required to produce 10.2 moles of phosphorous acid, #H_3PO_3# in the reaction #P_2O_3 + 3H_2O -> 2H_3PO_3#?</h1>
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<div class="markdown"><p><mathjax>#"561 g P"_2"O"_3"#</mathjax> are required to produce <mathjax>#"10.2 mol H"_3"PO"_3"#</mathjax>.</p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong><br/>
<mathjax>#"P"_2"O"_3 + "3H"_2"O"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2H"_3"PO"_3"#</mathjax></p>
<p>First use the balanced equation to determine <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#"H"_3"PO"_3"#</mathjax> and <mathjax>#"P"_2"O"_3"#</mathjax>. This ratio will be used to determine the moles <mathjax>#"P"_2"O"_3"#</mathjax> required to produce <mathjax>#10.2#</mathjax> moles <mathjax>#"H"_3"PO"_3"#</mathjax>.</p>
<p>Mole Ratio Between <mathjax>#"P"_2"O"_3"#</mathjax> and <mathjax>#"H"_3"PO"_3"#</mathjax> from the balanced equation.</p>
<p><mathjax>#(1"mol P"_2"O"_3)/(2"mol H"_3"PO"_3)#</mathjax> and <mathjax>#(2"mol H"_3"PO"_3)/(1"mol P"_2"O"_3)#</mathjax></p>
<p>Multiply the moles <mathjax>#"H"_3"PO"_3"#</mathjax> by the molar mass that cancels <mathjax>#"H"_3"PO"_3"#</mathjax> and leaves <mathjax>#"P"_2"O"_3"#</mathjax></p>
<p><mathjax>#10.2color(red)cancel(color(black)("mol H"_3"PO"_3))xx(1"mol P"_2"O"_3)/(2color(red)cancel(color(black)("mol H"_3"PO"_3)))="5.10 mol P"_2"O"_3#</mathjax></p>
<p>Now that the moles <mathjax>#"P"_2"O"_3"#</mathjax> required to produce <mathjax>#"10.2 mol H"_3"PO"_3"#</mathjax> are known, multiply the number of moles by its molar mass, <mathjax>#"109.945 g/mol"#</mathjax>. <a href="https://www.ncbi.nlm.nih.gov/pccompound?term=P2O3" rel="nofollow" target="_blank">https://www.ncbi.nlm.nih.gov/pccompound?term=P2O3</a> This will give the mass in grams needed for <mathjax>#"P"_2"O"_3"#</mathjax> to produce <mathjax>#"10.2 mol H"_3"PO"_3"#</mathjax>.</p>
<p><mathjax>#5.10color(red)cancel(color(black)("mol P"_2"O"_3))xx(109.945"g P"_2"O"_3)/(1color(red)cancel(color(black)("mol P"_2"O"_3)))="561 g P"_2"O"_3#</mathjax> rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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</article> | How many grams of diphosphorus trioxide, #P_2O3#, are required to produce 10.2 moles of phosphorous acid, #H_3PO_3# in the reaction #P_2O_3 + 3H_2O -> 2H_3PO_3#? | null |
3,395 | a98f6c1a-6ddd-11ea-a4b6-ccda262736ce | https://socratic.org/questions/58266570b72cff56ead3eda4 | 8.69 L | start physical_unit 5 5 volume l qc_end physical_unit 5 5 1 2 volume qc_end physical_unit 5 5 6 7 pressure qc_end physical_unit 5 5 10 11 temperature qc_end physical_unit 5 5 16 17 temperature qc_end physical_unit 5 5 19 20 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"8.69 L"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{780 mL}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{1.00 atm}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{310 K}"},{"type":"physical unit","value":"Temperature2 [OF] the gas [=] \\pu{295 K}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{65 mmHg}"}] | <h1 class="questionTitle" itemprop="name">A #780*mL# volume of gas #1.00*atm# pressure, and #310*K# temperature, is cooled to #295*K#, and #65*mm*Hg# pressure. What is the new volume?</h1> | null | 8.69 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For a given quantity of gas, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> holds that <mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>, temperature quoted in <mathjax>#"degrees Kelvin"#</mathjax>.</p>
<p>Thus <mathjax>#V_2=(P_1V_1)/T_1xxT_2/P_2#</mathjax></p>
<p><mathjax>#=(1.00*atmxx780*mL)/(310*K)xx(295*K)/((65*mm*Hg)/(760*mm*Hg*atm^-1)#</mathjax></p>
<p><mathjax>#=8.69xx10^3*mL=8.70*L#</mathjax></p>
<p>We use here the useful relationship that <mathjax>#760*mm*Hg-=1*atm#</mathjax>.</p>
<p>And thus a measurement of length on a mercury manometer, which is fairly easy to do in a lab, can be used to solve Gas law problems.</p>
<p>Please note that I used <mathjax>#22#</mathjax> <mathjax>#""^@C#</mathjax>, <mathjax>#-=#</mathjax> <mathjax>#295*K#</mathjax>, and <mathjax>#37#</mathjax> <mathjax>#""^@C#</mathjax>, <mathjax>#-=#</mathjax> <mathjax>#310*K#</mathjax>. You can change the calculation accordingly if I have got your starting conditions wrong. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>For a given quantity of gas, .........<mathjax>#V_2=8.70*L.#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For a given quantity of gas, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> holds that <mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>, temperature quoted in <mathjax>#"degrees Kelvin"#</mathjax>.</p>
<p>Thus <mathjax>#V_2=(P_1V_1)/T_1xxT_2/P_2#</mathjax></p>
<p><mathjax>#=(1.00*atmxx780*mL)/(310*K)xx(295*K)/((65*mm*Hg)/(760*mm*Hg*atm^-1)#</mathjax></p>
<p><mathjax>#=8.69xx10^3*mL=8.70*L#</mathjax></p>
<p>We use here the useful relationship that <mathjax>#760*mm*Hg-=1*atm#</mathjax>.</p>
<p>And thus a measurement of length on a mercury manometer, which is fairly easy to do in a lab, can be used to solve Gas law problems.</p>
<p>Please note that I used <mathjax>#22#</mathjax> <mathjax>#""^@C#</mathjax>, <mathjax>#-=#</mathjax> <mathjax>#295*K#</mathjax>, and <mathjax>#37#</mathjax> <mathjax>#""^@C#</mathjax>, <mathjax>#-=#</mathjax> <mathjax>#310*K#</mathjax>. You can change the calculation accordingly if I have got your starting conditions wrong. </p></div>
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<h1 class="questionTitle" itemprop="name">A #780*mL# volume of gas #1.00*atm# pressure, and #310*K# temperature, is cooled to #295*K#, and #65*mm*Hg# pressure. What is the new volume?</h1>
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anor277
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<div class="markdown"><p>For a given quantity of gas, .........<mathjax>#V_2=8.70*L.#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For a given quantity of gas, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> holds that <mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>, temperature quoted in <mathjax>#"degrees Kelvin"#</mathjax>.</p>
<p>Thus <mathjax>#V_2=(P_1V_1)/T_1xxT_2/P_2#</mathjax></p>
<p><mathjax>#=(1.00*atmxx780*mL)/(310*K)xx(295*K)/((65*mm*Hg)/(760*mm*Hg*atm^-1)#</mathjax></p>
<p><mathjax>#=8.69xx10^3*mL=8.70*L#</mathjax></p>
<p>We use here the useful relationship that <mathjax>#760*mm*Hg-=1*atm#</mathjax>.</p>
<p>And thus a measurement of length on a mercury manometer, which is fairly easy to do in a lab, can be used to solve Gas law problems.</p>
<p>Please note that I used <mathjax>#22#</mathjax> <mathjax>#""^@C#</mathjax>, <mathjax>#-=#</mathjax> <mathjax>#295*K#</mathjax>, and <mathjax>#37#</mathjax> <mathjax>#""^@C#</mathjax>, <mathjax>#-=#</mathjax> <mathjax>#310*K#</mathjax>. You can change the calculation accordingly if I have got your starting conditions wrong. </p></div>
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</article> | A #780*mL# volume of gas #1.00*atm# pressure, and #310*K# temperature, is cooled to #295*K#, and #65*mm*Hg# pressure. What is the new volume? | null |
3,396 | ab81a841-6ddd-11ea-993f-ccda262736ce | https://socratic.org/questions/calculate-the-volume-of-hydrogen-required-for-complete-hydrogenation-of-0-25-dm- | 0.50 dm^3 | start physical_unit 4 4 volume dm^3 qc_end c_other OTHER qc_end physical_unit 13 13 10 11 volume qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] hydrogen [IN] dm^3"}] | [{"type":"physical unit","value":"0.50 dm^3"}] | [{"type":"other","value":"Complete hydrogenation."},{"type":"physical unit","value":"Volume [OF] ethyne [=] \\pu{0.25 dm^3}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">Calculate the volume of hydrogen required for complete hydrogenation of 0.25 #dm^3# of ethyne at STP?</h1> | null | 0.50 dm^3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Like with any <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> problem, the key tool you have at your disposal is the <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a>. </p>
<p>However, an interesting thing takes place when you're dealing with gases that are <em>under the same conditions</em> for pressure and temperature. In such cases, <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio becomes the <strong>volume ratio</strong>. </p>
<p>The idea is that, since you're dealing with two ideal gases, you can use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation to write</p>
<p><mathjax>#P * V_1 = n_1 * RT#</mathjax> <mathjax>#->#</mathjax> for ethyne;</p>
<p><mathjax>#P * V_2 = n_2 * RT#</mathjax> <mathjax>#->#</mathjax> for hydrogen.</p>
<p>The pressure and the temperature are the same for both gases, since the reaction presumably takes place at <strong>STP</strong>. </p>
<p>If you divided these two equations, you'll get</p>
<p><mathjax>#(cancel(P) * V_1)/(cancel(P) * V_2) = (n_1 * cancel(RT))/(n_2 * cancel(RT))#</mathjax></p>
<p>This is equivalent to </p>
<p><mathjax>#n_1/n_2 = V_1/V_2#</mathjax></p>
<p>The mole ratio is equivalent to the volume ratio. </p>
<p>All you need now is the balanced chemical equation for the reaction, which looks like this </p>
<p><mathjax>#C_2H_(2(g)) + color(red)(2)H_(2(g)) -> C_2H_(6(g))#</mathjax></p>
<p>Notice that you have a <mathjax>#1:color(red)(2)#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between ethyne and hydrogen gas. This means that the reaction needs <strong>twice as many</strong> moles of hydrogen gas than of ethyne. </p>
<p>This translates into volumes as well. In other words, the volume of hydrogen gas must be <strong>twice as big</strong> as the volume of ethyne. </p>
<p><mathjax>#0.25cancel("dm"^3"ethyne") * (color(red)(2)" dm"^3 "hydrogen")/(1cancel("dm"^3 "ethyne")) = color(green)("0.50 dm"""^3"hydrogen")#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>You'd need <mathjax>#"50 dm"""^3#</mathjax> of hydrogen.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Like with any <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> problem, the key tool you have at your disposal is the <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a>. </p>
<p>However, an interesting thing takes place when you're dealing with gases that are <em>under the same conditions</em> for pressure and temperature. In such cases, <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio becomes the <strong>volume ratio</strong>. </p>
<p>The idea is that, since you're dealing with two ideal gases, you can use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation to write</p>
<p><mathjax>#P * V_1 = n_1 * RT#</mathjax> <mathjax>#->#</mathjax> for ethyne;</p>
<p><mathjax>#P * V_2 = n_2 * RT#</mathjax> <mathjax>#->#</mathjax> for hydrogen.</p>
<p>The pressure and the temperature are the same for both gases, since the reaction presumably takes place at <strong>STP</strong>. </p>
<p>If you divided these two equations, you'll get</p>
<p><mathjax>#(cancel(P) * V_1)/(cancel(P) * V_2) = (n_1 * cancel(RT))/(n_2 * cancel(RT))#</mathjax></p>
<p>This is equivalent to </p>
<p><mathjax>#n_1/n_2 = V_1/V_2#</mathjax></p>
<p>The mole ratio is equivalent to the volume ratio. </p>
<p>All you need now is the balanced chemical equation for the reaction, which looks like this </p>
<p><mathjax>#C_2H_(2(g)) + color(red)(2)H_(2(g)) -> C_2H_(6(g))#</mathjax></p>
<p>Notice that you have a <mathjax>#1:color(red)(2)#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between ethyne and hydrogen gas. This means that the reaction needs <strong>twice as many</strong> moles of hydrogen gas than of ethyne. </p>
<p>This translates into volumes as well. In other words, the volume of hydrogen gas must be <strong>twice as big</strong> as the volume of ethyne. </p>
<p><mathjax>#0.25cancel("dm"^3"ethyne") * (color(red)(2)" dm"^3 "hydrogen")/(1cancel("dm"^3 "ethyne")) = color(green)("0.50 dm"""^3"hydrogen")#</mathjax></p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">Calculate the volume of hydrogen required for complete hydrogenation of 0.25 #dm^3# of ethyne at STP?</h1>
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<div class="markdown"><p>You'd need <mathjax>#"50 dm"""^3#</mathjax> of hydrogen.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Like with any <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> problem, the key tool you have at your disposal is the <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a>. </p>
<p>However, an interesting thing takes place when you're dealing with gases that are <em>under the same conditions</em> for pressure and temperature. In such cases, <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio becomes the <strong>volume ratio</strong>. </p>
<p>The idea is that, since you're dealing with two ideal gases, you can use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation to write</p>
<p><mathjax>#P * V_1 = n_1 * RT#</mathjax> <mathjax>#->#</mathjax> for ethyne;</p>
<p><mathjax>#P * V_2 = n_2 * RT#</mathjax> <mathjax>#->#</mathjax> for hydrogen.</p>
<p>The pressure and the temperature are the same for both gases, since the reaction presumably takes place at <strong>STP</strong>. </p>
<p>If you divided these two equations, you'll get</p>
<p><mathjax>#(cancel(P) * V_1)/(cancel(P) * V_2) = (n_1 * cancel(RT))/(n_2 * cancel(RT))#</mathjax></p>
<p>This is equivalent to </p>
<p><mathjax>#n_1/n_2 = V_1/V_2#</mathjax></p>
<p>The mole ratio is equivalent to the volume ratio. </p>
<p>All you need now is the balanced chemical equation for the reaction, which looks like this </p>
<p><mathjax>#C_2H_(2(g)) + color(red)(2)H_(2(g)) -> C_2H_(6(g))#</mathjax></p>
<p>Notice that you have a <mathjax>#1:color(red)(2)#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between ethyne and hydrogen gas. This means that the reaction needs <strong>twice as many</strong> moles of hydrogen gas than of ethyne. </p>
<p>This translates into volumes as well. In other words, the volume of hydrogen gas must be <strong>twice as big</strong> as the volume of ethyne. </p>
<p><mathjax>#0.25cancel("dm"^3"ethyne") * (color(red)(2)" dm"^3 "hydrogen")/(1cancel("dm"^3 "ethyne")) = color(green)("0.50 dm"""^3"hydrogen")#</mathjax></p></div>
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</article> | Calculate the volume of hydrogen required for complete hydrogenation of 0.25 #dm^3# of ethyne at STP? | null |
3,397 | aa1fead2-6ddd-11ea-b135-ccda262736ce | https://socratic.org/questions/5-lithium-nitride-has-the-chemical-formula-li-3n-rubidium-is-in-the-same-chemica | Rb3N | start chemical_formula qc_end c_other OTHER qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Formula [OF] rubidium nitride [IN] default"}] | [{"type":"chemical equation","value":"Rb3N"}] | [{"type":"other","value":"Lithium nitride has the chemical formula Li3N."},{"type":"other","value":"Rubidium is in the same chemical family as lithium."}] | <h1 class="questionTitle" itemprop="name">5. Lithium nitride has the chemical formula #Li_3N#. Rubidium is in the same chemical family as lithium. What is the formula of rubidium nitride? </h1> | null | Rb3N | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Rubidium, an alkali metal, commonly forms <mathjax>#Rb^+#</mathjax> salts, just as lithium commonly forms <mathjax>#Li^+#</mathjax> salts. We would assume that the formula of rubidium nitride (if it exists!) is <mathjax>#Rb_3N#</mathjax>. </p></div>
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<div class="markdown"><p>We would assume <mathjax>#Rb_3N#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Rubidium, an alkali metal, commonly forms <mathjax>#Rb^+#</mathjax> salts, just as lithium commonly forms <mathjax>#Li^+#</mathjax> salts. We would assume that the formula of rubidium nitride (if it exists!) is <mathjax>#Rb_3N#</mathjax>. </p></div>
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<h1 class="questionTitle" itemprop="name">5. Lithium nitride has the chemical formula #Li_3N#. Rubidium is in the same chemical family as lithium. What is the formula of rubidium nitride? </h1>
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<div class="markdown"><p>We would assume <mathjax>#Rb_3N#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Rubidium, an alkali metal, commonly forms <mathjax>#Rb^+#</mathjax> salts, just as lithium commonly forms <mathjax>#Li^+#</mathjax> salts. We would assume that the formula of rubidium nitride (if it exists!) is <mathjax>#Rb_3N#</mathjax>. </p></div>
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</article> | 5. Lithium nitride has the chemical formula #Li_3N#. Rubidium is in the same chemical family as lithium. What is the formula of rubidium nitride? | null |
3,398 | a92b5248-6ddd-11ea-98a2-ccda262736ce | https://socratic.org/questions/how-do-you-write-out-the-equation-for-this-reaction-sodium-phosphate-reacts-with | Na3PO4 + Ca(NO3)2 -> NaNO3 + Ca3(PO4)2 | start chemical_equation qc_end substance 10 11 qc_end substance 14 15 qc_end substance 18 19 qc_end substance 21 22 qc_end end | [{"type":"other","value":"Chemical Equation [OF] this reaction"}] | [{"type":"chemical equation","value":"Na3PO4 + Ca(NO3)2 -> NaNO3 + Ca3(PO4)2"}] | [{"type":"substance name","value":"Sodium phosphate"},{"type":"substance name","value":"Calcium nitrate"},{"type":"substance name","value":"Sodium nitrate"},{"type":"substance name","value":"Calcium phosphate"}] | <h1 class="questionTitle" itemprop="name">How do you write out the equation for this reaction: Sodium phosphate reacts with calcium nitrate to produce sodium nitrate plus calcium phosphate?</h1> | null | Na3PO4 + Ca(NO3)2 -> NaNO3 + Ca3(PO4)2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium phosphate = <mathjax>#Na_"3"PO_"4"#</mathjax>. <br/>
Calcium nitrate = <mathjax>#Ca(NO_"3")_"2"#</mathjax>. </p>
<p>Sodium nitrate = <mathjax>#NaNO_"3"#</mathjax>. <br/>
Calcium phosphate = <mathjax>#Ca_"3"(PO_"4")_"2"#</mathjax>. </p>
<blockquote>
<p><mathjax>#Na_"3"PO_"4" (aq) + Ca(NO_"3")_"2" (aq) -> NaNO_"3" (aq) + Ca_"3"(PO_"4")_"2" (s)#</mathjax></p>
</blockquote>
<p>States might be wrong :/ </p>
<p>... but hope this helps :)</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#Na_"3"PO_"4" (l) + Ca(NO_"3")_"2" (s) -> NaNO_"3" (s) + Ca_"3"(PO_"4")_"2" (s)#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium phosphate = <mathjax>#Na_"3"PO_"4"#</mathjax>. <br/>
Calcium nitrate = <mathjax>#Ca(NO_"3")_"2"#</mathjax>. </p>
<p>Sodium nitrate = <mathjax>#NaNO_"3"#</mathjax>. <br/>
Calcium phosphate = <mathjax>#Ca_"3"(PO_"4")_"2"#</mathjax>. </p>
<blockquote>
<p><mathjax>#Na_"3"PO_"4" (aq) + Ca(NO_"3")_"2" (aq) -> NaNO_"3" (aq) + Ca_"3"(PO_"4")_"2" (s)#</mathjax></p>
</blockquote>
<p>States might be wrong :/ </p>
<p>... but hope this helps :)</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you write out the equation for this reaction: Sodium phosphate reacts with calcium nitrate to produce sodium nitrate plus calcium phosphate?</h1>
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<div class="markdown"><p><mathjax>#Na_"3"PO_"4" (l) + Ca(NO_"3")_"2" (s) -> NaNO_"3" (s) + Ca_"3"(PO_"4")_"2" (s)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium phosphate = <mathjax>#Na_"3"PO_"4"#</mathjax>. <br/>
Calcium nitrate = <mathjax>#Ca(NO_"3")_"2"#</mathjax>. </p>
<p>Sodium nitrate = <mathjax>#NaNO_"3"#</mathjax>. <br/>
Calcium phosphate = <mathjax>#Ca_"3"(PO_"4")_"2"#</mathjax>. </p>
<blockquote>
<p><mathjax>#Na_"3"PO_"4" (aq) + Ca(NO_"3")_"2" (aq) -> NaNO_"3" (aq) + Ca_"3"(PO_"4")_"2" (s)#</mathjax></p>
</blockquote>
<p>States might be wrong :/ </p>
<p>... but hope this helps :)</p></div>
</div>
</div>
</div>
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</article> | How do you write out the equation for this reaction: Sodium phosphate reacts with calcium nitrate to produce sodium nitrate plus calcium phosphate? | null |
3,399 | ac332699-6ddd-11ea-bc87-ccda262736ce | https://socratic.org/questions/a-350-ml-sample-of-air-collected-at-35-c-has-a-pressure-of-550-torr-what-pressur | 485 torr | start physical_unit 3 5 pressure torr qc_end physical_unit 3 5 1 2 volume qc_end physical_unit 3 5 8 9 temperature qc_end physical_unit 3 5 14 15 pressure qc_end physical_unit 3 5 29 30 volume qc_end physical_unit 3 5 32 33 temperature qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] air sample [IN] torr"}] | [{"type":"physical unit","value":"485 torr"}] | [{"type":"physical unit","value":"Volume1 [OF] air sample [=] \\pu{350 mL}"},{"type":"physical unit","value":"Temperature1 [OF] air sample [=] \\pu{35 ℃}"},{"type":"physical unit","value":"Pressure1 [OF] air sample [=] \\pu{550 torr}"},{"type":"physical unit","value":"Volume2 [OF] air sample [=] \\pu{425 mL}"},{"type":"physical unit","value":"Temperature2 [OF] air sample [=] \\pu{57 ℃}"}] | <h1 class="questionTitle" itemprop="name">A 350 mL sample of air collected at 35°C has a pressure of 550 torr. What pressure will the air exert if it is allowed to expand to 425 mL at 57°C?</h1> | null | 485 torr | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can solve this equation using the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a></strong>:</p>
<p><mathjax>#(P_1V_1)/(T_1) = (P_2V_2)/(T_2)#</mathjax></p>
<p>Remember, <em>always</em> use the absolute (Kelvin) temperature when working with gas equations.</p>
<p>The temperature conversions are</p>
<p><mathjax>#T_1 = ""^"o""C" + 273 = 35^"o" + 273 = color(red)(308#</mathjax> <mathjax>#color(red)("K"#</mathjax></p>
<p><mathjax>#T_2 = 57^"o""C" + 273 = color(green)(330#</mathjax> <mathjax>#color(green)("K"#</mathjax></p>
<p>Since we're trying to find the final pressure, let's rearrange this equation to solve for <mathjax>#P_2#</mathjax>:</p>
<p><mathjax>#P_2 = (P_1V_1T_2)/(T_1V_2)#</mathjax></p>
<p><strong>Our known values</strong>:</p>
<p><mathjax>#P_1 = 550#</mathjax> <mathjax>#"torr"#</mathjax></p>
<p><mathjax>#V_1 = 350#</mathjax> <mathjax>#"mL"#</mathjax></p>
<p><mathjax>#T_1 = color(red)(308#</mathjax> <mathjax>#color(red)("K"#</mathjax></p>
<p><mathjax>#V_2 = 425#</mathjax> "<mathjax>#mL"#</mathjax></p>
<p><mathjax>#T_2 = color(green)(330#</mathjax> <mathjax>#color(green)("K"#</mathjax></p>
<blockquote></blockquote>
<p>Let's plug these into the equation to find the final pressure:</p>
<p><mathjax>#P_2 = ((550"torr")(350cancel("mL"))(330cancel("K")))/((308cancel("K"))(425cancel("mL"))) = color(blue)(485#</mathjax> <mathjax>#color(blue)("torr"#</mathjax></p>
<p>The final pressure after it is subjected to these changes is thus <mathjax>#color(blue)(485#</mathjax> <mathjax>#sfcolor(blue)("torr"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#485#</mathjax> <mathjax>#"torr"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can solve this equation using the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a></strong>:</p>
<p><mathjax>#(P_1V_1)/(T_1) = (P_2V_2)/(T_2)#</mathjax></p>
<p>Remember, <em>always</em> use the absolute (Kelvin) temperature when working with gas equations.</p>
<p>The temperature conversions are</p>
<p><mathjax>#T_1 = ""^"o""C" + 273 = 35^"o" + 273 = color(red)(308#</mathjax> <mathjax>#color(red)("K"#</mathjax></p>
<p><mathjax>#T_2 = 57^"o""C" + 273 = color(green)(330#</mathjax> <mathjax>#color(green)("K"#</mathjax></p>
<p>Since we're trying to find the final pressure, let's rearrange this equation to solve for <mathjax>#P_2#</mathjax>:</p>
<p><mathjax>#P_2 = (P_1V_1T_2)/(T_1V_2)#</mathjax></p>
<p><strong>Our known values</strong>:</p>
<p><mathjax>#P_1 = 550#</mathjax> <mathjax>#"torr"#</mathjax></p>
<p><mathjax>#V_1 = 350#</mathjax> <mathjax>#"mL"#</mathjax></p>
<p><mathjax>#T_1 = color(red)(308#</mathjax> <mathjax>#color(red)("K"#</mathjax></p>
<p><mathjax>#V_2 = 425#</mathjax> "<mathjax>#mL"#</mathjax></p>
<p><mathjax>#T_2 = color(green)(330#</mathjax> <mathjax>#color(green)("K"#</mathjax></p>
<blockquote></blockquote>
<p>Let's plug these into the equation to find the final pressure:</p>
<p><mathjax>#P_2 = ((550"torr")(350cancel("mL"))(330cancel("K")))/((308cancel("K"))(425cancel("mL"))) = color(blue)(485#</mathjax> <mathjax>#color(blue)("torr"#</mathjax></p>
<p>The final pressure after it is subjected to these changes is thus <mathjax>#color(blue)(485#</mathjax> <mathjax>#sfcolor(blue)("torr"#</mathjax>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A 350 mL sample of air collected at 35°C has a pressure of 550 torr. What pressure will the air exert if it is allowed to expand to 425 mL at 57°C?</h1>
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Nathan L.
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<span class="dateCreated" datetime="2017-06-17T22:08:55" itemprop="dateCreated">
Jun 17, 2017
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<div class="markdown"><p><mathjax>#485#</mathjax> <mathjax>#"torr"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can solve this equation using the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a></strong>:</p>
<p><mathjax>#(P_1V_1)/(T_1) = (P_2V_2)/(T_2)#</mathjax></p>
<p>Remember, <em>always</em> use the absolute (Kelvin) temperature when working with gas equations.</p>
<p>The temperature conversions are</p>
<p><mathjax>#T_1 = ""^"o""C" + 273 = 35^"o" + 273 = color(red)(308#</mathjax> <mathjax>#color(red)("K"#</mathjax></p>
<p><mathjax>#T_2 = 57^"o""C" + 273 = color(green)(330#</mathjax> <mathjax>#color(green)("K"#</mathjax></p>
<p>Since we're trying to find the final pressure, let's rearrange this equation to solve for <mathjax>#P_2#</mathjax>:</p>
<p><mathjax>#P_2 = (P_1V_1T_2)/(T_1V_2)#</mathjax></p>
<p><strong>Our known values</strong>:</p>
<p><mathjax>#P_1 = 550#</mathjax> <mathjax>#"torr"#</mathjax></p>
<p><mathjax>#V_1 = 350#</mathjax> <mathjax>#"mL"#</mathjax></p>
<p><mathjax>#T_1 = color(red)(308#</mathjax> <mathjax>#color(red)("K"#</mathjax></p>
<p><mathjax>#V_2 = 425#</mathjax> "<mathjax>#mL"#</mathjax></p>
<p><mathjax>#T_2 = color(green)(330#</mathjax> <mathjax>#color(green)("K"#</mathjax></p>
<blockquote></blockquote>
<p>Let's plug these into the equation to find the final pressure:</p>
<p><mathjax>#P_2 = ((550"torr")(350cancel("mL"))(330cancel("K")))/((308cancel("K"))(425cancel("mL"))) = color(blue)(485#</mathjax> <mathjax>#color(blue)("torr"#</mathjax></p>
<p>The final pressure after it is subjected to these changes is thus <mathjax>#color(blue)(485#</mathjax> <mathjax>#sfcolor(blue)("torr"#</mathjax>.</p></div>
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</article> | A 350 mL sample of air collected at 35°C has a pressure of 550 torr. What pressure will the air exert if it is allowed to expand to 425 mL at 57°C? | null |
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