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500 | a874a04c-6ddd-11ea-99b6-ccda262736ce | https://socratic.org/questions/what-is-the-coefficient-for-h-when-the-redox-equation-below-is-balanced | 4 | start physical_unit 5 5 coefficient none qc_end chemical_equation 14 29 qc_end end | [{"type":"physical unit","value":"Coefficient [OF] H^+"}] | [{"type":"physical unit","value":"4"}] | [{"type":"chemical equation","value":"Ag(s) + ? NO3^-(aq) + ? H^+ -> ? Ag^+(aq) + ? NO(g) + ? H2O"}] | <h1 class="questionTitle" itemprop="name">What is the coefficient for #H^+# when the redox equation below is balanced?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#?Ag(s) + ?NO_3^(-)(aq) + ?H^(+)(aq) -> ?Ag^(+)(aq) + ?NO(g) + ?H_2O#</mathjax></p></div>
</h2>
</div>
</div> | 4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As usual we write separate <a href="https://socratic.org/chemistry/chemical-reactions/redox-reactions">redox reactions</a>:</p>
<p><mathjax>#"Oxidation:"#</mathjax></p>
<p><mathjax>#Ag(s) rarr Ag^(+) + e^-#</mathjax> <mathjax>#(i)#</mathjax></p>
<p><mathjax>#"Reduction:"#</mathjax></p>
<p><mathjax>#NO_3^(-) +4H^(+) + 3e^(-) rarr NO +2H_2O#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>And so, we add <mathjax>#3xx(i) + (ii):#</mathjax></p>
<p><mathjax>#3Ag(s) +NO_3^(-) +4H^(+)rarr 3Ag^(+) +NO +2H_2O#</mathjax></p>
<p>This is balanced with respect to mass and charge, so it is reasonable. </p>
<p>The nitrate anion is in fact a potent oxidant; however, in aqueous solution, often its reactivity is kinetically hindered. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#3Ag(s) +NO_3^(-) +4H^(+)rarr 3Ag^(+) +NO +2H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As usual we write separate <a href="https://socratic.org/chemistry/chemical-reactions/redox-reactions">redox reactions</a>:</p>
<p><mathjax>#"Oxidation:"#</mathjax></p>
<p><mathjax>#Ag(s) rarr Ag^(+) + e^-#</mathjax> <mathjax>#(i)#</mathjax></p>
<p><mathjax>#"Reduction:"#</mathjax></p>
<p><mathjax>#NO_3^(-) +4H^(+) + 3e^(-) rarr NO +2H_2O#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>And so, we add <mathjax>#3xx(i) + (ii):#</mathjax></p>
<p><mathjax>#3Ag(s) +NO_3^(-) +4H^(+)rarr 3Ag^(+) +NO +2H_2O#</mathjax></p>
<p>This is balanced with respect to mass and charge, so it is reasonable. </p>
<p>The nitrate anion is in fact a potent oxidant; however, in aqueous solution, often its reactivity is kinetically hindered. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the coefficient for #H^+# when the redox equation below is balanced?</h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#?Ag(s) + ?NO_3^(-)(aq) + ?H^(+)(aq) -> ?Ag^(+)(aq) + ?NO(g) + ?H_2O#</mathjax></p></div>
</h2>
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anor277
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Nov 14, 2016
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<div class="markdown"><p><mathjax>#3Ag(s) +NO_3^(-) +4H^(+)rarr 3Ag^(+) +NO +2H_2O#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As usual we write separate <a href="https://socratic.org/chemistry/chemical-reactions/redox-reactions">redox reactions</a>:</p>
<p><mathjax>#"Oxidation:"#</mathjax></p>
<p><mathjax>#Ag(s) rarr Ag^(+) + e^-#</mathjax> <mathjax>#(i)#</mathjax></p>
<p><mathjax>#"Reduction:"#</mathjax></p>
<p><mathjax>#NO_3^(-) +4H^(+) + 3e^(-) rarr NO +2H_2O#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>And so, we add <mathjax>#3xx(i) + (ii):#</mathjax></p>
<p><mathjax>#3Ag(s) +NO_3^(-) +4H^(+)rarr 3Ag^(+) +NO +2H_2O#</mathjax></p>
<p>This is balanced with respect to mass and charge, so it is reasonable. </p>
<p>The nitrate anion is in fact a potent oxidant; however, in aqueous solution, often its reactivity is kinetically hindered. </p></div>
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</article> | What is the coefficient for #H^+# when the redox equation below is balanced? |
#?Ag(s) + ?NO_3^(-)(aq) + ?H^(+)(aq) -> ?Ag^(+)(aq) + ?NO(g) + ?H_2O#
|
501 | a9358d76-6ddd-11ea-b36e-ccda262736ce | https://socratic.org/questions/a-30-0-g-sample-of-water-at-280-k-is-mixed-with-50-0-g-of-water-at-330-k-how-wou | 311.25 K | start physical_unit 27 28 temperature k qc_end physical_unit 3 5 1 2 mass qc_end physical_unit 3 5 7 8 temperature qc_end physical_unit 3 5 12 13 mass qc_end physical_unit 3 5 17 18 temperature qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Temperature3 [OF] the mixture [IN] K"}] | [{"type":"physical unit","value":"311.25 K"}] | [{"type":"physical unit","value":"Mass1 [OF] water sample [=] \\pu{30.0 g}"},{"type":"physical unit","value":"Temperature1 [OF] water sample [=] \\pu{280 K}"},{"type":"physical unit","value":"Mass2 [OF] water sample [=] \\pu{50.0 g}"},{"type":"physical unit","value":"Temperature2 [OF] water sample [=] \\pu{330 K}"},{"type":"other","value":"Assume no heat is lost to the surroundings."}] | <h1 class="questionTitle" itemprop="name">A 30-0 g sample of water at 280 K is mixed with 50.0 g of water at 330 K. How would you calculate the final temperature of the mixture assuming no heat is lost to the surroundings?
</h1> | null | 311.25 K | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If you assume that no heat is lost to the surroundings, then you can say that the heat <strong>lost</strong> by the <em>hotter</em> sample will be <strong>equal to</strong> the heat <strong>absorbed</strong> by the <em>cooler</em> sample. </p>
<p>Now, the equation that establishes a relationship between heat lost or gained and change in temperature looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(q = m * c * DeltaT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - heat lost or gained<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the change in temperature, defined as <em>final temperature</em> minus <em>initial temperature</em></p>
<p>Now, before doing any calculation, try to predict what will happen when the two samples are mixed. </p>
<p>Notice that you have more hot water than cold water, which means that you can expect the final temperature of the mixture to be <strong>closer</strong> to <mathjax>#"330 K"#</mathjax> than to <mathjax>#"280 K"#</mathjax>. </p>
<p>So, you can say that</p>
<blockquote>
<p><mathjax>#-q_"lost" = q_"gained"#</mathjax></p>
</blockquote>
<p>Here the minus signed is used because heat <strong>lost</strong> is <strong>negative</strong>. </p>
<p>This is equivalent to </p>
<blockquote>
<p><mathjax>#m_"hot" * color(red)(cancel(color(black)(c))) * DeltaT_"hot" = m_"cold" * color(red)(cancel(color(black)(c))) * DeltaT_"cold"#</mathjax></p>
</blockquote>
<p>Let's say that the final temperature is <mathjax>#color(blue)(T_"f")#</mathjax>. This means that you have</p>
<blockquote>
<p><mathjax>#-50.0color(red)(cancel(color(black)("g"))) * (color(blue)(T_"f") - 330)color(red)(cancel(color(black)("K"))) = 30.0color(red)(cancel(color(black)("g"))) * (color(blue)(T_"f") - 280)color(red)(cancel(color(black)("K")))#</mathjax></p>
<p><mathjax>#-50.0 * color(blue)(T_"f") + 16500 = 30.0 * color(blue)(T_"f") - 8400#</mathjax></p>
<p><mathjax>#80.0 * color(blue)(T_"f") = 24900#</mathjax></p>
<p><mathjax>#color(blue)(T_"f") = 24900/80.0 = "311.25 K"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the two temperatures, the answer will be </p>
<blockquote>
<p><mathjax>#T_"f" = color(green)("310 K")#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"310 K"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If you assume that no heat is lost to the surroundings, then you can say that the heat <strong>lost</strong> by the <em>hotter</em> sample will be <strong>equal to</strong> the heat <strong>absorbed</strong> by the <em>cooler</em> sample. </p>
<p>Now, the equation that establishes a relationship between heat lost or gained and change in temperature looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(q = m * c * DeltaT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - heat lost or gained<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the change in temperature, defined as <em>final temperature</em> minus <em>initial temperature</em></p>
<p>Now, before doing any calculation, try to predict what will happen when the two samples are mixed. </p>
<p>Notice that you have more hot water than cold water, which means that you can expect the final temperature of the mixture to be <strong>closer</strong> to <mathjax>#"330 K"#</mathjax> than to <mathjax>#"280 K"#</mathjax>. </p>
<p>So, you can say that</p>
<blockquote>
<p><mathjax>#-q_"lost" = q_"gained"#</mathjax></p>
</blockquote>
<p>Here the minus signed is used because heat <strong>lost</strong> is <strong>negative</strong>. </p>
<p>This is equivalent to </p>
<blockquote>
<p><mathjax>#m_"hot" * color(red)(cancel(color(black)(c))) * DeltaT_"hot" = m_"cold" * color(red)(cancel(color(black)(c))) * DeltaT_"cold"#</mathjax></p>
</blockquote>
<p>Let's say that the final temperature is <mathjax>#color(blue)(T_"f")#</mathjax>. This means that you have</p>
<blockquote>
<p><mathjax>#-50.0color(red)(cancel(color(black)("g"))) * (color(blue)(T_"f") - 330)color(red)(cancel(color(black)("K"))) = 30.0color(red)(cancel(color(black)("g"))) * (color(blue)(T_"f") - 280)color(red)(cancel(color(black)("K")))#</mathjax></p>
<p><mathjax>#-50.0 * color(blue)(T_"f") + 16500 = 30.0 * color(blue)(T_"f") - 8400#</mathjax></p>
<p><mathjax>#80.0 * color(blue)(T_"f") = 24900#</mathjax></p>
<p><mathjax>#color(blue)(T_"f") = 24900/80.0 = "311.25 K"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the two temperatures, the answer will be </p>
<blockquote>
<p><mathjax>#T_"f" = color(green)("310 K")#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A 30-0 g sample of water at 280 K is mixed with 50.0 g of water at 330 K. How would you calculate the final temperature of the mixture assuming no heat is lost to the surroundings?
</h1>
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Stefan V.
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<span class="dateCreated" datetime="2015-11-09T23:45:41" itemprop="dateCreated">
Nov 9, 2015
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<div>
<div class="markdown"><p><mathjax>#"310 K"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If you assume that no heat is lost to the surroundings, then you can say that the heat <strong>lost</strong> by the <em>hotter</em> sample will be <strong>equal to</strong> the heat <strong>absorbed</strong> by the <em>cooler</em> sample. </p>
<p>Now, the equation that establishes a relationship between heat lost or gained and change in temperature looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(q = m * c * DeltaT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - heat lost or gained<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the change in temperature, defined as <em>final temperature</em> minus <em>initial temperature</em></p>
<p>Now, before doing any calculation, try to predict what will happen when the two samples are mixed. </p>
<p>Notice that you have more hot water than cold water, which means that you can expect the final temperature of the mixture to be <strong>closer</strong> to <mathjax>#"330 K"#</mathjax> than to <mathjax>#"280 K"#</mathjax>. </p>
<p>So, you can say that</p>
<blockquote>
<p><mathjax>#-q_"lost" = q_"gained"#</mathjax></p>
</blockquote>
<p>Here the minus signed is used because heat <strong>lost</strong> is <strong>negative</strong>. </p>
<p>This is equivalent to </p>
<blockquote>
<p><mathjax>#m_"hot" * color(red)(cancel(color(black)(c))) * DeltaT_"hot" = m_"cold" * color(red)(cancel(color(black)(c))) * DeltaT_"cold"#</mathjax></p>
</blockquote>
<p>Let's say that the final temperature is <mathjax>#color(blue)(T_"f")#</mathjax>. This means that you have</p>
<blockquote>
<p><mathjax>#-50.0color(red)(cancel(color(black)("g"))) * (color(blue)(T_"f") - 330)color(red)(cancel(color(black)("K"))) = 30.0color(red)(cancel(color(black)("g"))) * (color(blue)(T_"f") - 280)color(red)(cancel(color(black)("K")))#</mathjax></p>
<p><mathjax>#-50.0 * color(blue)(T_"f") + 16500 = 30.0 * color(blue)(T_"f") - 8400#</mathjax></p>
<p><mathjax>#80.0 * color(blue)(T_"f") = 24900#</mathjax></p>
<p><mathjax>#color(blue)(T_"f") = 24900/80.0 = "311.25 K"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the two temperatures, the answer will be </p>
<blockquote>
<p><mathjax>#T_"f" = color(green)("310 K")#</mathjax></p>
</blockquote></div>
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</article> | A 30-0 g sample of water at 280 K is mixed with 50.0 g of water at 330 K. How would you calculate the final temperature of the mixture assuming no heat is lost to the surroundings?
| null |
502 | a9c964c9-6ddd-11ea-a78e-ccda262736ce | https://socratic.org/questions/an-acidic-solution-has-a-ph-of-4-02-what-is-the-h-3o-concentration-of-this-solut | 9.55 × 10^(-6) M | start physical_unit 11 11 concentration mol/l qc_end physical_unit 1 2 7 7 ph qc_end end | [{"type":"physical unit","value":"concentration [OF] H3O+ [IN] M"}] | [{"type":"physical unit","value":"9.55 × 10^(-6) M"}] | [{"type":"physical unit","value":"pH [OF] the acidic solution [=] \\pu{4.02}"}] | <h1 class="questionTitle" itemprop="name">An acidic solution has a pH of 4.02. What is the #H_3O^+# concentration of this solution in M? </h1> | null | 9.55 × 10^(-6) M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition, <mathjax>#pH=-log_(10)[H_3O^+]#</mathjax></p>
<p>And thus if <mathjax>#pH=4.02#</mathjax>, <mathjax>#[H_3O^+]=10^(-4.02)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#9.55xx10^-6*mol*L^-1#</mathjax>.</p>
<p>Back in the day, before the advent of cheap electronic calculators, A level students routinely used log tables to aid complex multiplications and divisions. See <a href="https://socratic.org/questions/how-do-you-calculate-the-poh-of-a-solution">here</a> and elsewhere.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#H_3O^+=9.55xx10^-6*mol*L^-1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition, <mathjax>#pH=-log_(10)[H_3O^+]#</mathjax></p>
<p>And thus if <mathjax>#pH=4.02#</mathjax>, <mathjax>#[H_3O^+]=10^(-4.02)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#9.55xx10^-6*mol*L^-1#</mathjax>.</p>
<p>Back in the day, before the advent of cheap electronic calculators, A level students routinely used log tables to aid complex multiplications and divisions. See <a href="https://socratic.org/questions/how-do-you-calculate-the-poh-of-a-solution">here</a> and elsewhere.</p></div>
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<h1 class="questionTitle" itemprop="name">An acidic solution has a pH of 4.02. What is the #H_3O^+# concentration of this solution in M? </h1>
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anor277
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<div class="markdown"><p><mathjax>#H_3O^+=9.55xx10^-6*mol*L^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition, <mathjax>#pH=-log_(10)[H_3O^+]#</mathjax></p>
<p>And thus if <mathjax>#pH=4.02#</mathjax>, <mathjax>#[H_3O^+]=10^(-4.02)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#9.55xx10^-6*mol*L^-1#</mathjax>.</p>
<p>Back in the day, before the advent of cheap electronic calculators, A level students routinely used log tables to aid complex multiplications and divisions. See <a href="https://socratic.org/questions/how-do-you-calculate-the-poh-of-a-solution">here</a> and elsewhere.</p></div>
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</article> | An acidic solution has a pH of 4.02. What is the #H_3O^+# concentration of this solution in M? | null |
503 | ab5c9682-6ddd-11ea-a2c3-ccda262736ce | https://socratic.org/questions/what-s-the-balanced-equation-for-magnesium-and-hydrochloric-acid | Mg(s) + 2 HCl(aq) -> MgCl(aq) + H2(g) | start chemical_equation qc_end substance 5 5 qc_end substance 7 8 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the balanced equation"}] | [{"type":"chemical equation","value":"Mg(s) + 2 HCl(aq) -> MgCl(aq) + H2(g)"}] | [{"type":"substance name","value":"Magnesium"},{"type":"substance name","value":"Hydrochloric acid"}] | <h1 class="questionTitle" itemprop="name">What's the balanced equation for magnesium and hydrochloric acid?</h1> | null | Mg(s) + 2 HCl(aq) -> MgCl(aq) + H2(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction between magnesium and hydrochloric acid combine to form a salt of magnesium chloride and release hydrogen gas. This single replacement reaction is a classic example of a metal reacting in an acid to release hydrogen gas. </p>
<p>I hope this was helpful.<br/>
SMARTERTEACHER</p>
<p>
<iframe src="https://www.youtube.com/embed/jwNx-7XDpPg?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p>
<p>You can watch this reaction in the video below.</p>
<p>
<iframe src="https://www.youtube.com/embed/ry33yyXdTg0?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>The balanced chemical equation is:<br/>
<mathjax>#Mg_((s))+ 2HCl_((aq))rarrMgCl_(2(aq)) + H_(2(g)#</mathjax> </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction between magnesium and hydrochloric acid combine to form a salt of magnesium chloride and release hydrogen gas. This single replacement reaction is a classic example of a metal reacting in an acid to release hydrogen gas. </p>
<p>I hope this was helpful.<br/>
SMARTERTEACHER</p>
<p>
<iframe src="https://www.youtube.com/embed/jwNx-7XDpPg?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p>
<p>You can watch this reaction in the video below.</p>
<p>
<iframe src="https://www.youtube.com/embed/ry33yyXdTg0?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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<h1 class="questionTitle" itemprop="name">What's the balanced equation for magnesium and hydrochloric acid?</h1>
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BRIAN M.
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<div class="markdown"><p>The balanced chemical equation is:<br/>
<mathjax>#Mg_((s))+ 2HCl_((aq))rarrMgCl_(2(aq)) + H_(2(g)#</mathjax> </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction between magnesium and hydrochloric acid combine to form a salt of magnesium chloride and release hydrogen gas. This single replacement reaction is a classic example of a metal reacting in an acid to release hydrogen gas. </p>
<p>I hope this was helpful.<br/>
SMARTERTEACHER</p>
<p>
<iframe src="https://www.youtube.com/embed/jwNx-7XDpPg?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p>
<p>You can watch this reaction in the video below.</p>
<p>
<iframe src="https://www.youtube.com/embed/ry33yyXdTg0?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
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</article> | What's the balanced equation for magnesium and hydrochloric acid? | null |
504 | ab5451b6-6ddd-11ea-91a2-ccda262736ce | https://socratic.org/questions/57ff49b411ef6b31dfbee8b6 | H2SO4(aq) + 2 NaOH(aq) -> Na2SO4(aq) + 2 H2O(l) | start chemical_equation qc_end substance 7 8 qc_end substance 10 11 qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Equation [OF] the neutralization"}] | [{"type":"chemical equation","value":"H2SO4(aq) + 2 NaOH(aq) -> Na2SO4(aq) + 2 H2O(l)"}] | [{"type":"substance name","value":"Sulfuric acid"},{"type":"substance name","value":"Sodium hydroxide"},{"type":"other","value":"In a stoichiometrically balanced chemical equation."}] | <h1 class="questionTitle" itemprop="name">How do we represent the neutralization of sulfuric acid by sodium hydroxide in a stoichiometrically balanced chemical equation? </h1> | null | H2SO4(aq) + 2 NaOH(aq) -> Na2SO4(aq) + 2 H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>When you write or consider a chemical equation, the two things you must ask yourself are (i) is mass balanced?, and (ii) is charge balanced? If they are not then you know that this does not reflect reality. Are these conditions satisfied here?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#H_2SO_4(aq) + 2NaOH(aq) rarr Na_2SO_4(aq) + 2H_2O(l)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>When you write or consider a chemical equation, the two things you must ask yourself are (i) is mass balanced?, and (ii) is charge balanced? If they are not then you know that this does not reflect reality. Are these conditions satisfied here?</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How do we represent the neutralization of sulfuric acid by sodium hydroxide in a stoichiometrically balanced chemical equation? </h1>
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anor277
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Oct 13, 2016
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<div class="markdown"><p><mathjax>#H_2SO_4(aq) + 2NaOH(aq) rarr Na_2SO_4(aq) + 2H_2O(l)#</mathjax></p></div>
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<div class="markdown"><p>When you write or consider a chemical equation, the two things you must ask yourself are (i) is mass balanced?, and (ii) is charge balanced? If they are not then you know that this does not reflect reality. Are these conditions satisfied here?</p></div>
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</article> | How do we represent the neutralization of sulfuric acid by sodium hydroxide in a stoichiometrically balanced chemical equation? | null |
505 | ac9cf4bc-6ddd-11ea-938d-ccda262736ce | https://socratic.org/questions/a-beaker-with-120ml-of-an-acetic-acid-buffer-with-a-ph-of-5-00-is-sitting-on-a-b | 0.29 | start physical_unit 9 9 ph none qc_end physical_unit 7 9 3 4 volume qc_end physical_unit 7 9 14 14 ph qc_end physical_unit 24 27 32 33 molarity qc_end physical_unit 43 44 37 38 volume qc_end physical_unit 43 44 41 42 molarity qc_end physical_unit 7 8 60 60 pka qc_end end | [{"type":"physical unit","value":"Changed pH [OF] the buffer"}] | [{"type":"physical unit","value":"0.29"}] | [{"type":"physical unit","value":"Volume [OF] acetic acid buffer [=] \\pu{120 mL}"},{"type":"physical unit","value":"pH [OF] acetic acid buffer [=] \\pu{5.00}"},{"type":"physical unit","value":"Molarity [OF] acid and conjugate base [=] \\pu{0.1 M}"},{"type":"physical unit","value":"Volume [OF] HCl solution [=] \\pu{6.60 mL}"},{"type":"physical unit","value":"Molarity [OF] HCl solution [=] \\pu{0.300 M}"},{"type":"physical unit","value":"pKa [OF] acetic acid [=] \\pu{4.76}"}] | <h1 class="questionTitle" itemprop="name">A beaker with 120mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.1M. A student adds 6.60mL of a 0.300M HCl solution to the beaker. How much will the pH change? </h1> | <div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The pKa of acetic acid is 4.76.</p></div>
</h2>
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</div> | 0.29 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>!! LONG ANSWER !!</strong></p>
<p>The idea here is that you need to use the <strong>Henderson-Hasselbalch equation</strong> to determine the <em>ratio</em> that exists between the concentration of the weak acid and of its conjugate base in the <strong><a href="https://socratic.org/chemistry/reactions-in-solution/buffer-theory">buffer solution</a></strong>. </p>
<p>Once you know that, you can use the <strong>total <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the acid and of the conjugate base to find the <em>number of moles</em> of these two chemical species present in the buffer. </p>
<p>So, the <em>Henderson-Hasselbalch equation</em> looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pH" = pK_a + log( (["conjugate base"])/(["weak acid"]))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you have acetic acid, <mathjax>#"CH"_3"COOH"#</mathjax>, as the weak acid and the acetate anion, <mathjax>#"CH"_3"COO"^(-)#</mathjax>, as its conjugate base. The <mathjax>#pK_a#</mathjax> of the acid is said to be equal to <mathjax>#4.76#</mathjax>, which means that you have </p>
<blockquote>
<p><mathjax>#"pH" = 4.76 + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#</mathjax></p>
</blockquote>
<p>The <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> is equal to <mathjax>#5#</mathjax>, and so </p>
<blockquote>
<p><mathjax>#5.00 = 4.76 + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#</mathjax></p>
<p><mathjax>#log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 0.24#</mathjax></p>
</blockquote>
<p>This will be equivalent to </p>
<blockquote>
<p><mathjax>#10^log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 10^(0.24)#</mathjax></p>
</blockquote>
<p>which will give you </p>
<blockquote>
<p><mathjax>#(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = 1.74#</mathjax></p>
</blockquote>
<p>This means that your buffer contains <mathjax>#1.74#</mathjax> <strong>times more</strong> conjugate base than weak acid </p>
<blockquote>
<p><mathjax>#["CH"_3"COO"^(-)] = 1.74 xx ["CH"_3"COOH"]#</mathjax></p>
</blockquote>
<p>Now, because both chemical species share <strong>the same volume</strong>, <mathjax>#"120 mL"#</mathjax>, this can be rewritten as </p>
<blockquote>
<p><mathjax>#n_(CH_3COO^(-))/color(red)(cancel(color(black)(120 * 10^(-3)"L"))) = 1.74 xx n_(CH_3COOH)/color(red)(cancel(color(black)(120 * 10^(-3)"L")))#</mathjax></p>
</blockquote>
<p>which is </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(n_(CH_3COO^(-)) = 1.74 xx n_(CH_3COOH))color(white)(a/a)|)))" " " "color(orange)((1))#</mathjax></p>
</blockquote>
<p>So, the buffer contains <mathjax>#1.74#</mathjax> <strong>times more</strong> <em>moles</em> of acetate anions that of acetic acid. </p>
<p>Now, the <strong>total molarity</strong> of the buffer is said to be equal to <mathjax>#"0.1 M"#</mathjax>. You thus have</p>
<blockquote>
<p><mathjax>#["CH"_3"COOH"] + ["CH"_3"COO"^(-)] = "0.10 M"#</mathjax></p>
</blockquote>
<p>Once again, use the volume of the buffer to write </p>
<blockquote>
<p><mathjax>#n_(CH_3COOH)/(120 * 10^(-3)color(red)(cancel(color(black)("L")))) + n_(CH_3COO^(-))/(120 * 10^(-3)color(red)(cancel(color(black)("L")))) = 0.1"moles"/color(red)(cancel(color(black)("L")))#</mathjax></p>
</blockquote>
<p>This will be equivalent to </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(n_(CH_3COO^(-)) + n_(CH_3COOH) = 0.012)color(white)(a/a)|)))" " " " color(orange)((2))#</mathjax></p>
</blockquote>
<p>Use equations <mathjax>#color(orange)((1))#</mathjax> and <mathjax>#color(orange)((2))#</mathjax> to find how many moles of acetate ions you have in the buffer</p>
<blockquote>
<p><mathjax>#1.74 * n_(CH_3COOH) + n_(CH_3COOH) = 0.012#</mathjax></p>
<p><mathjax>#n_(CH_3COOH) = 0.012/(1.74 + 1) = "0.004380 moles CH"_3"COOH"#</mathjax></p>
</blockquote>
<p>This means that you have </p>
<blockquote>
<p><mathjax>#n_(CH_3COO^(-)) = 1.74 * "0.004380 moles"#</mathjax></p>
<p><mathjax>#n_(CH_3COO^(-)) = "0.007621 moles CH"_3"COO"^(-)#</mathjax></p>
</blockquote>
<p>Now, hydrochloric acid, <mathjax>#"HCl"#</mathjax>, will react with the acetate anions to form acetic acid and chloride anions, <mathjax>#"Cl"^(-)#</mathjax> </p>
<blockquote>
<p><mathjax>#color(red)("H")"Cl"_ ((aq)) + "CH"_ 3"COO"_ ((aq))^(-) -> "CH"_ 3"COO"color(red)("H")_ ((aq)) + "Cl"_((aq))^(-)#</mathjax></p>
</blockquote>
<p>Notice that the reaction consumes hydrochloric acid and acetate ions in a <mathjax>#1:1#</mathjax> <strong><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a></strong>, and produces acetic acid in a <mathjax>#1:1#</mathjax> mole ratio. </p>
<p>Use the <em>molarity</em> and <em>volume</em> of the hydrochloric acid solution to determine how many moles of strong acid you have </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#</mathjax> </p>
</blockquote>
<p>In your case, this gets you </p>
<blockquote>
<p><mathjax>#n_(HCl) = "0.300 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(6.60 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))#</mathjax></p>
<p><mathjax>#n_(HCl) = "0.001980 moles HCl"#</mathjax></p>
</blockquote>
<p>The hydrochloric acid will be <strong>completely consumed</strong> by the reaction, and the resulting solution will contain </p>
<blockquote>
<p><mathjax>#n_(HCl) = "0 moles" ->#</mathjax> <em><strong>completely consumed</strong></em></p>
<p><mathjax>#n_(CH_3COO^(-)) = "0.007621 moles" - "0.001980 moles"#</mathjax> </p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#= "0.005641 moles CH"_3"COO"^(-)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p><mathjax>#n_(CH_3COOH) = "0.004380 moles" + "0.001980 moles"#</mathjax> </p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#= "0.006360 moles CH"_3"COOH"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>The <strong>total volume</strong> of the solution will now be </p>
<blockquote>
<p><mathjax>#V_"total" = "120 mL" + "6.60 mL" = "126.6 mL"#</mathjax></p>
</blockquote>
<p>The concentrations of acetic acid and acetate ions will be </p>
<blockquote>
<p><mathjax>#["CH"_3"COOH"] = "0.006360 moles"/(126.6 * 10^(-3)"L") = "0.05024 M"#</mathjax></p>
<p><mathjax>#["CH"_3"COO"^(-)] = "0.005641 moles"/(126.6 * 10^(-3)"L") = "0.04456 M"#</mathjax></p>
</blockquote>
<p>Use the Henderson-Hasselbalch equation to find the new pH of the solution </p>
<blockquote>
<p><mathjax>#"pH" = 4.76 + log( (0.04456 color(red)(cancel(color(black)("M"))))/(0.05024color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#"pH" = 4.71#</mathjax></p>
</blockquote>
<p>Therefore, the pH of the solution <strong>decreased</strong> by </p>
<blockquote>
<p><mathjax>#Delta_"pH" = |4.71 - 5.00| = color(green)(|bar(ul(color(white)(a/a)"0.29 units"color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#Delta_"pH" = 0.29#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>!! LONG ANSWER !!</strong></p>
<p>The idea here is that you need to use the <strong>Henderson-Hasselbalch equation</strong> to determine the <em>ratio</em> that exists between the concentration of the weak acid and of its conjugate base in the <strong><a href="https://socratic.org/chemistry/reactions-in-solution/buffer-theory">buffer solution</a></strong>. </p>
<p>Once you know that, you can use the <strong>total <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the acid and of the conjugate base to find the <em>number of moles</em> of these two chemical species present in the buffer. </p>
<p>So, the <em>Henderson-Hasselbalch equation</em> looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pH" = pK_a + log( (["conjugate base"])/(["weak acid"]))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you have acetic acid, <mathjax>#"CH"_3"COOH"#</mathjax>, as the weak acid and the acetate anion, <mathjax>#"CH"_3"COO"^(-)#</mathjax>, as its conjugate base. The <mathjax>#pK_a#</mathjax> of the acid is said to be equal to <mathjax>#4.76#</mathjax>, which means that you have </p>
<blockquote>
<p><mathjax>#"pH" = 4.76 + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#</mathjax></p>
</blockquote>
<p>The <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> is equal to <mathjax>#5#</mathjax>, and so </p>
<blockquote>
<p><mathjax>#5.00 = 4.76 + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#</mathjax></p>
<p><mathjax>#log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 0.24#</mathjax></p>
</blockquote>
<p>This will be equivalent to </p>
<blockquote>
<p><mathjax>#10^log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 10^(0.24)#</mathjax></p>
</blockquote>
<p>which will give you </p>
<blockquote>
<p><mathjax>#(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = 1.74#</mathjax></p>
</blockquote>
<p>This means that your buffer contains <mathjax>#1.74#</mathjax> <strong>times more</strong> conjugate base than weak acid </p>
<blockquote>
<p><mathjax>#["CH"_3"COO"^(-)] = 1.74 xx ["CH"_3"COOH"]#</mathjax></p>
</blockquote>
<p>Now, because both chemical species share <strong>the same volume</strong>, <mathjax>#"120 mL"#</mathjax>, this can be rewritten as </p>
<blockquote>
<p><mathjax>#n_(CH_3COO^(-))/color(red)(cancel(color(black)(120 * 10^(-3)"L"))) = 1.74 xx n_(CH_3COOH)/color(red)(cancel(color(black)(120 * 10^(-3)"L")))#</mathjax></p>
</blockquote>
<p>which is </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(n_(CH_3COO^(-)) = 1.74 xx n_(CH_3COOH))color(white)(a/a)|)))" " " "color(orange)((1))#</mathjax></p>
</blockquote>
<p>So, the buffer contains <mathjax>#1.74#</mathjax> <strong>times more</strong> <em>moles</em> of acetate anions that of acetic acid. </p>
<p>Now, the <strong>total molarity</strong> of the buffer is said to be equal to <mathjax>#"0.1 M"#</mathjax>. You thus have</p>
<blockquote>
<p><mathjax>#["CH"_3"COOH"] + ["CH"_3"COO"^(-)] = "0.10 M"#</mathjax></p>
</blockquote>
<p>Once again, use the volume of the buffer to write </p>
<blockquote>
<p><mathjax>#n_(CH_3COOH)/(120 * 10^(-3)color(red)(cancel(color(black)("L")))) + n_(CH_3COO^(-))/(120 * 10^(-3)color(red)(cancel(color(black)("L")))) = 0.1"moles"/color(red)(cancel(color(black)("L")))#</mathjax></p>
</blockquote>
<p>This will be equivalent to </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(n_(CH_3COO^(-)) + n_(CH_3COOH) = 0.012)color(white)(a/a)|)))" " " " color(orange)((2))#</mathjax></p>
</blockquote>
<p>Use equations <mathjax>#color(orange)((1))#</mathjax> and <mathjax>#color(orange)((2))#</mathjax> to find how many moles of acetate ions you have in the buffer</p>
<blockquote>
<p><mathjax>#1.74 * n_(CH_3COOH) + n_(CH_3COOH) = 0.012#</mathjax></p>
<p><mathjax>#n_(CH_3COOH) = 0.012/(1.74 + 1) = "0.004380 moles CH"_3"COOH"#</mathjax></p>
</blockquote>
<p>This means that you have </p>
<blockquote>
<p><mathjax>#n_(CH_3COO^(-)) = 1.74 * "0.004380 moles"#</mathjax></p>
<p><mathjax>#n_(CH_3COO^(-)) = "0.007621 moles CH"_3"COO"^(-)#</mathjax></p>
</blockquote>
<p>Now, hydrochloric acid, <mathjax>#"HCl"#</mathjax>, will react with the acetate anions to form acetic acid and chloride anions, <mathjax>#"Cl"^(-)#</mathjax> </p>
<blockquote>
<p><mathjax>#color(red)("H")"Cl"_ ((aq)) + "CH"_ 3"COO"_ ((aq))^(-) -> "CH"_ 3"COO"color(red)("H")_ ((aq)) + "Cl"_((aq))^(-)#</mathjax></p>
</blockquote>
<p>Notice that the reaction consumes hydrochloric acid and acetate ions in a <mathjax>#1:1#</mathjax> <strong><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a></strong>, and produces acetic acid in a <mathjax>#1:1#</mathjax> mole ratio. </p>
<p>Use the <em>molarity</em> and <em>volume</em> of the hydrochloric acid solution to determine how many moles of strong acid you have </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#</mathjax> </p>
</blockquote>
<p>In your case, this gets you </p>
<blockquote>
<p><mathjax>#n_(HCl) = "0.300 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(6.60 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))#</mathjax></p>
<p><mathjax>#n_(HCl) = "0.001980 moles HCl"#</mathjax></p>
</blockquote>
<p>The hydrochloric acid will be <strong>completely consumed</strong> by the reaction, and the resulting solution will contain </p>
<blockquote>
<p><mathjax>#n_(HCl) = "0 moles" ->#</mathjax> <em><strong>completely consumed</strong></em></p>
<p><mathjax>#n_(CH_3COO^(-)) = "0.007621 moles" - "0.001980 moles"#</mathjax> </p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#= "0.005641 moles CH"_3"COO"^(-)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p><mathjax>#n_(CH_3COOH) = "0.004380 moles" + "0.001980 moles"#</mathjax> </p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#= "0.006360 moles CH"_3"COOH"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>The <strong>total volume</strong> of the solution will now be </p>
<blockquote>
<p><mathjax>#V_"total" = "120 mL" + "6.60 mL" = "126.6 mL"#</mathjax></p>
</blockquote>
<p>The concentrations of acetic acid and acetate ions will be </p>
<blockquote>
<p><mathjax>#["CH"_3"COOH"] = "0.006360 moles"/(126.6 * 10^(-3)"L") = "0.05024 M"#</mathjax></p>
<p><mathjax>#["CH"_3"COO"^(-)] = "0.005641 moles"/(126.6 * 10^(-3)"L") = "0.04456 M"#</mathjax></p>
</blockquote>
<p>Use the Henderson-Hasselbalch equation to find the new pH of the solution </p>
<blockquote>
<p><mathjax>#"pH" = 4.76 + log( (0.04456 color(red)(cancel(color(black)("M"))))/(0.05024color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#"pH" = 4.71#</mathjax></p>
</blockquote>
<p>Therefore, the pH of the solution <strong>decreased</strong> by </p>
<blockquote>
<p><mathjax>#Delta_"pH" = |4.71 - 5.00| = color(green)(|bar(ul(color(white)(a/a)"0.29 units"color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A beaker with 120mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.1M. A student adds 6.60mL of a 0.300M HCl solution to the beaker. How much will the pH change? </h1>
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<div class="markdown"><p>The pKa of acetic acid is 4.76.</p></div>
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Stefan V.
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<span class="dateCreated" datetime="2016-05-08T03:24:26" itemprop="dateCreated">
May 8, 2016
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<div class="markdown"><p><mathjax>#Delta_"pH" = 0.29#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>!! LONG ANSWER !!</strong></p>
<p>The idea here is that you need to use the <strong>Henderson-Hasselbalch equation</strong> to determine the <em>ratio</em> that exists between the concentration of the weak acid and of its conjugate base in the <strong><a href="https://socratic.org/chemistry/reactions-in-solution/buffer-theory">buffer solution</a></strong>. </p>
<p>Once you know that, you can use the <strong>total <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the acid and of the conjugate base to find the <em>number of moles</em> of these two chemical species present in the buffer. </p>
<p>So, the <em>Henderson-Hasselbalch equation</em> looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pH" = pK_a + log( (["conjugate base"])/(["weak acid"]))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you have acetic acid, <mathjax>#"CH"_3"COOH"#</mathjax>, as the weak acid and the acetate anion, <mathjax>#"CH"_3"COO"^(-)#</mathjax>, as its conjugate base. The <mathjax>#pK_a#</mathjax> of the acid is said to be equal to <mathjax>#4.76#</mathjax>, which means that you have </p>
<blockquote>
<p><mathjax>#"pH" = 4.76 + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#</mathjax></p>
</blockquote>
<p>The <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> is equal to <mathjax>#5#</mathjax>, and so </p>
<blockquote>
<p><mathjax>#5.00 = 4.76 + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#</mathjax></p>
<p><mathjax>#log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 0.24#</mathjax></p>
</blockquote>
<p>This will be equivalent to </p>
<blockquote>
<p><mathjax>#10^log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 10^(0.24)#</mathjax></p>
</blockquote>
<p>which will give you </p>
<blockquote>
<p><mathjax>#(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = 1.74#</mathjax></p>
</blockquote>
<p>This means that your buffer contains <mathjax>#1.74#</mathjax> <strong>times more</strong> conjugate base than weak acid </p>
<blockquote>
<p><mathjax>#["CH"_3"COO"^(-)] = 1.74 xx ["CH"_3"COOH"]#</mathjax></p>
</blockquote>
<p>Now, because both chemical species share <strong>the same volume</strong>, <mathjax>#"120 mL"#</mathjax>, this can be rewritten as </p>
<blockquote>
<p><mathjax>#n_(CH_3COO^(-))/color(red)(cancel(color(black)(120 * 10^(-3)"L"))) = 1.74 xx n_(CH_3COOH)/color(red)(cancel(color(black)(120 * 10^(-3)"L")))#</mathjax></p>
</blockquote>
<p>which is </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(n_(CH_3COO^(-)) = 1.74 xx n_(CH_3COOH))color(white)(a/a)|)))" " " "color(orange)((1))#</mathjax></p>
</blockquote>
<p>So, the buffer contains <mathjax>#1.74#</mathjax> <strong>times more</strong> <em>moles</em> of acetate anions that of acetic acid. </p>
<p>Now, the <strong>total molarity</strong> of the buffer is said to be equal to <mathjax>#"0.1 M"#</mathjax>. You thus have</p>
<blockquote>
<p><mathjax>#["CH"_3"COOH"] + ["CH"_3"COO"^(-)] = "0.10 M"#</mathjax></p>
</blockquote>
<p>Once again, use the volume of the buffer to write </p>
<blockquote>
<p><mathjax>#n_(CH_3COOH)/(120 * 10^(-3)color(red)(cancel(color(black)("L")))) + n_(CH_3COO^(-))/(120 * 10^(-3)color(red)(cancel(color(black)("L")))) = 0.1"moles"/color(red)(cancel(color(black)("L")))#</mathjax></p>
</blockquote>
<p>This will be equivalent to </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(n_(CH_3COO^(-)) + n_(CH_3COOH) = 0.012)color(white)(a/a)|)))" " " " color(orange)((2))#</mathjax></p>
</blockquote>
<p>Use equations <mathjax>#color(orange)((1))#</mathjax> and <mathjax>#color(orange)((2))#</mathjax> to find how many moles of acetate ions you have in the buffer</p>
<blockquote>
<p><mathjax>#1.74 * n_(CH_3COOH) + n_(CH_3COOH) = 0.012#</mathjax></p>
<p><mathjax>#n_(CH_3COOH) = 0.012/(1.74 + 1) = "0.004380 moles CH"_3"COOH"#</mathjax></p>
</blockquote>
<p>This means that you have </p>
<blockquote>
<p><mathjax>#n_(CH_3COO^(-)) = 1.74 * "0.004380 moles"#</mathjax></p>
<p><mathjax>#n_(CH_3COO^(-)) = "0.007621 moles CH"_3"COO"^(-)#</mathjax></p>
</blockquote>
<p>Now, hydrochloric acid, <mathjax>#"HCl"#</mathjax>, will react with the acetate anions to form acetic acid and chloride anions, <mathjax>#"Cl"^(-)#</mathjax> </p>
<blockquote>
<p><mathjax>#color(red)("H")"Cl"_ ((aq)) + "CH"_ 3"COO"_ ((aq))^(-) -> "CH"_ 3"COO"color(red)("H")_ ((aq)) + "Cl"_((aq))^(-)#</mathjax></p>
</blockquote>
<p>Notice that the reaction consumes hydrochloric acid and acetate ions in a <mathjax>#1:1#</mathjax> <strong><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a></strong>, and produces acetic acid in a <mathjax>#1:1#</mathjax> mole ratio. </p>
<p>Use the <em>molarity</em> and <em>volume</em> of the hydrochloric acid solution to determine how many moles of strong acid you have </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#</mathjax> </p>
</blockquote>
<p>In your case, this gets you </p>
<blockquote>
<p><mathjax>#n_(HCl) = "0.300 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(6.60 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))#</mathjax></p>
<p><mathjax>#n_(HCl) = "0.001980 moles HCl"#</mathjax></p>
</blockquote>
<p>The hydrochloric acid will be <strong>completely consumed</strong> by the reaction, and the resulting solution will contain </p>
<blockquote>
<p><mathjax>#n_(HCl) = "0 moles" ->#</mathjax> <em><strong>completely consumed</strong></em></p>
<p><mathjax>#n_(CH_3COO^(-)) = "0.007621 moles" - "0.001980 moles"#</mathjax> </p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#= "0.005641 moles CH"_3"COO"^(-)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p><mathjax>#n_(CH_3COOH) = "0.004380 moles" + "0.001980 moles"#</mathjax> </p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#= "0.006360 moles CH"_3"COOH"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>The <strong>total volume</strong> of the solution will now be </p>
<blockquote>
<p><mathjax>#V_"total" = "120 mL" + "6.60 mL" = "126.6 mL"#</mathjax></p>
</blockquote>
<p>The concentrations of acetic acid and acetate ions will be </p>
<blockquote>
<p><mathjax>#["CH"_3"COOH"] = "0.006360 moles"/(126.6 * 10^(-3)"L") = "0.05024 M"#</mathjax></p>
<p><mathjax>#["CH"_3"COO"^(-)] = "0.005641 moles"/(126.6 * 10^(-3)"L") = "0.04456 M"#</mathjax></p>
</blockquote>
<p>Use the Henderson-Hasselbalch equation to find the new pH of the solution </p>
<blockquote>
<p><mathjax>#"pH" = 4.76 + log( (0.04456 color(red)(cancel(color(black)("M"))))/(0.05024color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#"pH" = 4.71#</mathjax></p>
</blockquote>
<p>Therefore, the pH of the solution <strong>decreased</strong> by </p>
<blockquote>
<p><mathjax>#Delta_"pH" = |4.71 - 5.00| = color(green)(|bar(ul(color(white)(a/a)"0.29 units"color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</article> | A beaker with 120mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.1M. A student adds 6.60mL of a 0.300M HCl solution to the beaker. How much will the pH change? |
The pKa of acetic acid is 4.76.
|
506 | a9d7fe2c-6ddd-11ea-8066-ccda262736ce | https://socratic.org/questions/how-much-energy-in-joules-must-be-absorbed-if-all-the-o-3-molecules-in-the-sampl | 8.85 × 10^(-3) joules | start physical_unit 11 12 heat_energy j qc_end physical_unit 38 38 40 41 wavelength qc_end c_other OTHER qc_end chemical_equation 58 64 qc_end physical_unit 15 17 66 67 volume qc_end physical_unit 15 17 72 73 temperature qc_end physical_unit 15 17 75 76 pressure qc_end end | [{"type":"physical unit","value":"Absorbed energy [OF] O3 molecules [IN] joules"}] | [{"type":"physical unit","value":"8.85 × 10^(-3) joules"}] | [{"type":"physical unit","value":"Wavelength [OF] radiation [=] \\pu{254 nm}"},{"type":"other","value":"Each photon absorbed causes one O3 molecule to dissociate."},{"type":"chemical equation","value":"O3 + hν -> O2 + O"},{"type":"physical unit","value":"Volume [OF] air sample [=] \\pu{1.85 L}"},{"type":"physical unit","value":"Temperature [OF] air sample [=] \\pu{22 ℃ }"},{"type":"physical unit","value":"Pressure [OF] air sample [=] \\pu{748 mmHg}"},{"type":"physical unit","value":"Molarity [OF] O3 in the air samole [=] \\pu{0.25 ppm}"}] | <h1 class="questionTitle" itemprop="name">How much energy, in joules, must be absorbed if all the #O_3# molecules in the sample of air are to dissociate?
Assume that each photon absorbed causes one #O_3# molecule to dissociate, and that the wavelength of the radiation is 254 nm.</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><em>Information provided with question</em></p>
<p>Ozone, <mathjax>#O_3#</mathjax>, absorbs ultraviolet radiation and dissociates into <mathjax>#O_2#</mathjax> molecules and <mathjax>#O#</mathjax> atoms:<br/>
<mathjax>#O_3+h\nu→O_2+O.#</mathjax><br/>
A 1.85-<mathjax>#L#</mathjax> sample of air at 22<mathjax>#°C#</mathjax> and 748 <mathjax>#mmHg#</mathjax> contains 0.25 ppm of <mathjax>#O_3#</mathjax>.</p>
<p><em>My thoughts</em></p>
<p>I assume "parts per million" (ppm) is "<mathjax>#(mL)/L#</mathjax>"?<br/>
Also, I don't really know what section I should be posting this in, sorry.</p></div>
</h2>
</div>
</div> | 8.85 × 10^(-3) joules | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a very difficult problem. What makes it difficult is the number of steps involved in getting to the answer.</p>
<p>The outline of how we are going to solve the problem is this:<br/>
1) figure out how much energy, in Joules, a single photon of 254 nm light has.<br/>
2) figure out how many ozone molecules you have in the container. This is even more complicated because "ppm" can refer to mg/kg or <mathjax>#mu#</mathjax>l/l. Fortunately, the convention for gas is <mathjax>#mu#</mathjax>l/l, so we will go with that.<br/>
3) multiply the two together to get the answer.</p>
<p>So, step 1: we will use E = hc/ <mathjax>#lambda#</mathjax> where E is in Joules, h is Plank's constant (<mathjax>#6.62607 xx 10^(-34) J*s#</mathjax>), c is the speed of light in meters/sec (299,792,458 m/sec), and <mathjax>#lambda#</mathjax> is 254 nm (<mathjax>#254 xx 10^(-9)#</mathjax> meters). Each photon thus has an energy of <mathjax>#7.82065 xx 10^(-19)#</mathjax> J.</p>
<p>Reality check: is this number reasonable?<br/>
(Yes! A Joule is a significant amount of heat--enough that you can feel it on your skin. But one photon has a very small amount of energy. so a very small number is what we are expecting for this step of the problem.)</p>
<p>Step 2: We will rearrange the ideal gas equation to determine how many moles of gas are present. </p>
<p><mathjax>#m_(gas) = (P_(atm) * V_l) / (R_((l atm)/(mol K)) * T_K)#</mathjax> </p>
<p>We will then take 0.25 ppm(vol) of that for the number of moles of ozone present.</p>
<p><mathjax>#m_(gas) = (((748 mmHg)/(760 mmHg))*1.85 liters)/(0.082057_((l atm)/(mol K)) * (22+273.15)K) = 7.5180 xx 10^(-2) m#</mathjax> </p>
<p>Reality check: we know at STP that 1 mole of gas occupies 22.414 liters. We have calculated a little less than 1/10 of a mole here, and that fits since the volume is a little less than 2 liters and the pressure is a little less than 1 atm. </p>
<p>So far, so good!</p>
<p>So how many molecules of ozone do we have?</p>
<p><mathjax>#"molecules " O_3 = 6.022 xx 10^(23) " molecules " O_3/("mole " O_3) xx 7.5180 xx 10^(-2)" mole " O_3 xx 0.25 xx 10^(-6)"(ppm of " O_3 " in the sample) " = 1.1318 xx 10^(16) " Ozone molecules"#</mathjax></p>
<p>Now to multiply the part 1 answer by the part 2 answer:</p>
<p><mathjax>#7.82065 xx 10^(-19)" J/photon" xx " 1 photon"/" Ozone molecule" xx 1.1318 xx 10^(16) " Ozone molecules" = 8.85166 xx 10^(-3) " joules"#</mathjax></p>
<p>For this question, the limiting number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> is 2 from the 0.25 ppm ozone concentration. Thus the final answer is</p>
<p>8.9 mJ.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>8.9 mJ</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a very difficult problem. What makes it difficult is the number of steps involved in getting to the answer.</p>
<p>The outline of how we are going to solve the problem is this:<br/>
1) figure out how much energy, in Joules, a single photon of 254 nm light has.<br/>
2) figure out how many ozone molecules you have in the container. This is even more complicated because "ppm" can refer to mg/kg or <mathjax>#mu#</mathjax>l/l. Fortunately, the convention for gas is <mathjax>#mu#</mathjax>l/l, so we will go with that.<br/>
3) multiply the two together to get the answer.</p>
<p>So, step 1: we will use E = hc/ <mathjax>#lambda#</mathjax> where E is in Joules, h is Plank's constant (<mathjax>#6.62607 xx 10^(-34) J*s#</mathjax>), c is the speed of light in meters/sec (299,792,458 m/sec), and <mathjax>#lambda#</mathjax> is 254 nm (<mathjax>#254 xx 10^(-9)#</mathjax> meters). Each photon thus has an energy of <mathjax>#7.82065 xx 10^(-19)#</mathjax> J.</p>
<p>Reality check: is this number reasonable?<br/>
(Yes! A Joule is a significant amount of heat--enough that you can feel it on your skin. But one photon has a very small amount of energy. so a very small number is what we are expecting for this step of the problem.)</p>
<p>Step 2: We will rearrange the ideal gas equation to determine how many moles of gas are present. </p>
<p><mathjax>#m_(gas) = (P_(atm) * V_l) / (R_((l atm)/(mol K)) * T_K)#</mathjax> </p>
<p>We will then take 0.25 ppm(vol) of that for the number of moles of ozone present.</p>
<p><mathjax>#m_(gas) = (((748 mmHg)/(760 mmHg))*1.85 liters)/(0.082057_((l atm)/(mol K)) * (22+273.15)K) = 7.5180 xx 10^(-2) m#</mathjax> </p>
<p>Reality check: we know at STP that 1 mole of gas occupies 22.414 liters. We have calculated a little less than 1/10 of a mole here, and that fits since the volume is a little less than 2 liters and the pressure is a little less than 1 atm. </p>
<p>So far, so good!</p>
<p>So how many molecules of ozone do we have?</p>
<p><mathjax>#"molecules " O_3 = 6.022 xx 10^(23) " molecules " O_3/("mole " O_3) xx 7.5180 xx 10^(-2)" mole " O_3 xx 0.25 xx 10^(-6)"(ppm of " O_3 " in the sample) " = 1.1318 xx 10^(16) " Ozone molecules"#</mathjax></p>
<p>Now to multiply the part 1 answer by the part 2 answer:</p>
<p><mathjax>#7.82065 xx 10^(-19)" J/photon" xx " 1 photon"/" Ozone molecule" xx 1.1318 xx 10^(16) " Ozone molecules" = 8.85166 xx 10^(-3) " joules"#</mathjax></p>
<p>For this question, the limiting number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> is 2 from the 0.25 ppm ozone concentration. Thus the final answer is</p>
<p>8.9 mJ.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How much energy, in joules, must be absorbed if all the #O_3# molecules in the sample of air are to dissociate?
Assume that each photon absorbed causes one #O_3# molecule to dissociate, and that the wavelength of the radiation is 254 nm.</h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><em>Information provided with question</em></p>
<p>Ozone, <mathjax>#O_3#</mathjax>, absorbs ultraviolet radiation and dissociates into <mathjax>#O_2#</mathjax> molecules and <mathjax>#O#</mathjax> atoms:<br/>
<mathjax>#O_3+h\nu→O_2+O.#</mathjax><br/>
A 1.85-<mathjax>#L#</mathjax> sample of air at 22<mathjax>#°C#</mathjax> and 748 <mathjax>#mmHg#</mathjax> contains 0.25 ppm of <mathjax>#O_3#</mathjax>.</p>
<p><em>My thoughts</em></p>
<p>I assume "parts per million" (ppm) is "<mathjax>#(mL)/L#</mathjax>"?<br/>
Also, I don't really know what section I should be posting this in, sorry.</p></div>
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Andy Wolff
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<span class="dateCreated" datetime="2016-11-20T03:57:58" itemprop="dateCreated">
Nov 20, 2016
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<div class="markdown"><p>8.9 mJ</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a very difficult problem. What makes it difficult is the number of steps involved in getting to the answer.</p>
<p>The outline of how we are going to solve the problem is this:<br/>
1) figure out how much energy, in Joules, a single photon of 254 nm light has.<br/>
2) figure out how many ozone molecules you have in the container. This is even more complicated because "ppm" can refer to mg/kg or <mathjax>#mu#</mathjax>l/l. Fortunately, the convention for gas is <mathjax>#mu#</mathjax>l/l, so we will go with that.<br/>
3) multiply the two together to get the answer.</p>
<p>So, step 1: we will use E = hc/ <mathjax>#lambda#</mathjax> where E is in Joules, h is Plank's constant (<mathjax>#6.62607 xx 10^(-34) J*s#</mathjax>), c is the speed of light in meters/sec (299,792,458 m/sec), and <mathjax>#lambda#</mathjax> is 254 nm (<mathjax>#254 xx 10^(-9)#</mathjax> meters). Each photon thus has an energy of <mathjax>#7.82065 xx 10^(-19)#</mathjax> J.</p>
<p>Reality check: is this number reasonable?<br/>
(Yes! A Joule is a significant amount of heat--enough that you can feel it on your skin. But one photon has a very small amount of energy. so a very small number is what we are expecting for this step of the problem.)</p>
<p>Step 2: We will rearrange the ideal gas equation to determine how many moles of gas are present. </p>
<p><mathjax>#m_(gas) = (P_(atm) * V_l) / (R_((l atm)/(mol K)) * T_K)#</mathjax> </p>
<p>We will then take 0.25 ppm(vol) of that for the number of moles of ozone present.</p>
<p><mathjax>#m_(gas) = (((748 mmHg)/(760 mmHg))*1.85 liters)/(0.082057_((l atm)/(mol K)) * (22+273.15)K) = 7.5180 xx 10^(-2) m#</mathjax> </p>
<p>Reality check: we know at STP that 1 mole of gas occupies 22.414 liters. We have calculated a little less than 1/10 of a mole here, and that fits since the volume is a little less than 2 liters and the pressure is a little less than 1 atm. </p>
<p>So far, so good!</p>
<p>So how many molecules of ozone do we have?</p>
<p><mathjax>#"molecules " O_3 = 6.022 xx 10^(23) " molecules " O_3/("mole " O_3) xx 7.5180 xx 10^(-2)" mole " O_3 xx 0.25 xx 10^(-6)"(ppm of " O_3 " in the sample) " = 1.1318 xx 10^(16) " Ozone molecules"#</mathjax></p>
<p>Now to multiply the part 1 answer by the part 2 answer:</p>
<p><mathjax>#7.82065 xx 10^(-19)" J/photon" xx " 1 photon"/" Ozone molecule" xx 1.1318 xx 10^(16) " Ozone molecules" = 8.85166 xx 10^(-3) " joules"#</mathjax></p>
<p>For this question, the limiting number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> is 2 from the 0.25 ppm ozone concentration. Thus the final answer is</p>
<p>8.9 mJ.</p></div>
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Ernest Z.
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Andy Wolff
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<span class="dateCreated" datetime="2016-11-29T00:12:44" itemprop="dateCreated">
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<div class="markdown"><p><strong>WARNING! Long answer!</strong> The energy absorbed is 8.9 mJ.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1: Calculate the number of molecules in the container.</strong></p>
<p>The <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> 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 the formula to get</p>
<blockquote>
<blockquote>
<p><mathjax>#n = (PV)/(RT)#</mathjax></p>
</blockquote>
</blockquote>
<p>In this problem,</p>
<p><mathjax>#P = 748 color(red)(cancel(color(black)("mmHg"))) × ("1 atm")/(760 color(red)(cancel(color(black)("mmHg")))) = "0.9842 atm"#</mathjax></p>
<p><mathjax>#V = "1.85 L"#</mathjax></p>
<p><mathjax>#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#</mathjax></p>
<p><mathjax>#T = "22 °C" = "295.15 K"#</mathjax></p>
<p><mathjax>#n = (0.9842 color(red)(cancel(color(black)("atm"))) × 1.85 color(red)(cancel(color(black)("L"))))/( "0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 295.15 color(red)(cancel(color(black)("K")))) = "0.075 18 mol"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the number of gas molecules</strong></p>
<p><mathjax>#"Number of gas molecules" = "0.075 18" color(red)(cancel(color(black)("mol"))) × (6.022 × 10^23 color(white)(l)"molecules")/(1 color(red)(cancel(color(black)("mol"))))#</mathjax></p>
<p><mathjax>#= 4.527 × 10^22 color(white)(l)"molecules"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the number of <mathjax>#"O"_3#</mathjax> molecules.</strong></p>
<p><mathjax>#"Number of"color(white)(l) "O"_3 color(white)(l)"molecules" = 4.527 × 10^22 color(red)(cancel(color(black)("gas molecules"))) × ("0.25 O"_3 color(white)(l)"molecule")/(10^6 color(red)(cancel(color(black)("gas molecules")))) = 1.13 × 10^16 color(white)(l)"O"_3 color(white)(l)"molecules"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4. Calculate the energy absorbed by 1 <mathjax>#"O"_3#</mathjax> molecule.</strong></p>
<p>The formula for the energy <mathjax>#E#</mathjax> of a quantum is</p>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) E = (hc)/λ color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
<p>where</p>
<p><mathjax>#h#</mathjax> = <a href="https://socratic.org/chemistry/the-bohr-model-of-the-atom/calculations-with-plancks-constant-and-frequency">Planck's constant</a><br/>
<mathjax>#c#</mathjax> = the speed of light<br/>
<mathjax>#λ#</mathjax> = the wavelength of the light</p>
<p><mathjax>#E = (6.626 × 10^"-34" "J"·color(red)(cancel(color(black)("s"))) × 2.998 × 10^8 color(red)(cancel(color(black)("m·s"^"-1"))))/(254 × 10^"-9" color(red)(cancel(color(black)("m")))) = 7.821 × 10^"-19"color(white)(l) "J"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 5. Calculate the energy absorbed by all the <mathjax>#"O"_3#</mathjax> molecules</strong></p>
<p><mathjax>#"Total energy" = 1.13 × 10^16 color(red)(cancel(color(black)("O"_3color(white)(l) "molecules"))) × (7.821 × 10^"-19"color(white)(l) "J")/(1 color(red)(cancel(color(black)("O"_3color(white)(l) "molecule")))) = 8.9 × 10^"-3"color(white)(l) "J" = "8.9 mJ"#</mathjax></p></div>
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</article> | How much energy, in joules, must be absorbed if all the #O_3# molecules in the sample of air are to dissociate?
Assume that each photon absorbed causes one #O_3# molecule to dissociate, and that the wavelength of the radiation is 254 nm. |
Information provided with question
Ozone, #O_3#, absorbs ultraviolet radiation and dissociates into #O_2# molecules and #O# atoms:
#O_3+h\nu→O_2+O.#
A 1.85-#L# sample of air at 22#°C# and 748 #mmHg# contains 0.25 ppm of #O_3#.
My thoughts
I assume "parts per million" (ppm) is "#(mL)/L#"?
Also, I don't really know what section I should be posting this in, sorry.
|
507 | ac34d426-6ddd-11ea-9552-ccda262736ce | https://socratic.org/questions/how-many-calories-are-required-to-raise-the-temperature-of-a-35-0-g-sample-of-ir | 37.8 cal | start physical_unit 13 15 energy cal qc_end physical_unit 13 15 11 12 mass qc_end physical_unit 13 15 17 18 temperature qc_end physical_unit 13 15 20 21 temperature qc_end physical_unit 15 15 28 31 specific_heat qc_end end | [{"type":"physical unit","value":"Required energy [OF] iron sample [IN] cal"}] | [{"type":"physical unit","value":"37.8 cal"}] | [{"type":"physical unit","value":"Mass [OF] iron sample [=] \\pu{35.0 g}"},{"type":"physical unit","value":"Temperature1 [OF] iron sample [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] iron sample [=] \\pu{35 ℃}"},{"type":"physical unit","value":"Specific heat [OF] iron [=] \\pu{0.108 cal/(g * ℃)}"}] | <h1 class="questionTitle" itemprop="name">How many calories are required to raise the temperature of a 35.0 g sample of iron from 25°C to 35°C? Iron has a specific heat of 0.108 cal/g°C?</h1> | null | 37.8 cal | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key to this problem lies in the value of the <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong> of iron. </p>
<blockquote>
<p><mathjax>#c_"iron" = "0.108 cal"#</mathjax> <mathjax>#color(blue)("g"^(-1))color(darkorange)(""^@"C"^(-1))#</mathjax></p>
</blockquote>
<p>This tells you that in order to increase the temperature of <em>one unit of mass</em> of iron, i.e. of <mathjax>#color(blue)("1 g")#</mathjax> of iron, by <em>one unit of temperature</em>, i.e. by <mathjax>#color(darkorange)(1^@"C")#</mathjax>, you need to provide it with <mathjax>#"0.108 cal"#</mathjax>. </p>
<p>Now, you can use the specific heat of iron to figure out the amount of heat needed to increase the temperature of <mathjax>#"35.0 g"#</mathjax> of iron </p>
<blockquote>
<p><mathjax>#35.0 color(red)(cancel(color(black)("g"))) * "0.108 cal"/(color(blue)(1)color(red)(cancel(color(blue)("g"))) * color(darkorange)(1^@"C")) = "3.78 cal"#</mathjax> <mathjax>#color(darkorange)(""^@"C"^(-1))#</mathjax></p>
</blockquote>
<p>This tells you that in order to increase the temperature of <mathjax>#"35.0 g"#</mathjax> of iron by <mathjax>#color(darkorange)(1^@"C")#</mathjax>, you need to provide it with <mathjax>#"3.78 cal"#</mathjax> of heat. </p>
<p>In your case, the temperature of the iron must increase by </p>
<blockquote>
<p><mathjax>#35^@"C" - 25^@"C" = 10^@"C"#</mathjax></p>
</blockquote>
<p>which means that you will need</p>
<blockquote>
<p><mathjax>#10color(red)(cancel(color(black)(""^@"C"))) * "3.78 cal"/(color(darkorange)(1)color(red)(cancel(color(darkorange)(""^@"C")))) = "37.8 cal"#</mathjax></p>
</blockquote>
<p>Now, you should round the answer to <em>one significant figure</em>, the number of <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> you have for the <strong>change in temperature</strong>, i.e. for <mathjax>#10^@"C"#</mathjax>, but I'll leave it rounded to two sig figs</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("heat needed = 38 cal")))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"38 cal"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key to this problem lies in the value of the <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong> of iron. </p>
<blockquote>
<p><mathjax>#c_"iron" = "0.108 cal"#</mathjax> <mathjax>#color(blue)("g"^(-1))color(darkorange)(""^@"C"^(-1))#</mathjax></p>
</blockquote>
<p>This tells you that in order to increase the temperature of <em>one unit of mass</em> of iron, i.e. of <mathjax>#color(blue)("1 g")#</mathjax> of iron, by <em>one unit of temperature</em>, i.e. by <mathjax>#color(darkorange)(1^@"C")#</mathjax>, you need to provide it with <mathjax>#"0.108 cal"#</mathjax>. </p>
<p>Now, you can use the specific heat of iron to figure out the amount of heat needed to increase the temperature of <mathjax>#"35.0 g"#</mathjax> of iron </p>
<blockquote>
<p><mathjax>#35.0 color(red)(cancel(color(black)("g"))) * "0.108 cal"/(color(blue)(1)color(red)(cancel(color(blue)("g"))) * color(darkorange)(1^@"C")) = "3.78 cal"#</mathjax> <mathjax>#color(darkorange)(""^@"C"^(-1))#</mathjax></p>
</blockquote>
<p>This tells you that in order to increase the temperature of <mathjax>#"35.0 g"#</mathjax> of iron by <mathjax>#color(darkorange)(1^@"C")#</mathjax>, you need to provide it with <mathjax>#"3.78 cal"#</mathjax> of heat. </p>
<p>In your case, the temperature of the iron must increase by </p>
<blockquote>
<p><mathjax>#35^@"C" - 25^@"C" = 10^@"C"#</mathjax></p>
</blockquote>
<p>which means that you will need</p>
<blockquote>
<p><mathjax>#10color(red)(cancel(color(black)(""^@"C"))) * "3.78 cal"/(color(darkorange)(1)color(red)(cancel(color(darkorange)(""^@"C")))) = "37.8 cal"#</mathjax></p>
</blockquote>
<p>Now, you should round the answer to <em>one significant figure</em>, the number of <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> you have for the <strong>change in temperature</strong>, i.e. for <mathjax>#10^@"C"#</mathjax>, but I'll leave it rounded to two sig figs</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("heat needed = 38 cal")))#</mathjax></p>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">How many calories are required to raise the temperature of a 35.0 g sample of iron from 25°C to 35°C? Iron has a specific heat of 0.108 cal/g°C?</h1>
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<div class="markdown"><p><mathjax>#"38 cal"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key to this problem lies in the value of the <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong> of iron. </p>
<blockquote>
<p><mathjax>#c_"iron" = "0.108 cal"#</mathjax> <mathjax>#color(blue)("g"^(-1))color(darkorange)(""^@"C"^(-1))#</mathjax></p>
</blockquote>
<p>This tells you that in order to increase the temperature of <em>one unit of mass</em> of iron, i.e. of <mathjax>#color(blue)("1 g")#</mathjax> of iron, by <em>one unit of temperature</em>, i.e. by <mathjax>#color(darkorange)(1^@"C")#</mathjax>, you need to provide it with <mathjax>#"0.108 cal"#</mathjax>. </p>
<p>Now, you can use the specific heat of iron to figure out the amount of heat needed to increase the temperature of <mathjax>#"35.0 g"#</mathjax> of iron </p>
<blockquote>
<p><mathjax>#35.0 color(red)(cancel(color(black)("g"))) * "0.108 cal"/(color(blue)(1)color(red)(cancel(color(blue)("g"))) * color(darkorange)(1^@"C")) = "3.78 cal"#</mathjax> <mathjax>#color(darkorange)(""^@"C"^(-1))#</mathjax></p>
</blockquote>
<p>This tells you that in order to increase the temperature of <mathjax>#"35.0 g"#</mathjax> of iron by <mathjax>#color(darkorange)(1^@"C")#</mathjax>, you need to provide it with <mathjax>#"3.78 cal"#</mathjax> of heat. </p>
<p>In your case, the temperature of the iron must increase by </p>
<blockquote>
<p><mathjax>#35^@"C" - 25^@"C" = 10^@"C"#</mathjax></p>
</blockquote>
<p>which means that you will need</p>
<blockquote>
<p><mathjax>#10color(red)(cancel(color(black)(""^@"C"))) * "3.78 cal"/(color(darkorange)(1)color(red)(cancel(color(darkorange)(""^@"C")))) = "37.8 cal"#</mathjax></p>
</blockquote>
<p>Now, you should round the answer to <em>one significant figure</em>, the number of <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> you have for the <strong>change in temperature</strong>, i.e. for <mathjax>#10^@"C"#</mathjax>, but I'll leave it rounded to two sig figs</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("heat needed = 38 cal")))#</mathjax></p>
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</article> | How many calories are required to raise the temperature of a 35.0 g sample of iron from 25°C to 35°C? Iron has a specific heat of 0.108 cal/g°C? | null |
508 | ab1923e9-6ddd-11ea-a159-ccda262736ce | https://socratic.org/questions/a-solution-contains-oh-4-0-times-10-5-m-what-is-the-concentration-of-h-3o | 2.50 × 10^(-10) M | start physical_unit 14 14 concentration mol/l qc_end physical_unit 1 1 5 8 [oh-] qc_end end | [{"type":"physical unit","value":"Concentration [OF] [H3O+] [IN] M"}] | [{"type":"physical unit","value":"2.50 × 10^(-10) M"}] | [{"type":"physical unit","value":"[OH-] [OF] the solution [=] \\pu{4.0 × 10^(-5) M}"}] | <h1 class="questionTitle" itemprop="name">A solution contains #[OH^-] = 4.0 times 10^-5# #M#, what is the concentration of #[H_3O^+]#?</h1> | null | 2.50 × 10^(-10) M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In aqueous solution, it is a fact that the following equilibrium operates under standard conditions.........</p>
<p><mathjax>#2H_2O(l) rightleftharpoons H_3O^(+) +HO^(-)#</mathjax></p>
<p>Under standard conditions............</p>
<p><mathjax>#K_"rxn"=K_w=[H_3O^+][HO^-]=10^-14#</mathjax>. And typically we would use logarithms to reduce this expression to.......</p>
<p><mathjax>#pH+pOH=14#</mathjax>, where the <mathjax>#pH#</mathjax> function means <mathjax>#-log_10[H_3O^+]#</mathjax> etc. </p>
<p>So <mathjax>#pOH=-log_10[4.0xx10^-5]=4.40#</mathjax>, and <mathjax>#pH=9.60#</mathjax>.</p>
<p>And thus <mathjax>#[H_3O^+]=10^(-9.60)=2.50xx10^-10*mol*L^-1.#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#[H_3O^+]=2.50xx10^-10*mol*L^-1#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In aqueous solution, it is a fact that the following equilibrium operates under standard conditions.........</p>
<p><mathjax>#2H_2O(l) rightleftharpoons H_3O^(+) +HO^(-)#</mathjax></p>
<p>Under standard conditions............</p>
<p><mathjax>#K_"rxn"=K_w=[H_3O^+][HO^-]=10^-14#</mathjax>. And typically we would use logarithms to reduce this expression to.......</p>
<p><mathjax>#pH+pOH=14#</mathjax>, where the <mathjax>#pH#</mathjax> function means <mathjax>#-log_10[H_3O^+]#</mathjax> etc. </p>
<p>So <mathjax>#pOH=-log_10[4.0xx10^-5]=4.40#</mathjax>, and <mathjax>#pH=9.60#</mathjax>.</p>
<p>And thus <mathjax>#[H_3O^+]=10^(-9.60)=2.50xx10^-10*mol*L^-1.#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A solution contains #[OH^-] = 4.0 times 10^-5# #M#, what is the concentration of #[H_3O^+]#?</h1>
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anor277
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<div class="markdown"><p><mathjax>#[H_3O^+]=2.50xx10^-10*mol*L^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In aqueous solution, it is a fact that the following equilibrium operates under standard conditions.........</p>
<p><mathjax>#2H_2O(l) rightleftharpoons H_3O^(+) +HO^(-)#</mathjax></p>
<p>Under standard conditions............</p>
<p><mathjax>#K_"rxn"=K_w=[H_3O^+][HO^-]=10^-14#</mathjax>. And typically we would use logarithms to reduce this expression to.......</p>
<p><mathjax>#pH+pOH=14#</mathjax>, where the <mathjax>#pH#</mathjax> function means <mathjax>#-log_10[H_3O^+]#</mathjax> etc. </p>
<p>So <mathjax>#pOH=-log_10[4.0xx10^-5]=4.40#</mathjax>, and <mathjax>#pH=9.60#</mathjax>.</p>
<p>And thus <mathjax>#[H_3O^+]=10^(-9.60)=2.50xx10^-10*mol*L^-1.#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#[H_3O^+]=2.5*10^(-10) M#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Given that no hydrolysis happened in the solution (inferred from question) Calculate <mathjax>#[H_3O^+]#</mathjax> from <mathjax>#k_w#</mathjax> directly.<br/>
<mathjax>#[H_3O^+]#</mathjax><br/>
<mathjax>#=k_w/([OH^-])#</mathjax><br/>
<mathjax>#=10^-14/(4.0*10^-5)#</mathjax><br/>
<mathjax>#=2.5*10^-10M#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#2.5 xx 10 ^-10#</mathjax> <mathjax>#"M"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><img alt="My own work (Mert METİN)" src="https://useruploads.socratic.org/010qHTHRUiX2eGsLU1gu_photo5852947402972047700.jpg"/> </p></div>
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</article> | A solution contains #[OH^-] = 4.0 times 10^-5# #M#, what is the concentration of #[H_3O^+]#? | null |
509 | ad0ef3e6-6ddd-11ea-a9be-ccda262736ce | https://socratic.org/questions/how-do-i-calculate-the-molarity-of-a-solution-prepared-by-diluting-87-44-ml-of-0 | 0.43 M | start physical_unit 17 18 molarity mol/l qc_end physical_unit 17 18 15 16 molarity qc_end physical_unit 8 8 12 13 volume qc_end physical_unit 8 8 20 21 volume qc_end end | [{"type":"physical unit","value":"Molarity2 [OF] potassium chloride solution [IN] M"}] | [{"type":"physical unit","value":"0.43 M"}] | [{"type":"physical unit","value":"Molarity1 [OF] potassium chloride solution [=] \\pu{0.743 M}"},{"type":"physical unit","value":"Volume1 [OF] potassium chloride solution [=] \\pu{87.44 mL}"},{"type":"physical unit","value":"Volume2 [OF] potassium chloride solution [=] \\pu{150.00 mL}"}] | <h1 class="questionTitle" itemprop="name">How do I calculate the molarity of a solution prepared by diluting 87.44 mL of 0.743 M potassium chloride to 150.00 mL?</h1> | null | 0.43 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And we have initially, <mathjax>#87.44xx10^-3*Lxx0.743*mol*L^-1=0.0650*mol#</mathjax>, and we divide this molar quantity by the NEW volume to get the NEW concentration....</p>
<p><mathjax>#(87.44xx10^-3*Lxx0.743*mol*L^-1)/(150*mLxx10^-3*L*mL^-1)=??*mol*L^-1.#</mathjax></p>
<p>Alternatively, we could use the old expression.......</p>
<p><mathjax>#C_1V_1=C_2V_2#</mathjax>, i.e. </p>
<p><mathjax>#C_2=(C_1V_1)/(V_2)=(0.743*mol*L^(-1)xx87.44*cancel(mL))/(150.00*cancel(mL))#</mathjax></p>
<p>Which of course is the same quotient.........which explicity gives an answer with units of <mathjax>#mol*L^-1#</mathjax>, as required...........</p></div>
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<div class="markdown"><p><mathjax>#"Concentration"="Moles of solute"/"Volume of solution"#</mathjax></p>
<p>Here, <mathjax>#"concentration"~=0.4*mol*L^-1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And we have initially, <mathjax>#87.44xx10^-3*Lxx0.743*mol*L^-1=0.0650*mol#</mathjax>, and we divide this molar quantity by the NEW volume to get the NEW concentration....</p>
<p><mathjax>#(87.44xx10^-3*Lxx0.743*mol*L^-1)/(150*mLxx10^-3*L*mL^-1)=??*mol*L^-1.#</mathjax></p>
<p>Alternatively, we could use the old expression.......</p>
<p><mathjax>#C_1V_1=C_2V_2#</mathjax>, i.e. </p>
<p><mathjax>#C_2=(C_1V_1)/(V_2)=(0.743*mol*L^(-1)xx87.44*cancel(mL))/(150.00*cancel(mL))#</mathjax></p>
<p>Which of course is the same quotient.........which explicity gives an answer with units of <mathjax>#mol*L^-1#</mathjax>, as required...........</p></div>
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<h1 class="questionTitle" itemprop="name">How do I calculate the molarity of a solution prepared by diluting 87.44 mL of 0.743 M potassium chloride to 150.00 mL?</h1>
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<div class="markdown"><p><mathjax>#"Concentration"="Moles of solute"/"Volume of solution"#</mathjax></p>
<p>Here, <mathjax>#"concentration"~=0.4*mol*L^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And we have initially, <mathjax>#87.44xx10^-3*Lxx0.743*mol*L^-1=0.0650*mol#</mathjax>, and we divide this molar quantity by the NEW volume to get the NEW concentration....</p>
<p><mathjax>#(87.44xx10^-3*Lxx0.743*mol*L^-1)/(150*mLxx10^-3*L*mL^-1)=??*mol*L^-1.#</mathjax></p>
<p>Alternatively, we could use the old expression.......</p>
<p><mathjax>#C_1V_1=C_2V_2#</mathjax>, i.e. </p>
<p><mathjax>#C_2=(C_1V_1)/(V_2)=(0.743*mol*L^(-1)xx87.44*cancel(mL))/(150.00*cancel(mL))#</mathjax></p>
<p>Which of course is the same quotient.........which explicity gives an answer with units of <mathjax>#mol*L^-1#</mathjax>, as required...........</p></div>
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</article> | How do I calculate the molarity of a solution prepared by diluting 87.44 mL of 0.743 M potassium chloride to 150.00 mL? | null |
510 | acb24938-6ddd-11ea-b6f1-ccda262736ce | https://socratic.org/questions/57c7758711ef6b056dca6307 | 0.04 kg | start physical_unit 4 5 mass kg qc_end physical_unit 4 5 8 9 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] C atom [IN] kg"}] | [{"type":"physical unit","value":"0.04 kg"}] | [{"type":"physical unit","value":"Mole [OF] C atom [=] \\pu{3.6 mols}"}] | <h1 class="questionTitle" itemprop="name">How many #"kg"# of #"C"# atom are in #"3.6 mols"# of #"C"# atom?</h1> | null | 0.04 kg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>3.6 mol C * <mathjax>#(12.01 g C)/(1 mol C)#</mathjax> = 43.236 g C</p>
<p>43.236 g C * <mathjax>#(1 kg)/(1000 g)#</mathjax> = 0.043 kg C</p>
<p>You only need the 6.022 when converting to molecules.</p></div>
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<div class="markdown"><p>0.043 kg C.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>3.6 mol C * <mathjax>#(12.01 g C)/(1 mol C)#</mathjax> = 43.236 g C</p>
<p>43.236 g C * <mathjax>#(1 kg)/(1000 g)#</mathjax> = 0.043 kg C</p>
<p>You only need the 6.022 when converting to molecules.</p></div>
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<h1 class="questionTitle" itemprop="name">How many #"kg"# of #"C"# atom are in #"3.6 mols"# of #"C"# atom?</h1>
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<div class="markdown"><p>0.043 kg C.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>3.6 mol C * <mathjax>#(12.01 g C)/(1 mol C)#</mathjax> = 43.236 g C</p>
<p>43.236 g C * <mathjax>#(1 kg)/(1000 g)#</mathjax> = 0.043 kg C</p>
<p>You only need the 6.022 when converting to molecules.</p></div>
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</article> | How many #"kg"# of #"C"# atom are in #"3.6 mols"# of #"C"# atom? | null |
511 | ac799e8a-6ddd-11ea-a20b-ccda262736ce | https://socratic.org/questions/the-ph-at-one-half-the-equivalence-point-in-an-acid-base-titration-was-found-to- | 2.1 × 10^(-6) | start physical_unit 23 25 ka none qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Value of Ka [OF] this unknown acid"}] | [{"type":"physical unit","value":"2.1 × 10^(-6)"}] | [{"type":"physical unit","value":"pH [OF] the solution [=] \\pu{5.67}"},{"type":"other","value":"One-half the equivalence point in an acid-base titration."}] | <h1 class="questionTitle" itemprop="name">The #"pH"# at one-half the equivalence point in an acid-base titration was found to be #5.67#. What is the value of #K_a# for this unknown acid?</h1> | null | 2.1 × 10^(-6) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that at the <strong>half equivalence point</strong>, the <mathjax>#"pH"#</mathjax> of the solution will be equal to the <mathjax>#"p"K_a#</mathjax> of the weak acid. </p>
<p>Assuming that you're titrating a weak monoprotic acid <mathjax>#"HA"#</mathjax> with a strong base that I'll represent as <mathjax>#"OH"^(-)#</mathjax>, you know that at the <strong>equivalence point</strong>, the strong base will completely neutralize the weak acid.</p>
<blockquote>
<p><mathjax>#"HA"_ ((aq)) + "OH"_ ((aq))^(-) -> "A"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>So when you're adding <strong>equal numbers of moles</strong> of weak acid and of strong base, all the moles of the weak will be consumed and you'll be left with <mathjax>#"A"^(-)#</mathjax>, the <strong>conjugate base</strong> of the weak acid. </p>
<p>Now, at the <strong>half equivalence point</strong>, you're adding enough moles of the strong base to neutralize <strong>half of the moles</strong> of the weak acid present in the solution. </p>
<p>The reaction will consume <strong>half</strong> of the moles of the weak acid and produce just as many moles of the conjugate base <mathjax>#->#</mathjax> the weak acid, the strong base, and the conjugate base are all in <mathjax>#1:1#</mathjax> <strong><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong>, meaning that what you consume from the weak acid and the strong base, you produce as the conjugate base. </p>
<p>And so at the <strong>half equivalence point</strong>, the solution will contain <strong>equal numbers of moles</strong> of the weak acid and of its conjugate base, which implies that you're now dealing with a <strong>buffer solution</strong>. </p>
<p><img alt="https://secondaryscience4all.wordpress.com/2014/08/10/ph-curves-titrations-and-indicators/" src="https://useruploads.socratic.org/mLYaYwnSRSWjnlT85VHk_slide211.jpg"/> </p>
<p>As you know, the <mathjax>#"pH"#</mathjax> of a weak acid-conjugate base buffer can be calcualted using the <strong>Henderson - Hasselbalch equation</strong></p>
<blockquote>
<p><mathjax>#"pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))#</mathjax></p>
</blockquote>
<p>At the half equivalence point, you have</p>
<blockquote>
<p><mathjax>#["HA"] = ["A"^(-)]#</mathjax></p>
</blockquote>
<p>which implies that</p>
<blockquote>
<p><mathjax>#log( (["HA"])/(["A"^(-)])) = log(1) = 0#</mathjax></p>
</blockquote>
<p>Therefore, you can say that at the half-equivalence point, the <mathjax>#"pH"#</mathjax> of the solution is <strong>equal</strong> to the <mathjax>#"p"K_a#</mathjax> of the weak acid. </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("At the half equivalence point: " -> " pH" = "p"K_a)))#</mathjax></p>
</blockquote>
<p>The <mathjax>#"p"K_a#</mathjax> is given by the <strong>acid dissociation constant</strong> of the weak acid, <mathjax>#K_a#</mathjax>. </p>
<blockquote>
<p><mathjax>#"p"K_a = - log(K_a)#</mathjax></p>
</blockquote>
<p>which implies that you have</p>
<blockquote>
<p><mathjax>#K_a = 10^(-"p"K_a)#</mathjax></p>
</blockquote>
<p>At the <strong>half equivalence point</strong>, you will have</p>
<blockquote>
<p><mathjax>#K_a = 10^(-"pH")#</mathjax></p>
</blockquote>
<p>Plug in your value to find</p>
<blockquote>
<p><mathjax>#K_a = 10^(-5.67) = color(darkgreen)(ul(color(black)(2.1 * 10^(-6))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of decimal places you have for the <mathjax>#"pH"#</mathjax> of the solution. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#K_a = 2.1 * 10^(-6)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that at the <strong>half equivalence point</strong>, the <mathjax>#"pH"#</mathjax> of the solution will be equal to the <mathjax>#"p"K_a#</mathjax> of the weak acid. </p>
<p>Assuming that you're titrating a weak monoprotic acid <mathjax>#"HA"#</mathjax> with a strong base that I'll represent as <mathjax>#"OH"^(-)#</mathjax>, you know that at the <strong>equivalence point</strong>, the strong base will completely neutralize the weak acid.</p>
<blockquote>
<p><mathjax>#"HA"_ ((aq)) + "OH"_ ((aq))^(-) -> "A"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>So when you're adding <strong>equal numbers of moles</strong> of weak acid and of strong base, all the moles of the weak will be consumed and you'll be left with <mathjax>#"A"^(-)#</mathjax>, the <strong>conjugate base</strong> of the weak acid. </p>
<p>Now, at the <strong>half equivalence point</strong>, you're adding enough moles of the strong base to neutralize <strong>half of the moles</strong> of the weak acid present in the solution. </p>
<p>The reaction will consume <strong>half</strong> of the moles of the weak acid and produce just as many moles of the conjugate base <mathjax>#->#</mathjax> the weak acid, the strong base, and the conjugate base are all in <mathjax>#1:1#</mathjax> <strong><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong>, meaning that what you consume from the weak acid and the strong base, you produce as the conjugate base. </p>
<p>And so at the <strong>half equivalence point</strong>, the solution will contain <strong>equal numbers of moles</strong> of the weak acid and of its conjugate base, which implies that you're now dealing with a <strong>buffer solution</strong>. </p>
<p><img alt="https://secondaryscience4all.wordpress.com/2014/08/10/ph-curves-titrations-and-indicators/" src="https://useruploads.socratic.org/mLYaYwnSRSWjnlT85VHk_slide211.jpg"/> </p>
<p>As you know, the <mathjax>#"pH"#</mathjax> of a weak acid-conjugate base buffer can be calcualted using the <strong>Henderson - Hasselbalch equation</strong></p>
<blockquote>
<p><mathjax>#"pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))#</mathjax></p>
</blockquote>
<p>At the half equivalence point, you have</p>
<blockquote>
<p><mathjax>#["HA"] = ["A"^(-)]#</mathjax></p>
</blockquote>
<p>which implies that</p>
<blockquote>
<p><mathjax>#log( (["HA"])/(["A"^(-)])) = log(1) = 0#</mathjax></p>
</blockquote>
<p>Therefore, you can say that at the half-equivalence point, the <mathjax>#"pH"#</mathjax> of the solution is <strong>equal</strong> to the <mathjax>#"p"K_a#</mathjax> of the weak acid. </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("At the half equivalence point: " -> " pH" = "p"K_a)))#</mathjax></p>
</blockquote>
<p>The <mathjax>#"p"K_a#</mathjax> is given by the <strong>acid dissociation constant</strong> of the weak acid, <mathjax>#K_a#</mathjax>. </p>
<blockquote>
<p><mathjax>#"p"K_a = - log(K_a)#</mathjax></p>
</blockquote>
<p>which implies that you have</p>
<blockquote>
<p><mathjax>#K_a = 10^(-"p"K_a)#</mathjax></p>
</blockquote>
<p>At the <strong>half equivalence point</strong>, you will have</p>
<blockquote>
<p><mathjax>#K_a = 10^(-"pH")#</mathjax></p>
</blockquote>
<p>Plug in your value to find</p>
<blockquote>
<p><mathjax>#K_a = 10^(-5.67) = color(darkgreen)(ul(color(black)(2.1 * 10^(-6))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of decimal places you have for the <mathjax>#"pH"#</mathjax> of the solution. </p></div>
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<h1 class="questionTitle" itemprop="name">The #"pH"# at one-half the equivalence point in an acid-base titration was found to be #5.67#. What is the value of #K_a# for this unknown acid?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#K_a = 2.1 * 10^(-6)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that at the <strong>half equivalence point</strong>, the <mathjax>#"pH"#</mathjax> of the solution will be equal to the <mathjax>#"p"K_a#</mathjax> of the weak acid. </p>
<p>Assuming that you're titrating a weak monoprotic acid <mathjax>#"HA"#</mathjax> with a strong base that I'll represent as <mathjax>#"OH"^(-)#</mathjax>, you know that at the <strong>equivalence point</strong>, the strong base will completely neutralize the weak acid.</p>
<blockquote>
<p><mathjax>#"HA"_ ((aq)) + "OH"_ ((aq))^(-) -> "A"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>So when you're adding <strong>equal numbers of moles</strong> of weak acid and of strong base, all the moles of the weak will be consumed and you'll be left with <mathjax>#"A"^(-)#</mathjax>, the <strong>conjugate base</strong> of the weak acid. </p>
<p>Now, at the <strong>half equivalence point</strong>, you're adding enough moles of the strong base to neutralize <strong>half of the moles</strong> of the weak acid present in the solution. </p>
<p>The reaction will consume <strong>half</strong> of the moles of the weak acid and produce just as many moles of the conjugate base <mathjax>#->#</mathjax> the weak acid, the strong base, and the conjugate base are all in <mathjax>#1:1#</mathjax> <strong><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong>, meaning that what you consume from the weak acid and the strong base, you produce as the conjugate base. </p>
<p>And so at the <strong>half equivalence point</strong>, the solution will contain <strong>equal numbers of moles</strong> of the weak acid and of its conjugate base, which implies that you're now dealing with a <strong>buffer solution</strong>. </p>
<p><img alt="https://secondaryscience4all.wordpress.com/2014/08/10/ph-curves-titrations-and-indicators/" src="https://useruploads.socratic.org/mLYaYwnSRSWjnlT85VHk_slide211.jpg"/> </p>
<p>As you know, the <mathjax>#"pH"#</mathjax> of a weak acid-conjugate base buffer can be calcualted using the <strong>Henderson - Hasselbalch equation</strong></p>
<blockquote>
<p><mathjax>#"pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))#</mathjax></p>
</blockquote>
<p>At the half equivalence point, you have</p>
<blockquote>
<p><mathjax>#["HA"] = ["A"^(-)]#</mathjax></p>
</blockquote>
<p>which implies that</p>
<blockquote>
<p><mathjax>#log( (["HA"])/(["A"^(-)])) = log(1) = 0#</mathjax></p>
</blockquote>
<p>Therefore, you can say that at the half-equivalence point, the <mathjax>#"pH"#</mathjax> of the solution is <strong>equal</strong> to the <mathjax>#"p"K_a#</mathjax> of the weak acid. </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("At the half equivalence point: " -> " pH" = "p"K_a)))#</mathjax></p>
</blockquote>
<p>The <mathjax>#"p"K_a#</mathjax> is given by the <strong>acid dissociation constant</strong> of the weak acid, <mathjax>#K_a#</mathjax>. </p>
<blockquote>
<p><mathjax>#"p"K_a = - log(K_a)#</mathjax></p>
</blockquote>
<p>which implies that you have</p>
<blockquote>
<p><mathjax>#K_a = 10^(-"p"K_a)#</mathjax></p>
</blockquote>
<p>At the <strong>half equivalence point</strong>, you will have</p>
<blockquote>
<p><mathjax>#K_a = 10^(-"pH")#</mathjax></p>
</blockquote>
<p>Plug in your value to find</p>
<blockquote>
<p><mathjax>#K_a = 10^(-5.67) = color(darkgreen)(ul(color(black)(2.1 * 10^(-6))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of decimal places you have for the <mathjax>#"pH"#</mathjax> of the solution. </p></div>
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</article> | The #"pH"# at one-half the equivalence point in an acid-base titration was found to be #5.67#. What is the value of #K_a# for this unknown acid? | null |
512 | a8b9c215-6ddd-11ea-9f2c-ccda262736ce | https://socratic.org/questions/following-reaction-has-a-reaction-yield-of-75-in-order-to-obtain-25-g-of-carbon- | 0.25 moles | start physical_unit 23 23 mass mol qc_end physical_unit 3 4 7 7 percent_yield qc_end physical_unit 15 16 12 13 mass qc_end chemical_equation 26 32 qc_end end | [{"type":"physical unit","value":"Amount [OF] propane [IN] moles"}] | [{"type":"physical unit","value":"0.25 moles"}] | [{"type":"physical unit","value":"Yield [OF] a reaction [=] \\pu{75%}"},{"type":"physical unit","value":"Mass [OF] carbon dioxide [=] \\pu{25 g}"},{"type":"chemical equation","value":"C3H8 + O2 -> CO2 + H2O"}] | <h1 class="questionTitle" itemprop="name">Following reaction has a reaction yield of #75%#. In order to obtain #"25 g"# of carbon dioxide, what is the necessary amount of propane in moles?
#"C"_3"H"_8 + "O"_2 -> "CO"_2 + "H"_2"O"# (unbalanced)</h1> | null | 0.25 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing the balanced chemical equation that describes this combustion reaction</p>
<blockquote>
<p><mathjax>#"C"_ 3"H"_ (8(g)) + 5"O"_ (2(g)) -> 3"CO"_ (2(g)) + 4"H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>Now, the balanced chemical equation tells you that in order for the reaction to produce <mathjax>#3#</mathjax> <strong>moles</strong> of carbon dioxide, it must consume <mathjax>#1#</mathjax> <strong>mole</strong> of propane. </p>
<p>This is the <strong>theoretical yield</strong> of the reaction, i.e. what you would get at <mathjax>#100%#</mathjax> <strong>yield</strong>. </p>
<p>In your case, the reaction is known to have a <strong><a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a></strong> equal to <mathjax>#75%#</mathjax>. This essentially means that for every <mathjax>#100#</mathjax> <strong>moles</strong> of carbon dioxide that the reaction could <em>theoretically produce</em>, you only get <mathjax>#75#</mathjax> <strong>moles</strong>. </p>
<p>Use the <strong>molar mass</strong> of carbon dioxide to convert the mass of the sample to <em>moles</em></p>
<blockquote>
<p><mathjax>#25 color(red)(cancel(color(black)("g"))) * "1 mole CO"_2/(44.01color(red)(cancel(color(black)("g")))) = "0.568 moles CO"_2#</mathjax></p>
</blockquote>
<p>Now, at <mathjax>#100%#</mathjax> yield, the reaction must consume </p>
<blockquote>
<p><mathjax>#0.568 color(red)(cancel(color(black)("moles CO"_2))) * ("1 mole C"_3"H"_8)/(3color(red)(cancel(color(black)("moles CO"_2)))) = "0.189 moles C"_3"H"_8#</mathjax></p>
</blockquote>
<p>in order to produce <mathjax>#0.568#</mathjax> <strong>moles</strong> of carbon dioxide. However, you know that the reaction has a percent yield equal to <mathjax>#75%#</mathjax>, which means that the reaction produced only <mathjax>#75%#</mathjax> of the number of moles of carbon dioxide that it would have produced at theoretical yield. </p>
<p>This means that in order to get <mathjax>#0.568#</mathjax> <strong>moles</strong> of carbon dioxide, the reaction must consume enough moles of propane to theoretically produce </p>
<blockquote>
<p><mathjax>#0.568 color(red)(cancel(color(black)("moles CO"_2))) * overbrace("100 moles CO"_2/(75 color(red)(cancel(color(black)("moles CO"_2)))))^(color(blue)("= 75% yield")) = "0.757 moles CO"_2#</mathjax></p>
</blockquote>
<p>You can thus say that the number of moles of propane needed is equal to</p>
<blockquote>
<p><mathjax>#0.757 color(red)(cancel(color(black)("moles CO"_2))) * ("1 mole C"_3"H"_8)/(3color(red)(cancel(color(black)("moles CO"_2)))) = color(darkgreen)(ul(color(black)("0.25 moles C"_3"H"_8)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the mass of carbon dioxide. </p>
<p>So, you can say that, for this reaction, you have</p>
<blockquote>
<p><mathjax>#"0.189 moles C"_3"H"_8 " "stackrel(color(white)(acolor(blue)("at 100% yield")aaa))(->) " " "0.568 moles CO"_2#</mathjax></p>
<p><mathjax>#"0.25 moles C"_3"H"_8 " "stackrel(color(white)(acolor(blue)("at 100% yield")aaa))(->) " " "0.757 moles CO"_2#</mathjax></p>
<p><mathjax>#"0.25 moles C"_3"H"_8 " "stackrel(color(white)(acolor(blue)("at 75% yield")aaa))(->) " " "0.568 moles CO"_2#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.25 moles C"_3"H"_8#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing the balanced chemical equation that describes this combustion reaction</p>
<blockquote>
<p><mathjax>#"C"_ 3"H"_ (8(g)) + 5"O"_ (2(g)) -> 3"CO"_ (2(g)) + 4"H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>Now, the balanced chemical equation tells you that in order for the reaction to produce <mathjax>#3#</mathjax> <strong>moles</strong> of carbon dioxide, it must consume <mathjax>#1#</mathjax> <strong>mole</strong> of propane. </p>
<p>This is the <strong>theoretical yield</strong> of the reaction, i.e. what you would get at <mathjax>#100%#</mathjax> <strong>yield</strong>. </p>
<p>In your case, the reaction is known to have a <strong><a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a></strong> equal to <mathjax>#75%#</mathjax>. This essentially means that for every <mathjax>#100#</mathjax> <strong>moles</strong> of carbon dioxide that the reaction could <em>theoretically produce</em>, you only get <mathjax>#75#</mathjax> <strong>moles</strong>. </p>
<p>Use the <strong>molar mass</strong> of carbon dioxide to convert the mass of the sample to <em>moles</em></p>
<blockquote>
<p><mathjax>#25 color(red)(cancel(color(black)("g"))) * "1 mole CO"_2/(44.01color(red)(cancel(color(black)("g")))) = "0.568 moles CO"_2#</mathjax></p>
</blockquote>
<p>Now, at <mathjax>#100%#</mathjax> yield, the reaction must consume </p>
<blockquote>
<p><mathjax>#0.568 color(red)(cancel(color(black)("moles CO"_2))) * ("1 mole C"_3"H"_8)/(3color(red)(cancel(color(black)("moles CO"_2)))) = "0.189 moles C"_3"H"_8#</mathjax></p>
</blockquote>
<p>in order to produce <mathjax>#0.568#</mathjax> <strong>moles</strong> of carbon dioxide. However, you know that the reaction has a percent yield equal to <mathjax>#75%#</mathjax>, which means that the reaction produced only <mathjax>#75%#</mathjax> of the number of moles of carbon dioxide that it would have produced at theoretical yield. </p>
<p>This means that in order to get <mathjax>#0.568#</mathjax> <strong>moles</strong> of carbon dioxide, the reaction must consume enough moles of propane to theoretically produce </p>
<blockquote>
<p><mathjax>#0.568 color(red)(cancel(color(black)("moles CO"_2))) * overbrace("100 moles CO"_2/(75 color(red)(cancel(color(black)("moles CO"_2)))))^(color(blue)("= 75% yield")) = "0.757 moles CO"_2#</mathjax></p>
</blockquote>
<p>You can thus say that the number of moles of propane needed is equal to</p>
<blockquote>
<p><mathjax>#0.757 color(red)(cancel(color(black)("moles CO"_2))) * ("1 mole C"_3"H"_8)/(3color(red)(cancel(color(black)("moles CO"_2)))) = color(darkgreen)(ul(color(black)("0.25 moles C"_3"H"_8)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the mass of carbon dioxide. </p>
<p>So, you can say that, for this reaction, you have</p>
<blockquote>
<p><mathjax>#"0.189 moles C"_3"H"_8 " "stackrel(color(white)(acolor(blue)("at 100% yield")aaa))(->) " " "0.568 moles CO"_2#</mathjax></p>
<p><mathjax>#"0.25 moles C"_3"H"_8 " "stackrel(color(white)(acolor(blue)("at 100% yield")aaa))(->) " " "0.757 moles CO"_2#</mathjax></p>
<p><mathjax>#"0.25 moles C"_3"H"_8 " "stackrel(color(white)(acolor(blue)("at 75% yield")aaa))(->) " " "0.568 moles CO"_2#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Following reaction has a reaction yield of #75%#. In order to obtain #"25 g"# of carbon dioxide, what is the necessary amount of propane in moles?
#"C"_3"H"_8 + "O"_2 -> "CO"_2 + "H"_2"O"# (unbalanced)</h1>
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Stefan V.
</a></div>
<div class="answerInfoBottom clearfix weak-text">
<span class="dateCreated" datetime="2017-12-24T00:02:06" itemprop="dateCreated">
Dec 24, 2017
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<div class="markdown"><p><mathjax>#"0.25 moles C"_3"H"_8#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing the balanced chemical equation that describes this combustion reaction</p>
<blockquote>
<p><mathjax>#"C"_ 3"H"_ (8(g)) + 5"O"_ (2(g)) -> 3"CO"_ (2(g)) + 4"H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>Now, the balanced chemical equation tells you that in order for the reaction to produce <mathjax>#3#</mathjax> <strong>moles</strong> of carbon dioxide, it must consume <mathjax>#1#</mathjax> <strong>mole</strong> of propane. </p>
<p>This is the <strong>theoretical yield</strong> of the reaction, i.e. what you would get at <mathjax>#100%#</mathjax> <strong>yield</strong>. </p>
<p>In your case, the reaction is known to have a <strong><a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a></strong> equal to <mathjax>#75%#</mathjax>. This essentially means that for every <mathjax>#100#</mathjax> <strong>moles</strong> of carbon dioxide that the reaction could <em>theoretically produce</em>, you only get <mathjax>#75#</mathjax> <strong>moles</strong>. </p>
<p>Use the <strong>molar mass</strong> of carbon dioxide to convert the mass of the sample to <em>moles</em></p>
<blockquote>
<p><mathjax>#25 color(red)(cancel(color(black)("g"))) * "1 mole CO"_2/(44.01color(red)(cancel(color(black)("g")))) = "0.568 moles CO"_2#</mathjax></p>
</blockquote>
<p>Now, at <mathjax>#100%#</mathjax> yield, the reaction must consume </p>
<blockquote>
<p><mathjax>#0.568 color(red)(cancel(color(black)("moles CO"_2))) * ("1 mole C"_3"H"_8)/(3color(red)(cancel(color(black)("moles CO"_2)))) = "0.189 moles C"_3"H"_8#</mathjax></p>
</blockquote>
<p>in order to produce <mathjax>#0.568#</mathjax> <strong>moles</strong> of carbon dioxide. However, you know that the reaction has a percent yield equal to <mathjax>#75%#</mathjax>, which means that the reaction produced only <mathjax>#75%#</mathjax> of the number of moles of carbon dioxide that it would have produced at theoretical yield. </p>
<p>This means that in order to get <mathjax>#0.568#</mathjax> <strong>moles</strong> of carbon dioxide, the reaction must consume enough moles of propane to theoretically produce </p>
<blockquote>
<p><mathjax>#0.568 color(red)(cancel(color(black)("moles CO"_2))) * overbrace("100 moles CO"_2/(75 color(red)(cancel(color(black)("moles CO"_2)))))^(color(blue)("= 75% yield")) = "0.757 moles CO"_2#</mathjax></p>
</blockquote>
<p>You can thus say that the number of moles of propane needed is equal to</p>
<blockquote>
<p><mathjax>#0.757 color(red)(cancel(color(black)("moles CO"_2))) * ("1 mole C"_3"H"_8)/(3color(red)(cancel(color(black)("moles CO"_2)))) = color(darkgreen)(ul(color(black)("0.25 moles C"_3"H"_8)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the mass of carbon dioxide. </p>
<p>So, you can say that, for this reaction, you have</p>
<blockquote>
<p><mathjax>#"0.189 moles C"_3"H"_8 " "stackrel(color(white)(acolor(blue)("at 100% yield")aaa))(->) " " "0.568 moles CO"_2#</mathjax></p>
<p><mathjax>#"0.25 moles C"_3"H"_8 " "stackrel(color(white)(acolor(blue)("at 100% yield")aaa))(->) " " "0.757 moles CO"_2#</mathjax></p>
<p><mathjax>#"0.25 moles C"_3"H"_8 " "stackrel(color(white)(acolor(blue)("at 75% yield")aaa))(->) " " "0.568 moles CO"_2#</mathjax></p>
</blockquote></div>
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</article> | Following reaction has a reaction yield of #75%#. In order to obtain #"25 g"# of carbon dioxide, what is the necessary amount of propane in moles?
#"C"_3"H"_8 + "O"_2 -> "CO"_2 + "H"_2"O"# (unbalanced) | null |
513 | ac583853-6ddd-11ea-9450-ccda262736ce | https://socratic.org/questions/how-many-moles-are-in-200-grams-of-fluorine | 10.53 moles | start physical_unit 8 8 mole mol qc_end physical_unit 8 8 5 6 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] fluorine [IN] moles "}] | [{"type":"physical unit","value":"10.53 moles "}] | [{"type":"physical unit","value":"Mass [OF] fluorine [=] \\pu{200 grams}"}] | <h1 class="questionTitle" itemprop="name">How many moles are in 200 grams of fluorine?</h1> | null | 10.53 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Fluorine has <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of 18.998403 g/mol rounding of we consider 19 g/mol</p>
<p>So,</p>
<p>In 19 g fluorine is 1 mole <br/>
so in 200g fluorine <mathjax>#200/19#</mathjax> moles are there</p>
<p>which equals 10.5263 moles</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>10.5263 moles</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Fluorine has <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of 18.998403 g/mol rounding of we consider 19 g/mol</p>
<p>So,</p>
<p>In 19 g fluorine is 1 mole <br/>
so in 200g fluorine <mathjax>#200/19#</mathjax> moles are there</p>
<p>which equals 10.5263 moles</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles are in 200 grams of fluorine?</h1>
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<div class="markdown"><p>10.5263 moles</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Fluorine has <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of 18.998403 g/mol rounding of we consider 19 g/mol</p>
<p>So,</p>
<p>In 19 g fluorine is 1 mole <br/>
so in 200g fluorine <mathjax>#200/19#</mathjax> moles are there</p>
<p>which equals 10.5263 moles</p></div>
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<div class="markdown"><p>Assume that:<br/>
n = number of moles<br/>
m = mass of substance<br/>
M = molar mass (same as atomic weight on periodic table)</p>
<p>200 grams of fluorine (m) has been provided for you.</p>
<p>The molar mass (M) of fluorine is 19.0 g/mol.</p>
<p>Therefore, the number of moles (n) for fluorine is:<br/>
[n = 200 grams <mathjax>#-:#</mathjax> 19.0 g/mol] = 10.5 moles.<br/>
Note: the answer is rounded to 3 <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p>
<p>There are 10.5 moles in 200 grams of fluorine.</p></div>
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</article> | How many moles are in 200 grams of fluorine? | null |
514 | aa36d447-6ddd-11ea-b47a-ccda262736ce | https://socratic.org/questions/for-the-reaction-n-2-g-2o-2-s-rightleftharpoons-2no-2-g-k-c-8-3x10-10-at-25-c-wh | 30120481927.71 M | start physical_unit 23 24 concentration mol/l qc_end chemical_equation 3 9 qc_end physical_unit 1 2 12 14 equilibrium_constant_k qc_end physical_unit 1 2 16 17 temperature qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Concentration [OF] N2 gas [IN] M"}] | [{"type":"physical unit","value":"30120481927.71 M"}] | [{"type":"chemical equation","value":"N2(g) + 2 O2(s) <=> 2 NO2(g)"},{"type":"physical unit","value":"Kc [OF] the reaction [=] \\pu{8.3 × 10^(-10)}"},{"type":"physical unit","value":"Temperature [OF] the reaction [=] \\pu{25 ℃}"},{"type":"other","value":"The concentration of NO2 is five times the concentration of O2 gas."}] | <h1 class="questionTitle" itemprop="name">For the reaction: #N_2(g) + 2O_2(s) rightleftharpoons 2NO_2(g)#, #K_c# - #8.3x10^-10# at 25 °C. What is the concentration of #N_2# gas at equilibrium when the concentration of #NO_2# is five times the concentration of #O_2# gas?</h1> | null | 30120481927.71 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Kc = [NO_2]^2/ ([N2][O2]^2) #</mathjax></p>
<p>The Kc is measured at the equilibrum </p>
<p><mathjax>#8.3 * 10^-10 = x^2/([N_2][x]^2)#</mathjax></p>
<p>Conc of <mathjax>#N_2 "when" NO_2 = O_2#</mathjax> = 1204819277.108434mol</p>
<p><mathjax>#8.3 * 10^-10 = (5x^2)/(N_2 * x^2)#</mathjax></p>
<p>Let's solve for <mathjax>#N_2#</mathjax>.</p>
<p><mathjax>#8.3(10^−10)= (5x)^2 /(x^2N_2)#</mathjax></p>
<p>Step 1: Multiply both sides by <mathjax>#N_2#</mathjax></p>
<p><mathjax>#0N_2=25#</mathjax></p>
<p>Step 2: Divide both sides by 0.</p>
<p><mathjax>#(0N_2)/0 = 25/0#</mathjax></p>
<p><mathjax>#N_2=30120481927.710842mol#</mathjax></p>
<p>Answer:<br/>
<mathjax>#N_2=30120481927.710842mol#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#N_2=30120481927.710842mol#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Kc = [NO_2]^2/ ([N2][O2]^2) #</mathjax></p>
<p>The Kc is measured at the equilibrum </p>
<p><mathjax>#8.3 * 10^-10 = x^2/([N_2][x]^2)#</mathjax></p>
<p>Conc of <mathjax>#N_2 "when" NO_2 = O_2#</mathjax> = 1204819277.108434mol</p>
<p><mathjax>#8.3 * 10^-10 = (5x^2)/(N_2 * x^2)#</mathjax></p>
<p>Let's solve for <mathjax>#N_2#</mathjax>.</p>
<p><mathjax>#8.3(10^−10)= (5x)^2 /(x^2N_2)#</mathjax></p>
<p>Step 1: Multiply both sides by <mathjax>#N_2#</mathjax></p>
<p><mathjax>#0N_2=25#</mathjax></p>
<p>Step 2: Divide both sides by 0.</p>
<p><mathjax>#(0N_2)/0 = 25/0#</mathjax></p>
<p><mathjax>#N_2=30120481927.710842mol#</mathjax></p>
<p>Answer:<br/>
<mathjax>#N_2=30120481927.710842mol#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">For the reaction: #N_2(g) + 2O_2(s) rightleftharpoons 2NO_2(g)#, #K_c# - #8.3x10^-10# at 25 °C. What is the concentration of #N_2# gas at equilibrium when the concentration of #NO_2# is five times the concentration of #O_2# gas?</h1>
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<div class="markdown"><p><mathjax>#N_2=30120481927.710842mol#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Kc = [NO_2]^2/ ([N2][O2]^2) #</mathjax></p>
<p>The Kc is measured at the equilibrum </p>
<p><mathjax>#8.3 * 10^-10 = x^2/([N_2][x]^2)#</mathjax></p>
<p>Conc of <mathjax>#N_2 "when" NO_2 = O_2#</mathjax> = 1204819277.108434mol</p>
<p><mathjax>#8.3 * 10^-10 = (5x^2)/(N_2 * x^2)#</mathjax></p>
<p>Let's solve for <mathjax>#N_2#</mathjax>.</p>
<p><mathjax>#8.3(10^−10)= (5x)^2 /(x^2N_2)#</mathjax></p>
<p>Step 1: Multiply both sides by <mathjax>#N_2#</mathjax></p>
<p><mathjax>#0N_2=25#</mathjax></p>
<p>Step 2: Divide both sides by 0.</p>
<p><mathjax>#(0N_2)/0 = 25/0#</mathjax></p>
<p><mathjax>#N_2=30120481927.710842mol#</mathjax></p>
<p>Answer:<br/>
<mathjax>#N_2=30120481927.710842mol#</mathjax></p></div>
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</article> | For the reaction: #N_2(g) + 2O_2(s) rightleftharpoons 2NO_2(g)#, #K_c# - #8.3x10^-10# at 25 °C. What is the concentration of #N_2# gas at equilibrium when the concentration of #NO_2# is five times the concentration of #O_2# gas? | null |
515 | aa5f39b5-6ddd-11ea-a501-ccda262736ce | https://socratic.org/questions/what-is-the-balanced-net-ionic-equation-for-the-reaction-in-acidic-solution-of-f | 5 Fe^2+ + MnO4- + 8 H+ -> Mn^2+ + 5 Fe^3+ + 4 H2O | start chemical_equation qc_end chemical_equation 14 14 qc_end chemical_equation 16 16 qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reaction"}] | [{"type":"chemical equation","value":"5 Fe^2+ + MnO4- + 8 H+ -> Mn^2+ + 5 Fe^3+ + 4 H2O"}] | [{"type":"chemical equation","value":"Fe^2+"},{"type":"chemical equation","value":"KMnO4"},{"type":"other","value":"The reaction is in acidic solution."}] | <h1 class="questionTitle" itemprop="name">What is the balanced net ionic equation for the reaction in acidic solution of #Fe^(2+)# and #KMnO_4#? </h1> | null | 5 Fe^2+ + MnO4- + 8 H+ -> Mn^2+ + 5 Fe^3+ + 4 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Potassium is a spectator ion so assume the following reaction ... </p>
<p><mathjax>#Fe^(+2) + MnO_4^-#</mathjax> => <mathjax>#Mn^(+2) + Fe^(+3)#</mathjax></p>
<p>Oxidation Half Rxn: <mathjax>#Fe^(+2)#</mathjax> => <mathjax>#Fe^(+3) + e^-#</mathjax><br/>
Reduction Half Rxn: <mathjax>#MnO_4^-#</mathjax> + <mathjax>#5e^-#</mathjax> => <mathjax>#Mn^(+2)#</mathjax></p>
<p>Balancing with respect to charge transfer ...<br/>
<mathjax>#5Fe^(+2) #</mathjax> =><mathjax>#5Fe^(+3) + 5e^-#</mathjax><br/>
<mathjax>#MnO_4^-#</mathjax> + <mathjax>#5e^ -#</mathjax> => <mathjax>#Mn^(+2)#</mathjax> </p>
<p>Balancing with respect to mass ... (Rxn in acidic medium, add one water for each oxygen needed to the side needing oxygen and follow with the appropriate number of <mathjax>#H^+#</mathjax> needed to the side needing hydrogen. </p>
<p><mathjax>#5Fe^(+2)#</mathjax> + <mathjax>#MnO_4^(-) + 8H^+#</mathjax> => <mathjax>#Mn^(+2) + 5Fe^(+3) + 4H_2O#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#5Fe^(+2)#</mathjax> + <mathjax>#MnO_4^(-) + 8H^+#</mathjax> => <mathjax>#Mn^(+2) + 5Fe^(+3) + 4H_2O#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Potassium is a spectator ion so assume the following reaction ... </p>
<p><mathjax>#Fe^(+2) + MnO_4^-#</mathjax> => <mathjax>#Mn^(+2) + Fe^(+3)#</mathjax></p>
<p>Oxidation Half Rxn: <mathjax>#Fe^(+2)#</mathjax> => <mathjax>#Fe^(+3) + e^-#</mathjax><br/>
Reduction Half Rxn: <mathjax>#MnO_4^-#</mathjax> + <mathjax>#5e^-#</mathjax> => <mathjax>#Mn^(+2)#</mathjax></p>
<p>Balancing with respect to charge transfer ...<br/>
<mathjax>#5Fe^(+2) #</mathjax> =><mathjax>#5Fe^(+3) + 5e^-#</mathjax><br/>
<mathjax>#MnO_4^-#</mathjax> + <mathjax>#5e^ -#</mathjax> => <mathjax>#Mn^(+2)#</mathjax> </p>
<p>Balancing with respect to mass ... (Rxn in acidic medium, add one water for each oxygen needed to the side needing oxygen and follow with the appropriate number of <mathjax>#H^+#</mathjax> needed to the side needing hydrogen. </p>
<p><mathjax>#5Fe^(+2)#</mathjax> + <mathjax>#MnO_4^(-) + 8H^+#</mathjax> => <mathjax>#Mn^(+2) + 5Fe^(+3) + 4H_2O#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the balanced net ionic equation for the reaction in acidic solution of #Fe^(2+)# and #KMnO_4#? </h1>
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<div class="markdown"><p><mathjax>#5Fe^(+2)#</mathjax> + <mathjax>#MnO_4^(-) + 8H^+#</mathjax> => <mathjax>#Mn^(+2) + 5Fe^(+3) + 4H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Potassium is a spectator ion so assume the following reaction ... </p>
<p><mathjax>#Fe^(+2) + MnO_4^-#</mathjax> => <mathjax>#Mn^(+2) + Fe^(+3)#</mathjax></p>
<p>Oxidation Half Rxn: <mathjax>#Fe^(+2)#</mathjax> => <mathjax>#Fe^(+3) + e^-#</mathjax><br/>
Reduction Half Rxn: <mathjax>#MnO_4^-#</mathjax> + <mathjax>#5e^-#</mathjax> => <mathjax>#Mn^(+2)#</mathjax></p>
<p>Balancing with respect to charge transfer ...<br/>
<mathjax>#5Fe^(+2) #</mathjax> =><mathjax>#5Fe^(+3) + 5e^-#</mathjax><br/>
<mathjax>#MnO_4^-#</mathjax> + <mathjax>#5e^ -#</mathjax> => <mathjax>#Mn^(+2)#</mathjax> </p>
<p>Balancing with respect to mass ... (Rxn in acidic medium, add one water for each oxygen needed to the side needing oxygen and follow with the appropriate number of <mathjax>#H^+#</mathjax> needed to the side needing hydrogen. </p>
<p><mathjax>#5Fe^(+2)#</mathjax> + <mathjax>#MnO_4^(-) + 8H^+#</mathjax> => <mathjax>#Mn^(+2) + 5Fe^(+3) + 4H_2O#</mathjax></p></div>
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</article> | What is the balanced net ionic equation for the reaction in acidic solution of #Fe^(2+)# and #KMnO_4#? | null |
516 | a96ff4de-6ddd-11ea-8816-ccda262736ce | https://socratic.org/questions/what-is-the-substance-that-forms-hydronium-ions-in-water-arrhenius | H3O+ | start chemical_formula qc_end substance 6 7 qc_end substance 9 9 qc_end end | [{"type":"other","value":"Chemical Formula [OF] the substance [IN] default"}] | [{"type":"chemical equation","value":"H3O+"}] | [{"type":"substance name","value":"Hydronium ions"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">What is the substance that forms hydronium ions in water (Arrhenius)?</h1> | null | H3O+ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>According to Arrhenius' concept of acids, an acid is such a compound which when dissolved in water, releases proton(s) or <mathjax>#H^+#</mathjax>. </p>
<p>Now as an Arrhenius acid releases <mathjax>#H^+#</mathjax>, the proton combines with a <mathjax>#H_2O#</mathjax> molecule to produce a <mathjax>#H_3O^+#</mathjax> ion which is an hydronium ion. </p>
<p>Examples of Arrhenius Acids - </p>
<p><mathjax>#HCl, H_2SO_4, HNO_3, HBr, HI, HClO_4#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>Substances which form hydronium ions (<mathjax>#H_3O^+#</mathjax>) in water are known as Arrhenius Acids. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>According to Arrhenius' concept of acids, an acid is such a compound which when dissolved in water, releases proton(s) or <mathjax>#H^+#</mathjax>. </p>
<p>Now as an Arrhenius acid releases <mathjax>#H^+#</mathjax>, the proton combines with a <mathjax>#H_2O#</mathjax> molecule to produce a <mathjax>#H_3O^+#</mathjax> ion which is an hydronium ion. </p>
<p>Examples of Arrhenius Acids - </p>
<p><mathjax>#HCl, H_2SO_4, HNO_3, HBr, HI, HClO_4#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the substance that forms hydronium ions in water (Arrhenius)?</h1>
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<div class="markdown"><p>Substances which form hydronium ions (<mathjax>#H_3O^+#</mathjax>) in water are known as Arrhenius Acids. </p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>According to Arrhenius' concept of acids, an acid is such a compound which when dissolved in water, releases proton(s) or <mathjax>#H^+#</mathjax>. </p>
<p>Now as an Arrhenius acid releases <mathjax>#H^+#</mathjax>, the proton combines with a <mathjax>#H_2O#</mathjax> molecule to produce a <mathjax>#H_3O^+#</mathjax> ion which is an hydronium ion. </p>
<p>Examples of Arrhenius Acids - </p>
<p><mathjax>#HCl, H_2SO_4, HNO_3, HBr, HI, HClO_4#</mathjax></p></div>
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</article> | What is the substance that forms hydronium ions in water (Arrhenius)? | null |
517 | aa767106-6ddd-11ea-b4ae-ccda262736ce | https://socratic.org/questions/a-0-01-m-acetic-acid-solution-is-1-ionised-an-another-acetic-acid-is-10-ionised- | 9 × 10^(-5) M | start physical_unit 10 12 concentration mol/l qc_end physical_unit 10 12 14 14 percent_ionization qc_end end | [{"type":"physical unit","value":"Concentration [OF] another acetic acid [IN] M"}] | [{"type":"physical unit","value":"9 × 10^(-5) M"}] | [{"type":"physical unit","value":"Molarity [OF] an acetic acid solution [=] \\pu{0.01 M}"},{"type":"physical unit","value":"Degree of ionization [OF] an acetic acid solution [=] \\pu{1%}"},{"type":"physical unit","value":"Degree of ionization [OF] another acetic acid [=] \\pu{10%}"}] | <h1 class="questionTitle" itemprop="name">A 0.01 M acetic acid solution is 1% ionised. An another acetic acid is 10% ionised. What will be the concentration of another acetic acid?</h1> | null | 9 × 10^(-5) M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Let <mathjax>#c#</mathjax> be the initial concentration of acetic acid.</p>
<p>The equation for the equilibrium is</p>
<p><mathjax>#color(white)(m)"HA" + "H"_2"O" ⇌ "A"^"-" + "H"_3"O"^"+"#</mathjax><br/>
<mathjax>#c-αc color(white)(mmmmm) αc color(white)(mml) αc#</mathjax></p>
<p>The equilibrium concentration of <mathjax>#"HA"#</mathjax> will decrease by <mathjax>#αc#</mathjax>, and those of <mathjax>#"A"^"-"#</mathjax> and <mathjax>#"H"_3"O"^"+"#</mathjax> will be <mathjax>#αc#</mathjax>.</p>
<p><mathjax>#K = (αc)^2/(c-αc) = (α^2c^2)/(c(1-α)) = (α^2c)/(1-α)#</mathjax></p>
<blockquote></blockquote>
<p><strong>For the first solution</strong></p>
<p><mathjax>#K = ((0.01)^2× 0.01)/(1-0.01) = (1 × 10^"-6")/0.99 = 1.0 × 10^"-6"#</mathjax></p>
<blockquote></blockquote>
<p><strong>For the second solution</strong>,</p>
<p><mathjax>#α^2c = K(1-α)#</mathjax></p>
<p><mathjax>#c= (K(1-α))/α^2#</mathjax></p>
<p><mathjax>#c = (1.0 × 10^"-6"(1-0.10))/(0.10)^2 = 9 × 10^"-5"color(white)(l) "mol/L"#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The concentration will be <mathjax>#9 × 10^"-5"color(white)(l) "mol/L"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Let <mathjax>#c#</mathjax> be the initial concentration of acetic acid.</p>
<p>The equation for the equilibrium is</p>
<p><mathjax>#color(white)(m)"HA" + "H"_2"O" ⇌ "A"^"-" + "H"_3"O"^"+"#</mathjax><br/>
<mathjax>#c-αc color(white)(mmmmm) αc color(white)(mml) αc#</mathjax></p>
<p>The equilibrium concentration of <mathjax>#"HA"#</mathjax> will decrease by <mathjax>#αc#</mathjax>, and those of <mathjax>#"A"^"-"#</mathjax> and <mathjax>#"H"_3"O"^"+"#</mathjax> will be <mathjax>#αc#</mathjax>.</p>
<p><mathjax>#K = (αc)^2/(c-αc) = (α^2c^2)/(c(1-α)) = (α^2c)/(1-α)#</mathjax></p>
<blockquote></blockquote>
<p><strong>For the first solution</strong></p>
<p><mathjax>#K = ((0.01)^2× 0.01)/(1-0.01) = (1 × 10^"-6")/0.99 = 1.0 × 10^"-6"#</mathjax></p>
<blockquote></blockquote>
<p><strong>For the second solution</strong>,</p>
<p><mathjax>#α^2c = K(1-α)#</mathjax></p>
<p><mathjax>#c= (K(1-α))/α^2#</mathjax></p>
<p><mathjax>#c = (1.0 × 10^"-6"(1-0.10))/(0.10)^2 = 9 × 10^"-5"color(white)(l) "mol/L"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A 0.01 M acetic acid solution is 1% ionised. An another acetic acid is 10% ionised. What will be the concentration of another acetic acid?</h1>
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<div class="markdown"><p>The concentration will be <mathjax>#9 × 10^"-5"color(white)(l) "mol/L"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Let <mathjax>#c#</mathjax> be the initial concentration of acetic acid.</p>
<p>The equation for the equilibrium is</p>
<p><mathjax>#color(white)(m)"HA" + "H"_2"O" ⇌ "A"^"-" + "H"_3"O"^"+"#</mathjax><br/>
<mathjax>#c-αc color(white)(mmmmm) αc color(white)(mml) αc#</mathjax></p>
<p>The equilibrium concentration of <mathjax>#"HA"#</mathjax> will decrease by <mathjax>#αc#</mathjax>, and those of <mathjax>#"A"^"-"#</mathjax> and <mathjax>#"H"_3"O"^"+"#</mathjax> will be <mathjax>#αc#</mathjax>.</p>
<p><mathjax>#K = (αc)^2/(c-αc) = (α^2c^2)/(c(1-α)) = (α^2c)/(1-α)#</mathjax></p>
<blockquote></blockquote>
<p><strong>For the first solution</strong></p>
<p><mathjax>#K = ((0.01)^2× 0.01)/(1-0.01) = (1 × 10^"-6")/0.99 = 1.0 × 10^"-6"#</mathjax></p>
<blockquote></blockquote>
<p><strong>For the second solution</strong>,</p>
<p><mathjax>#α^2c = K(1-α)#</mathjax></p>
<p><mathjax>#c= (K(1-α))/α^2#</mathjax></p>
<p><mathjax>#c = (1.0 × 10^"-6"(1-0.10))/(0.10)^2 = 9 × 10^"-5"color(white)(l) "mol/L"#</mathjax></p></div>
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</article> | A 0.01 M acetic acid solution is 1% ionised. An another acetic acid is 10% ionised. What will be the concentration of another acetic acid? | null |
518 | ab2386d4-6ddd-11ea-8f08-ccda262736ce | https://socratic.org/questions/how-many-moles-of-ethylene-c-2h-4-can-react-with-12-9-liters-of-oxygen-gas-at-1- | 0.21 moles | start physical_unit 5 5 mole mol qc_end physical_unit 12 13 9 10 volume qc_end physical_unit 12 13 15 16 pressure qc_end physical_unit 12 13 18 19 temperature qc_end end | [{"type":"physical unit","value":"Mole [OF] C2H4 [IN] moles"}] | [{"type":"physical unit","value":"0.21 moles"}] | [{"type":"physical unit","value":"Volume [OF] oxygen gas [=] \\pu{12.9 liters}"},{"type":"physical unit","value":"Pressure [OF] oxygen gas [=] \\pu{1.2 atmospheres}"},{"type":"physical unit","value":"Temperature [OF] oxygen gas [=] \\pu{297 Kelvin}"}] | <h1 class="questionTitle" itemprop="name">How many moles of ethylene (#C_2H_4#) can react with 12.9 liters of oxygen gas at 1.2 atmospheres and 297 Kelvin? </h1> | null | 0.21 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by using the ideal gas equation to figure out the the number of moles <mathjax>#O_2#</mathjax> used.</p>
<p><mathjax># P*V= n*R*T#</mathjax></p>
<p>Where</p>
<p><mathjax>#P" "#</mathjax> is the pressure expressed in atm.</p>
<p><mathjax>#V" "#</mathjax> is the volume expressed in L.</p>
<p><mathjax>#n" " #</mathjax> is the number of moles.</p>
<p><mathjax>#R" "#</mathjax> is the universal gas constant, it has a value of <mathjax>#0.0821\ L* atm. mol^-1*K^-1#</mathjax></p>
<p><mathjax>#T" " #</mathjax>is the Kelvin temperature.</p>
<p>Now rearrange the ideal gas equation and solve for n.</p>
<p><mathjax>#n = (P*V)/(R*T)#</mathjax></p>
<p><mathjax>#n = (12.9 \ Lxx1.2 \ atm)/(0.0821\ L* atm. mol^-1*K^-1 xx 297 \ K)#</mathjax></p>
<p><mathjax>#n = (12.9 \ cancel(L)xx1.2 \ cancel( atm))/(0.0821\ cancel(L) *cancel(atm) *mol^-1*cancel(K^-1) xx 297 \ cancel(K))#</mathjax></p>
<p><mathjax>#n = 0.635 \ mol.#</mathjax></p>
<p>In the second step, write a balanced chemical equation for the reaction and use the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the equation to figure out the moles of ethylene used.</p>
<p><mathjax>#C_2H_4 + 3O_2 -> 2CO_2 + 2H_2O#</mathjax></p>
<p><mathjax>#0.635 \ mol. O_2 xx (1\ mol. \ C_2H_4)/(3 \ mol. \ O_2) #</mathjax></p>
<p><mathjax>#0.635 \ cancel (mol. O_2) xx (1\ mol. \ C_2H_4)/(3 \ cancel( mol. \ O_2))#</mathjax></p>
<p><mathjax>#0.212 \ mol. C_2H_4#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#0.212 \ mol. C_2H_4#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by using the ideal gas equation to figure out the the number of moles <mathjax>#O_2#</mathjax> used.</p>
<p><mathjax># P*V= n*R*T#</mathjax></p>
<p>Where</p>
<p><mathjax>#P" "#</mathjax> is the pressure expressed in atm.</p>
<p><mathjax>#V" "#</mathjax> is the volume expressed in L.</p>
<p><mathjax>#n" " #</mathjax> is the number of moles.</p>
<p><mathjax>#R" "#</mathjax> is the universal gas constant, it has a value of <mathjax>#0.0821\ L* atm. mol^-1*K^-1#</mathjax></p>
<p><mathjax>#T" " #</mathjax>is the Kelvin temperature.</p>
<p>Now rearrange the ideal gas equation and solve for n.</p>
<p><mathjax>#n = (P*V)/(R*T)#</mathjax></p>
<p><mathjax>#n = (12.9 \ Lxx1.2 \ atm)/(0.0821\ L* atm. mol^-1*K^-1 xx 297 \ K)#</mathjax></p>
<p><mathjax>#n = (12.9 \ cancel(L)xx1.2 \ cancel( atm))/(0.0821\ cancel(L) *cancel(atm) *mol^-1*cancel(K^-1) xx 297 \ cancel(K))#</mathjax></p>
<p><mathjax>#n = 0.635 \ mol.#</mathjax></p>
<p>In the second step, write a balanced chemical equation for the reaction and use the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the equation to figure out the moles of ethylene used.</p>
<p><mathjax>#C_2H_4 + 3O_2 -> 2CO_2 + 2H_2O#</mathjax></p>
<p><mathjax>#0.635 \ mol. O_2 xx (1\ mol. \ C_2H_4)/(3 \ mol. \ O_2) #</mathjax></p>
<p><mathjax>#0.635 \ cancel (mol. O_2) xx (1\ mol. \ C_2H_4)/(3 \ cancel( mol. \ O_2))#</mathjax></p>
<p><mathjax>#0.212 \ mol. C_2H_4#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of ethylene (#C_2H_4#) can react with 12.9 liters of oxygen gas at 1.2 atmospheres and 297 Kelvin? </h1>
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Sam
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<div class="markdown"><p><mathjax>#0.212 \ mol. C_2H_4#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by using the ideal gas equation to figure out the the number of moles <mathjax>#O_2#</mathjax> used.</p>
<p><mathjax># P*V= n*R*T#</mathjax></p>
<p>Where</p>
<p><mathjax>#P" "#</mathjax> is the pressure expressed in atm.</p>
<p><mathjax>#V" "#</mathjax> is the volume expressed in L.</p>
<p><mathjax>#n" " #</mathjax> is the number of moles.</p>
<p><mathjax>#R" "#</mathjax> is the universal gas constant, it has a value of <mathjax>#0.0821\ L* atm. mol^-1*K^-1#</mathjax></p>
<p><mathjax>#T" " #</mathjax>is the Kelvin temperature.</p>
<p>Now rearrange the ideal gas equation and solve for n.</p>
<p><mathjax>#n = (P*V)/(R*T)#</mathjax></p>
<p><mathjax>#n = (12.9 \ Lxx1.2 \ atm)/(0.0821\ L* atm. mol^-1*K^-1 xx 297 \ K)#</mathjax></p>
<p><mathjax>#n = (12.9 \ cancel(L)xx1.2 \ cancel( atm))/(0.0821\ cancel(L) *cancel(atm) *mol^-1*cancel(K^-1) xx 297 \ cancel(K))#</mathjax></p>
<p><mathjax>#n = 0.635 \ mol.#</mathjax></p>
<p>In the second step, write a balanced chemical equation for the reaction and use the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the equation to figure out the moles of ethylene used.</p>
<p><mathjax>#C_2H_4 + 3O_2 -> 2CO_2 + 2H_2O#</mathjax></p>
<p><mathjax>#0.635 \ mol. O_2 xx (1\ mol. \ C_2H_4)/(3 \ mol. \ O_2) #</mathjax></p>
<p><mathjax>#0.635 \ cancel (mol. O_2) xx (1\ mol. \ C_2H_4)/(3 \ cancel( mol. \ O_2))#</mathjax></p>
<p><mathjax>#0.212 \ mol. C_2H_4#</mathjax></p></div>
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</article> | How many moles of ethylene (#C_2H_4#) can react with 12.9 liters of oxygen gas at 1.2 atmospheres and 297 Kelvin? | null |
519 | ac5cb666-6ddd-11ea-8842-ccda262736ce | https://socratic.org/questions/what-is-the-specific-heat-capacity-of-lead | 0.13 J/(g * K) | start physical_unit 7 7 specific_heat_capacity j/(g_·_k) qc_end substance 7 7 qc_end end | [{"type":"physical unit","value":"Specific heat capacity [OF] lead [IN] J/(g * K)"}] | [{"type":"physical unit","value":"0.13 J/(g * K)"}] | [{"type":"substance name","value":"Lead"}] | <h1 class="questionTitle" itemprop="name">What is the specific heat capacity of lead?</h1> | null | 0.13 J/(g * K) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There really isn't a way to find this yourself without your own lab. I had to google it also.</p>
<p><a href="http://socratic.org/chemistry/thermochemistry/specific-heat">Specific heat</a> describes the amount of energy (In Joules) it would take to raise a set mass (In grams) of a substance by one Kelvin.</p></div>
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<div class="markdown"><p><mathjax>#0.128 J/(g*K)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There really isn't a way to find this yourself without your own lab. I had to google it also.</p>
<p><a href="http://socratic.org/chemistry/thermochemistry/specific-heat">Specific heat</a> describes the amount of energy (In Joules) it would take to raise a set mass (In grams) of a substance by one Kelvin.</p></div>
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<div class="markdown"><p><mathjax>#0.128 J/(g*K)#</mathjax></p></div>
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<div class="markdown"><p>There really isn't a way to find this yourself without your own lab. I had to google it also.</p>
<p><a href="http://socratic.org/chemistry/thermochemistry/specific-heat">Specific heat</a> describes the amount of energy (In Joules) it would take to raise a set mass (In grams) of a substance by one Kelvin.</p></div>
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</article> | What is the specific heat capacity of lead? | null |
520 | a96c230b-6ddd-11ea-a5ef-ccda262736ce | https://socratic.org/questions/how-many-moles-of-oxygen-gas-are-produced-when-0-500-mol-na-2o-2-reacts-with-wat | 0.25 moles | start physical_unit 4 5 mole mol qc_end physical_unit 11 11 9 10 mole qc_end substance 14 14 qc_end end | [{"type":"physical unit","value":"Mole [OF] oxygen gas [IN] moles"}] | [{"type":"physical unit","value":"0.25 moles"}] | [{"type":"physical unit","value":"Mole [OF] Na2O2 [=] \\pu{0.500 mol}"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">How many moles of oxygen gas are produced when 0.500 mol #Na_2O_2# reacts with water?</h1> | null | 0.25 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium peroxide can be thermolyzed to give dioxygen gas...</p>
<p><mathjax>#Na_2O_2(s) +Delta rarr Na_2O(s) + 1/2O_2(g)uarr#</mathjax></p>
<p>But with water, we simply get an acid base reaction....</p>
<p><mathjax>#Na_2O_2(s) +2H_2O(l) rarr 2NaOH(aq) + H_2O_2(aq)#</mathjax>...</p></div>
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<div class="markdown"><p>As far as I know NONE.....</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium peroxide can be thermolyzed to give dioxygen gas...</p>
<p><mathjax>#Na_2O_2(s) +Delta rarr Na_2O(s) + 1/2O_2(g)uarr#</mathjax></p>
<p>But with water, we simply get an acid base reaction....</p>
<p><mathjax>#Na_2O_2(s) +2H_2O(l) rarr 2NaOH(aq) + H_2O_2(aq)#</mathjax>...</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of oxygen gas are produced when 0.500 mol #Na_2O_2# reacts with water?</h1>
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<div class="markdown"><p>As far as I know NONE.....</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium peroxide can be thermolyzed to give dioxygen gas...</p>
<p><mathjax>#Na_2O_2(s) +Delta rarr Na_2O(s) + 1/2O_2(g)uarr#</mathjax></p>
<p>But with water, we simply get an acid base reaction....</p>
<p><mathjax>#Na_2O_2(s) +2H_2O(l) rarr 2NaOH(aq) + H_2O_2(aq)#</mathjax>...</p></div>
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</article> | How many moles of oxygen gas are produced when 0.500 mol #Na_2O_2# reacts with water? | null |
521 | ac83c6ec-6ddd-11ea-bf24-ccda262736ce | https://socratic.org/questions/an-organic-compound-is-known-to-be-composed-of-46-60-carbon-6-800-hydrogen-and-4 | C4H7O3 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the organic compound [IN] empirical"}] | [{"type":"chemical equation","value":"C4H7O3"}] | [{"type":"physical unit","value":"Percent [OF] carbon in the organic compound [=] \\pu{46.60%}"},{"type":"physical unit","value":"Percent [OF] hydrogen in the organic compound [=] \\pu{6.800%}"},{"type":"physical unit","value":"Percent [OF] oxygen in the organic compound [=] \\pu{46.60%}"}] | <h1 class="questionTitle" itemprop="name">An organic compound is known to be composed of 46.60% carbon, 6.800% hydrogen, and 46.60% oxygen. What is its empirical formula? </h1> | null | C4H7O3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Step 1:</em></p>
<p>For this type of problem, if the percentages add up to 100% (and it almost always is), then it safe to assume 100g sample. </p>
<p>By assuming that the sample is 100g, we can then assume that the corresponding percentages are also the weights. </p>
<p>Despite the fact that, in your case, the sum is only 99.8%, we'll just go on with the assumption of 100g sample shall we?</p>
<p>Thus, mass of C = 46.60 g; mass of H = 6.80g; mass of O = 46.60g</p>
<p><em>Step 2:</em></p>
<p>Since chemical formulas deal more with number of moles rather than weights, we have to convert these masses by multiplying them with their respective atomic masses.</p>
<p>mole C = 46.60 <mathjax>#cancel g#</mathjax> x <mathjax>#(1mol)/(12.0107 cancel "g") C#</mathjax> = 3.8799 mol C</p>
<p>mole H = 6.80 <mathjax>#cancel g#</mathjax> x <mathjax>#(1mol)/(1.00794 cancel "g") H#</mathjax> = 6.7464 mol H</p>
<p>mole O = 46.60 <mathjax>#cancel g#</mathjax> x <mathjax>#(1mol)/(15.9994 cancel "g") O#</mathjax> = 2.9126 mol O</p>
<p><em>Step 3</em>:</p>
<p>Now that we have the number of moles of each element, we have to get their respective ratios by dividing everything with the <em>smallest</em> number of mole.</p>
<p><mathjax>#C#</mathjax> = <mathjax>#(3.8799 cancel "mol")/(2.9126 cancel "mol")#</mathjax> = <mathjax>#1.3321#</mathjax></p>
<p><mathjax>#H#</mathjax> = <mathjax>#(6.7464 cancel "mol")/(2.9126 cancel "mol")#</mathjax> = <mathjax>#2.3163#</mathjax></p>
<p><mathjax>#O#</mathjax> = <mathjax>#(2.9126 cancel "mol")/(2.9126 cancel "mol")#</mathjax> = <mathjax>#1.000#</mathjax></p>
<p><em>Step 4</em>:</p>
<p>Round off the answers to the nearest whole number. For the above answers, you can multiply each number by a factor to get the lowest whole number multiple. In this case, the factor is 3. Thus,</p>
<p><mathjax>#C#</mathjax> = 1.3321 x 3 = <mathjax>#3.99~~4#</mathjax></p>
<p><mathjax>#H#</mathjax> = 2.3163 x 3 = <mathjax>#6.95~~7#</mathjax></p>
<p><mathjax>#O#</mathjax> = 1.000 x 3 = <mathjax>#3#</mathjax></p>
<p>Therefore, answer is <mathjax>#C_4H_7O_3#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#C_4H_7O_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Step 1:</em></p>
<p>For this type of problem, if the percentages add up to 100% (and it almost always is), then it safe to assume 100g sample. </p>
<p>By assuming that the sample is 100g, we can then assume that the corresponding percentages are also the weights. </p>
<p>Despite the fact that, in your case, the sum is only 99.8%, we'll just go on with the assumption of 100g sample shall we?</p>
<p>Thus, mass of C = 46.60 g; mass of H = 6.80g; mass of O = 46.60g</p>
<p><em>Step 2:</em></p>
<p>Since chemical formulas deal more with number of moles rather than weights, we have to convert these masses by multiplying them with their respective atomic masses.</p>
<p>mole C = 46.60 <mathjax>#cancel g#</mathjax> x <mathjax>#(1mol)/(12.0107 cancel "g") C#</mathjax> = 3.8799 mol C</p>
<p>mole H = 6.80 <mathjax>#cancel g#</mathjax> x <mathjax>#(1mol)/(1.00794 cancel "g") H#</mathjax> = 6.7464 mol H</p>
<p>mole O = 46.60 <mathjax>#cancel g#</mathjax> x <mathjax>#(1mol)/(15.9994 cancel "g") O#</mathjax> = 2.9126 mol O</p>
<p><em>Step 3</em>:</p>
<p>Now that we have the number of moles of each element, we have to get their respective ratios by dividing everything with the <em>smallest</em> number of mole.</p>
<p><mathjax>#C#</mathjax> = <mathjax>#(3.8799 cancel "mol")/(2.9126 cancel "mol")#</mathjax> = <mathjax>#1.3321#</mathjax></p>
<p><mathjax>#H#</mathjax> = <mathjax>#(6.7464 cancel "mol")/(2.9126 cancel "mol")#</mathjax> = <mathjax>#2.3163#</mathjax></p>
<p><mathjax>#O#</mathjax> = <mathjax>#(2.9126 cancel "mol")/(2.9126 cancel "mol")#</mathjax> = <mathjax>#1.000#</mathjax></p>
<p><em>Step 4</em>:</p>
<p>Round off the answers to the nearest whole number. For the above answers, you can multiply each number by a factor to get the lowest whole number multiple. In this case, the factor is 3. Thus,</p>
<p><mathjax>#C#</mathjax> = 1.3321 x 3 = <mathjax>#3.99~~4#</mathjax></p>
<p><mathjax>#H#</mathjax> = 2.3163 x 3 = <mathjax>#6.95~~7#</mathjax></p>
<p><mathjax>#O#</mathjax> = 1.000 x 3 = <mathjax>#3#</mathjax></p>
<p>Therefore, answer is <mathjax>#C_4H_7O_3#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">An organic compound is known to be composed of 46.60% carbon, 6.800% hydrogen, and 46.60% oxygen. What is its empirical formula? </h1>
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Nikka C.
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Nov 3, 2015
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<div class="markdown"><p><mathjax>#C_4H_7O_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Step 1:</em></p>
<p>For this type of problem, if the percentages add up to 100% (and it almost always is), then it safe to assume 100g sample. </p>
<p>By assuming that the sample is 100g, we can then assume that the corresponding percentages are also the weights. </p>
<p>Despite the fact that, in your case, the sum is only 99.8%, we'll just go on with the assumption of 100g sample shall we?</p>
<p>Thus, mass of C = 46.60 g; mass of H = 6.80g; mass of O = 46.60g</p>
<p><em>Step 2:</em></p>
<p>Since chemical formulas deal more with number of moles rather than weights, we have to convert these masses by multiplying them with their respective atomic masses.</p>
<p>mole C = 46.60 <mathjax>#cancel g#</mathjax> x <mathjax>#(1mol)/(12.0107 cancel "g") C#</mathjax> = 3.8799 mol C</p>
<p>mole H = 6.80 <mathjax>#cancel g#</mathjax> x <mathjax>#(1mol)/(1.00794 cancel "g") H#</mathjax> = 6.7464 mol H</p>
<p>mole O = 46.60 <mathjax>#cancel g#</mathjax> x <mathjax>#(1mol)/(15.9994 cancel "g") O#</mathjax> = 2.9126 mol O</p>
<p><em>Step 3</em>:</p>
<p>Now that we have the number of moles of each element, we have to get their respective ratios by dividing everything with the <em>smallest</em> number of mole.</p>
<p><mathjax>#C#</mathjax> = <mathjax>#(3.8799 cancel "mol")/(2.9126 cancel "mol")#</mathjax> = <mathjax>#1.3321#</mathjax></p>
<p><mathjax>#H#</mathjax> = <mathjax>#(6.7464 cancel "mol")/(2.9126 cancel "mol")#</mathjax> = <mathjax>#2.3163#</mathjax></p>
<p><mathjax>#O#</mathjax> = <mathjax>#(2.9126 cancel "mol")/(2.9126 cancel "mol")#</mathjax> = <mathjax>#1.000#</mathjax></p>
<p><em>Step 4</em>:</p>
<p>Round off the answers to the nearest whole number. For the above answers, you can multiply each number by a factor to get the lowest whole number multiple. In this case, the factor is 3. Thus,</p>
<p><mathjax>#C#</mathjax> = 1.3321 x 3 = <mathjax>#3.99~~4#</mathjax></p>
<p><mathjax>#H#</mathjax> = 2.3163 x 3 = <mathjax>#6.95~~7#</mathjax></p>
<p><mathjax>#O#</mathjax> = 1.000 x 3 = <mathjax>#3#</mathjax></p>
<p>Therefore, answer is <mathjax>#C_4H_7O_3#</mathjax></p></div>
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</article> | An organic compound is known to be composed of 46.60% carbon, 6.800% hydrogen, and 46.60% oxygen. What is its empirical formula? | null |
522 | ac8c9ea2-6ddd-11ea-80e6-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-1-0-10-3-m-sodium-hydroxide-solution | 11.00 | start physical_unit 10 12 ph none qc_end physical_unit 10 12 6 9 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] sodium hydroxide solution"}] | [{"type":"physical unit","value":"11.00"}] | [{"type":"physical unit","value":"Molarity [OF] sodium hydroxide solution [=] \\pu{1.0 × 10^(-3) M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a #1.0 * 10^-3# #M# sodium hydroxide solution?</h1> | null | 11.00 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We examine the equilibrium:</p>
<p><mathjax>#2H_2O(l) rightleftharpoons H_3O^+ + ""^(-)OH#</mathjax></p>
<p>Under standard conditions, <mathjax>#[H_3O^+][HO^-]=10^-14#</mathjax>.</p>
<p>And if we take <mathjax>#log_10#</mathjax> of both sides:</p>
<p><mathjax>#log_10[H_3O^+]+log_10[HO^-]=-14#</mathjax></p>
<p>Om rearrangement:</p>
<p><mathjax>#14=-log_10[H_3O^+]-log_10[HO^-]#</mathjax></p>
<p>But by definition, <mathjax>#-log_10[H_3O^+]=pH#</mathjax>, and <mathjax>#log_10[HO^-]=pOH#</mathjax>.</p>
<p>And thus our definining relationship: <mathjax>#pH+pOH=14#</mathjax>.</p>
<p>Since (finally!) <mathjax>#pOH=-log_10[HO^-]=-log_10(1xx10^-3)=-(-3)=3#</mathjax>.</p>
<p>And if <mathjax>#pOH=3#</mathjax>, <mathjax>#pH=14-3=??#</mathjax></p>
<p>In A level, you do have to remember the result:</p>
<p><mathjax>#pH+pOH=14#</mathjax>. With this is mind, these problems become trivial. </p></div>
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<div>
<div class="markdown"><p><mathjax>#pH=11#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We examine the equilibrium:</p>
<p><mathjax>#2H_2O(l) rightleftharpoons H_3O^+ + ""^(-)OH#</mathjax></p>
<p>Under standard conditions, <mathjax>#[H_3O^+][HO^-]=10^-14#</mathjax>.</p>
<p>And if we take <mathjax>#log_10#</mathjax> of both sides:</p>
<p><mathjax>#log_10[H_3O^+]+log_10[HO^-]=-14#</mathjax></p>
<p>Om rearrangement:</p>
<p><mathjax>#14=-log_10[H_3O^+]-log_10[HO^-]#</mathjax></p>
<p>But by definition, <mathjax>#-log_10[H_3O^+]=pH#</mathjax>, and <mathjax>#log_10[HO^-]=pOH#</mathjax>.</p>
<p>And thus our definining relationship: <mathjax>#pH+pOH=14#</mathjax>.</p>
<p>Since (finally!) <mathjax>#pOH=-log_10[HO^-]=-log_10(1xx10^-3)=-(-3)=3#</mathjax>.</p>
<p>And if <mathjax>#pOH=3#</mathjax>, <mathjax>#pH=14-3=??#</mathjax></p>
<p>In A level, you do have to remember the result:</p>
<p><mathjax>#pH+pOH=14#</mathjax>. With this is mind, these problems become trivial. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a #1.0 * 10^-3# #M# sodium hydroxide solution?</h1>
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<div class="markdown"><p><mathjax>#pH=11#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We examine the equilibrium:</p>
<p><mathjax>#2H_2O(l) rightleftharpoons H_3O^+ + ""^(-)OH#</mathjax></p>
<p>Under standard conditions, <mathjax>#[H_3O^+][HO^-]=10^-14#</mathjax>.</p>
<p>And if we take <mathjax>#log_10#</mathjax> of both sides:</p>
<p><mathjax>#log_10[H_3O^+]+log_10[HO^-]=-14#</mathjax></p>
<p>Om rearrangement:</p>
<p><mathjax>#14=-log_10[H_3O^+]-log_10[HO^-]#</mathjax></p>
<p>But by definition, <mathjax>#-log_10[H_3O^+]=pH#</mathjax>, and <mathjax>#log_10[HO^-]=pOH#</mathjax>.</p>
<p>And thus our definining relationship: <mathjax>#pH+pOH=14#</mathjax>.</p>
<p>Since (finally!) <mathjax>#pOH=-log_10[HO^-]=-log_10(1xx10^-3)=-(-3)=3#</mathjax>.</p>
<p>And if <mathjax>#pOH=3#</mathjax>, <mathjax>#pH=14-3=??#</mathjax></p>
<p>In A level, you do have to remember the result:</p>
<p><mathjax>#pH+pOH=14#</mathjax>. With this is mind, these problems become trivial. </p></div>
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</article> | What is the pH of a #1.0 * 10^-3# #M# sodium hydroxide solution? | null |
523 | a9b4b4da-6ddd-11ea-ac17-ccda262736ce | https://socratic.org/questions/if-i-have-5-6-liters-of-gas-in-a-piston-at-a-pressure-of-1-5-atm-and-compress-th | 1.75 atm | start physical_unit 6 6 pressure atm qc_end physical_unit 6 6 3 4 volume qc_end physical_unit 6 6 14 15 pressure qc_end physical_unit 6 6 24 25 volume qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] gas [IN] atm"}] | [{"type":"physical unit","value":"1.75 atm"}] | [{"type":"physical unit","value":"Volume1 [OF] gas [=] \\pu{5.6 liters}"},{"type":"physical unit","value":"Pressure1 [OF] gas [=] \\pu{1.5 atm}"},{"type":"physical unit","value":"Volume2 [OF] gas [=] \\pu{4.8 L}"}] | <h1 class="questionTitle" itemprop="name">If I have 5.6 liters of gas in a piston at a pressure of 1.5 atm and compress the gas until its volume is 4.8 L, what will the new pressure inside the piston be? </h1> | null | 1.75 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At constant <mathjax>#T#</mathjax>, <mathjax>#P_1V_1=P_2V_2#</mathjax>.......</p>
<p>We solve for <mathjax>#P_2=(P_1V_1)/V_2#</mathjax>, and clearly the <mathjax>#"RHS"#</mathjax> will have the required units of pressure:</p>
<p><mathjax>#P_2=(1.5*atmxx5.6*L)/(4.8*L)=1.75*atm#</mathjax></p>
<p>That <mathjax>#P_2>P_1#</mathjax> is quite reasonable, because we are compressing the piston. </p></div>
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<div class="markdown"><p><mathjax>#P_2=1.75*atm.........#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At constant <mathjax>#T#</mathjax>, <mathjax>#P_1V_1=P_2V_2#</mathjax>.......</p>
<p>We solve for <mathjax>#P_2=(P_1V_1)/V_2#</mathjax>, and clearly the <mathjax>#"RHS"#</mathjax> will have the required units of pressure:</p>
<p><mathjax>#P_2=(1.5*atmxx5.6*L)/(4.8*L)=1.75*atm#</mathjax></p>
<p>That <mathjax>#P_2>P_1#</mathjax> is quite reasonable, because we are compressing the piston. </p></div>
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<h1 class="questionTitle" itemprop="name">If I have 5.6 liters of gas in a piston at a pressure of 1.5 atm and compress the gas until its volume is 4.8 L, what will the new pressure inside the piston be? </h1>
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<div class="markdown"><p><mathjax>#P_2=1.75*atm.........#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At constant <mathjax>#T#</mathjax>, <mathjax>#P_1V_1=P_2V_2#</mathjax>.......</p>
<p>We solve for <mathjax>#P_2=(P_1V_1)/V_2#</mathjax>, and clearly the <mathjax>#"RHS"#</mathjax> will have the required units of pressure:</p>
<p><mathjax>#P_2=(1.5*atmxx5.6*L)/(4.8*L)=1.75*atm#</mathjax></p>
<p>That <mathjax>#P_2>P_1#</mathjax> is quite reasonable, because we are compressing the piston. </p></div>
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</article> | If I have 5.6 liters of gas in a piston at a pressure of 1.5 atm and compress the gas until its volume is 4.8 L, what will the new pressure inside the piston be? | null |
524 | ad0b7170-6ddd-11ea-9555-ccda262736ce | https://socratic.org/questions/a-if-11-08-ml-of-0-1111-m-naoh-were-required-to-neutralize-25-00-ml-of-a-monopro | 0.05 M | start physical_unit 15 16 concentration mol/l qc_end physical_unit 6 6 4 5 concentration qc_end physical_unit 15 16 11 12 volume qc_end end | [{"type":"physical unit","value":"Concentration [OF] monoprotic acid [IN] M"}] | [{"type":"physical unit","value":"0.05 M"}] | [{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{11.08 mL}"},{"type":"physical unit","value":"Concentration [OF] NaOH solution [=] \\pu{0.1111 M}"},{"type":"physical unit","value":"Volume [OF] monoprotic acid [=] \\pu{25.00 mL }"}] | <h1 class="questionTitle" itemprop="name">If 11.08 mL of 0.1111 M NaOH were required to neutralize 25.00 mL of a monoprotic acid, what was the concentration of the acid?
</h1> | null | 0.05 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that you're dealing with a <em>monoprotic acid</em> here, so you can represent this unknown acid as <mathjax>#"HA"#</mathjax> and write the balanced chemical equation that describes the <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction like this</p>
<blockquote>
<p><mathjax>#"NaOH"_ ((aq)) + "HA"_ ((aq)) -> "NaA"_ ((aq)) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>So <mathjax>#1#</mathjax> <strong>mole</strong> of sodium hydroxide will neutralize <mathjax>#1#</mathjax> <strong>mole</strong> of the acid. </p>
<p>Now, you know that you can express the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present in a solution in terms of the <strong>volume</strong> of the solution and of its <strong>molar concentration</strong>. </p>
<blockquote>
<p><mathjax>#n = c * V#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#n#</mathjax> is the <strong>number of moles</strong> of solute</li>
<li><mathjax>#c#</mathjax> is the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the solution</li>
<li><mathjax>#V#</mathjax> is the <strong>volume</strong> of the solution, expressed in <strong>liters</strong></li>
</ul>
</blockquote>
<p>So, you know that this neutralization reaction required </p>
<blockquote>
<p><mathjax>#"11.08 mL" = 11.08 * 10^(-3) quad "L"#</mathjax> </p>
</blockquote>
<p>of a </p>
<blockquote>
<p><mathjax>#"0.1111 M" = "0.1111 mol L"^(-1)#</mathjax> </p>
</blockquote>
<p>sodium hydroxide solution, which means that you have</p>
<blockquote>
<p><mathjax>#n_ ("NaOH") = "0.1111 mol" color(red)(cancel(color(black)("L"^(-1)))) * 11.08 * 10^(-3) color(red)(cancel(color(black)("L")))#</mathjax></p>
<p><mathjax>#n_ ("NaOH") = 1.231 * 10^(-3) quad "moles"#</mathjax></p>
</blockquote>
<p>You also now that this number of moles of sodium hydroxide neutralized <mathjax>#"25.00 mL"#</mathjax> of a solution of the unknown acid. </p>
<p>Moreover, you know that the number of moles of sodium hydroxide consumed in the reaction must be <strong>equal</strong> to the number of moles of acid consumed in the reaction. </p>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#n_ ("HA") = 1.231 * 10^(-3) quad "moles"#</mathjax></p>
</blockquote>
<p>and--don't forget that the volume must be expressed in <strong>liters</strong>!</p>
<blockquote>
<p><mathjax>#1.231 * color(red)(cancel(color(black)(10^(-3)))) quad "moles" = c_ ("HA") * 25.00 * color(red)(cancel(color(black)(10^(-3)))) quad "L"#</mathjax></p>
</blockquote>
<p>The concentration of the acid solution is equal to</p>
<blockquote>
<p><mathjax>#c_ "HA" = "1.231 moles"/"25.00 L" = color(darkgreen)(ul(color(black)("0.04924 mol L"^(-1))))#</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></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.04924 mol L"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that you're dealing with a <em>monoprotic acid</em> here, so you can represent this unknown acid as <mathjax>#"HA"#</mathjax> and write the balanced chemical equation that describes the <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction like this</p>
<blockquote>
<p><mathjax>#"NaOH"_ ((aq)) + "HA"_ ((aq)) -> "NaA"_ ((aq)) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>So <mathjax>#1#</mathjax> <strong>mole</strong> of sodium hydroxide will neutralize <mathjax>#1#</mathjax> <strong>mole</strong> of the acid. </p>
<p>Now, you know that you can express the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present in a solution in terms of the <strong>volume</strong> of the solution and of its <strong>molar concentration</strong>. </p>
<blockquote>
<p><mathjax>#n = c * V#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#n#</mathjax> is the <strong>number of moles</strong> of solute</li>
<li><mathjax>#c#</mathjax> is the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the solution</li>
<li><mathjax>#V#</mathjax> is the <strong>volume</strong> of the solution, expressed in <strong>liters</strong></li>
</ul>
</blockquote>
<p>So, you know that this neutralization reaction required </p>
<blockquote>
<p><mathjax>#"11.08 mL" = 11.08 * 10^(-3) quad "L"#</mathjax> </p>
</blockquote>
<p>of a </p>
<blockquote>
<p><mathjax>#"0.1111 M" = "0.1111 mol L"^(-1)#</mathjax> </p>
</blockquote>
<p>sodium hydroxide solution, which means that you have</p>
<blockquote>
<p><mathjax>#n_ ("NaOH") = "0.1111 mol" color(red)(cancel(color(black)("L"^(-1)))) * 11.08 * 10^(-3) color(red)(cancel(color(black)("L")))#</mathjax></p>
<p><mathjax>#n_ ("NaOH") = 1.231 * 10^(-3) quad "moles"#</mathjax></p>
</blockquote>
<p>You also now that this number of moles of sodium hydroxide neutralized <mathjax>#"25.00 mL"#</mathjax> of a solution of the unknown acid. </p>
<p>Moreover, you know that the number of moles of sodium hydroxide consumed in the reaction must be <strong>equal</strong> to the number of moles of acid consumed in the reaction. </p>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#n_ ("HA") = 1.231 * 10^(-3) quad "moles"#</mathjax></p>
</blockquote>
<p>and--don't forget that the volume must be expressed in <strong>liters</strong>!</p>
<blockquote>
<p><mathjax>#1.231 * color(red)(cancel(color(black)(10^(-3)))) quad "moles" = c_ ("HA") * 25.00 * color(red)(cancel(color(black)(10^(-3)))) quad "L"#</mathjax></p>
</blockquote>
<p>The concentration of the acid solution is equal to</p>
<blockquote>
<p><mathjax>#c_ "HA" = "1.231 moles"/"25.00 L" = color(darkgreen)(ul(color(black)("0.04924 mol L"^(-1))))#</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></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If 11.08 mL of 0.1111 M NaOH were required to neutralize 25.00 mL of a monoprotic acid, what was the concentration of the acid?
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<div class="markdown"><p><mathjax>#"0.04924 mol L"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that you're dealing with a <em>monoprotic acid</em> here, so you can represent this unknown acid as <mathjax>#"HA"#</mathjax> and write the balanced chemical equation that describes the <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction like this</p>
<blockquote>
<p><mathjax>#"NaOH"_ ((aq)) + "HA"_ ((aq)) -> "NaA"_ ((aq)) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>So <mathjax>#1#</mathjax> <strong>mole</strong> of sodium hydroxide will neutralize <mathjax>#1#</mathjax> <strong>mole</strong> of the acid. </p>
<p>Now, you know that you can express the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present in a solution in terms of the <strong>volume</strong> of the solution and of its <strong>molar concentration</strong>. </p>
<blockquote>
<p><mathjax>#n = c * V#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#n#</mathjax> is the <strong>number of moles</strong> of solute</li>
<li><mathjax>#c#</mathjax> is the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the solution</li>
<li><mathjax>#V#</mathjax> is the <strong>volume</strong> of the solution, expressed in <strong>liters</strong></li>
</ul>
</blockquote>
<p>So, you know that this neutralization reaction required </p>
<blockquote>
<p><mathjax>#"11.08 mL" = 11.08 * 10^(-3) quad "L"#</mathjax> </p>
</blockquote>
<p>of a </p>
<blockquote>
<p><mathjax>#"0.1111 M" = "0.1111 mol L"^(-1)#</mathjax> </p>
</blockquote>
<p>sodium hydroxide solution, which means that you have</p>
<blockquote>
<p><mathjax>#n_ ("NaOH") = "0.1111 mol" color(red)(cancel(color(black)("L"^(-1)))) * 11.08 * 10^(-3) color(red)(cancel(color(black)("L")))#</mathjax></p>
<p><mathjax>#n_ ("NaOH") = 1.231 * 10^(-3) quad "moles"#</mathjax></p>
</blockquote>
<p>You also now that this number of moles of sodium hydroxide neutralized <mathjax>#"25.00 mL"#</mathjax> of a solution of the unknown acid. </p>
<p>Moreover, you know that the number of moles of sodium hydroxide consumed in the reaction must be <strong>equal</strong> to the number of moles of acid consumed in the reaction. </p>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#n_ ("HA") = 1.231 * 10^(-3) quad "moles"#</mathjax></p>
</blockquote>
<p>and--don't forget that the volume must be expressed in <strong>liters</strong>!</p>
<blockquote>
<p><mathjax>#1.231 * color(red)(cancel(color(black)(10^(-3)))) quad "moles" = c_ ("HA") * 25.00 * color(red)(cancel(color(black)(10^(-3)))) quad "L"#</mathjax></p>
</blockquote>
<p>The concentration of the acid solution is equal to</p>
<blockquote>
<p><mathjax>#c_ "HA" = "1.231 moles"/"25.00 L" = color(darkgreen)(ul(color(black)("0.04924 mol L"^(-1))))#</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></div>
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</article> | If 11.08 mL of 0.1111 M NaOH were required to neutralize 25.00 mL of a monoprotic acid, what was the concentration of the acid?
| null |
525 | a8aa2d5c-6ddd-11ea-869f-ccda262736ce | https://socratic.org/questions/581cc13e11ef6b302576a3dc | Na2CO3 + 2 HCl -> 2 NaCl + H2O + CO2 | start chemical_equation qc_end substance 10 11 qc_end substance 13 14 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reaction"}] | [{"type":"chemical equation","value":"Na2CO3 + 2 HCl -> 2 NaCl + H2O + CO2"}] | [{"type":"substance name","value":"Sodium carbonate"},{"type":"substance name","value":"Hydrochloric acid"}] | <h1 class="questionTitle" itemprop="name">Can you represent the stoichiometric equation for the reaction between sodium carbonate and hydrochloric acid...?</h1> | null | Na2CO3 + 2 HCl -> 2 NaCl + H2O + CO2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Or as a net ionic equation:</p>
<p><mathjax>#CO_3^(2-) + 2H_3O^(+) rarr CO_2(g)uarr+ 3H_2O(l)#</mathjax></p>
<p>With all these equations, not only must mass be balanced, charge must also be balanced as well. It is all too easy to make a mistake. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#Na_2CO_3(aq) + 2HCl(aq) rarr 2NaCl(aq) + H_2O(l) + CO_2(g)uarr#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Or as a net ionic equation:</p>
<p><mathjax>#CO_3^(2-) + 2H_3O^(+) rarr CO_2(g)uarr+ 3H_2O(l)#</mathjax></p>
<p>With all these equations, not only must mass be balanced, charge must also be balanced as well. It is all too easy to make a mistake. </p></div>
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<h1 class="questionTitle" itemprop="name">Can you represent the stoichiometric equation for the reaction between sodium carbonate and hydrochloric acid...?</h1>
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<div class="markdown"><p><mathjax>#Na_2CO_3(aq) + 2HCl(aq) rarr 2NaCl(aq) + H_2O(l) + CO_2(g)uarr#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Or as a net ionic equation:</p>
<p><mathjax>#CO_3^(2-) + 2H_3O^(+) rarr CO_2(g)uarr+ 3H_2O(l)#</mathjax></p>
<p>With all these equations, not only must mass be balanced, charge must also be balanced as well. It is all too easy to make a mistake. </p></div>
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</article> | Can you represent the stoichiometric equation for the reaction between sodium carbonate and hydrochloric acid...? | null |
526 | acd864ca-6ddd-11ea-9050-ccda262736ce | https://socratic.org/questions/57f29d5fb72cff3262ccaf22 | 142.14 moles | start physical_unit 4 4 mole mol qc_end physical_unit 4 4 7 8 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] iron [IN] moles"}] | [{"type":"physical unit","value":"142.14 moles"}] | [{"type":"physical unit","value":"Mass [OF] iron [=] \\pu{50 lb}"}] | <h1 class="questionTitle" itemprop="name">How many moles of iron in a #50*lb# mass of iron?</h1> | null | 142.14 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#b.#</mathjax> If you have a <mathjax>#50*lb#</mathjax> mass, AND <mathjax>#1*lb-=454*g#</mathjax> (I don't know, I don't use these units, especially as the States has its own special variety), then..........</p>
<p><mathjax>#"Mass of iron in grams"-=50*cancel(lb)xx454*g*cancel(lb^-1)=??lbs#</mathjax></p>
<p><mathjax>#"Moles of iron"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(50*lbxx454*g*lb^-1)/((2xx55.85+3xx15.999)*g*mol^-1)=142*mol#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#a.#</mathjax> <mathjax>#"Formula mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(2xx55.85+3xx15.999)*g*mol^-1#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#b.#</mathjax> If you have a <mathjax>#50*lb#</mathjax> mass, AND <mathjax>#1*lb-=454*g#</mathjax> (I don't know, I don't use these units, especially as the States has its own special variety), then..........</p>
<p><mathjax>#"Mass of iron in grams"-=50*cancel(lb)xx454*g*cancel(lb^-1)=??lbs#</mathjax></p>
<p><mathjax>#"Moles of iron"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(50*lbxx454*g*lb^-1)/((2xx55.85+3xx15.999)*g*mol^-1)=142*mol#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#a.#</mathjax> <mathjax>#"Formula mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(2xx55.85+3xx15.999)*g*mol^-1#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#b.#</mathjax> If you have a <mathjax>#50*lb#</mathjax> mass, AND <mathjax>#1*lb-=454*g#</mathjax> (I don't know, I don't use these units, especially as the States has its own special variety), then..........</p>
<p><mathjax>#"Mass of iron in grams"-=50*cancel(lb)xx454*g*cancel(lb^-1)=??lbs#</mathjax></p>
<p><mathjax>#"Moles of iron"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(50*lbxx454*g*lb^-1)/((2xx55.85+3xx15.999)*g*mol^-1)=142*mol#</mathjax>.</p></div>
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</article> | How many moles of iron in a #50*lb# mass of iron? | null |
527 | a9db1858-6ddd-11ea-b2e1-ccda262736ce | https://socratic.org/questions/how-many-grams-of-o-are-contained-in-8-52-g-k-2co-3 | 3.12 grams | start physical_unit 4 4 mass g qc_end physical_unit 10 10 8 9 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] O [IN] grams"}] | [{"type":"physical unit","value":"3.12 grams"}] | [{"type":"physical unit","value":"Mass [OF] K2CO3 [=] \\pu{8.52 g}"}] | <h1 class="questionTitle" itemprop="name">How many grams of #O# are contained in 8.52 g #K_2CO_3#?</h1> | null | 3.12 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Moles of <mathjax>#K_2CO_3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(8.52*g)/(131.21*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#mol#</mathjax>.</p>
<p>But there are 3 mol of oxygen in each mole of <mathjax>#K_2CO_3#</mathjax>.</p>
<p>So moles of oxygen <mathjax>#=#</mathjax> <mathjax>#3xx(8.52*g)/(131.21*g*mol^-1) = ?? mol#</mathjax>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>In 1 mole of potassium carbonate there are clearly 3 moles of oxygen atoms (i.e. 48.00 g), almost half the mass of 1 mol of the salt.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Moles of <mathjax>#K_2CO_3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(8.52*g)/(131.21*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#mol#</mathjax>.</p>
<p>But there are 3 mol of oxygen in each mole of <mathjax>#K_2CO_3#</mathjax>.</p>
<p>So moles of oxygen <mathjax>#=#</mathjax> <mathjax>#3xx(8.52*g)/(131.21*g*mol^-1) = ?? mol#</mathjax>. </p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of #O# are contained in 8.52 g #K_2CO_3#?</h1>
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<div class="markdown"><p>In 1 mole of potassium carbonate there are clearly 3 moles of oxygen atoms (i.e. 48.00 g), almost half the mass of 1 mol of the salt.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Moles of <mathjax>#K_2CO_3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(8.52*g)/(131.21*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#mol#</mathjax>.</p>
<p>But there are 3 mol of oxygen in each mole of <mathjax>#K_2CO_3#</mathjax>.</p>
<p>So moles of oxygen <mathjax>#=#</mathjax> <mathjax>#3xx(8.52*g)/(131.21*g*mol^-1) = ?? mol#</mathjax>. </p></div>
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</article> | How many grams of #O# are contained in 8.52 g #K_2CO_3#? | null |
528 | aa403666-6ddd-11ea-b52d-ccda262736ce | https://socratic.org/questions/aluminum-reacts-with-excess-hydrochloric-acid-to-form-aqueous-aluminum-chloride- | 0.03 grams | start physical_unit 30 31 mass g qc_end physical_unit 15 16 12 13 volume qc_end physical_unit 18 18 41 42 partial_pressure qc_end c_other OTHER qc_end substance 8 10 qc_end end | [{"type":"physical unit","value":"Mass [OF] aluminum reacted [IN] grams"}] | [{"type":"physical unit","value":"0.03 grams"}] | [{"type":"physical unit","value":"Pressure [OF] the reaction [=] \\pu{751 mmHg}"},{"type":"physical unit","value":"Temperature [OF] the reaction [=] \\pu{27 degrees}"},{"type":"physical unit","value":"Volume [OF] hydrogen gas [=] \\pu{46.5 mL}"},{"type":"physical unit","value":"Partial pressure [OF] water [=] \\pu{26.8 mmHg}"},{"type":"other","value":"Excess hydrochloric acid."},{"type":"substance name","value":"Aqueous aluminum chloride"}] | <h1 class="questionTitle" itemprop="name">Aluminum reacts with excess hydrochloric acid to form aqueous aluminum chloride and 46.5 mL of hydrogen gas over water at 27 degrees Celsius and 751 mmHg. How many grams of aluminum reacted? The partial pressure of water at 27 C is 26.8 mmHg.</h1> | null | 0.03 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the number of grams of <mathjax>#"Al"#</mathjax> that reacted, given some <mathjax>#"H"_2 (g)#</mathjax> product characteristics. </p>
<p>Let's first write the chemical equation for this reaction:</p>
<p><mathjax>#2"Al"(s) + 6"HCl"(aq) rarr2 "AlCl"_3(aq) + 3"H"_2 (g)#</mathjax></p>
<p>The total pressure of the gaseous system is given as <mathjax>#751#</mathjax> <mathjax>#"mm Hg"#</mathjax>, and the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of water vapor is <mathjax>#26.8#</mathjax> <mathjax>#"mm Hg"#</mathjax> at <mathjax>#27^"o""C"#</mathjax>. The pressure of <em>hydrogen gas</em> is thus</p>
<p><mathjax>#P_"total" = P_ ("H"_2"O") + P_ ("H"_2)#</mathjax></p>
<p><mathjax>#P_ ("H"_2) = 751#</mathjax> <mathjax>#"mm Hg"#</mathjax> <mathjax>#- 26.8#</mathjax> <mathjax>#"mm Hg"#</mathjax> <mathjax>#= color(red)(724#</mathjax> <mathjax>#color(red)("mm Hg"#</mathjax></p>
<p>This pressure in atmospheres is </p>
<p><mathjax>#724cancel("mm Hg")((1color(white)(l)"atm")/(760cancel("mm Hg"))) = color(red)(0.953#</mathjax> <mathjax>#color(red)("atm"#</mathjax></p>
<p>We'll now use the ideal gas equation to find the number of moles of <mathjax>#"H"_2#</mathjax> formed:</p>
<p>(<mathjax>#T = 27^"o""C" + 273 = 300#</mathjax> <mathjax>#"K"#</mathjax>)</p>
<p><mathjax>#n = (PV)/(RT) = ((color(red)(0.953)cancel(color(red)("atm")))(0.0465cancel("L")))/((0.082057(cancel("L")•cancel("atm"))/("mol"•cancel("K")))(300cancel("K"))) = color(green)(0.00180#</mathjax> <mathjax>#color(green)("mol H"_2#</mathjax></p>
<p>(volume converted to liters here)</p>
<p>Using the coefficients of the chemical equation, we'll now find the relative number of moles of <mathjax>#"Al"#</mathjax> that react:</p>
<p><mathjax>#color(green)(0.00180)cancel(color(green)("mol H"_2))((2color(white)(l)"mol Al")/(3cancel("mol H"_2))) = color(purple)(0.00120#</mathjax> <mathjax>#color(purple)("mol Al"#</mathjax></p>
<p>Lastly, we'll use the <em>molar mass</em> of aluminum (<mathjax>#26.98#</mathjax> <mathjax>#"g/mol"#</mathjax>) to find the number of grams that reacted:</p>
<p><mathjax>#color(purple)(0.00120)cancel(color(purple)("mol Al"))((26.98color(white)(l)"g Al")/(1cancel("mol Al"))) = color(blue)(0.0324#</mathjax> <mathjax>#color(blue)("g Al"#</mathjax></p>
<p>Thus, <mathjax>#color(blue)(0.0324#</mathjax> <mathjax>#sfcolor(blue)("grams of aluminum"#</mathjax> reacted. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#0.0324#</mathjax> <mathjax>#"g Al"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the number of grams of <mathjax>#"Al"#</mathjax> that reacted, given some <mathjax>#"H"_2 (g)#</mathjax> product characteristics. </p>
<p>Let's first write the chemical equation for this reaction:</p>
<p><mathjax>#2"Al"(s) + 6"HCl"(aq) rarr2 "AlCl"_3(aq) + 3"H"_2 (g)#</mathjax></p>
<p>The total pressure of the gaseous system is given as <mathjax>#751#</mathjax> <mathjax>#"mm Hg"#</mathjax>, and the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of water vapor is <mathjax>#26.8#</mathjax> <mathjax>#"mm Hg"#</mathjax> at <mathjax>#27^"o""C"#</mathjax>. The pressure of <em>hydrogen gas</em> is thus</p>
<p><mathjax>#P_"total" = P_ ("H"_2"O") + P_ ("H"_2)#</mathjax></p>
<p><mathjax>#P_ ("H"_2) = 751#</mathjax> <mathjax>#"mm Hg"#</mathjax> <mathjax>#- 26.8#</mathjax> <mathjax>#"mm Hg"#</mathjax> <mathjax>#= color(red)(724#</mathjax> <mathjax>#color(red)("mm Hg"#</mathjax></p>
<p>This pressure in atmospheres is </p>
<p><mathjax>#724cancel("mm Hg")((1color(white)(l)"atm")/(760cancel("mm Hg"))) = color(red)(0.953#</mathjax> <mathjax>#color(red)("atm"#</mathjax></p>
<p>We'll now use the ideal gas equation to find the number of moles of <mathjax>#"H"_2#</mathjax> formed:</p>
<p>(<mathjax>#T = 27^"o""C" + 273 = 300#</mathjax> <mathjax>#"K"#</mathjax>)</p>
<p><mathjax>#n = (PV)/(RT) = ((color(red)(0.953)cancel(color(red)("atm")))(0.0465cancel("L")))/((0.082057(cancel("L")•cancel("atm"))/("mol"•cancel("K")))(300cancel("K"))) = color(green)(0.00180#</mathjax> <mathjax>#color(green)("mol H"_2#</mathjax></p>
<p>(volume converted to liters here)</p>
<p>Using the coefficients of the chemical equation, we'll now find the relative number of moles of <mathjax>#"Al"#</mathjax> that react:</p>
<p><mathjax>#color(green)(0.00180)cancel(color(green)("mol H"_2))((2color(white)(l)"mol Al")/(3cancel("mol H"_2))) = color(purple)(0.00120#</mathjax> <mathjax>#color(purple)("mol Al"#</mathjax></p>
<p>Lastly, we'll use the <em>molar mass</em> of aluminum (<mathjax>#26.98#</mathjax> <mathjax>#"g/mol"#</mathjax>) to find the number of grams that reacted:</p>
<p><mathjax>#color(purple)(0.00120)cancel(color(purple)("mol Al"))((26.98color(white)(l)"g Al")/(1cancel("mol Al"))) = color(blue)(0.0324#</mathjax> <mathjax>#color(blue)("g Al"#</mathjax></p>
<p>Thus, <mathjax>#color(blue)(0.0324#</mathjax> <mathjax>#sfcolor(blue)("grams of aluminum"#</mathjax> reacted. </p></div>
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<h1 class="questionTitle" itemprop="name">Aluminum reacts with excess hydrochloric acid to form aqueous aluminum chloride and 46.5 mL of hydrogen gas over water at 27 degrees Celsius and 751 mmHg. How many grams of aluminum reacted? The partial pressure of water at 27 C is 26.8 mmHg.</h1>
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Nathan L.
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Jul 8, 2017
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<div class="markdown"><p><mathjax>#0.0324#</mathjax> <mathjax>#"g Al"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the number of grams of <mathjax>#"Al"#</mathjax> that reacted, given some <mathjax>#"H"_2 (g)#</mathjax> product characteristics. </p>
<p>Let's first write the chemical equation for this reaction:</p>
<p><mathjax>#2"Al"(s) + 6"HCl"(aq) rarr2 "AlCl"_3(aq) + 3"H"_2 (g)#</mathjax></p>
<p>The total pressure of the gaseous system is given as <mathjax>#751#</mathjax> <mathjax>#"mm Hg"#</mathjax>, and the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of water vapor is <mathjax>#26.8#</mathjax> <mathjax>#"mm Hg"#</mathjax> at <mathjax>#27^"o""C"#</mathjax>. The pressure of <em>hydrogen gas</em> is thus</p>
<p><mathjax>#P_"total" = P_ ("H"_2"O") + P_ ("H"_2)#</mathjax></p>
<p><mathjax>#P_ ("H"_2) = 751#</mathjax> <mathjax>#"mm Hg"#</mathjax> <mathjax>#- 26.8#</mathjax> <mathjax>#"mm Hg"#</mathjax> <mathjax>#= color(red)(724#</mathjax> <mathjax>#color(red)("mm Hg"#</mathjax></p>
<p>This pressure in atmospheres is </p>
<p><mathjax>#724cancel("mm Hg")((1color(white)(l)"atm")/(760cancel("mm Hg"))) = color(red)(0.953#</mathjax> <mathjax>#color(red)("atm"#</mathjax></p>
<p>We'll now use the ideal gas equation to find the number of moles of <mathjax>#"H"_2#</mathjax> formed:</p>
<p>(<mathjax>#T = 27^"o""C" + 273 = 300#</mathjax> <mathjax>#"K"#</mathjax>)</p>
<p><mathjax>#n = (PV)/(RT) = ((color(red)(0.953)cancel(color(red)("atm")))(0.0465cancel("L")))/((0.082057(cancel("L")•cancel("atm"))/("mol"•cancel("K")))(300cancel("K"))) = color(green)(0.00180#</mathjax> <mathjax>#color(green)("mol H"_2#</mathjax></p>
<p>(volume converted to liters here)</p>
<p>Using the coefficients of the chemical equation, we'll now find the relative number of moles of <mathjax>#"Al"#</mathjax> that react:</p>
<p><mathjax>#color(green)(0.00180)cancel(color(green)("mol H"_2))((2color(white)(l)"mol Al")/(3cancel("mol H"_2))) = color(purple)(0.00120#</mathjax> <mathjax>#color(purple)("mol Al"#</mathjax></p>
<p>Lastly, we'll use the <em>molar mass</em> of aluminum (<mathjax>#26.98#</mathjax> <mathjax>#"g/mol"#</mathjax>) to find the number of grams that reacted:</p>
<p><mathjax>#color(purple)(0.00120)cancel(color(purple)("mol Al"))((26.98color(white)(l)"g Al")/(1cancel("mol Al"))) = color(blue)(0.0324#</mathjax> <mathjax>#color(blue)("g Al"#</mathjax></p>
<p>Thus, <mathjax>#color(blue)(0.0324#</mathjax> <mathjax>#sfcolor(blue)("grams of aluminum"#</mathjax> reacted. </p></div>
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</article> | Aluminum reacts with excess hydrochloric acid to form aqueous aluminum chloride and 46.5 mL of hydrogen gas over water at 27 degrees Celsius and 751 mmHg. How many grams of aluminum reacted? The partial pressure of water at 27 C is 26.8 mmHg. | null |
529 | ad1926ba-6ddd-11ea-aae0-ccda262736ce | https://socratic.org/questions/a-canister-contains-25-kpa-of-co-2-7-50-kpa-of-n-2-and-52-5-kpa-of-o-2-what-is-t | 85.00 kPa | start physical_unit 22 23 total_pressure kpa qc_end physical_unit 6 6 3 4 partial_pressure qc_end physical_unit 10 10 7 8 partial_pressure qc_end physical_unit 15 15 12 13 partial_pressure qc_end end | [{"type":"physical unit","value":"Total pressure [OF] the canister [IN] kPa"}] | [{"type":"physical unit","value":"85.00 kPa"}] | [{"type":"physical unit","value":"Partial pressure [OF] CO2 [=] \\pu{25 kPa}"},{"type":"physical unit","value":"Partial pressure [OF] N2 [=] \\pu{7.50 kPa}"},{"type":"physical unit","value":"Partial pressure [OF] O2 [=] \\pu{52.5 kPa}"}] | <h1 class="questionTitle" itemprop="name">A canister contains 25 kPa of #CO_2#, 7.50 kPa of #N_2#, and 52.5 kPa of #O_2#. What is the total pressure within the canister?</h1> | null | 85.00 kPa | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And this follows Dalton's law of partial pressures.....<mathjax>#"in a gaseous"#</mathjax><br/>
<mathjax>#"mixture the partial pressure exerted by each gas is the same "#</mathjax><br/>
<mathjax>#"as the pressure it would exert if it alone occupied the container."#</mathjax></p>
<p><mathjax>#"The total pressure is the sum of the individual partial pressures."#</mathjax></p>
<p>This is an experimental gas law that has copious experimental support. </p>
<p>ANd thus <mathjax>#P_"Total"=P_(CO_2)+P_(N_2)+P_(O_2)#</mathjax></p>
<p><mathjax>#={25+7.5+52.5}*kPa#</mathjax></p>
<p><mathjax>#=85*kPa#</mathjax></p>
<p>I think you can see from where the examiner pulled those figures.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>This is simply the sum of the individual partial pressures...........and you can do the sum in your head.......to get <mathjax>#85*kPa#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And this follows Dalton's law of partial pressures.....<mathjax>#"in a gaseous"#</mathjax><br/>
<mathjax>#"mixture the partial pressure exerted by each gas is the same "#</mathjax><br/>
<mathjax>#"as the pressure it would exert if it alone occupied the container."#</mathjax></p>
<p><mathjax>#"The total pressure is the sum of the individual partial pressures."#</mathjax></p>
<p>This is an experimental gas law that has copious experimental support. </p>
<p>ANd thus <mathjax>#P_"Total"=P_(CO_2)+P_(N_2)+P_(O_2)#</mathjax></p>
<p><mathjax>#={25+7.5+52.5}*kPa#</mathjax></p>
<p><mathjax>#=85*kPa#</mathjax></p>
<p>I think you can see from where the examiner pulled those figures.</p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">A canister contains 25 kPa of #CO_2#, 7.50 kPa of #N_2#, and 52.5 kPa of #O_2#. What is the total pressure within the canister?</h1>
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anor277
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<span class="dateCreated" datetime="2017-08-07T05:19:02" itemprop="dateCreated">
Aug 7, 2017
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<div class="markdown"><p>This is simply the sum of the individual partial pressures...........and you can do the sum in your head.......to get <mathjax>#85*kPa#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And this follows Dalton's law of partial pressures.....<mathjax>#"in a gaseous"#</mathjax><br/>
<mathjax>#"mixture the partial pressure exerted by each gas is the same "#</mathjax><br/>
<mathjax>#"as the pressure it would exert if it alone occupied the container."#</mathjax></p>
<p><mathjax>#"The total pressure is the sum of the individual partial pressures."#</mathjax></p>
<p>This is an experimental gas law that has copious experimental support. </p>
<p>ANd thus <mathjax>#P_"Total"=P_(CO_2)+P_(N_2)+P_(O_2)#</mathjax></p>
<p><mathjax>#={25+7.5+52.5}*kPa#</mathjax></p>
<p><mathjax>#=85*kPa#</mathjax></p>
<p>I think you can see from where the examiner pulled those figures.</p></div>
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</article> | A canister contains 25 kPa of #CO_2#, 7.50 kPa of #N_2#, and 52.5 kPa of #O_2#. What is the total pressure within the canister? | null |
530 | aaa29126-6ddd-11ea-97cb-ccda262736ce | https://socratic.org/questions/what-is-the-grams-in-1-26-10-4-mol-of-hc-2h-3o-2 | 7.57 × 10^(-3) grams | start physical_unit 10 10 mass g qc_end physical_unit 10 10 5 8 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] HC2H3O2 [IN] grams"}] | [{"type":"physical unit","value":"7.57 × 10^(-3) grams"}] | [{"type":"physical unit","value":"Mole [OF] HC2H3O2 [=] \\pu{1.26 × 10^(-4) mol}"}] | <h1 class="questionTitle" itemprop="name">What is the grams in #1.26 * 10^-4# mol of #HC_2H_3O_2#?</h1> | null | 7.57 × 10^(-3) grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#HC_2H_3O_2#</mathjax> = <mathjax>#60.05#</mathjax> g/mol</p>
<p>The number of mols can be expressed by <mathjax>#n#</mathjax> = <mathjax>#m/M#</mathjax> </p>
<p>We may rearrange this into <mathjax>#m#</mathjax> = <mathjax>#n*M#</mathjax> </p>
<p><mathjax>#m#</mathjax> = <mathjax>#0.000126*60.05#</mathjax> <br/>
<mathjax>#m#</mathjax> = <mathjax>#0.0075663#</mathjax> grams</p></div>
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<div class="markdown"><p><mathjax>#0.0075663#</mathjax> grams of <mathjax>#HC_2H_3O_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#HC_2H_3O_2#</mathjax> = <mathjax>#60.05#</mathjax> g/mol</p>
<p>The number of mols can be expressed by <mathjax>#n#</mathjax> = <mathjax>#m/M#</mathjax> </p>
<p>We may rearrange this into <mathjax>#m#</mathjax> = <mathjax>#n*M#</mathjax> </p>
<p><mathjax>#m#</mathjax> = <mathjax>#0.000126*60.05#</mathjax> <br/>
<mathjax>#m#</mathjax> = <mathjax>#0.0075663#</mathjax> grams</p></div>
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<h1 class="questionTitle" itemprop="name">What is the grams in #1.26 * 10^-4# mol of #HC_2H_3O_2#?</h1>
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<div class="markdown"><p><mathjax>#0.0075663#</mathjax> grams of <mathjax>#HC_2H_3O_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#HC_2H_3O_2#</mathjax> = <mathjax>#60.05#</mathjax> g/mol</p>
<p>The number of mols can be expressed by <mathjax>#n#</mathjax> = <mathjax>#m/M#</mathjax> </p>
<p>We may rearrange this into <mathjax>#m#</mathjax> = <mathjax>#n*M#</mathjax> </p>
<p><mathjax>#m#</mathjax> = <mathjax>#0.000126*60.05#</mathjax> <br/>
<mathjax>#m#</mathjax> = <mathjax>#0.0075663#</mathjax> grams</p></div>
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</article> | What is the grams in #1.26 * 10^-4# mol of #HC_2H_3O_2#? | null |
531 | ab0c64e4-6ddd-11ea-a64f-ccda262736ce | https://socratic.org/questions/how-many-milliliters-of-12-0-m-hydrochloric-acid-hcl-is-required-to-prepare-0-75 | 15.63 mL | start physical_unit 19 19 volume ml qc_end physical_unit 8 8 4 5 molarity qc_end physical_unit 19 19 13 14 volume qc_end physical_unit 8 8 17 18 molarity qc_end end | [{"type":"physical unit","value":"Volume1 [OF] HCl solution [IN] mL"}] | [{"type":"physical unit","value":"15.63 mL"}] | [{"type":"physical unit","value":"Molarity1 [OF] HCl solution [=] \\pu{12.0 M}"},{"type":"physical unit","value":"Volume2 [OF] HCl solution [=] \\pu{0.75 L}"},{"type":"physical unit","value":"Molarity2 [OF] HCl solution [=] \\pu{0.25 M}"}] | <h1 class="questionTitle" itemprop="name">How many milliliters of 12.0 M hydrochloric acid (#HCl#) is required to prepare 0.75 L of a 0.25 M solution of #HCl#?</h1> | null | 15.63 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We have to use the titration formula:<br/>
<mathjax>#M_aV_a=M_bV_b#</mathjax><br/>
Now just plug it in.<br/>
<mathjax>#(12)(x)=(.75)(.25)#</mathjax><br/>
<mathjax>#12x = .1875#</mathjax><br/>
<mathjax>#x= 0.015625#</mathjax> liters<br/>
To convert liter to mL, multiply it by 1000, since 1 L= 1000 mL<br/>
Therefore, the answer rounded is about 16 mL</p></div>
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<div class="markdown"><p>16mL</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We have to use the titration formula:<br/>
<mathjax>#M_aV_a=M_bV_b#</mathjax><br/>
Now just plug it in.<br/>
<mathjax>#(12)(x)=(.75)(.25)#</mathjax><br/>
<mathjax>#12x = .1875#</mathjax><br/>
<mathjax>#x= 0.015625#</mathjax> liters<br/>
To convert liter to mL, multiply it by 1000, since 1 L= 1000 mL<br/>
Therefore, the answer rounded is about 16 mL</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many milliliters of 12.0 M hydrochloric acid (#HCl#) is required to prepare 0.75 L of a 0.25 M solution of #HCl#?</h1>
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<div class="markdown"><p>16mL</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We have to use the titration formula:<br/>
<mathjax>#M_aV_a=M_bV_b#</mathjax><br/>
Now just plug it in.<br/>
<mathjax>#(12)(x)=(.75)(.25)#</mathjax><br/>
<mathjax>#12x = .1875#</mathjax><br/>
<mathjax>#x= 0.015625#</mathjax> liters<br/>
To convert liter to mL, multiply it by 1000, since 1 L= 1000 mL<br/>
Therefore, the answer rounded is about 16 mL</p></div>
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</article> | How many milliliters of 12.0 M hydrochloric acid (#HCl#) is required to prepare 0.75 L of a 0.25 M solution of #HCl#? | null |
532 | a8f000a7-6ddd-11ea-a6b0-ccda262736ce | https://socratic.org/questions/what-is-the-mole-ratio-of-fe-3o-4-to-fe | 0.24 | start physical_unit 6 8 mole_fraction none qc_end chemical_equation 6 6 qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Mole ratio [OF] Fe3O4 to Fe"}] | [{"type":"physical unit","value":"0.24"}] | [{"type":"chemical equation","value":"Fe3O4"},{"type":"chemical equation","value":"Fe"}] | <h1 class="questionTitle" itemprop="name">What is the mole ratio of #Fe_3O_4# to #Fe#?</h1> | null | 0.24 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I think that by using simple arithmetic you should very quickly be able to tell me that I had 36 iron atoms, and 48 oxygen atoms....do you agree?</p>
<p>Here we have a <mathjax>#"mole"#</mathjax> of <mathjax>#Fe_3O_4#</mathjax> units, i.e. a mass of <mathjax>#231.4*g#</mathjax>...but a mole specifies Avogadro's number of particles....we write <mathjax>#N_A=6.022xx10^23*mol^-1#</mathjax>.. And so we can use the <mathjax>#"mole"#</mathjax> as a collective number precisely as we do the <mathjax>#"dozen"#</mathjax>.</p>
<p>And so we gots <mathjax>#1*mol#</mathjax> <mathjax>#Fe_3O_4#</mathjax>...we gots <mathjax>#3*mol#</mathjax> of iron atoms, and <mathjax>#4*mol#</mathjax> of oxygen atoms...</p>
<p>If I got one mole of <mathjax>#P_4O_10#</mathjax>, can you tell me the numbers and the masses of the constituent atoms?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Suppose instead I had a dozen formula units of <mathjax>#Fe_3O_4#</mathjax>...</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I think that by using simple arithmetic you should very quickly be able to tell me that I had 36 iron atoms, and 48 oxygen atoms....do you agree?</p>
<p>Here we have a <mathjax>#"mole"#</mathjax> of <mathjax>#Fe_3O_4#</mathjax> units, i.e. a mass of <mathjax>#231.4*g#</mathjax>...but a mole specifies Avogadro's number of particles....we write <mathjax>#N_A=6.022xx10^23*mol^-1#</mathjax>.. And so we can use the <mathjax>#"mole"#</mathjax> as a collective number precisely as we do the <mathjax>#"dozen"#</mathjax>.</p>
<p>And so we gots <mathjax>#1*mol#</mathjax> <mathjax>#Fe_3O_4#</mathjax>...we gots <mathjax>#3*mol#</mathjax> of iron atoms, and <mathjax>#4*mol#</mathjax> of oxygen atoms...</p>
<p>If I got one mole of <mathjax>#P_4O_10#</mathjax>, can you tell me the numbers and the masses of the constituent atoms?</p></div>
</div>
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<div class="markdown"><p>Suppose instead I had a dozen formula units of <mathjax>#Fe_3O_4#</mathjax>...</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I think that by using simple arithmetic you should very quickly be able to tell me that I had 36 iron atoms, and 48 oxygen atoms....do you agree?</p>
<p>Here we have a <mathjax>#"mole"#</mathjax> of <mathjax>#Fe_3O_4#</mathjax> units, i.e. a mass of <mathjax>#231.4*g#</mathjax>...but a mole specifies Avogadro's number of particles....we write <mathjax>#N_A=6.022xx10^23*mol^-1#</mathjax>.. And so we can use the <mathjax>#"mole"#</mathjax> as a collective number precisely as we do the <mathjax>#"dozen"#</mathjax>.</p>
<p>And so we gots <mathjax>#1*mol#</mathjax> <mathjax>#Fe_3O_4#</mathjax>...we gots <mathjax>#3*mol#</mathjax> of iron atoms, and <mathjax>#4*mol#</mathjax> of oxygen atoms...</p>
<p>If I got one mole of <mathjax>#P_4O_10#</mathjax>, can you tell me the numbers and the masses of the constituent atoms?</p></div>
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</article> | What is the mole ratio of #Fe_3O_4# to #Fe#? | null |
533 | a918d8af-6ddd-11ea-a1f5-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-formula-of-baking-soda | NaHCO3 | start chemical_formula qc_end substance 6 7 qc_end end | [{"type":"other","value":"Chemical Formula [OF] baking soda [IN] default"}] | [{"type":"chemical equation","value":"NaHCO3"}] | [{"type":"substance name","value":"Baking soda"}] | <h1 class="questionTitle" itemprop="name">What is the chemical formula of baking soda?</h1> | null | NaHCO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>See <a href="https://socratic.org/questions/what-is-the-chemical-name-of-baking-soda#344264">here.</a> </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>Baking soda is largely <mathjax>#NaHCO_3#</mathjax>; some other salts are present. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>See <a href="https://socratic.org/questions/what-is-the-chemical-name-of-baking-soda#344264">here.</a> </p></div>
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<div class="markdown"><p>Baking soda is largely <mathjax>#NaHCO_3#</mathjax>; some other salts are present. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>See <a href="https://socratic.org/questions/what-is-the-chemical-name-of-baking-soda#344264">here.</a> </p></div>
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</article> | What is the chemical formula of baking soda? | null |
534 | abd6a4f4-6ddd-11ea-a395-ccda262736ce | https://socratic.org/questions/5a0f0a6a11ef6b622cf5fdb9 | 110.00 g | start physical_unit 16 16 mass g qc_end physical_unit 5 5 19 20 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] solute [IN] g"}] | [{"type":"physical unit","value":"110.00 g"}] | [{"type":"physical unit","value":"m/m [OF] solute in solution [=] \\pu{5.5%}"},{"type":"physical unit","value":"Mass [OF] solution [=] \\pu{2000 g}"}] | <h1 class="questionTitle" itemprop="name">If we have #5.5%# #"m/m"# solution with respect to a solution....what is the mass of solute in a #2000*g# mass of solution...?</h1> | null | 110.00 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We are given <mathjax>#"mass of solute"/"mass of solution"xx100%=5.5%#</mathjax></p>
<p>And thus <mathjax>#"mass of solute"=0.055xx"mass of solution"#</mathjax></p>
<p><mathjax>#=0.055xx2000*g=110*g#</mathjax></p>
<p>Now the volume of this solution would be as near as dammit to <mathjax>#1890*mL#</mathjax> (why?). Would you tell us the concentration of potassium chloride in <mathjax>#mol*L^-1#</mathjax>?</p></div>
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<div class="markdown"><p>Is it not <mathjax>#110*g#</mathjax>.....</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We are given <mathjax>#"mass of solute"/"mass of solution"xx100%=5.5%#</mathjax></p>
<p>And thus <mathjax>#"mass of solute"=0.055xx"mass of solution"#</mathjax></p>
<p><mathjax>#=0.055xx2000*g=110*g#</mathjax></p>
<p>Now the volume of this solution would be as near as dammit to <mathjax>#1890*mL#</mathjax> (why?). Would you tell us the concentration of potassium chloride in <mathjax>#mol*L^-1#</mathjax>?</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If we have #5.5%# #"m/m"# solution with respect to a solution....what is the mass of solute in a #2000*g# mass of solution...?</h1>
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Nov 20, 2017
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<div class="markdown"><p>Is it not <mathjax>#110*g#</mathjax>.....</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We are given <mathjax>#"mass of solute"/"mass of solution"xx100%=5.5%#</mathjax></p>
<p>And thus <mathjax>#"mass of solute"=0.055xx"mass of solution"#</mathjax></p>
<p><mathjax>#=0.055xx2000*g=110*g#</mathjax></p>
<p>Now the volume of this solution would be as near as dammit to <mathjax>#1890*mL#</mathjax> (why?). Would you tell us the concentration of potassium chloride in <mathjax>#mol*L^-1#</mathjax>?</p></div>
</div>
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<a href="https://socratic.org/answers/508805" itemprop="url">Answer link</a>
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</article> | If we have #5.5%# #"m/m"# solution with respect to a solution....what is the mass of solute in a #2000*g# mass of solution...? | null |
535 | a9e1c88a-6ddd-11ea-9986-ccda262736ce | https://socratic.org/questions/how-do-you-balance-na-2so-4-ca-no-3-2-caso-4-nano-3 | Na2SO4(aq) + Ca(NO3)2(aq) -> CaSO4(s) + 2 NaNO3(aq) | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"Na2SO4(aq) + Ca(NO3)2(aq) -> CaSO4(s) + 2 NaNO3(aq)"}] | [{"type":"chemical equation","value":"Na2SO4 + Ca(NO3)2 -> CaSO4 + NaNO3"}] | <h1 class="questionTitle" itemprop="name">How do you balance #Na_2SO_4 + Ca(NO_3)_2 -> CaSO_4 + NaNO_3#?</h1> | null | Na2SO4(aq) + Ca(NO3)2(aq) -> CaSO4(s) + 2 NaNO3(aq) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Every time you're dealing with a <a href="https://socratic.org/chemistry/chemical-reactions/double-replacement-reactions">double replacement reaction</a>, you can make your life easier by looking at <strong>ions</strong>, rather than at <strong>atoms</strong>, when trying to balance the chemical equation. </p>
<p>A double replacement reaction is characterized by the reaction of the <strong>soluble <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a></strong> and the formation of an <strong>insoluble solid</strong> that precipitates out of solution. </p>
<p><em>Sodium sulfate</em>, <mathjax>#"Na"_2"SO"_4#</mathjax>, and <em>calcium nitrate</em>, <mathjax>#"Ca"("NO"_3)_2#</mathjax>, are both soluble <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a>, which means that they exist as <em>ions</em> in aqueous solution</p>
<blockquote>
<p><mathjax>#"Na"_2"SO"_text(4(aq]) -> color(red)(2)"Na"_text((aq])^(+) + "SO"_text(4(aq])^(2-)#</mathjax></p>
<p><mathjax>#"Ca"("NO"_3)_text(2(aq]) -> "Ca"_text((aq])^(2+) + color(blue)(2)"NO"_text(3(aq])^(-)#</mathjax></p>
</blockquote>
<p>The reaction produces <em>calcium sulfate</em>, an <strong>insoluble solid</strong> that precipitates out of solution, and <em>sodium nitrate</em>, <mathjax>#"NaNO"_3#</mathjax>, another soluble ionic compound that exists as ions in solution</p>
<blockquote>
<p><mathjax>#"NaNO"_text(3(aq]) -> "Na"_text((aq])^(+) + "NO"_text(3(aq])^(-)#</mathjax></p>
</blockquote>
<p>Now, the idea here is that <strong>all the ions</strong> present on the reactants' side <strong>must be present</strong>, either as dissociated ions or as part of the solid, on the products' side. </p>
<p>Notice that the one calcium cation, <mathjax>#"Ca"^(2+)#</mathjax>, and the one sulfate anion, <mathjax>#"SO"_4^(2-)#</mathjax>, will group to form the insoluble calcium sulfate in a <mathjax>#1:1#</mathjax> mole ratio, so you don't need to balance these two ions out.</p>
<p>On the other hand, notice that you have <mathjax>#color(red)(2)#</mathjax> sodium cations on the reactants side, but only <mathjax>#1#</mathjax> on the products' side. Likewise, you have <mathjax>#color(blue)(2)#</mathjax> nitrate anions on the reactants' side, and once again only one on the products' side. </p>
<p>This tells you that will have to multiply the <strong>sodium nitrate</strong> by <mathjax>#color(green)(2)#</mathjax> to balance these two ions out. </p>
<p>The complete chemical equation for this reaction will look like this </p>
<blockquote>
<p><mathjax>#"Na"_2"SO"_text(4(aq]) + "Ca"("NO"_3)_text(2(aq]) -> "CaSO"_text(4(s]) darr + color(green)(2)"NaNO"_text(3(aq])#</mathjax></p>
</blockquote>
<p>The <strong>net ionic equation</strong>, which doesn't include <em>spectator ions</em>, i.e. the ions that exist <em>dissociated</em> on both sides of the equation, will look like this </p>
<blockquote>
<p><mathjax>#"SO"_text(4(aq])^(2-) + "Ca"_text((aq])^(2+) -> "CaSO"_text(4(s]) darr#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Na"_2"SO"_text(4(aq]) + "Ca"("NO"_3)_text(2(aq]) -> "CaSO"_text(4(s]) darr + 2"NaNO"_text(3(aq])#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Every time you're dealing with a <a href="https://socratic.org/chemistry/chemical-reactions/double-replacement-reactions">double replacement reaction</a>, you can make your life easier by looking at <strong>ions</strong>, rather than at <strong>atoms</strong>, when trying to balance the chemical equation. </p>
<p>A double replacement reaction is characterized by the reaction of the <strong>soluble <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a></strong> and the formation of an <strong>insoluble solid</strong> that precipitates out of solution. </p>
<p><em>Sodium sulfate</em>, <mathjax>#"Na"_2"SO"_4#</mathjax>, and <em>calcium nitrate</em>, <mathjax>#"Ca"("NO"_3)_2#</mathjax>, are both soluble <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a>, which means that they exist as <em>ions</em> in aqueous solution</p>
<blockquote>
<p><mathjax>#"Na"_2"SO"_text(4(aq]) -> color(red)(2)"Na"_text((aq])^(+) + "SO"_text(4(aq])^(2-)#</mathjax></p>
<p><mathjax>#"Ca"("NO"_3)_text(2(aq]) -> "Ca"_text((aq])^(2+) + color(blue)(2)"NO"_text(3(aq])^(-)#</mathjax></p>
</blockquote>
<p>The reaction produces <em>calcium sulfate</em>, an <strong>insoluble solid</strong> that precipitates out of solution, and <em>sodium nitrate</em>, <mathjax>#"NaNO"_3#</mathjax>, another soluble ionic compound that exists as ions in solution</p>
<blockquote>
<p><mathjax>#"NaNO"_text(3(aq]) -> "Na"_text((aq])^(+) + "NO"_text(3(aq])^(-)#</mathjax></p>
</blockquote>
<p>Now, the idea here is that <strong>all the ions</strong> present on the reactants' side <strong>must be present</strong>, either as dissociated ions or as part of the solid, on the products' side. </p>
<p>Notice that the one calcium cation, <mathjax>#"Ca"^(2+)#</mathjax>, and the one sulfate anion, <mathjax>#"SO"_4^(2-)#</mathjax>, will group to form the insoluble calcium sulfate in a <mathjax>#1:1#</mathjax> mole ratio, so you don't need to balance these two ions out.</p>
<p>On the other hand, notice that you have <mathjax>#color(red)(2)#</mathjax> sodium cations on the reactants side, but only <mathjax>#1#</mathjax> on the products' side. Likewise, you have <mathjax>#color(blue)(2)#</mathjax> nitrate anions on the reactants' side, and once again only one on the products' side. </p>
<p>This tells you that will have to multiply the <strong>sodium nitrate</strong> by <mathjax>#color(green)(2)#</mathjax> to balance these two ions out. </p>
<p>The complete chemical equation for this reaction will look like this </p>
<blockquote>
<p><mathjax>#"Na"_2"SO"_text(4(aq]) + "Ca"("NO"_3)_text(2(aq]) -> "CaSO"_text(4(s]) darr + color(green)(2)"NaNO"_text(3(aq])#</mathjax></p>
</blockquote>
<p>The <strong>net ionic equation</strong>, which doesn't include <em>spectator ions</em>, i.e. the ions that exist <em>dissociated</em> on both sides of the equation, will look like this </p>
<blockquote>
<p><mathjax>#"SO"_text(4(aq])^(2-) + "Ca"_text((aq])^(2+) -> "CaSO"_text(4(s]) darr#</mathjax></p>
</blockquote></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you balance #Na_2SO_4 + Ca(NO_3)_2 -> CaSO_4 + NaNO_3#?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"Na"_2"SO"_text(4(aq]) + "Ca"("NO"_3)_text(2(aq]) -> "CaSO"_text(4(s]) darr + 2"NaNO"_text(3(aq])#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Every time you're dealing with a <a href="https://socratic.org/chemistry/chemical-reactions/double-replacement-reactions">double replacement reaction</a>, you can make your life easier by looking at <strong>ions</strong>, rather than at <strong>atoms</strong>, when trying to balance the chemical equation. </p>
<p>A double replacement reaction is characterized by the reaction of the <strong>soluble <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a></strong> and the formation of an <strong>insoluble solid</strong> that precipitates out of solution. </p>
<p><em>Sodium sulfate</em>, <mathjax>#"Na"_2"SO"_4#</mathjax>, and <em>calcium nitrate</em>, <mathjax>#"Ca"("NO"_3)_2#</mathjax>, are both soluble <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a>, which means that they exist as <em>ions</em> in aqueous solution</p>
<blockquote>
<p><mathjax>#"Na"_2"SO"_text(4(aq]) -> color(red)(2)"Na"_text((aq])^(+) + "SO"_text(4(aq])^(2-)#</mathjax></p>
<p><mathjax>#"Ca"("NO"_3)_text(2(aq]) -> "Ca"_text((aq])^(2+) + color(blue)(2)"NO"_text(3(aq])^(-)#</mathjax></p>
</blockquote>
<p>The reaction produces <em>calcium sulfate</em>, an <strong>insoluble solid</strong> that precipitates out of solution, and <em>sodium nitrate</em>, <mathjax>#"NaNO"_3#</mathjax>, another soluble ionic compound that exists as ions in solution</p>
<blockquote>
<p><mathjax>#"NaNO"_text(3(aq]) -> "Na"_text((aq])^(+) + "NO"_text(3(aq])^(-)#</mathjax></p>
</blockquote>
<p>Now, the idea here is that <strong>all the ions</strong> present on the reactants' side <strong>must be present</strong>, either as dissociated ions or as part of the solid, on the products' side. </p>
<p>Notice that the one calcium cation, <mathjax>#"Ca"^(2+)#</mathjax>, and the one sulfate anion, <mathjax>#"SO"_4^(2-)#</mathjax>, will group to form the insoluble calcium sulfate in a <mathjax>#1:1#</mathjax> mole ratio, so you don't need to balance these two ions out.</p>
<p>On the other hand, notice that you have <mathjax>#color(red)(2)#</mathjax> sodium cations on the reactants side, but only <mathjax>#1#</mathjax> on the products' side. Likewise, you have <mathjax>#color(blue)(2)#</mathjax> nitrate anions on the reactants' side, and once again only one on the products' side. </p>
<p>This tells you that will have to multiply the <strong>sodium nitrate</strong> by <mathjax>#color(green)(2)#</mathjax> to balance these two ions out. </p>
<p>The complete chemical equation for this reaction will look like this </p>
<blockquote>
<p><mathjax>#"Na"_2"SO"_text(4(aq]) + "Ca"("NO"_3)_text(2(aq]) -> "CaSO"_text(4(s]) darr + color(green)(2)"NaNO"_text(3(aq])#</mathjax></p>
</blockquote>
<p>The <strong>net ionic equation</strong>, which doesn't include <em>spectator ions</em>, i.e. the ions that exist <em>dissociated</em> on both sides of the equation, will look like this </p>
<blockquote>
<p><mathjax>#"SO"_text(4(aq])^(2-) + "Ca"_text((aq])^(2+) -> "CaSO"_text(4(s]) darr#</mathjax></p>
</blockquote></div>
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</article> | How do you balance #Na_2SO_4 + Ca(NO_3)_2 -> CaSO_4 + NaNO_3#? | null |
536 | aa158b30-6ddd-11ea-8fec-ccda262736ce | https://socratic.org/questions/how-many-moles-of-cabr-2-are-in-5-0-grams-of-cabr-2 | 0.25 moles | start physical_unit 4 4 mole mol qc_end physical_unit 4 4 7 8 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] CaBr2 [IN] moles"}] | [{"type":"physical unit","value":"0.25 moles"}] | [{"type":"physical unit","value":"Mass [OF] CaBr2 [=] \\pu{5.0 grams}"}] | <h1 class="questionTitle" itemprop="name">How many moles of #CaBr_2# are in 5.0 grams of #CaBr_2#?</h1> | null | 0.25 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Molar mass of Calcium Bromide is 200 g <mathjax>#mol^-1#</mathjax></p>
<p>Mass of Calcium Bromide = 5.0 g </p>
<p>moles of Ca<mathjax>#Br_2#</mathjax> = mass of Ca<mathjax>#Br_2#</mathjax> / molar mass of Ca<mathjax>#Br_2#</mathjax> </p>
<p>moles of Ca<mathjax>#Br_2#</mathjax> = 5.0 g / 200 g <mathjax>#mol^-1#</mathjax></p>
<p>moles of Ca<mathjax>#Br_2#</mathjax> = 0.25 mol </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>0.25 mol</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Molar mass of Calcium Bromide is 200 g <mathjax>#mol^-1#</mathjax></p>
<p>Mass of Calcium Bromide = 5.0 g </p>
<p>moles of Ca<mathjax>#Br_2#</mathjax> = mass of Ca<mathjax>#Br_2#</mathjax> / molar mass of Ca<mathjax>#Br_2#</mathjax> </p>
<p>moles of Ca<mathjax>#Br_2#</mathjax> = 5.0 g / 200 g <mathjax>#mol^-1#</mathjax></p>
<p>moles of Ca<mathjax>#Br_2#</mathjax> = 0.25 mol </p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of #CaBr_2# are in 5.0 grams of #CaBr_2#?</h1>
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<div class="markdown"><p>0.25 mol</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Molar mass of Calcium Bromide is 200 g <mathjax>#mol^-1#</mathjax></p>
<p>Mass of Calcium Bromide = 5.0 g </p>
<p>moles of Ca<mathjax>#Br_2#</mathjax> = mass of Ca<mathjax>#Br_2#</mathjax> / molar mass of Ca<mathjax>#Br_2#</mathjax> </p>
<p>moles of Ca<mathjax>#Br_2#</mathjax> = 5.0 g / 200 g <mathjax>#mol^-1#</mathjax></p>
<p>moles of Ca<mathjax>#Br_2#</mathjax> = 0.25 mol </p></div>
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</article> | How many moles of #CaBr_2# are in 5.0 grams of #CaBr_2#? | null |
537 | a912e563-6ddd-11ea-be29-ccda262736ce | https://socratic.org/questions/the-density-of-water-is-1-00-g-ml-at-4-c-how-many-water-molecules-are-present-in | 8.23 × 10^22 | start physical_unit 12 13 number none qc_end physical_unit 3 3 5 6 density qc_end physical_unit 3 3 8 9 temperature qc_end physical_unit 3 3 17 18 volume qc_end end | [{"type":"physical unit","value":"Number [OF] water molecules"}] | [{"type":"physical unit","value":"8.23 × 10^22"}] | [{"type":"physical unit","value":"Density [OF] water [=] \\pu{1.00 g/mL }"},{"type":"physical unit","value":"Temperature [OF] water [=] \\pu{4 ℃}"},{"type":"physical unit","value":"Volume [OF] water [=] \\pu{2.46 mL}"}] | <h1 class="questionTitle" itemprop="name">The density of water is 1.00 g/mL at 4°C. How many water molecules are present in 2.46 mL of water at this temperature?</h1> | null | 8.23 × 10^22 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to use the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of water at <mathjax>#4^@"C"#</mathjax> and the volume of the sample to find its <em>mass</em>, then use water's molar mass to find how many <em>moles</em> of water you get in this sample. </p>
<p>So, density is defined as mass per unit of volume. In this case, a density of <mathjax>#"`1.00 g/mL"#</mathjax> tells you that <strong>every milliliter</strong> of water has a mass of <mathjax>#"1.00 g"#</mathjax>.</p>
<p>This means that <mathjax>#"2.46 mL"#</mathjax> will have a mass of </p>
<blockquote>
<p><mathjax>#2.46color(red)(cancel(color(black)("mL"))) * "1.00 g"/(1color(red)(cancel(color(black)("mL")))) = "2.46 g"#</mathjax></p>
</blockquote>
<p>Now, water has <em>molar mass</em> of <mathjax>#"18.015 g/mol"#</mathjax>. This tells you that <strong>every mole</strong> of water has mass of <mathjax>#"18.015 g"#</mathjax>. In your case, the sample will contain</p>
<blockquote>
<p><mathjax>#2.46color(red)(cancel(color(black)("g"))) * "1 mole water"/(18.015color(red)(cancel(color(black)("g")))) = "0.1366 moles water"#</mathjax></p>
</blockquote>
<p>Finally, the relationship between <em>number of moles</em> and <em>number of molecules</em> is given by <strong>Avogadro's number</strong>, which tells you that <strong>every mole</strong> of a substance contains exactly <mathjax>#6.022 * 20^(23)#</mathjax> molecules of that substance. </p>
<p>In this case, you will have</p>
<blockquote>
<p><mathjax>#0.1366color(red)(cancel(color(black)("moles"))) * (6.022 * 10^(23)"molecules")/(1color(red)(cancel(color(black)("mole")))) = color(green)(8.23 * 10^(22)"molecules")#</mathjax></p>
</blockquote>
<p>So, <mathjax>#"2.46 mL"#</mathjax> of water at a temperature of <mathjax>#4^@"C"#</mathjax> wil contain a total of <mathjax>#8.23 * 10^(22)#</mathjax> molecules of water. </p></div>
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<div class="answerSummary">
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<div class="markdown"><p><mathjax>#8.23 * 10^(22)#</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 use the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of water at <mathjax>#4^@"C"#</mathjax> and the volume of the sample to find its <em>mass</em>, then use water's molar mass to find how many <em>moles</em> of water you get in this sample. </p>
<p>So, density is defined as mass per unit of volume. In this case, a density of <mathjax>#"`1.00 g/mL"#</mathjax> tells you that <strong>every milliliter</strong> of water has a mass of <mathjax>#"1.00 g"#</mathjax>.</p>
<p>This means that <mathjax>#"2.46 mL"#</mathjax> will have a mass of </p>
<blockquote>
<p><mathjax>#2.46color(red)(cancel(color(black)("mL"))) * "1.00 g"/(1color(red)(cancel(color(black)("mL")))) = "2.46 g"#</mathjax></p>
</blockquote>
<p>Now, water has <em>molar mass</em> of <mathjax>#"18.015 g/mol"#</mathjax>. This tells you that <strong>every mole</strong> of water has mass of <mathjax>#"18.015 g"#</mathjax>. In your case, the sample will contain</p>
<blockquote>
<p><mathjax>#2.46color(red)(cancel(color(black)("g"))) * "1 mole water"/(18.015color(red)(cancel(color(black)("g")))) = "0.1366 moles water"#</mathjax></p>
</blockquote>
<p>Finally, the relationship between <em>number of moles</em> and <em>number of molecules</em> is given by <strong>Avogadro's number</strong>, which tells you that <strong>every mole</strong> of a substance contains exactly <mathjax>#6.022 * 20^(23)#</mathjax> molecules of that substance. </p>
<p>In this case, you will have</p>
<blockquote>
<p><mathjax>#0.1366color(red)(cancel(color(black)("moles"))) * (6.022 * 10^(23)"molecules")/(1color(red)(cancel(color(black)("mole")))) = color(green)(8.23 * 10^(22)"molecules")#</mathjax></p>
</blockquote>
<p>So, <mathjax>#"2.46 mL"#</mathjax> of water at a temperature of <mathjax>#4^@"C"#</mathjax> wil contain a total of <mathjax>#8.23 * 10^(22)#</mathjax> molecules of water. </p></div>
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<h1 class="questionTitle" itemprop="name">The density of water is 1.00 g/mL at 4°C. How many water molecules are present in 2.46 mL of water at this temperature?</h1>
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<div class="markdown"><p><mathjax>#8.23 * 10^(22)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to use the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of water at <mathjax>#4^@"C"#</mathjax> and the volume of the sample to find its <em>mass</em>, then use water's molar mass to find how many <em>moles</em> of water you get in this sample. </p>
<p>So, density is defined as mass per unit of volume. In this case, a density of <mathjax>#"`1.00 g/mL"#</mathjax> tells you that <strong>every milliliter</strong> of water has a mass of <mathjax>#"1.00 g"#</mathjax>.</p>
<p>This means that <mathjax>#"2.46 mL"#</mathjax> will have a mass of </p>
<blockquote>
<p><mathjax>#2.46color(red)(cancel(color(black)("mL"))) * "1.00 g"/(1color(red)(cancel(color(black)("mL")))) = "2.46 g"#</mathjax></p>
</blockquote>
<p>Now, water has <em>molar mass</em> of <mathjax>#"18.015 g/mol"#</mathjax>. This tells you that <strong>every mole</strong> of water has mass of <mathjax>#"18.015 g"#</mathjax>. In your case, the sample will contain</p>
<blockquote>
<p><mathjax>#2.46color(red)(cancel(color(black)("g"))) * "1 mole water"/(18.015color(red)(cancel(color(black)("g")))) = "0.1366 moles water"#</mathjax></p>
</blockquote>
<p>Finally, the relationship between <em>number of moles</em> and <em>number of molecules</em> is given by <strong>Avogadro's number</strong>, which tells you that <strong>every mole</strong> of a substance contains exactly <mathjax>#6.022 * 20^(23)#</mathjax> molecules of that substance. </p>
<p>In this case, you will have</p>
<blockquote>
<p><mathjax>#0.1366color(red)(cancel(color(black)("moles"))) * (6.022 * 10^(23)"molecules")/(1color(red)(cancel(color(black)("mole")))) = color(green)(8.23 * 10^(22)"molecules")#</mathjax></p>
</blockquote>
<p>So, <mathjax>#"2.46 mL"#</mathjax> of water at a temperature of <mathjax>#4^@"C"#</mathjax> wil contain a total of <mathjax>#8.23 * 10^(22)#</mathjax> molecules of water. </p></div>
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</article> | The density of water is 1.00 g/mL at 4°C. How many water molecules are present in 2.46 mL of water at this temperature? | null |
538 | aafac438-6ddd-11ea-ab66-ccda262736ce | https://socratic.org/questions/what-is-the-equation-for-gaseous-ammonia-reacts-with-gaseous-oxygen-to-form-gase | 2 NH3(g) + 5/2 O2(g) -> 2 NO(g) + 3 H2O(g) | start chemical_equation qc_end substance 5 6 qc_end substance 9 10 qc_end substance 13 15 qc_end substance 17 18 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 NH3(g) + 5/2 O2(g) -> 2 NO(g) + 3 H2O(g)"}] | [{"type":"substance name","value":"Gaseous ammonia"},{"type":"substance name","value":"Gaseous oxygen"},{"type":"substance name","value":"Gaseous nitrogen monoxide"},{"type":"substance name","value":"Gaseous water"}] | <h1 class="questionTitle" itemprop="name">What is the equation for: Gaseous ammonia reacts with gaseous oxygen to form gaseous nitrogen monoxide and gaseous water?</h1> | null | 2 NH3(g) + 5/2 O2(g) -> 2 NO(g) + 3 H2O(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Is this balanced? How do you know?</p>
<p>Is this reaction a redox reaction? Could oxidation to <mathjax>#NO_2(g)#</mathjax> occur?</p></div>
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<div class="markdown"><p><mathjax>#2NH_3(g) + 5/2O_2(g) rarr 2NO(g) + 3H_2O(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Is this balanced? How do you know?</p>
<p>Is this reaction a redox reaction? Could oxidation to <mathjax>#NO_2(g)#</mathjax> occur?</p></div>
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<h1 class="questionTitle" itemprop="name">What is the equation for: Gaseous ammonia reacts with gaseous oxygen to form gaseous nitrogen monoxide and gaseous water?</h1>
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<div class="markdown"><p><mathjax>#2NH_3(g) + 5/2O_2(g) rarr 2NO(g) + 3H_2O(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Is this balanced? How do you know?</p>
<p>Is this reaction a redox reaction? Could oxidation to <mathjax>#NO_2(g)#</mathjax> occur?</p></div>
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</article> | What is the equation for: Gaseous ammonia reacts with gaseous oxygen to form gaseous nitrogen monoxide and gaseous water? | null |
539 | ac865d1e-6ddd-11ea-9c9d-ccda262736ce | https://socratic.org/questions/500-ml-of-hydrogen-at-27-c-and-3-atmospheres-pressure-is-compressed-to-100-ml-at | 18.65 atmospheres | start physical_unit 3 3 pressure atm qc_end physical_unit 3 3 8 9 pressure qc_end physical_unit 3 3 0 1 volume qc_end physical_unit 3 3 5 6 temperature qc_end physical_unit 3 3 17 18 temperature qc_end physical_unit 3 3 14 15 volume qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] hydrogen [IN] atmospheres"}] | [{"type":"physical unit","value":"18.65 atmospheres"}] | [{"type":"physical unit","value":"Pressure1 [OF] hydrogen [=] \\pu{3 atmospheres}"},{"type":"physical unit","value":"Volume1 [OF] hydrogen [=] \\pu{500 mL}"},{"type":"physical unit","value":"Temperature1 [OF] hydrogen [=] \\pu{27 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] hydrogen [=] \\pu{100 ℃}"},{"type":"physical unit","value":"Volume2 [OF] hydrogen [=] \\pu{100 mL}"}] | <h1 class="questionTitle" itemprop="name">500 mL of hydrogen at 27°C and 3 atmospheres pressure is compressed to 100 mL at 100°C. What is the new pressure in atmospheres?</h1> | null | 18.65 atmospheres | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>This appears to be a <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a></strong> problem.</p>
<p><mathjax>#(P_1V_1)/T_1 = (P_2V_2)/T_2#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#P_1 = "3 atm"; V_1 = "500 mL"; T_1 = "(27 + 273.15) K = 300.15 K"#</mathjax></p>
<p><mathjax>#P_2 = "?";color(white)(mml) V_2 = "100 mL"; T_2 = "(100 + 273.15) K = 373.15 K"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#P_2 = P_1 × V_1/V_2 × T_2/T_1 = "3 atm" × (500 color(red)(cancel(color(black)("mL"))))/(100 color(red)(cancel(color(black)("mL")))) × (373.15 color(red)(cancel(color(black)("K"))))/(300.15 color(red)(cancel(color(black)("K")))) = "18.6 atm"#</mathjax></p>
<p>The new pressure is <mathjax>#"18.6 atm"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The new pressure is <mathjax>#"18.6 atm"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>This appears to be a <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a></strong> problem.</p>
<p><mathjax>#(P_1V_1)/T_1 = (P_2V_2)/T_2#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#P_1 = "3 atm"; V_1 = "500 mL"; T_1 = "(27 + 273.15) K = 300.15 K"#</mathjax></p>
<p><mathjax>#P_2 = "?";color(white)(mml) V_2 = "100 mL"; T_2 = "(100 + 273.15) K = 373.15 K"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#P_2 = P_1 × V_1/V_2 × T_2/T_1 = "3 atm" × (500 color(red)(cancel(color(black)("mL"))))/(100 color(red)(cancel(color(black)("mL")))) × (373.15 color(red)(cancel(color(black)("K"))))/(300.15 color(red)(cancel(color(black)("K")))) = "18.6 atm"#</mathjax></p>
<p>The new pressure is <mathjax>#"18.6 atm"#</mathjax>.</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">500 mL of hydrogen at 27°C and 3 atmospheres pressure is compressed to 100 mL at 100°C. What is the new pressure in atmospheres?</h1>
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<div class="markdown"><p>The new pressure is <mathjax>#"18.6 atm"#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>This appears to be a <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a></strong> problem.</p>
<p><mathjax>#(P_1V_1)/T_1 = (P_2V_2)/T_2#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#P_1 = "3 atm"; V_1 = "500 mL"; T_1 = "(27 + 273.15) K = 300.15 K"#</mathjax></p>
<p><mathjax>#P_2 = "?";color(white)(mml) V_2 = "100 mL"; T_2 = "(100 + 273.15) K = 373.15 K"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#P_2 = P_1 × V_1/V_2 × T_2/T_1 = "3 atm" × (500 color(red)(cancel(color(black)("mL"))))/(100 color(red)(cancel(color(black)("mL")))) × (373.15 color(red)(cancel(color(black)("K"))))/(300.15 color(red)(cancel(color(black)("K")))) = "18.6 atm"#</mathjax></p>
<p>The new pressure is <mathjax>#"18.6 atm"#</mathjax>.</p></div>
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</article> | 500 mL of hydrogen at 27°C and 3 atmospheres pressure is compressed to 100 mL at 100°C. What is the new pressure in atmospheres? | null |
540 | acabc10c-6ddd-11ea-9a84-ccda262736ce | https://socratic.org/questions/5904bb2db72cff212c680fad | H2O | start chemical_formula qc_end physical_unit 13 13 10 11 mass qc_end physical_unit 18 18 15 16 mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] the hydride [IN] empirical"}] | [{"type":"chemical equation","value":"H2O"}] | [{"type":"physical unit","value":"Mass [OF] hydrogen [=] \\pu{4 g}"},{"type":"physical unit","value":"Mass [OF] oxygen [=] \\pu{28 g}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula of a hydride prepared from #4*g# of hydrogen, and #28*g# of oxygen?</h1> | null | H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that you have 4 grams of Hydrogen and 32 grams of Oxygen in your molecule so the first step is converting both weights given into moles. You use each molecules molecular weight (1.0079 g/mol for Hydrogen and 15.999 g/mol for Oxygen) to convert between moles and grams. These values are found on a periodic table! </p>
<p><mathjax>#("4g H")/1 * ("1mol H")/("1.0079g H") = "4mol H"#</mathjax></p>
<p><mathjax>#("32g O")/1 * ("1mol O")/("15.999g O") = "2mol O"#</mathjax></p>
<p>After you know how many moles of each compound you have, look at the ratio and see if you can reduce like you would a fraction.</p>
<p><mathjax>#("4mol H")/("2mol O") = ("2mol H")/("1mol O") = H_2O#</mathjax> </p>
<p>This gives you your final answer!</p></div>
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<div>
<div class="markdown"><p>Yep!</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that you have 4 grams of Hydrogen and 32 grams of Oxygen in your molecule so the first step is converting both weights given into moles. You use each molecules molecular weight (1.0079 g/mol for Hydrogen and 15.999 g/mol for Oxygen) to convert between moles and grams. These values are found on a periodic table! </p>
<p><mathjax>#("4g H")/1 * ("1mol H")/("1.0079g H") = "4mol H"#</mathjax></p>
<p><mathjax>#("32g O")/1 * ("1mol O")/("15.999g O") = "2mol O"#</mathjax></p>
<p>After you know how many moles of each compound you have, look at the ratio and see if you can reduce like you would a fraction.</p>
<p><mathjax>#("4mol H")/("2mol O") = ("2mol H")/("1mol O") = H_2O#</mathjax> </p>
<p>This gives you your final answer!</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the empirical formula of a hydride prepared from #4*g# of hydrogen, and #28*g# of oxygen?</h1>
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<div class="markdown"><p>Yep!</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that you have 4 grams of Hydrogen and 32 grams of Oxygen in your molecule so the first step is converting both weights given into moles. You use each molecules molecular weight (1.0079 g/mol for Hydrogen and 15.999 g/mol for Oxygen) to convert between moles and grams. These values are found on a periodic table! </p>
<p><mathjax>#("4g H")/1 * ("1mol H")/("1.0079g H") = "4mol H"#</mathjax></p>
<p><mathjax>#("32g O")/1 * ("1mol O")/("15.999g O") = "2mol O"#</mathjax></p>
<p>After you know how many moles of each compound you have, look at the ratio and see if you can reduce like you would a fraction.</p>
<p><mathjax>#("4mol H")/("2mol O") = ("2mol H")/("1mol O") = H_2O#</mathjax> </p>
<p>This gives you your final answer!</p></div>
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anor277
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<div class="markdown"><p>The empirical formula is <mathjax>#H_2O#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The empirical formula is the simplest, whole number ratio defining constituent atoms in a species. And how do we get this? Well, we divide the elemental masses thru by the atomic masses of each element...........</p>
<p><mathjax>#"Moles of hydrogen"=(4.0*g)/(1.0*g*mol^-1)=4*mol#</mathjax></p>
<p><mathjax>#"Moles of oxygen"=(32.0*g)/(16.0*g*mol^-1)=2*mol#</mathjax></p>
<p>If we divide thru by the smallest molar quantity (that of oxygen) we get <mathjax>#H_2O#</mathjax> as the empirical formula. </p></div>
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</article> | What is the empirical formula of a hydride prepared from #4*g# of hydrogen, and #28*g# of oxygen? | null |
541 | abcbd41f-6ddd-11ea-b2f7-ccda262736ce | https://socratic.org/questions/the-formula-for-sugar-is-c-12h-22o-11-how-many-moles-of-carbon-are-in-2-moles-of | 24.00 moles | start physical_unit 10 10 mole mol qc_end physical_unit 5 5 13 14 mole qc_end end | [{"type":"physical unit","value":"Mole [OF] carbon [IN] moles"}] | [{"type":"physical unit","value":"24.00 moles"}] | [{"type":"physical unit","value":"Mole [OF] C12H22O11 [=] \\pu{2 moles}"}] | <h1 class="questionTitle" itemprop="name">The formula for sugar is #C_12H_22O_11#. How many moles of carbon are in 2 moles of sugar?</h1> | null | 24.00 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Agreed? And this in 2 moles of sugar there should be how many moles of carbon? What is the mass of this quantity of carbon?</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>Well in one mole of sugar, CLEARLY there are <mathjax>#12*"moles of carbon atoms"#</mathjax>.</p></div>
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<div class="markdown"><p>Agreed? And this in 2 moles of sugar there should be how many moles of carbon? What is the mass of this quantity of carbon?</p></div>
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<div class="markdown"><p>Well in one mole of sugar, CLEARLY there are <mathjax>#12*"moles of carbon atoms"#</mathjax>.</p></div>
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<div class="markdown"><p>Agreed? And this in 2 moles of sugar there should be how many moles of carbon? What is the mass of this quantity of carbon?</p></div>
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</article> | The formula for sugar is #C_12H_22O_11#. How many moles of carbon are in 2 moles of sugar? | null |
542 | ace66852-6ddd-11ea-8c74-ccda262736ce | https://socratic.org/questions/59299731b72cff0244159da5 | 2 S2O3^2- -> S4O6^2- + 2 e- | start chemical_equation qc_end chemical_equation 9 9 qc_end chemical_equation 12 12 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the oxidation"}] | [{"type":"chemical equation","value":"2 S2O3^2- -> S4O6^2- + 2 e-"}] | [{"type":"chemical equation","value":"S2O3^2-"},{"type":"chemical equation","value":"S4O6^2-"}] | <h1 class="questionTitle" itemprop="name">How do we represent the oxidation of thiosulfate anion, #S_2O_3^(2-)# to tetrathionite, #S_4O_6^(2-)#...?</h1> | null | 2 S2O3^2- -> S4O6^2- + 2 e- | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I always considered <mathjax>#"thiosulfate dianion"#</mathjax>, <mathjax>#S_2O_3^(2-)#</mathjax>, to be the same as sulfate dianion, <mathjax>#SO_4^(2-)#</mathjax>, EXCEPT that one of the oxygen atoms has been replaced by congeneric sulfur. And this sulfur has PRECISELY the same oxidation state as the oxygen it replaces, i.e. <mathjax>#stackrel(-II)S#</mathjax>. The AVERAGE oxidation states is still.................................</p>
<p><mathjax>#(S(-II)+S(+VI))/2=S(+II)#</mathjax></p>
<p>Anyway, let's try to balance the beast.</p>
<p><mathjax>#2S_2O_3^(2-) rarr stackrel()S_4O_6^(2-)+2e^(-)#</mathjax> <mathjax>#;stackrel(II)Srarrstackrel(II*1/2)S#</mathjax></p>
<p>Do you agree that it is balanced with respect to mass and charge?</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#"Thiosulfate to tetrathionite..........."#</mathjax></p>
<p><mathjax>#2S_2O_3^(2-) rarr S_4O_6^(2-)+2e^(-)#</mathjax> <mathjax>#;stackrel(II)Srarrstackrel(II*1/2)S#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I always considered <mathjax>#"thiosulfate dianion"#</mathjax>, <mathjax>#S_2O_3^(2-)#</mathjax>, to be the same as sulfate dianion, <mathjax>#SO_4^(2-)#</mathjax>, EXCEPT that one of the oxygen atoms has been replaced by congeneric sulfur. And this sulfur has PRECISELY the same oxidation state as the oxygen it replaces, i.e. <mathjax>#stackrel(-II)S#</mathjax>. The AVERAGE oxidation states is still.................................</p>
<p><mathjax>#(S(-II)+S(+VI))/2=S(+II)#</mathjax></p>
<p>Anyway, let's try to balance the beast.</p>
<p><mathjax>#2S_2O_3^(2-) rarr stackrel()S_4O_6^(2-)+2e^(-)#</mathjax> <mathjax>#;stackrel(II)Srarrstackrel(II*1/2)S#</mathjax></p>
<p>Do you agree that it is balanced with respect to mass and charge?</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How do we represent the oxidation of thiosulfate anion, #S_2O_3^(2-)# to tetrathionite, #S_4O_6^(2-)#...?</h1>
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<div class="markdown"><p><mathjax>#"Thiosulfate to tetrathionite..........."#</mathjax></p>
<p><mathjax>#2S_2O_3^(2-) rarr S_4O_6^(2-)+2e^(-)#</mathjax> <mathjax>#;stackrel(II)Srarrstackrel(II*1/2)S#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I always considered <mathjax>#"thiosulfate dianion"#</mathjax>, <mathjax>#S_2O_3^(2-)#</mathjax>, to be the same as sulfate dianion, <mathjax>#SO_4^(2-)#</mathjax>, EXCEPT that one of the oxygen atoms has been replaced by congeneric sulfur. And this sulfur has PRECISELY the same oxidation state as the oxygen it replaces, i.e. <mathjax>#stackrel(-II)S#</mathjax>. The AVERAGE oxidation states is still.................................</p>
<p><mathjax>#(S(-II)+S(+VI))/2=S(+II)#</mathjax></p>
<p>Anyway, let's try to balance the beast.</p>
<p><mathjax>#2S_2O_3^(2-) rarr stackrel()S_4O_6^(2-)+2e^(-)#</mathjax> <mathjax>#;stackrel(II)Srarrstackrel(II*1/2)S#</mathjax></p>
<p>Do you agree that it is balanced with respect to mass and charge?</p></div>
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</article> | How do we represent the oxidation of thiosulfate anion, #S_2O_3^(2-)# to tetrathionite, #S_4O_6^(2-)#...? | null |
543 | abdcfe8b-6ddd-11ea-9d35-ccda262736ce | https://socratic.org/questions/how-do-you-balance-ca-p-ca-2p-3 | 2 Ca + 3 P -> Ca2P3 | start chemical_equation qc_end chemical_equation 4 8 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 Ca + 3 P -> Ca2P3"}] | [{"type":"chemical equation","value":"Ca + P -> Ca2P3"}] | <h1 class="questionTitle" itemprop="name">How do you balance #Ca + P -> Ca_2P_3#?</h1> | null | 2 Ca + 3 P -> Ca2P3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Assume:<br/>
<mathjax>#color(white)("XXX")color(red)(x)Ca+color(red)(y)P rarr color(red)(z)Ca_2P_3#</mathjax></p>
<p>Based on <mathjax>#Ca#</mathjax><br/>
[1]<mathjax>#color(white)("XXX")x-2z=0#</mathjax><br/>
Based on <mathjax>#P#</mathjax><br/>
[2]<mathjax>#color(white)("XXX")y-3z=0#</mathjax></p>
<p>In order to find the ratio of <mathjax>#(x:y:z)#</mathjax> we will arbitrarily set one of our variables to <mathjax>#1#</mathjax>; later we may need to adjust our ratios to get integer results.</p>
<p>In order to see how this works in the general case I have chosen to use <mathjax>#x=1#</mathjax></p>
<p>From [1] with <mathjax>#x=1#</mathjax><br/>
[3]<mathjax>#color(white)("XXX")z=1/2#</mathjax><br/>
From [2] with <mathjax>#z=1/2#</mathjax><br/>
[4]<mathjax>#color(white)("XXX")y=3/2#</mathjax></p>
<p>So the ratio of components is<br/>
<mathjax>#color(white)("XXX")(1,3/2,1/2)#</mathjax></p>
<p>Since we want integer factors we will multiply all terms by <mathjax>#2#</mathjax> to clear all fractions, giving:<br/>
<mathjax>#color(white)("XXX")(x:y:z) = (2:3:1)#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#color(red)(2)Ca+color(red)(3)P rarr color(red)(1)Ca_2P_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Assume:<br/>
<mathjax>#color(white)("XXX")color(red)(x)Ca+color(red)(y)P rarr color(red)(z)Ca_2P_3#</mathjax></p>
<p>Based on <mathjax>#Ca#</mathjax><br/>
[1]<mathjax>#color(white)("XXX")x-2z=0#</mathjax><br/>
Based on <mathjax>#P#</mathjax><br/>
[2]<mathjax>#color(white)("XXX")y-3z=0#</mathjax></p>
<p>In order to find the ratio of <mathjax>#(x:y:z)#</mathjax> we will arbitrarily set one of our variables to <mathjax>#1#</mathjax>; later we may need to adjust our ratios to get integer results.</p>
<p>In order to see how this works in the general case I have chosen to use <mathjax>#x=1#</mathjax></p>
<p>From [1] with <mathjax>#x=1#</mathjax><br/>
[3]<mathjax>#color(white)("XXX")z=1/2#</mathjax><br/>
From [2] with <mathjax>#z=1/2#</mathjax><br/>
[4]<mathjax>#color(white)("XXX")y=3/2#</mathjax></p>
<p>So the ratio of components is<br/>
<mathjax>#color(white)("XXX")(1,3/2,1/2)#</mathjax></p>
<p>Since we want integer factors we will multiply all terms by <mathjax>#2#</mathjax> to clear all fractions, giving:<br/>
<mathjax>#color(white)("XXX")(x:y:z) = (2:3:1)#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How do you balance #Ca + P -> Ca_2P_3#?</h1>
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Alan P.
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Dec 25, 2015
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<div class="markdown"><p><mathjax>#color(red)(2)Ca+color(red)(3)P rarr color(red)(1)Ca_2P_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Assume:<br/>
<mathjax>#color(white)("XXX")color(red)(x)Ca+color(red)(y)P rarr color(red)(z)Ca_2P_3#</mathjax></p>
<p>Based on <mathjax>#Ca#</mathjax><br/>
[1]<mathjax>#color(white)("XXX")x-2z=0#</mathjax><br/>
Based on <mathjax>#P#</mathjax><br/>
[2]<mathjax>#color(white)("XXX")y-3z=0#</mathjax></p>
<p>In order to find the ratio of <mathjax>#(x:y:z)#</mathjax> we will arbitrarily set one of our variables to <mathjax>#1#</mathjax>; later we may need to adjust our ratios to get integer results.</p>
<p>In order to see how this works in the general case I have chosen to use <mathjax>#x=1#</mathjax></p>
<p>From [1] with <mathjax>#x=1#</mathjax><br/>
[3]<mathjax>#color(white)("XXX")z=1/2#</mathjax><br/>
From [2] with <mathjax>#z=1/2#</mathjax><br/>
[4]<mathjax>#color(white)("XXX")y=3/2#</mathjax></p>
<p>So the ratio of components is<br/>
<mathjax>#color(white)("XXX")(1,3/2,1/2)#</mathjax></p>
<p>Since we want integer factors we will multiply all terms by <mathjax>#2#</mathjax> to clear all fractions, giving:<br/>
<mathjax>#color(white)("XXX")(x:y:z) = (2:3:1)#</mathjax></p></div>
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</article> | How do you balance #Ca + P -> Ca_2P_3#? | null |
544 | ab069877-6ddd-11ea-929e-ccda262736ce | https://socratic.org/questions/how-many-moles-of-tin-metal-will-be-produced-when-0-74-moles-of-tin-iv-oxide-is- | 0.74 moles | start physical_unit 4 5 mole mol qc_end chemical_equation 21 29 qc_end physical_unit 13 15 10 11 mole qc_end end | [{"type":"physical unit","value":"Mole [OF] tin metal [IN] moles"}] | [{"type":"physical unit","value":"0.74 moles"}] | [{"type":"chemical equation","value":"SnO2 + 2 H2 -> Sn + 2 H2O"},{"type":"physical unit","value":"Mole [OF] tin IV oxide [=] \\pu{0.74 moles}"}] | <h1 class="questionTitle" itemprop="name">How many moles of tin metal will be produced when 0.74 moles of tin IV oxide is heated in the reaction #SnO_2 + 2H_2 -> Sn + 2H_2O#?</h1> | null | 0.74 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, why clearly?</p>
<p>Well, to begin, you already have the stoichiometric equation:</p>
<p><mathjax>#SnO_2(s) +2H_2(g) rarr Sn(s) + 2H_2O(l)#</mathjax></p>
<p>This is a so-called <mathjax>#"redox reaction"#</mathjax>: stannic oxide is reduced to tin metal; and dihydrogen is oxidized to water.</p>
<p>There is 1:1 equivalence of stannic oxide to tin metal. And thus 1:1 molar equivalence. Since stannic oxide has a molar mass of <mathjax>#150.71*g*mol^-1#</mathjax>, normally you would be asked how much metal would be produced. So, how much metal in grams?</p></div>
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<div class="markdown"><p>Clearly, <mathjax>#0.74*mol#</mathjax> of tin metal will result in the presence of stoichiometric dihydrogen. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, why clearly?</p>
<p>Well, to begin, you already have the stoichiometric equation:</p>
<p><mathjax>#SnO_2(s) +2H_2(g) rarr Sn(s) + 2H_2O(l)#</mathjax></p>
<p>This is a so-called <mathjax>#"redox reaction"#</mathjax>: stannic oxide is reduced to tin metal; and dihydrogen is oxidized to water.</p>
<p>There is 1:1 equivalence of stannic oxide to tin metal. And thus 1:1 molar equivalence. Since stannic oxide has a molar mass of <mathjax>#150.71*g*mol^-1#</mathjax>, normally you would be asked how much metal would be produced. So, how much metal in grams?</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of tin metal will be produced when 0.74 moles of tin IV oxide is heated in the reaction #SnO_2 + 2H_2 -> Sn + 2H_2O#?</h1>
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<div class="markdown"><p>Clearly, <mathjax>#0.74*mol#</mathjax> of tin metal will result in the presence of stoichiometric dihydrogen. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So, why clearly?</p>
<p>Well, to begin, you already have the stoichiometric equation:</p>
<p><mathjax>#SnO_2(s) +2H_2(g) rarr Sn(s) + 2H_2O(l)#</mathjax></p>
<p>This is a so-called <mathjax>#"redox reaction"#</mathjax>: stannic oxide is reduced to tin metal; and dihydrogen is oxidized to water.</p>
<p>There is 1:1 equivalence of stannic oxide to tin metal. And thus 1:1 molar equivalence. Since stannic oxide has a molar mass of <mathjax>#150.71*g*mol^-1#</mathjax>, normally you would be asked how much metal would be produced. So, how much metal in grams?</p></div>
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</article> | How many moles of tin metal will be produced when 0.74 moles of tin IV oxide is heated in the reaction #SnO_2 + 2H_2 -> Sn + 2H_2O#? | null |
545 | abe47e17-6ddd-11ea-b2e4-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-solution-with-h-7-0-times-10-12 | 11.15 | start physical_unit 6 6 ph none qc_end physical_unit 6 6 10 13 [h+] qc_end end | [{"type":"physical unit","value":"pH [OF] the solution"}] | [{"type":"physical unit","value":"11.15"}] | [{"type":"physical unit","value":"[H+] [OF] the solution [=] \\pu{7.0 × 10^(-12) M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a solution with #[H^+] = 7.0 times 10^-12 "M"?#</h1> | null | 11.15 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#pH=-log[H^+]=-log[7.0 * 10^-12] = 11.15.#</mathjax> Just plug and play at this point.</p></div>
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<div class="markdown"><p><a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> = 11.15</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#pH=-log[H^+]=-log[7.0 * 10^-12] = 11.15.#</mathjax> Just plug and play at this point.</p></div>
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<div class="markdown"><p><a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> = 11.15</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#pH=-log[H^+]=-log[7.0 * 10^-12] = 11.15.#</mathjax> Just plug and play at this point.</p></div>
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</article> | What is the pH of a solution with #[H^+] = 7.0 times 10^-12 "M"?# | null |
546 | aa3b19fa-6ddd-11ea-b0d1-ccda262736ce | https://socratic.org/questions/asample-of-a-compound-contains-0-150-mol-c-0-105-mole-h-0-0300-mol-o-and-0-0450- | C20H14O4N6 | start chemical_formula qc_end physical_unit 7 7 5 6 mole qc_end physical_unit 10 10 8 9 mole qc_end physical_unit 13 13 11 12 mole qc_end physical_unit 17 17 15 16 mole qc_end physical_unit 22 23 28 29 molecular_weight qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] default"}] | [{"type":"chemical equation","value":"C20H14O4N6"}] | [{"type":"physical unit","value":"Mole [OF] C [=] \\pu{0.150 mol}"},{"type":"physical unit","value":"Mole [OF] H [=] \\pu{0.105 mole}"},{"type":"physical unit","value":"Mole [OF] O [=] \\pu{0.0300 mol}"},{"type":"physical unit","value":"Mole [OF] N [=] \\pu{0.0450 mol}"},{"type":"physical unit","value":"molecular mass [OF] the compound [=] \\pu{402 g/mol}"}] | <h1 class="questionTitle" itemprop="name">Asample of a compound contains 0.150 mol C, 0.105 mole H, 0.0300 mol O, and 0.0450 mol N. The molecular mass of the compound is found to be 402 g/mol. What is the compound's chemical formula? </h1> | null | C20H14O4N6 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Calculate the empirical formula.</strong></p>
<p>The <strong>empirical formula</strong> is the simplest whole-number ratio of atoms in a compound.</p>
<p>The ratio of atoms is the same as the ratio of moles, so our job is to calculate the molar ratio of <mathjax>#"C"#</mathjax> to <mathjax>#"H"#</mathjax> to <mathjax>#"O"#</mathjax> to <mathjax>#"N"#</mathjax>.</p>
<blockquote></blockquote>
<p>I like to summarize the calculations in a table.</p>
<p><mathjax>#"Element"color(white)(m)"Moles"color(white)(Xll) "Ratio"color(white)(mll)×2 color(white)(m) "Integers"#</mathjax><br/>
<mathjax>#stackrel(——————————————————)(color(white)(m)"C" color(white)(XXXl)0.150 color(white)(Xml)5.00
color(white)(Xml)10.0color(white)(mm) 10#</mathjax><br/>
<mathjax>#color(white)(m)"H" color(white)(XXXl)0.105 color(white)(mml)"3.50 color(white)(Xml)7.00color(white)(mmll) 7#</mathjax><br/>
<mathjax>#color(white)(m)"O" color(white)(XXXl)0.0300 color(white)(mm)"1 color(white)(Xmmll)2 color(white)(mmmm)2#</mathjax><br/>
<mathjax>#color(white)(m)"N" color(white)(XXXl)0.0450 color(white)(mm)"1.50 color(white)(Xml)3.00 color(white)(mml)3#</mathjax></p>
<p>The empirical formula of the compound is <mathjax>#"C"_10"H"_7"O"_2"N"_3#</mathjax>.</p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the empirical formula mass</strong></p>
<p>The molecular mass of <mathjax>#"C"_10"H"_7"O"_2"N"_3#</mathjax> is 201.18 u.</p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the molecular mass.</strong></p>
<p>The molecular mass must be an integral multiple multiple of the empirical formula mass.</p>
<p><mathjax>#"MM" = n × "EFM"#</mathjax></p>
<p><mathjax>#n = "MM"/"EFM" = (402 color(red)(cancel(color(black)("u"))))/(201.18 color(red)(cancel(color(black)("u")))) = 2.00 ≈ 2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4. Calculate the Molecular Formula</strong></p>
<p>The molecular formula must be twice the empirical formula.</p>
<p>The molecular formula is <mathjax>#"C"_20"H"_14"O"_4"N"_6#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The chemical formula is <mathjax>#"C"_20"H"_14"O"_4"N"_6#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Calculate the empirical formula.</strong></p>
<p>The <strong>empirical formula</strong> is the simplest whole-number ratio of atoms in a compound.</p>
<p>The ratio of atoms is the same as the ratio of moles, so our job is to calculate the molar ratio of <mathjax>#"C"#</mathjax> to <mathjax>#"H"#</mathjax> to <mathjax>#"O"#</mathjax> to <mathjax>#"N"#</mathjax>.</p>
<blockquote></blockquote>
<p>I like to summarize the calculations in a table.</p>
<p><mathjax>#"Element"color(white)(m)"Moles"color(white)(Xll) "Ratio"color(white)(mll)×2 color(white)(m) "Integers"#</mathjax><br/>
<mathjax>#stackrel(——————————————————)(color(white)(m)"C" color(white)(XXXl)0.150 color(white)(Xml)5.00
color(white)(Xml)10.0color(white)(mm) 10#</mathjax><br/>
<mathjax>#color(white)(m)"H" color(white)(XXXl)0.105 color(white)(mml)"3.50 color(white)(Xml)7.00color(white)(mmll) 7#</mathjax><br/>
<mathjax>#color(white)(m)"O" color(white)(XXXl)0.0300 color(white)(mm)"1 color(white)(Xmmll)2 color(white)(mmmm)2#</mathjax><br/>
<mathjax>#color(white)(m)"N" color(white)(XXXl)0.0450 color(white)(mm)"1.50 color(white)(Xml)3.00 color(white)(mml)3#</mathjax></p>
<p>The empirical formula of the compound is <mathjax>#"C"_10"H"_7"O"_2"N"_3#</mathjax>.</p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the empirical formula mass</strong></p>
<p>The molecular mass of <mathjax>#"C"_10"H"_7"O"_2"N"_3#</mathjax> is 201.18 u.</p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the molecular mass.</strong></p>
<p>The molecular mass must be an integral multiple multiple of the empirical formula mass.</p>
<p><mathjax>#"MM" = n × "EFM"#</mathjax></p>
<p><mathjax>#n = "MM"/"EFM" = (402 color(red)(cancel(color(black)("u"))))/(201.18 color(red)(cancel(color(black)("u")))) = 2.00 ≈ 2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4. Calculate the Molecular Formula</strong></p>
<p>The molecular formula must be twice the empirical formula.</p>
<p>The molecular formula is <mathjax>#"C"_20"H"_14"O"_4"N"_6#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Asample of a compound contains 0.150 mol C, 0.105 mole H, 0.0300 mol O, and 0.0450 mol N. The molecular mass of the compound is found to be 402 g/mol. What is the compound's chemical formula? </h1>
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Ernest Z.
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<div class="markdown"><p>The chemical formula is <mathjax>#"C"_20"H"_14"O"_4"N"_6#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Calculate the empirical formula.</strong></p>
<p>The <strong>empirical formula</strong> is the simplest whole-number ratio of atoms in a compound.</p>
<p>The ratio of atoms is the same as the ratio of moles, so our job is to calculate the molar ratio of <mathjax>#"C"#</mathjax> to <mathjax>#"H"#</mathjax> to <mathjax>#"O"#</mathjax> to <mathjax>#"N"#</mathjax>.</p>
<blockquote></blockquote>
<p>I like to summarize the calculations in a table.</p>
<p><mathjax>#"Element"color(white)(m)"Moles"color(white)(Xll) "Ratio"color(white)(mll)×2 color(white)(m) "Integers"#</mathjax><br/>
<mathjax>#stackrel(——————————————————)(color(white)(m)"C" color(white)(XXXl)0.150 color(white)(Xml)5.00
color(white)(Xml)10.0color(white)(mm) 10#</mathjax><br/>
<mathjax>#color(white)(m)"H" color(white)(XXXl)0.105 color(white)(mml)"3.50 color(white)(Xml)7.00color(white)(mmll) 7#</mathjax><br/>
<mathjax>#color(white)(m)"O" color(white)(XXXl)0.0300 color(white)(mm)"1 color(white)(Xmmll)2 color(white)(mmmm)2#</mathjax><br/>
<mathjax>#color(white)(m)"N" color(white)(XXXl)0.0450 color(white)(mm)"1.50 color(white)(Xml)3.00 color(white)(mml)3#</mathjax></p>
<p>The empirical formula of the compound is <mathjax>#"C"_10"H"_7"O"_2"N"_3#</mathjax>.</p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the empirical formula mass</strong></p>
<p>The molecular mass of <mathjax>#"C"_10"H"_7"O"_2"N"_3#</mathjax> is 201.18 u.</p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the molecular mass.</strong></p>
<p>The molecular mass must be an integral multiple multiple of the empirical formula mass.</p>
<p><mathjax>#"MM" = n × "EFM"#</mathjax></p>
<p><mathjax>#n = "MM"/"EFM" = (402 color(red)(cancel(color(black)("u"))))/(201.18 color(red)(cancel(color(black)("u")))) = 2.00 ≈ 2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4. Calculate the Molecular Formula</strong></p>
<p>The molecular formula must be twice the empirical formula.</p>
<p>The molecular formula is <mathjax>#"C"_20"H"_14"O"_4"N"_6#</mathjax></p></div>
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</article> | Asample of a compound contains 0.150 mol C, 0.105 mole H, 0.0300 mol O, and 0.0450 mol N. The molecular mass of the compound is found to be 402 g/mol. What is the compound's chemical formula? | null |
547 | a8a831e2-6ddd-11ea-a2f1-ccda262736ce | https://socratic.org/questions/58e1711911ef6b0999f06efa | 7.47 × 10^(-5) mol/L | start physical_unit 4 5 concentration mol/l qc_end physical_unit 28 28 21 22 volume qc_end physical_unit 28 28 24 27 concentration qc_end physical_unit 4 5 30 31 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Concentration [OF] Vitamin C [IN] mol/L"}] | [{"type":"physical unit","value":"7.47 × 10^(-5) mol/L"}] | [{"type":"physical unit","value":"Volume [OF] K3Fe(CN)6 [=] \\pu{43.9 mL}"},{"type":"physical unit","value":"Concentration [OF] K3Fe(CN)6 [=] \\pu{1.12 × 10^(-4) mol/L}"},{"type":"physical unit","value":"Volume [OF] Vitamin C [=] \\pu{32.9 mL}"},{"type":"other","value":"The hexacyanoferrate(III) ion oxidizes Vitamin C (ascorbic acid, H2Asc) to dehydroascorbic acid (Asc)."}] | <h1 class="questionTitle" itemprop="name">The hexacyanoferrate(III) ion oxidizes Vitamin C (ascorbic acid, #"H"_2"Asc"#) to dehydroascorbic acid (#"Asc"#). What is the concentration of Vitamin C if 43.9 mL of #1.12 × 10^"-4"color(white)(l) "mol/L K"_3"Fe(CN)"_6# oxidizes 32.9 mL of Vitamin C?</h1> | null | 7.47 × 10^(-5) mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The equations for the redox reaction are:</p>
<p><mathjax>#2×["Fe(CN)"_6^"3-" + "e"^"-" → "Fe(CN)"_6^"4-"]#</mathjax><br/>
<mathjax>#1×["AscH"_2 → "Asc" + "2H"^"+" + "2e"^"-"#</mathjax><br/>
<mathjax>#stackrel(————————————————————)("2Fe(CN)"_6^"3-" +"AscH"_2 → "2Fe(CN)"_6^"4-" + "Asc" + "2H"^"+")#</mathjax></p>
<blockquote></blockquote>
<p><strong>1.</strong> Calculate the moles of <mathjax>#"Fe(CN)"_6^"3-"#</mathjax></p>
<p><mathjax>#"Moles of Fe(CN)"_6^"3-" = "0.0439" color(red)(cancel(color(black)("L Fe(CN)"_6^"3-"))) ×(1.12 × 10^"-4" color(white)(l)"mol Fe(CN)"_6^"3-")/(1 color(red)(cancel(color(black)("L Fe(CN)"_6^"3-")))) = 4.917 × 10^"-6"color(white)(l) "mol Fe(CN)"_6^"3-"#</mathjax></p>
<blockquote></blockquote>
<p><strong>2.</strong> Calculate the moles of <mathjax>#"AscH"_2#</mathjax></p>
<p><mathjax>#"Moles of AscH"_2 = 4.917 × 10^"-6"color(red)(cancel(color(black)("mol Fe(CN)"_6^"3-"))) × ("1 mol AscH"_2)/(2 color(red)(cancel(color(black)("mol Fe(CN)"_6^"3-")))) = 2.458 × 10^"-6"color(white)(l)"mol AscH"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>3.</strong> Calculate the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the <mathjax>#"AscH"_2#</mathjax></p>
<p><mathjax>#"Molarity" = "moles"/"litres" = (2.458 × 10^"-6"color(white)(l) "mol")/("0.0329 L") = 7.47 × 10^"-5"color(white)(l) "mol/L"#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>The concentration is <mathjax>#7.47 × 10^"-5"color(white)(l) "mol/L"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The equations for the redox reaction are:</p>
<p><mathjax>#2×["Fe(CN)"_6^"3-" + "e"^"-" → "Fe(CN)"_6^"4-"]#</mathjax><br/>
<mathjax>#1×["AscH"_2 → "Asc" + "2H"^"+" + "2e"^"-"#</mathjax><br/>
<mathjax>#stackrel(————————————————————)("2Fe(CN)"_6^"3-" +"AscH"_2 → "2Fe(CN)"_6^"4-" + "Asc" + "2H"^"+")#</mathjax></p>
<blockquote></blockquote>
<p><strong>1.</strong> Calculate the moles of <mathjax>#"Fe(CN)"_6^"3-"#</mathjax></p>
<p><mathjax>#"Moles of Fe(CN)"_6^"3-" = "0.0439" color(red)(cancel(color(black)("L Fe(CN)"_6^"3-"))) ×(1.12 × 10^"-4" color(white)(l)"mol Fe(CN)"_6^"3-")/(1 color(red)(cancel(color(black)("L Fe(CN)"_6^"3-")))) = 4.917 × 10^"-6"color(white)(l) "mol Fe(CN)"_6^"3-"#</mathjax></p>
<blockquote></blockquote>
<p><strong>2.</strong> Calculate the moles of <mathjax>#"AscH"_2#</mathjax></p>
<p><mathjax>#"Moles of AscH"_2 = 4.917 × 10^"-6"color(red)(cancel(color(black)("mol Fe(CN)"_6^"3-"))) × ("1 mol AscH"_2)/(2 color(red)(cancel(color(black)("mol Fe(CN)"_6^"3-")))) = 2.458 × 10^"-6"color(white)(l)"mol AscH"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>3.</strong> Calculate the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the <mathjax>#"AscH"_2#</mathjax></p>
<p><mathjax>#"Molarity" = "moles"/"litres" = (2.458 × 10^"-6"color(white)(l) "mol")/("0.0329 L") = 7.47 × 10^"-5"color(white)(l) "mol/L"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">The hexacyanoferrate(III) ion oxidizes Vitamin C (ascorbic acid, #"H"_2"Asc"#) to dehydroascorbic acid (#"Asc"#). What is the concentration of Vitamin C if 43.9 mL of #1.12 × 10^"-4"color(white)(l) "mol/L K"_3"Fe(CN)"_6# oxidizes 32.9 mL of Vitamin C?</h1>
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<div class="markdown"><p>The concentration is <mathjax>#7.47 × 10^"-5"color(white)(l) "mol/L"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<p>The equations for the redox reaction are:</p>
<p><mathjax>#2×["Fe(CN)"_6^"3-" + "e"^"-" → "Fe(CN)"_6^"4-"]#</mathjax><br/>
<mathjax>#1×["AscH"_2 → "Asc" + "2H"^"+" + "2e"^"-"#</mathjax><br/>
<mathjax>#stackrel(————————————————————)("2Fe(CN)"_6^"3-" +"AscH"_2 → "2Fe(CN)"_6^"4-" + "Asc" + "2H"^"+")#</mathjax></p>
<blockquote></blockquote>
<p><strong>1.</strong> Calculate the moles of <mathjax>#"Fe(CN)"_6^"3-"#</mathjax></p>
<p><mathjax>#"Moles of Fe(CN)"_6^"3-" = "0.0439" color(red)(cancel(color(black)("L Fe(CN)"_6^"3-"))) ×(1.12 × 10^"-4" color(white)(l)"mol Fe(CN)"_6^"3-")/(1 color(red)(cancel(color(black)("L Fe(CN)"_6^"3-")))) = 4.917 × 10^"-6"color(white)(l) "mol Fe(CN)"_6^"3-"#</mathjax></p>
<blockquote></blockquote>
<p><strong>2.</strong> Calculate the moles of <mathjax>#"AscH"_2#</mathjax></p>
<p><mathjax>#"Moles of AscH"_2 = 4.917 × 10^"-6"color(red)(cancel(color(black)("mol Fe(CN)"_6^"3-"))) × ("1 mol AscH"_2)/(2 color(red)(cancel(color(black)("mol Fe(CN)"_6^"3-")))) = 2.458 × 10^"-6"color(white)(l)"mol AscH"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>3.</strong> Calculate the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the <mathjax>#"AscH"_2#</mathjax></p>
<p><mathjax>#"Molarity" = "moles"/"litres" = (2.458 × 10^"-6"color(white)(l) "mol")/("0.0329 L") = 7.47 × 10^"-5"color(white)(l) "mol/L"#</mathjax></p></div>
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</article> | The hexacyanoferrate(III) ion oxidizes Vitamin C (ascorbic acid, #"H"_2"Asc"#) to dehydroascorbic acid (#"Asc"#). What is the concentration of Vitamin C if 43.9 mL of #1.12 × 10^"-4"color(white)(l) "mol/L K"_3"Fe(CN)"_6# oxidizes 32.9 mL of Vitamin C? | null |
548 | ad1b4008-6ddd-11ea-8ac7-ccda262736ce | https://socratic.org/questions/ph-of-0-08mol-dm-3-of-hocl-is-2-85-calculate-its-ionization-constant | 2.54 × 10^(-5) | start physical_unit 18 19 equilibrium_constant_k none qc_end physical_unit 6 6 12 12 ph qc_end physical_unit 8 8 4 5 molarity qc_end end | [{"type":"physical unit","value":"Ionization constant [OF] the acid"}] | [{"type":"physical unit","value":"2.54 × 10^(-5)"}] | [{"type":"physical unit","value":"pH [OF] HClO solution [=] \\pu{2.85}"},{"type":"physical unit","value":"Molarity [OF] HClO solution [=] \\pu{0.08 mol/dm^3}"}] | <h1 class="questionTitle" itemprop="name">The #"pH"# of a #"0.08 mol dm"^(-3)# solution of #"HClO"# is equal to #2.85#. Calculate the ionization constant of the acid?</h1> | null | 2.54 × 10^(-5) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the <mathjax>#"pH"#</mathjax> of the solution will give you the <strong>equilibrium concentration</strong> of hydronium cations, which, in turn, will give you the equilibrium concentration of the hypochlorite anions and of the hypochlorous acid.</p>
<blockquote>
<p><mathjax>#"HClO"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "ClO"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Now, you know that when <mathjax>#1#</mathjax> <strong>mole</strong> of hypochlorous acid <strong>dissociates</strong>, it produces <mathjax>#1#</mathjax> <strong>mole</strong> of hypochlorite anions and <mathjax>#1#</mathjax> <strong>mole</strong> of hydronium cations. </p>
<p>This means that if the solution contains</p>
<blockquote>
<p><mathjax>#["H"_ 3"O"^(+)] = 10^(-"pH") quad "M"#</mathjax></p>
</blockquote>
<p>then it must also contain</p>
<blockquote>
<p><mathjax>#["ClO"^(-)] = 10^(-"pH") quad "M"#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#["HClO"] = (0.08 - 10^(-"pH")) quad "M"#</mathjax></p>
<blockquote>
<p>This is what you'd expect the <strong>equilibrium concentration</strong> of hypochlorous acid to be equal to if its dissociation produced <mathjax>#10^(-"pH")#</mathjax> <mathjax>#"M"#</mathjax> of hydronium cations. </p>
<p>In other words, in order for the dissociation of the acid to produce <mathjax>#["H"_3"O"^(+)] = 10^(-"pH")#</mathjax> <mathjax>#"M"#</mathjax>, the concentration of the acid must <strong>decrease</strong> by <mathjax>#["H"_3"O"^(+)] = 10^(-"pH")#</mathjax> <mathjax>#"M"#</mathjax>.</p>
</blockquote>
</blockquote>
<p>So now that you know the equilibrium concentrations for all three chemical species that are of interest here, you can plug them into the expression of the <strong>acid dissociation constant</strong>, <mathjax>#K_a#</mathjax>.</p>
<blockquote>
<p><mathjax>#K_a = (["ClO"^(-)] * ["H"_ 3"O"^(+)])/(["HClO"])#</mathjax></p>
</blockquote>
<p>You will end up with</p>
<blockquote>
<p><mathjax>#K_a = (10^(-"pH") * 10^(-"pH"))/(0.08 - 10^(-"pH"))#</mathjax></p>
</blockquote>
<p>which gets you</p>
<blockquote>
<p><mathjax>#K_a = (10^(-2.85))^2/(0.08 - 10^(-2.85)) = color(darkgreen)(ul(color(black)(2.5 * 10^(-5))))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that you have only one significant figure for the initial concentration of the acid. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2.5 * 10^(-5)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the <mathjax>#"pH"#</mathjax> of the solution will give you the <strong>equilibrium concentration</strong> of hydronium cations, which, in turn, will give you the equilibrium concentration of the hypochlorite anions and of the hypochlorous acid.</p>
<blockquote>
<p><mathjax>#"HClO"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "ClO"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Now, you know that when <mathjax>#1#</mathjax> <strong>mole</strong> of hypochlorous acid <strong>dissociates</strong>, it produces <mathjax>#1#</mathjax> <strong>mole</strong> of hypochlorite anions and <mathjax>#1#</mathjax> <strong>mole</strong> of hydronium cations. </p>
<p>This means that if the solution contains</p>
<blockquote>
<p><mathjax>#["H"_ 3"O"^(+)] = 10^(-"pH") quad "M"#</mathjax></p>
</blockquote>
<p>then it must also contain</p>
<blockquote>
<p><mathjax>#["ClO"^(-)] = 10^(-"pH") quad "M"#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#["HClO"] = (0.08 - 10^(-"pH")) quad "M"#</mathjax></p>
<blockquote>
<p>This is what you'd expect the <strong>equilibrium concentration</strong> of hypochlorous acid to be equal to if its dissociation produced <mathjax>#10^(-"pH")#</mathjax> <mathjax>#"M"#</mathjax> of hydronium cations. </p>
<p>In other words, in order for the dissociation of the acid to produce <mathjax>#["H"_3"O"^(+)] = 10^(-"pH")#</mathjax> <mathjax>#"M"#</mathjax>, the concentration of the acid must <strong>decrease</strong> by <mathjax>#["H"_3"O"^(+)] = 10^(-"pH")#</mathjax> <mathjax>#"M"#</mathjax>.</p>
</blockquote>
</blockquote>
<p>So now that you know the equilibrium concentrations for all three chemical species that are of interest here, you can plug them into the expression of the <strong>acid dissociation constant</strong>, <mathjax>#K_a#</mathjax>.</p>
<blockquote>
<p><mathjax>#K_a = (["ClO"^(-)] * ["H"_ 3"O"^(+)])/(["HClO"])#</mathjax></p>
</blockquote>
<p>You will end up with</p>
<blockquote>
<p><mathjax>#K_a = (10^(-"pH") * 10^(-"pH"))/(0.08 - 10^(-"pH"))#</mathjax></p>
</blockquote>
<p>which gets you</p>
<blockquote>
<p><mathjax>#K_a = (10^(-2.85))^2/(0.08 - 10^(-2.85)) = color(darkgreen)(ul(color(black)(2.5 * 10^(-5))))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that you have only one significant figure for the initial concentration of the acid. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">The #"pH"# of a #"0.08 mol dm"^(-3)# solution of #"HClO"# is equal to #2.85#. Calculate the ionization constant of the acid?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2018-01-07T00:51:47" itemprop="dateCreated">
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<div class="markdown"><p><mathjax>#2.5 * 10^(-5)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the <mathjax>#"pH"#</mathjax> of the solution will give you the <strong>equilibrium concentration</strong> of hydronium cations, which, in turn, will give you the equilibrium concentration of the hypochlorite anions and of the hypochlorous acid.</p>
<blockquote>
<p><mathjax>#"HClO"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "ClO"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Now, you know that when <mathjax>#1#</mathjax> <strong>mole</strong> of hypochlorous acid <strong>dissociates</strong>, it produces <mathjax>#1#</mathjax> <strong>mole</strong> of hypochlorite anions and <mathjax>#1#</mathjax> <strong>mole</strong> of hydronium cations. </p>
<p>This means that if the solution contains</p>
<blockquote>
<p><mathjax>#["H"_ 3"O"^(+)] = 10^(-"pH") quad "M"#</mathjax></p>
</blockquote>
<p>then it must also contain</p>
<blockquote>
<p><mathjax>#["ClO"^(-)] = 10^(-"pH") quad "M"#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#["HClO"] = (0.08 - 10^(-"pH")) quad "M"#</mathjax></p>
<blockquote>
<p>This is what you'd expect the <strong>equilibrium concentration</strong> of hypochlorous acid to be equal to if its dissociation produced <mathjax>#10^(-"pH")#</mathjax> <mathjax>#"M"#</mathjax> of hydronium cations. </p>
<p>In other words, in order for the dissociation of the acid to produce <mathjax>#["H"_3"O"^(+)] = 10^(-"pH")#</mathjax> <mathjax>#"M"#</mathjax>, the concentration of the acid must <strong>decrease</strong> by <mathjax>#["H"_3"O"^(+)] = 10^(-"pH")#</mathjax> <mathjax>#"M"#</mathjax>.</p>
</blockquote>
</blockquote>
<p>So now that you know the equilibrium concentrations for all three chemical species that are of interest here, you can plug them into the expression of the <strong>acid dissociation constant</strong>, <mathjax>#K_a#</mathjax>.</p>
<blockquote>
<p><mathjax>#K_a = (["ClO"^(-)] * ["H"_ 3"O"^(+)])/(["HClO"])#</mathjax></p>
</blockquote>
<p>You will end up with</p>
<blockquote>
<p><mathjax>#K_a = (10^(-"pH") * 10^(-"pH"))/(0.08 - 10^(-"pH"))#</mathjax></p>
</blockquote>
<p>which gets you</p>
<blockquote>
<p><mathjax>#K_a = (10^(-2.85))^2/(0.08 - 10^(-2.85)) = color(darkgreen)(ul(color(black)(2.5 * 10^(-5))))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that you have only one significant figure for the initial concentration of the acid. </p></div>
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</article> | The #"pH"# of a #"0.08 mol dm"^(-3)# solution of #"HClO"# is equal to #2.85#. Calculate the ionization constant of the acid? | null |
549 | ac887f6f-6ddd-11ea-9831-ccda262736ce | https://socratic.org/questions/what-is-the-molar-mass-of-lithium-oxide-li-2o | 29.88 g/mol | start physical_unit 8 8 molar_mass g/mol qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Molar mass [OF] Li2O [IN] g/mol"}] | [{"type":"physical unit","value":"29.88 g/mol"}] | [{"type":"chemical equation","value":"Li2O"}] | <h1 class="questionTitle" itemprop="name">What is the molar mass of lithium oxide, #Li_2O#? </h1> | null | 29.88 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#color(purple)"Let's use this equation to answer the question:"#</mathjax></p>
<p><strong>atomic weight of element <mathjax>#xx#</mathjax>number of atoms given by subscript <mathjax>#=#</mathjax> molar mass</strong></p>
<p>First, you want to have your handy dandy periodic table available so you can determine the atomic weight of Li and O:</p>
<p>Li has an atomic weight of 6.94 g/mol<br/>
O has an atomic wight of 16.00 g/mol</p>
<p>Based on the chemical formula, we have 2 atoms of Li, so the atomic weight of Li has to be multiplied by 2 to obtain 13.88 g/mol.</p>
<p>You have 1 atom of O, so the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> would just be 16.00 g/mol</p>
<p>Now you want to add the mass of each atom together to obtain the molar mass of the entire compound:</p>
<p><mathjax>#color(red) ("13.88 g/mol + 16.00 g/mol = 29.88 g/mol")#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#color(red)("29.88 g/(mol))#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#color(purple)"Let's use this equation to answer the question:"#</mathjax></p>
<p><strong>atomic weight of element <mathjax>#xx#</mathjax>number of atoms given by subscript <mathjax>#=#</mathjax> molar mass</strong></p>
<p>First, you want to have your handy dandy periodic table available so you can determine the atomic weight of Li and O:</p>
<p>Li has an atomic weight of 6.94 g/mol<br/>
O has an atomic wight of 16.00 g/mol</p>
<p>Based on the chemical formula, we have 2 atoms of Li, so the atomic weight of Li has to be multiplied by 2 to obtain 13.88 g/mol.</p>
<p>You have 1 atom of O, so the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> would just be 16.00 g/mol</p>
<p>Now you want to add the mass of each atom together to obtain the molar mass of the entire compound:</p>
<p><mathjax>#color(red) ("13.88 g/mol + 16.00 g/mol = 29.88 g/mol")#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the molar mass of lithium oxide, #Li_2O#? </h1>
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<div class="markdown"><p><mathjax>#color(red)("29.88 g/(mol))#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#color(purple)"Let's use this equation to answer the question:"#</mathjax></p>
<p><strong>atomic weight of element <mathjax>#xx#</mathjax>number of atoms given by subscript <mathjax>#=#</mathjax> molar mass</strong></p>
<p>First, you want to have your handy dandy periodic table available so you can determine the atomic weight of Li and O:</p>
<p>Li has an atomic weight of 6.94 g/mol<br/>
O has an atomic wight of 16.00 g/mol</p>
<p>Based on the chemical formula, we have 2 atoms of Li, so the atomic weight of Li has to be multiplied by 2 to obtain 13.88 g/mol.</p>
<p>You have 1 atom of O, so the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> would just be 16.00 g/mol</p>
<p>Now you want to add the mass of each atom together to obtain the molar mass of the entire compound:</p>
<p><mathjax>#color(red) ("13.88 g/mol + 16.00 g/mol = 29.88 g/mol")#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#Li_2O#</mathjax> is composed of Lithium and Oxygen atoms.</p>
<p>If you refer to your periodic table, the atomic weight is the molar mass of each element.</p>
<p>The molar mass of Lithium is 6.9 g/mol and Oxygen 16.0 g/mol.</p>
<p>Since there are 2 atoms of Lithium - if you look closely in <mathjax>#Li_2O#</mathjax>,<br/>
the molar mass of lithium <mathjax>#Li_2O#</mathjax> is 6.9 <mathjax>#xx#</mathjax> 2 = 13.8 g/mol.</p>
<p>Add the molar masses of Lithium and Oxygen together:<br/>
13.8 g/mol + 16.0 g/mol therefore the molar mass of Lithium oxide is 29.8 g/mol.</p></div>
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</article> | What is the molar mass of lithium oxide, #Li_2O#? | null |
550 | abb16692-6ddd-11ea-87c2-ccda262736ce | https://socratic.org/questions/how-many-moles-of-gas-are-contained-in-890-0-ml-at-21-0-c-and-750-0-mmhg-pressur | 0.04 moles | start physical_unit 4 4 mole mol qc_end physical_unit 4 4 8 9 volume qc_end physical_unit 4 4 11 12 temperature qc_end physical_unit 4 4 14 15 pressure qc_end end | [{"type":"physical unit","value":"Mole [OF] the gas [IN] moles"}] | [{"type":"physical unit","value":"0.04 moles"}] | [{"type":"physical unit","value":"Volume [OF] the gas [=] \\pu{890.0 mL}"},{"type":"physical unit","value":"Temperature [OF] the gas [=] \\pu{21.0 ℃}"},{"type":"physical unit","value":"Pressure [OF] the gas [=] \\pu{750.0 mmHg}"}] | <h1 class="questionTitle" itemprop="name">How many moles of gas are contained in 890.0 mL at 21.0 °C and 750.0 mmHg pressure? </h1> | null | 0.04 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We solve the Ideal Gas equation using <mathjax>#R=0.0821*L*atm*K^-1*mol^-1#</mathjax>...and we know that <mathjax>#760*mm*Hg-=1*atm#</mathjax>...</p>
<p><mathjax>#n=(PV)/(RT)=((750*mm*Hg)/(760*mm*Hg*atm^-1)xx0.8900*L)/(0.0821*(L*atm)/(K*mol)xx294.15*K)#</mathjax></p>
<p><mathjax>#-=0.0364*mol#</mathjax>.....</p></div>
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<div class="markdown"><p>There are <mathjax>#0.0364*mol#</mathjax>... with respect to the gas...</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We solve the Ideal Gas equation using <mathjax>#R=0.0821*L*atm*K^-1*mol^-1#</mathjax>...and we know that <mathjax>#760*mm*Hg-=1*atm#</mathjax>...</p>
<p><mathjax>#n=(PV)/(RT)=((750*mm*Hg)/(760*mm*Hg*atm^-1)xx0.8900*L)/(0.0821*(L*atm)/(K*mol)xx294.15*K)#</mathjax></p>
<p><mathjax>#-=0.0364*mol#</mathjax>.....</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of gas are contained in 890.0 mL at 21.0 °C and 750.0 mmHg pressure? </h1>
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<div class="markdown"><p>There are <mathjax>#0.0364*mol#</mathjax>... with respect to the gas...</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We solve the Ideal Gas equation using <mathjax>#R=0.0821*L*atm*K^-1*mol^-1#</mathjax>...and we know that <mathjax>#760*mm*Hg-=1*atm#</mathjax>...</p>
<p><mathjax>#n=(PV)/(RT)=((750*mm*Hg)/(760*mm*Hg*atm^-1)xx0.8900*L)/(0.0821*(L*atm)/(K*mol)xx294.15*K)#</mathjax></p>
<p><mathjax>#-=0.0364*mol#</mathjax>.....</p></div>
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</article> | How many moles of gas are contained in 890.0 mL at 21.0 °C and 750.0 mmHg pressure? | null |
551 | aae7195c-6ddd-11ea-b1fa-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-solution-with-oh-1-10-4-m | 10.00 | start physical_unit 6 6 ph none qc_end physical_unit 6 6 10 13 [oh-] qc_end end | [{"type":"physical unit","value":"pH [OF] the solution"}] | [{"type":"physical unit","value":"10.00"}] | [{"type":"physical unit","value":"[OH-] [OF] the solution [=] \\pu{1 × 10^(-4) M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a solution with #[OH^-] = 1*10^-4# #M#?</h1> | null | 10.00 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First solve for pOH using the equation pOH=-log[OH-] = 4<br/>
Then plug the pOH in the equation, pH + pOH =14 <br/>
then solve for pH. pH =14 - 4 = 10</p></div>
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<div class="markdown"><p><a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> =10</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First solve for pOH using the equation pOH=-log[OH-] = 4<br/>
Then plug the pOH in the equation, pH + pOH =14 <br/>
then solve for pH. pH =14 - 4 = 10</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a solution with #[OH^-] = 1*10^-4# #M#?</h1>
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<div class="markdown"><p><a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> =10</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>First solve for pOH using the equation pOH=-log[OH-] = 4<br/>
Then plug the pOH in the equation, pH + pOH =14 <br/>
then solve for pH. pH =14 - 4 = 10</p></div>
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</article> | What is the pH of a solution with #[OH^-] = 1*10^-4# #M#? | null |
552 | ab45536e-6ddd-11ea-8c08-ccda262736ce | https://socratic.org/questions/what-is-the-molecular-formula-if-the-empirical-formula-is-c-3h-7-and-the-experim | C6H14 | start chemical_formula qc_end physical_unit 10 11 20 21 molar_mass qc_end chemical_equation 13 13 qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] molecular"}] | [{"type":"chemical equation","value":"C6H14"}] | [{"type":"physical unit","value":"experimental molar mass [OF] the compound [=] \\pu{86 g/mole}"},{"type":"chemical equation","value":"C3H7"}] | <h1 class="questionTitle" itemprop="name">What is the molecular formula if the empirical formula is #C_3H_7#, and the experimental molar mass is 86 g/mole?</h1> | null | C6H14 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Brute force method:</p>
<ol>
<li>find molar mass of the empirical formula you are given</li>
<li>set it equal to 86g/mol</li>
<li>multiply it until you get 86g/mol</li>
</ol>
<p>in this case:<br/>
C=12.01<br/>
N=14<br/>
H=1.008</p>
<p>the empirical formula adds up to 43g/mol<br/>
multiply by two, we get 86g/mol</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#C_6H_14#</mathjax> </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Brute force method:</p>
<ol>
<li>find molar mass of the empirical formula you are given</li>
<li>set it equal to 86g/mol</li>
<li>multiply it until you get 86g/mol</li>
</ol>
<p>in this case:<br/>
C=12.01<br/>
N=14<br/>
H=1.008</p>
<p>the empirical formula adds up to 43g/mol<br/>
multiply by two, we get 86g/mol</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the molecular formula if the empirical formula is #C_3H_7#, and the experimental molar mass is 86 g/mole?</h1>
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Layth A.
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<span class="dateCreated" datetime="2016-05-31T04:22:15" itemprop="dateCreated">
May 31, 2016
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<div>
<div class="markdown"><p><mathjax>#C_6H_14#</mathjax> </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Brute force method:</p>
<ol>
<li>find molar mass of the empirical formula you are given</li>
<li>set it equal to 86g/mol</li>
<li>multiply it until you get 86g/mol</li>
</ol>
<p>in this case:<br/>
C=12.01<br/>
N=14<br/>
H=1.008</p>
<p>the empirical formula adds up to 43g/mol<br/>
multiply by two, we get 86g/mol</p></div>
</div>
</div>
</div>
</div>
</div>
<div>
<a href="https://socratic.org/answers/271701" itemprop="url">Answer link</a>
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</article> | What is the molecular formula if the empirical formula is #C_3H_7#, and the experimental molar mass is 86 g/mole? | null |
553 | aafff44e-6ddd-11ea-8978-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>
</div>
</div>
</div> | <article>
<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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Mar 18, 2017
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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 |
554 | ab690e32-6ddd-11ea-ac1b-ccda262736ce | https://socratic.org/questions/what-volume-in-ml-of-a-3-50m-lif-stock-solution-would-you-use-to-make-1-5l-of-0- | 214.3 mL | start physical_unit 8 10 volume ml qc_end physical_unit 8 10 6 7 molarity qc_end physical_unit 21 22 19 20 molarity qc_end physical_unit 21 22 16 17 volume qc_end end | [{"type":"physical unit","value":"Volume2 [OF] LiF stock solution [IN] mL"}] | [{"type":"physical unit","value":"214.3 mL"}] | [{"type":"physical unit","value":"Molarity2 [OF] LiF stock solution [=] \\pu{3.50 M}"},{"type":"physical unit","value":"Molarity1 [OF] LiF solution [=] \\pu{0.500 M}"},{"type":"physical unit","value":"Volume1 [OF] LiF solution [=] \\pu{1.5 L}"}] | <h1 class="questionTitle" itemprop="name">What volume (in mL) of a 3.50M #LiF# stock solution would you use to make 1.5L of 0.500M #LiF# solution? </h1> | null | 214.3 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of lithium fluoride"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.5*Lxx0.500*mol*L^-1=0.75*mol#</mathjax>.</p>
<p>And thus we need a volume of the stock solution that contains <mathjax>#0.75*mol#</mathjax> lithium fluoride. </p>
<p><mathjax>#(0.75*cancel(mol))/(3.50*cancel(mol)*cancel(L^-1))xx10^3*mL*cancel(L^-1)=214.3*mL#</mathjax></p>
<p>This solution would be slightly basic, i.e. <mathjax>#pH>7#</mathjax>. Why?</p></div>
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<div class="markdown"><p>A volume a bit over <mathjax>#214*mL#</mathjax> is required. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Moles of lithium fluoride"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.5*Lxx0.500*mol*L^-1=0.75*mol#</mathjax>.</p>
<p>And thus we need a volume of the stock solution that contains <mathjax>#0.75*mol#</mathjax> lithium fluoride. </p>
<p><mathjax>#(0.75*cancel(mol))/(3.50*cancel(mol)*cancel(L^-1))xx10^3*mL*cancel(L^-1)=214.3*mL#</mathjax></p>
<p>This solution would be slightly basic, i.e. <mathjax>#pH>7#</mathjax>. Why?</p></div>
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<h1 class="questionTitle" itemprop="name">What volume (in mL) of a 3.50M #LiF# stock solution would you use to make 1.5L of 0.500M #LiF# solution? </h1>
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<div class="markdown"><p>A volume a bit over <mathjax>#214*mL#</mathjax> is required. </p></div>
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<div class="markdown"><p><mathjax>#"Moles of lithium fluoride"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.5*Lxx0.500*mol*L^-1=0.75*mol#</mathjax>.</p>
<p>And thus we need a volume of the stock solution that contains <mathjax>#0.75*mol#</mathjax> lithium fluoride. </p>
<p><mathjax>#(0.75*cancel(mol))/(3.50*cancel(mol)*cancel(L^-1))xx10^3*mL*cancel(L^-1)=214.3*mL#</mathjax></p>
<p>This solution would be slightly basic, i.e. <mathjax>#pH>7#</mathjax>. Why?</p></div>
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</article> | What volume (in mL) of a 3.50M #LiF# stock solution would you use to make 1.5L of 0.500M #LiF# solution? | null |
555 | ac3c4076-6ddd-11ea-be84-ccda262736ce | https://socratic.org/questions/a-compound-with-an-empirical-formula-of-c-4h-4o-and-a-molar-mass-of-136-grams-pe | C8H8O2 | start chemical_formula qc_end c_other OTHER qc_end physical_unit 23 24 13 16 molar_mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] this compound [IN] molecular"}] | [{"type":"chemical equation","value":"C8H8O2"}] | [{"type":"other","value":"A compound with an empirical formula of C4H4O."},{"type":"physical unit","value":"Molar mass [OF] this compound [=] \\pu{136 grams per mole}"}] | <h1 class="questionTitle" itemprop="name">A compound with an empirical formula of #C_4H_4O# and a molar mass of 136 grams per mole. What is the molecular formula of this compound?</h1> | null | C8H8O2 | <div class="answerDescription">
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<div class="markdown"><p>Well, you have the reduced formula, and you have the molar mass. There's not much to do other than to define what the terms "empirical formula" and "molecular formula" mean. That would help make the connection.</p>
<p><strong>Molecular Formula</strong> - The combination of atomic symbols and subscripts in such a way that they <strong><em>exactly</em></strong> represent a molecule's proportions of each element that composes the molecule.</p>
<p><strong>Empirical Formula</strong> - The most reduced form of the molecular formula, i.e. <em>the smallest subscript is odd.</em></p>
<p>To determine what to do from here, just figure out what the <em>molar mass</em> of the <strong><em>given</em></strong> empirical formula is.</p>
<blockquote>
<p><mathjax>#M_"C" ~~ "12.011 g/mol"#</mathjax><br/>
<mathjax>#M_"H" ~~ "1.0079 g/mol"#</mathjax><br/>
<mathjax>#M_"O" ~~ "15.999 g/mol"#</mathjax></p>
<p><mathjax>#=> M_("C"_4"H"_4"O") ~~ "68.0746 g/mol"#</mathjax></p>
</blockquote>
<p>Now we should see that this is about <strong>half</strong> of the given "true" molar mass, so the molecular formula is:</p>
<blockquote>
<p><mathjax>#color(blue)("C"_8"H"_8"O"_2)#</mathjax></p>
</blockquote>
<p>Granted, this could be any number of <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> (like methyl benzoate, <em>o</em>-methylbenzoic acid, <em>p</em>-methylbenzoic acid, etc), but you aren't supposed to figure anything out beyond this.</p></div>
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<div class="markdown"><p>Well, you have the reduced formula, and you have the molar mass. There's not much to do other than to define what the terms "empirical formula" and "molecular formula" mean. That would help make the connection.</p>
<p><strong>Molecular Formula</strong> - The combination of atomic symbols and subscripts in such a way that they <strong><em>exactly</em></strong> represent a molecule's proportions of each element that composes the molecule.</p>
<p><strong>Empirical Formula</strong> - The most reduced form of the molecular formula, i.e. <em>the smallest subscript is odd.</em></p>
<p>To determine what to do from here, just figure out what the <em>molar mass</em> of the <strong><em>given</em></strong> empirical formula is.</p>
<blockquote>
<p><mathjax>#M_"C" ~~ "12.011 g/mol"#</mathjax><br/>
<mathjax>#M_"H" ~~ "1.0079 g/mol"#</mathjax><br/>
<mathjax>#M_"O" ~~ "15.999 g/mol"#</mathjax></p>
<p><mathjax>#=> M_("C"_4"H"_4"O") ~~ "68.0746 g/mol"#</mathjax></p>
</blockquote>
<p>Now we should see that this is about <strong>half</strong> of the given "true" molar mass, so the molecular formula is:</p>
<blockquote>
<p><mathjax>#color(blue)("C"_8"H"_8"O"_2)#</mathjax></p>
</blockquote>
<p>Granted, this could be any number of <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> (like methyl benzoate, <em>o</em>-methylbenzoic acid, <em>p</em>-methylbenzoic acid, etc), but you aren't supposed to figure anything out beyond this.</p></div>
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<h1 class="questionTitle" itemprop="name">A compound with an empirical formula of #C_4H_4O# and a molar mass of 136 grams per mole. What is the molecular formula of this compound?</h1>
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<div class="markdown"><p>Well, you have the reduced formula, and you have the molar mass. There's not much to do other than to define what the terms "empirical formula" and "molecular formula" mean. That would help make the connection.</p>
<p><strong>Molecular Formula</strong> - The combination of atomic symbols and subscripts in such a way that they <strong><em>exactly</em></strong> represent a molecule's proportions of each element that composes the molecule.</p>
<p><strong>Empirical Formula</strong> - The most reduced form of the molecular formula, i.e. <em>the smallest subscript is odd.</em></p>
<p>To determine what to do from here, just figure out what the <em>molar mass</em> of the <strong><em>given</em></strong> empirical formula is.</p>
<blockquote>
<p><mathjax>#M_"C" ~~ "12.011 g/mol"#</mathjax><br/>
<mathjax>#M_"H" ~~ "1.0079 g/mol"#</mathjax><br/>
<mathjax>#M_"O" ~~ "15.999 g/mol"#</mathjax></p>
<p><mathjax>#=> M_("C"_4"H"_4"O") ~~ "68.0746 g/mol"#</mathjax></p>
</blockquote>
<p>Now we should see that this is about <strong>half</strong> of the given "true" molar mass, so the molecular formula is:</p>
<blockquote>
<p><mathjax>#color(blue)("C"_8"H"_8"O"_2)#</mathjax></p>
</blockquote>
<p>Granted, this could be any number of <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> (like methyl benzoate, <em>o</em>-methylbenzoic acid, <em>p</em>-methylbenzoic acid, etc), but you aren't supposed to figure anything out beyond this.</p></div>
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<div class="markdown"><p>The molecular formula is <mathjax>#"C"_8"H"_8"O"_2"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The empirical formula mass of <mathjax>#"C"_4"H"_4"O"#</mathjax> is <mathjax>#(4xx12)+(4xx1)+(1xx16)="68 g/mol"#</mathjax>.</p>
<p>To determine the molecular formula, divide the molecular molar mass by the empirical formula mass.</p>
<p><mathjax>#136/68=2#</mathjax></p>
<p>Multiply the subscripts in the empirical formula by 2 to get the molecular formula.</p>
<p><mathjax>#"Molecular Formula"="C"_8"H"_8"O"_2"#</mathjax></p></div>
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</article> | A compound with an empirical formula of #C_4H_4O# and a molar mass of 136 grams per mole. What is the molecular formula of this compound? | null |
556 | aad22d2f-6ddd-11ea-a3a8-ccda262736ce | https://socratic.org/questions/what-mass-of-ca-oh-2-is-contained-in-1500-ml-of-0-0250m-ca-oh-2-solution | 2.78 g | start physical_unit 3 3 mass g qc_end physical_unit 12 13 7 8 volume qc_end physical_unit 12 13 10 11 molarity qc_end end | [{"type":"physical unit","value":"Mass [OF] Ca(OH)2 [IN] g"}] | [{"type":"physical unit","value":"2.78 g"}] | [{"type":"physical unit","value":"Volume [OF] Ca(OH)2 solution [=] \\pu{1500 mL}"},{"type":"physical unit","value":"Molarity [OF] Ca(OH)2 solution [=] \\pu{0.0250 M}"}] | <h1 class="questionTitle" itemprop="name">What mass of #Ca(OH)_2# is contained in 1500 mL of 0.0250M #Ca(OH)_2# solution?</h1> | null | 2.78 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#text(Molarity)= (text(Mass of solute)) / (text(Gram molecular mass of solute) × text(Volume of solution in litres))#</mathjax></p>
<p><mathjax>#text(Mass of)#</mathjax> <mathjax>#Ca(OH)_2= (0.0250M × 74.093g/(mol) × 1.5L)#</mathjax></p>
<p>Mass of <mathjax>#Ca(OH)_2#</mathjax> = <mathjax>#2.778g#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>2.778g</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#text(Molarity)= (text(Mass of solute)) / (text(Gram molecular mass of solute) × text(Volume of solution in litres))#</mathjax></p>
<p><mathjax>#text(Mass of)#</mathjax> <mathjax>#Ca(OH)_2= (0.0250M × 74.093g/(mol) × 1.5L)#</mathjax></p>
<p>Mass of <mathjax>#Ca(OH)_2#</mathjax> = <mathjax>#2.778g#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What mass of #Ca(OH)_2# is contained in 1500 mL of 0.0250M #Ca(OH)_2# solution?</h1>
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Junaid Mirza
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<div class="markdown"><p>2.778g</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#text(Molarity)= (text(Mass of solute)) / (text(Gram molecular mass of solute) × text(Volume of solution in litres))#</mathjax></p>
<p><mathjax>#text(Mass of)#</mathjax> <mathjax>#Ca(OH)_2= (0.0250M × 74.093g/(mol) × 1.5L)#</mathjax></p>
<p>Mass of <mathjax>#Ca(OH)_2#</mathjax> = <mathjax>#2.778g#</mathjax></p></div>
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</article> | What mass of #Ca(OH)_2# is contained in 1500 mL of 0.0250M #Ca(OH)_2# solution? | null |
557 | aa136792-6ddd-11ea-9245-ccda262736ce | https://socratic.org/questions/coal-gasification-is-a-process-that-converts-coal-into-methane-gas-if-this-react | 709.45 grams | start physical_unit 9 10 mass g qc_end physical_unit 12 13 19 19 percent_yield qc_end physical_unit 30 30 27 28 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] methane gas [IN] grams"}] | [{"type":"physical unit","value":"709.45 grams"}] | [{"type":"physical unit","value":"Percent yield [OF] this reaction [=] \\pu{85.0%}"},{"type":"physical unit","value":"Mass [OF] carbon [=] \\pu{1250 g}"}] | <h1 class="questionTitle" itemprop="name"> Coal gasification is a process that converts coal into methane gas. If this reaction has a percent yield of 85.0%, how much methane can be obtained from 1250 g of carbon?</h1> | null | 709.45 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing the balanced chemical equation that describes this reaction</p>
<blockquote>
<p><mathjax>#color(blue)(2)"C"_ ((s)) + 2"H"_ 2"O"_ ((g)) -> "CO"_ (2(g)) + "CH"_ (4(g))#</mathjax></p>
</blockquote>
<p>Notice that the reaction produces <mathjax>#1#</mathjax> <strong>mole</strong> of methane <strong>for every</strong> <mathjax>#color(blue)(2)#</mathjax> <strong>moles</strong> of carbon that take part in the reaction. </p>
<p>This represents the reaction's <strong>theoretical yield</strong>, i.e. what you get if the reaction has a <mathjax>#100%#</mathjax> <strong>yield</strong>. </p>
<p>You can convert this mole ratio to a <strong>gram ratio</strong> by using the <strong>molar masses</strong> of carbon and methane</p>
<blockquote>
<p><mathjax>#2 color(red)(cancel(color(black)("moles C"))) * "12.011 g"/(1color(red)(cancel(color(black)("mole C")))) = "24.022 g"#</mathjax></p>
<p><mathjax>#1 color(red)(cancel(color(black)("mole CH"_4))) * "16.04 g"/(1color(red)(cancel(color(black)("mole CH"_4)))) = "16.04 g"#</mathjax></p>
</blockquote>
<p>You can thus say that the <strong>theoretical yield</strong> of the reaction will have it produce <mathjax>#"16.04 g"#</mathjax> of methane <strong>for every</strong> <mathjax>#"24.022 g"#</mathjax> of carbon that react. </p>
<p>This means that <mathjax>#"1250 g"#</mathjax> of carbon will <em>theoretically</em> produce</p>
<blockquote>
<p><mathjax>#1250 color(red)(cancel(color(black)("g C"))) * "16.04 g CH"_4/(24.022color(red)(cancel(color(black)("g C")))) = "834.65 g CH"_4#</mathjax></p>
</blockquote>
<p>Now, the reaction is said to have an <mathjax>#85.0%#</mathjax> yield, which basically means that <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of methane that <em>could</em> be produced, you only get <mathjax>#"85.0 g"#</mathjax>. </p>
<p>As a result, the <strong>actual yield</strong> of the reaction will be</p>
<blockquote>
<p><mathjax>#834.65 color(red)(cancel(color(black)("g CH"_4))) * "85.0 g produced"/(100color(red)(cancel(color(black)("g CH"_4)))) = color(darkgreen)(ul(color(black)("709 g CH"_4)))#</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>#"709 g CH"_4#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing the balanced chemical equation that describes this reaction</p>
<blockquote>
<p><mathjax>#color(blue)(2)"C"_ ((s)) + 2"H"_ 2"O"_ ((g)) -> "CO"_ (2(g)) + "CH"_ (4(g))#</mathjax></p>
</blockquote>
<p>Notice that the reaction produces <mathjax>#1#</mathjax> <strong>mole</strong> of methane <strong>for every</strong> <mathjax>#color(blue)(2)#</mathjax> <strong>moles</strong> of carbon that take part in the reaction. </p>
<p>This represents the reaction's <strong>theoretical yield</strong>, i.e. what you get if the reaction has a <mathjax>#100%#</mathjax> <strong>yield</strong>. </p>
<p>You can convert this mole ratio to a <strong>gram ratio</strong> by using the <strong>molar masses</strong> of carbon and methane</p>
<blockquote>
<p><mathjax>#2 color(red)(cancel(color(black)("moles C"))) * "12.011 g"/(1color(red)(cancel(color(black)("mole C")))) = "24.022 g"#</mathjax></p>
<p><mathjax>#1 color(red)(cancel(color(black)("mole CH"_4))) * "16.04 g"/(1color(red)(cancel(color(black)("mole CH"_4)))) = "16.04 g"#</mathjax></p>
</blockquote>
<p>You can thus say that the <strong>theoretical yield</strong> of the reaction will have it produce <mathjax>#"16.04 g"#</mathjax> of methane <strong>for every</strong> <mathjax>#"24.022 g"#</mathjax> of carbon that react. </p>
<p>This means that <mathjax>#"1250 g"#</mathjax> of carbon will <em>theoretically</em> produce</p>
<blockquote>
<p><mathjax>#1250 color(red)(cancel(color(black)("g C"))) * "16.04 g CH"_4/(24.022color(red)(cancel(color(black)("g C")))) = "834.65 g CH"_4#</mathjax></p>
</blockquote>
<p>Now, the reaction is said to have an <mathjax>#85.0%#</mathjax> yield, which basically means that <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of methane that <em>could</em> be produced, you only get <mathjax>#"85.0 g"#</mathjax>. </p>
<p>As a result, the <strong>actual yield</strong> of the reaction will be</p>
<blockquote>
<p><mathjax>#834.65 color(red)(cancel(color(black)("g CH"_4))) * "85.0 g produced"/(100color(red)(cancel(color(black)("g CH"_4)))) = color(darkgreen)(ul(color(black)("709 g CH"_4)))#</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"> Coal gasification is a process that converts coal into methane gas. If this reaction has a percent yield of 85.0%, how much methane can be obtained from 1250 g of carbon?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"709 g CH"_4#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing the balanced chemical equation that describes this reaction</p>
<blockquote>
<p><mathjax>#color(blue)(2)"C"_ ((s)) + 2"H"_ 2"O"_ ((g)) -> "CO"_ (2(g)) + "CH"_ (4(g))#</mathjax></p>
</blockquote>
<p>Notice that the reaction produces <mathjax>#1#</mathjax> <strong>mole</strong> of methane <strong>for every</strong> <mathjax>#color(blue)(2)#</mathjax> <strong>moles</strong> of carbon that take part in the reaction. </p>
<p>This represents the reaction's <strong>theoretical yield</strong>, i.e. what you get if the reaction has a <mathjax>#100%#</mathjax> <strong>yield</strong>. </p>
<p>You can convert this mole ratio to a <strong>gram ratio</strong> by using the <strong>molar masses</strong> of carbon and methane</p>
<blockquote>
<p><mathjax>#2 color(red)(cancel(color(black)("moles C"))) * "12.011 g"/(1color(red)(cancel(color(black)("mole C")))) = "24.022 g"#</mathjax></p>
<p><mathjax>#1 color(red)(cancel(color(black)("mole CH"_4))) * "16.04 g"/(1color(red)(cancel(color(black)("mole CH"_4)))) = "16.04 g"#</mathjax></p>
</blockquote>
<p>You can thus say that the <strong>theoretical yield</strong> of the reaction will have it produce <mathjax>#"16.04 g"#</mathjax> of methane <strong>for every</strong> <mathjax>#"24.022 g"#</mathjax> of carbon that react. </p>
<p>This means that <mathjax>#"1250 g"#</mathjax> of carbon will <em>theoretically</em> produce</p>
<blockquote>
<p><mathjax>#1250 color(red)(cancel(color(black)("g C"))) * "16.04 g CH"_4/(24.022color(red)(cancel(color(black)("g C")))) = "834.65 g CH"_4#</mathjax></p>
</blockquote>
<p>Now, the reaction is said to have an <mathjax>#85.0%#</mathjax> yield, which basically means that <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of methane that <em>could</em> be produced, you only get <mathjax>#"85.0 g"#</mathjax>. </p>
<p>As a result, the <strong>actual yield</strong> of the reaction will be</p>
<blockquote>
<p><mathjax>#834.65 color(red)(cancel(color(black)("g CH"_4))) * "85.0 g produced"/(100color(red)(cancel(color(black)("g CH"_4)))) = color(darkgreen)(ul(color(black)("709 g CH"_4)))#</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> | Coal gasification is a process that converts coal into methane gas. If this reaction has a percent yield of 85.0%, how much methane can be obtained from 1250 g of carbon? | null |
558 | aad1b82f-6ddd-11ea-a885-ccda262736ce | https://socratic.org/questions/an-imaginary-ideal-gas-has-a-density-of-3-g-l-at-stp-what-is-the-molar-mas-of-th | 68.09 g/mol | start physical_unit 18 19 molar_mass g/mol qc_end physical_unit 1 3 8 9 density qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Molar mass [OF] this gas [IN] g/mol"}] | [{"type":"physical unit","value":"68.09 g/mol"}] | [{"type":"physical unit","value":"Density [OF] the imaginary ideal gas [=] \\pu{3 g/L}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">An imaginary ideal gas has a density of 3 g/L at STP. What is the molar mas of this gas?</h1> | null | 68.09 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that you're dealing with a sample of an unknown gas that has a <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of <mathjax>#"3 g/L"#</mathjax> at <strong>STP</strong>, or Standard Temperature and Pressure. </p>
<p>In order to find the gas' molar mass, you need to know two things</p>
<blockquote>
<ul>
<li><em>the mass of this sample of gas</em></li>
<li><em>the number of moles of gas present in this sample</em></li>
</ul>
</blockquote>
<p>Now, STP conditions imply a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>. </p>
<p>You will have to use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal as law</a></p>
<blockquote>
<p><mathjax>#color(blue)(PV = nRT)#</mathjax></p>
</blockquote>
<p>in order to try to find a relationship between the gas' <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> and its molar mass. </p>
<p>You know that molar mass is defined as <em>mass per mole</em>, so you can say that</p>
<blockquote>
<p><mathjax>#M_"m" = m/n implies n = m/M_"m"#</mathjax></p>
</blockquote>
<p>Replace the number of moles in the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation to get</p>
<blockquote>
<p><mathjax>#PV = m/M_"m" * RT#</mathjax></p>
</blockquote>
<p>You also know that density is defined as <em>mass per unit of volume</em></p>
<blockquote>
<p><mathjax>#rho = m/V#</mathjax></p>
</blockquote>
<p>Notice what happens if you divide both sides of the ideal gas law equation by <mathjax>#V#</mathjax></p>
<blockquote>
<p><mathjax>#(P * color(red)(cancel(color(black)(V))))/color(red)(cancel(color(black)(V))) = m/M_"m" * (RT)/V#</mathjax></p>
<p><mathjax>#P = underbrace(m/V)_(color(green)("density")) * (RT)/M_"m" = rho * (RT)/M_"m"#</mathjax></p>
</blockquote>
<p>Finally, isolate the molar mass on one side of the equation to get</p>
<blockquote>
<p><mathjax>#M_"m" = rho * (RT)/P#</mathjax></p>
</blockquote>
<p>Now all you have to do is plug in your values - remember that </p>
<blockquote>
<p><mathjax>#R = 0.082("atm" * "L")/("mol" * "K")#</mathjax></p>
</blockquote>
<p>and that the pressure must be expressed in <em>atm</em> and the temperature in <em>Kelvin</em>!</p>
<blockquote>
<p><mathjax>#M_"m" = 3"g"/color(red)(cancel(color(black)("L"))) * (0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 0)color(red)(cancel(color(black)("K"))))/(100/101.325color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#M_"m" = "68.085 g/mol"#</mathjax></p>
</blockquote>
<p>SInce you only gave one <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a> for the density of the gas, the answer will be </p>
<blockquote>
<p><mathjax>#M_"m" = color(green)("70 g/mol")#</mathjax></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/AHY8yvrkQxc?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>#"70 g/mol"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that you're dealing with a sample of an unknown gas that has a <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of <mathjax>#"3 g/L"#</mathjax> at <strong>STP</strong>, or Standard Temperature and Pressure. </p>
<p>In order to find the gas' molar mass, you need to know two things</p>
<blockquote>
<ul>
<li><em>the mass of this sample of gas</em></li>
<li><em>the number of moles of gas present in this sample</em></li>
</ul>
</blockquote>
<p>Now, STP conditions imply a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>. </p>
<p>You will have to use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal as law</a></p>
<blockquote>
<p><mathjax>#color(blue)(PV = nRT)#</mathjax></p>
</blockquote>
<p>in order to try to find a relationship between the gas' <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> and its molar mass. </p>
<p>You know that molar mass is defined as <em>mass per mole</em>, so you can say that</p>
<blockquote>
<p><mathjax>#M_"m" = m/n implies n = m/M_"m"#</mathjax></p>
</blockquote>
<p>Replace the number of moles in the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation to get</p>
<blockquote>
<p><mathjax>#PV = m/M_"m" * RT#</mathjax></p>
</blockquote>
<p>You also know that density is defined as <em>mass per unit of volume</em></p>
<blockquote>
<p><mathjax>#rho = m/V#</mathjax></p>
</blockquote>
<p>Notice what happens if you divide both sides of the ideal gas law equation by <mathjax>#V#</mathjax></p>
<blockquote>
<p><mathjax>#(P * color(red)(cancel(color(black)(V))))/color(red)(cancel(color(black)(V))) = m/M_"m" * (RT)/V#</mathjax></p>
<p><mathjax>#P = underbrace(m/V)_(color(green)("density")) * (RT)/M_"m" = rho * (RT)/M_"m"#</mathjax></p>
</blockquote>
<p>Finally, isolate the molar mass on one side of the equation to get</p>
<blockquote>
<p><mathjax>#M_"m" = rho * (RT)/P#</mathjax></p>
</blockquote>
<p>Now all you have to do is plug in your values - remember that </p>
<blockquote>
<p><mathjax>#R = 0.082("atm" * "L")/("mol" * "K")#</mathjax></p>
</blockquote>
<p>and that the pressure must be expressed in <em>atm</em> and the temperature in <em>Kelvin</em>!</p>
<blockquote>
<p><mathjax>#M_"m" = 3"g"/color(red)(cancel(color(black)("L"))) * (0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 0)color(red)(cancel(color(black)("K"))))/(100/101.325color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#M_"m" = "68.085 g/mol"#</mathjax></p>
</blockquote>
<p>SInce you only gave one <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a> for the density of the gas, the answer will be </p>
<blockquote>
<p><mathjax>#M_"m" = color(green)("70 g/mol")#</mathjax></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/AHY8yvrkQxc?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">An imaginary ideal gas has a density of 3 g/L at STP. What is the molar mas of this gas?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2015-10-25T02:32:47" itemprop="dateCreated">
Oct 25, 2015
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<div class="markdown"><p><mathjax>#"70 g/mol"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that you're dealing with a sample of an unknown gas that has a <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of <mathjax>#"3 g/L"#</mathjax> at <strong>STP</strong>, or Standard Temperature and Pressure. </p>
<p>In order to find the gas' molar mass, you need to know two things</p>
<blockquote>
<ul>
<li><em>the mass of this sample of gas</em></li>
<li><em>the number of moles of gas present in this sample</em></li>
</ul>
</blockquote>
<p>Now, STP conditions imply a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>. </p>
<p>You will have to use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal as law</a></p>
<blockquote>
<p><mathjax>#color(blue)(PV = nRT)#</mathjax></p>
</blockquote>
<p>in order to try to find a relationship between the gas' <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> and its molar mass. </p>
<p>You know that molar mass is defined as <em>mass per mole</em>, so you can say that</p>
<blockquote>
<p><mathjax>#M_"m" = m/n implies n = m/M_"m"#</mathjax></p>
</blockquote>
<p>Replace the number of moles in the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation to get</p>
<blockquote>
<p><mathjax>#PV = m/M_"m" * RT#</mathjax></p>
</blockquote>
<p>You also know that density is defined as <em>mass per unit of volume</em></p>
<blockquote>
<p><mathjax>#rho = m/V#</mathjax></p>
</blockquote>
<p>Notice what happens if you divide both sides of the ideal gas law equation by <mathjax>#V#</mathjax></p>
<blockquote>
<p><mathjax>#(P * color(red)(cancel(color(black)(V))))/color(red)(cancel(color(black)(V))) = m/M_"m" * (RT)/V#</mathjax></p>
<p><mathjax>#P = underbrace(m/V)_(color(green)("density")) * (RT)/M_"m" = rho * (RT)/M_"m"#</mathjax></p>
</blockquote>
<p>Finally, isolate the molar mass on one side of the equation to get</p>
<blockquote>
<p><mathjax>#M_"m" = rho * (RT)/P#</mathjax></p>
</blockquote>
<p>Now all you have to do is plug in your values - remember that </p>
<blockquote>
<p><mathjax>#R = 0.082("atm" * "L")/("mol" * "K")#</mathjax></p>
</blockquote>
<p>and that the pressure must be expressed in <em>atm</em> and the temperature in <em>Kelvin</em>!</p>
<blockquote>
<p><mathjax>#M_"m" = 3"g"/color(red)(cancel(color(black)("L"))) * (0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 0)color(red)(cancel(color(black)("K"))))/(100/101.325color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#M_"m" = "68.085 g/mol"#</mathjax></p>
</blockquote>
<p>SInce you only gave one <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a> for the density of the gas, the answer will be </p>
<blockquote>
<p><mathjax>#M_"m" = color(green)("70 g/mol")#</mathjax></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/AHY8yvrkQxc?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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</article> | An imaginary ideal gas has a density of 3 g/L at STP. What is the molar mas of this gas? | null |
559 | ac0c1776-6ddd-11ea-b07b-ccda262736ce | https://socratic.org/questions/what-is-the-total-number-of-joules-released-when-a-5-00-gram-sample-of-water-cha | 1700 joules | start physical_unit 12 14 heat_energy j qc_end physical_unit 12 14 10 11 mass qc_end physical_unit 12 14 22 23 temperature qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Released energy [OF] water sample [IN] joules"}] | [{"type":"physical unit","value":"1700 joules"}] | [{"type":"physical unit","value":"Mass [OF] water sample [=] \\pu{5.00 grams}"},{"type":"physical unit","value":"Temperature [OF] water sample [=] \\pu{0.0 ℃}"},{"type":"other","value":"Water changes from liquid to a solid."}] | <h1 class="questionTitle" itemprop="name">What is the total number of joules released when a 5.00-gram sample of water changes from liquid to a solid at 0.0°C? </h1> | null | 1700 joules | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>here you have a change of phase from liquid to solid where we can evaluate the heat released <mathjax>#Q#</mathjax> (in Joules) using:</p>
<p><mathjax>#Q=mL_s#</mathjax></p>
<p>where:<br/>
<mathjax>#m=#</mathjax> mass;<br/>
<mathjax>#L_s=#</mathjax> latent heat of solidification of water that from literature is: <mathjax>#3.5xx10^5J/(kg)#</mathjax></p>
<p>so for <mathjax>#5g=0.005kg#</mathjax> of water we get:</p>
<p><mathjax>#Q=0.005*3.4xx10^5=1700J#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>I found: <mathjax>#1700J#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>here you have a change of phase from liquid to solid where we can evaluate the heat released <mathjax>#Q#</mathjax> (in Joules) using:</p>
<p><mathjax>#Q=mL_s#</mathjax></p>
<p>where:<br/>
<mathjax>#m=#</mathjax> mass;<br/>
<mathjax>#L_s=#</mathjax> latent heat of solidification of water that from literature is: <mathjax>#3.5xx10^5J/(kg)#</mathjax></p>
<p>so for <mathjax>#5g=0.005kg#</mathjax> of water we get:</p>
<p><mathjax>#Q=0.005*3.4xx10^5=1700J#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the total number of joules released when a 5.00-gram sample of water changes from liquid to a solid at 0.0°C? </h1>
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Gió
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<span class="dateCreated" datetime="2016-12-09T17:34:31" itemprop="dateCreated">
Dec 9, 2016
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<div class="markdown"><p>I found: <mathjax>#1700J#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>here you have a change of phase from liquid to solid where we can evaluate the heat released <mathjax>#Q#</mathjax> (in Joules) using:</p>
<p><mathjax>#Q=mL_s#</mathjax></p>
<p>where:<br/>
<mathjax>#m=#</mathjax> mass;<br/>
<mathjax>#L_s=#</mathjax> latent heat of solidification of water that from literature is: <mathjax>#3.5xx10^5J/(kg)#</mathjax></p>
<p>so for <mathjax>#5g=0.005kg#</mathjax> of water we get:</p>
<p><mathjax>#Q=0.005*3.4xx10^5=1700J#</mathjax></p></div>
</div>
</div>
</div>
</div>
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</article> | What is the total number of joules released when a 5.00-gram sample of water changes from liquid to a solid at 0.0°C? | null |
560 | ab30664c-6ddd-11ea-b98b-ccda262736ce | https://socratic.org/questions/an-experiment-shows-that-a-248-ml-gas-sample-has-a-mass-of-0-433-g-at-a-pressure | 101.65 g/mol | start physical_unit 33 34 molar_mass g/mol qc_end physical_unit 7 8 5 6 volume qc_end physical_unit 7 8 13 14 mass qc_end physical_unit 7 8 19 20 pressure qc_end physical_unit 7 8 25 26 temperature qc_end end | [{"type":"physical unit","value":"Molar mass [OF] the gas [IN] g/mol"}] | [{"type":"physical unit","value":"101.65 g/mol"}] | [{"type":"physical unit","value":"Volume [OF] gas sample [=] \\pu{248 mL}"},{"type":"physical unit","value":"Mass [OF] gas sample [=] \\pu{0.433 g}"},{"type":"physical unit","value":"Pressure [OF] gas sample [=] \\pu{745 mmHg}"},{"type":"physical unit","value":"Temperature [OF] gas sample [=] \\pu{28 ℃}"}] | <h1 class="questionTitle" itemprop="name">An experiment shows that a 248-mL gas sample has a mass of 0.433 g at a pressure of 745 mm Hg and a temperature of 28°C. What is the molar mass of the gas?</h1> | null | 101.65 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#PV=nRT="mass"/"molar mass"RT#</mathjax></p>
<p>And thus <mathjax>#"molar mass"=(RT)/(PV)#</mathjax></p>
<p>And so here....<mathjax>#"molar mass"=(0.0821*(L*atm)/(K*mol)xx301.1*K)/((745*mm*Hg)/(760*mm*Hg*atm^-1)xx0.248*L)#</mathjax></p>
<p><mathjax>#=??*g*mol^-1#</mathjax></p>
<p>The key idea of this problem was the recognition that <mathjax>#760*mm*Hg-=1*atm#</mathjax>...and thus a unit of length was proportional to a unit of pressure....</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#"Molar mass"~=100*g*mol^-1#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#PV=nRT="mass"/"molar mass"RT#</mathjax></p>
<p>And thus <mathjax>#"molar mass"=(RT)/(PV)#</mathjax></p>
<p>And so here....<mathjax>#"molar mass"=(0.0821*(L*atm)/(K*mol)xx301.1*K)/((745*mm*Hg)/(760*mm*Hg*atm^-1)xx0.248*L)#</mathjax></p>
<p><mathjax>#=??*g*mol^-1#</mathjax></p>
<p>The key idea of this problem was the recognition that <mathjax>#760*mm*Hg-=1*atm#</mathjax>...and thus a unit of length was proportional to a unit of pressure....</p></div>
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<h1 class="questionTitle" itemprop="name">An experiment shows that a 248-mL gas sample has a mass of 0.433 g at a pressure of 745 mm Hg and a temperature of 28°C. What is the molar mass of the gas?</h1>
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anor277
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May 31, 2018
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<div class="markdown"><p><mathjax>#"Molar mass"~=100*g*mol^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#PV=nRT="mass"/"molar mass"RT#</mathjax></p>
<p>And thus <mathjax>#"molar mass"=(RT)/(PV)#</mathjax></p>
<p>And so here....<mathjax>#"molar mass"=(0.0821*(L*atm)/(K*mol)xx301.1*K)/((745*mm*Hg)/(760*mm*Hg*atm^-1)xx0.248*L)#</mathjax></p>
<p><mathjax>#=??*g*mol^-1#</mathjax></p>
<p>The key idea of this problem was the recognition that <mathjax>#760*mm*Hg-=1*atm#</mathjax>...and thus a unit of length was proportional to a unit of pressure....</p></div>
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</article> | An experiment shows that a 248-mL gas sample has a mass of 0.433 g at a pressure of 745 mm Hg and a temperature of 28°C. What is the molar mass of the gas? | null |
561 | ab5737db-6ddd-11ea-96e3-ccda262736ce | https://socratic.org/questions/if-49-5-moles-of-an-ideal-gas-occupies-37-5-liters-at-479-k-what-is-the-pressure | 51.9 atm | start physical_unit 18 19 pressure atm qc_end physical_unit 5 6 1 2 mole qc_end physical_unit 5 6 8 9 volume qc_end physical_unit 5 6 11 12 temperature qc_end end | [{"type":"physical unit","value":"Pressure [OF] the gas [IN] atm"}] | [{"type":"physical unit","value":"51.9 atm"}] | [{"type":"physical unit","value":"Mole [OF] the ideal gas [=] \\pu{49.5 moles}"},{"type":"physical unit","value":"Volume [OF] the ideal gas [=] \\pu{37.5 liters}"},{"type":"physical unit","value":"Temperature [OF] the ideal gas [=] \\pu{479 K}"}] | <h1 class="questionTitle" itemprop="name">If 49.5 moles of an ideal gas occupies 37.5 liters at 479 K, what is the pressure of the gas?</h1> | null | 51.9 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>This looks like a good time to apply the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a>:</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(PV = nRT)|)#</mathjax>,</p>
</blockquote>
<p>where </p>
<ul>
<li><mathjax>#P#</mathjax> is the pressure</li>
<li><mathjax>#V#</mathjax> is the volume</li>
<li><mathjax>#n#</mathjax> is the number of moles</li>
<li><mathjax>#R#</mathjax> is the gas constant</li>
<li><mathjax>#T#</mathjax> is the temperature</li>
</ul>
<p>We can rearrange the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to get</p>
<p><mathjax>#P = (nRT)/V#</mathjax></p>
<p><mathjax>#n = "49.5 mol"#</mathjax><br/>
<mathjax>#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "479 K"#</mathjax><br/>
<mathjax>#V = "37.5 L"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#P = (nRT)/V = (49.5 color(red)(cancel(color(black)("mol"))) × "0.082 06" color(red)(cancel(color(black)("L")))"·atm·"color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 479 color(red)(cancel(color(black)("K"))))/(37.5 color(red)(cancel(color(black)("L")))) = "51.9 atm"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The pressure of the gas is <strong>51.9 atm</strong>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>This looks like a good time to apply the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a>:</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(PV = nRT)|)#</mathjax>,</p>
</blockquote>
<p>where </p>
<ul>
<li><mathjax>#P#</mathjax> is the pressure</li>
<li><mathjax>#V#</mathjax> is the volume</li>
<li><mathjax>#n#</mathjax> is the number of moles</li>
<li><mathjax>#R#</mathjax> is the gas constant</li>
<li><mathjax>#T#</mathjax> is the temperature</li>
</ul>
<p>We can rearrange the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to get</p>
<p><mathjax>#P = (nRT)/V#</mathjax></p>
<p><mathjax>#n = "49.5 mol"#</mathjax><br/>
<mathjax>#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "479 K"#</mathjax><br/>
<mathjax>#V = "37.5 L"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#P = (nRT)/V = (49.5 color(red)(cancel(color(black)("mol"))) × "0.082 06" color(red)(cancel(color(black)("L")))"·atm·"color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 479 color(red)(cancel(color(black)("K"))))/(37.5 color(red)(cancel(color(black)("L")))) = "51.9 atm"#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">If 49.5 moles of an ideal gas occupies 37.5 liters at 479 K, what is the pressure of the gas?</h1>
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Ernest Z.
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<div class="markdown"><p>The pressure of the gas is <strong>51.9 atm</strong>.</p></div>
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</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>This looks like a good time to apply the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a>:</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(PV = nRT)|)#</mathjax>,</p>
</blockquote>
<p>where </p>
<ul>
<li><mathjax>#P#</mathjax> is the pressure</li>
<li><mathjax>#V#</mathjax> is the volume</li>
<li><mathjax>#n#</mathjax> is the number of moles</li>
<li><mathjax>#R#</mathjax> is the gas constant</li>
<li><mathjax>#T#</mathjax> is the temperature</li>
</ul>
<p>We can rearrange the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to get</p>
<p><mathjax>#P = (nRT)/V#</mathjax></p>
<p><mathjax>#n = "49.5 mol"#</mathjax><br/>
<mathjax>#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "479 K"#</mathjax><br/>
<mathjax>#V = "37.5 L"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#P = (nRT)/V = (49.5 color(red)(cancel(color(black)("mol"))) × "0.082 06" color(red)(cancel(color(black)("L")))"·atm·"color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 479 color(red)(cancel(color(black)("K"))))/(37.5 color(red)(cancel(color(black)("L")))) = "51.9 atm"#</mathjax></p></div>
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</article> | If 49.5 moles of an ideal gas occupies 37.5 liters at 479 K, what is the pressure of the gas? | null |
562 | a92cb1da-6ddd-11ea-81c3-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-equation-for-the-combustion-of-c-3h-6 | 2 C3H6 + 9 O2 -> 6 CO2 + 6 H2O | start chemical_equation qc_end chemical_equation 9 9 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the combustion"}] | [{"type":"chemical equation","value":"2 C3H6 + 9 O2 -> 6 CO2 + 6 H2O"}] | [{"type":"chemical equation","value":"C3H6"}] | <h1 class="questionTitle" itemprop="name">What is the chemical equation for the combustion of #C_3H_6#?</h1> | null | 2 C3H6 + 9 O2 -> 6 CO2 + 6 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Propene</em>, <mathjax>#"C"_3"H"_6#</mathjax>, is a <strong>hydrocarbon</strong>, which is the term used to designate a compound that consists of only carbon and hydrogen. </p>
<p>When a <em>hydrocarbon</em> undergoes <em>complete combustion</em>, only <strong>two products</strong> are formed </p>
<blockquote>
<ul>
<li><em>carbon dioxide</em>, <mathjax>#"CO"_2#</mathjax></li>
<li><em>water</em>, <mathjax>#"H"_2"O"#</mathjax></li>
</ul>
</blockquote>
<p>This tells you that <strong>all the carbon</strong> that was a part of the hydrocarbon will now be a part of the <em>carbon dioxide</em> and <strong>all the hydrogen</strong> that was a part of the hydrocarbon will now be a part of <em>water</em>.</p>
<p>So, you know that <strong>one mole</strong> of propene contains </p>
<blockquote>
<ul>
<li><em><strong>three</strong> moles of carbon</em></li>
<li><em><strong>six</strong> moles of hydrogen</em></li>
</ul>
</blockquote>
<p>This tells you that the carbon dioxide produced by the reaction <strong>must contain</strong> three moles of carbon and the water <strong>must contain</strong> six moles of hydrogen. </p>
<p>Since <strong>one mole</strong> of carbon dioxide contains <strong>one mole</strong> of carbon, you can conclude that you need to have <strong>three moles</strong> of <mathjax>#"CO"_2#</mathjax> in the products. </p>
<p>Likewise, since <strong>one mole</strong> of water contains <strong>two moles</strong> of hydrogen, you can say that you need to have <strong>three moles</strong> of water in the products. </p>
<p>Therefore, you can write</p>
<blockquote>
<p><mathjax>#"C"_3"H"_text(6(g]) + "O"_text(2(g]) -> 3"CO"_text(2(g]) + 3"H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>Now, notice that the oxygen atoms are not balanced. More specifically, you have <mathjax>#2#</mathjax> oxygen atoms on the reactants' side and a total of <mathjax>##9#</mathjax> oxygen atoms on the products' side. </p>
<p>Here is where a <em>fractional coefficient</em> comes in handy. Multiply the oxygen molecule by <mathjax>#9/2#</mathjax> to get</p>
<blockquote>
<p><mathjax>#"C"_3"H"_text(6(g]) + 9/2"O"_text(2(g]) -> 3"CO"_text(2(g]) + 3"H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>Finally, to get rid of this fractional coefficient, multiply <strong>all the compounds</strong> by <mathjax>#2#</mathjax> to get the balanced </p>
<p>chemical equation for the combustion of propene</p>
<blockquote>
<p><mathjax>#2"C"_3"H"_text(6(g]) + 9"O"_text(2(g]) -> 6"CO"_text(2(g]) + 6"H"_2"O"_text((l])#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2"C"_3"H"_text(6(g]) + 9"O"_text(2(g]) -> 6"CO"_text(2(g]) + 6"H"_2"O"_text((l])#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Propene</em>, <mathjax>#"C"_3"H"_6#</mathjax>, is a <strong>hydrocarbon</strong>, which is the term used to designate a compound that consists of only carbon and hydrogen. </p>
<p>When a <em>hydrocarbon</em> undergoes <em>complete combustion</em>, only <strong>two products</strong> are formed </p>
<blockquote>
<ul>
<li><em>carbon dioxide</em>, <mathjax>#"CO"_2#</mathjax></li>
<li><em>water</em>, <mathjax>#"H"_2"O"#</mathjax></li>
</ul>
</blockquote>
<p>This tells you that <strong>all the carbon</strong> that was a part of the hydrocarbon will now be a part of the <em>carbon dioxide</em> and <strong>all the hydrogen</strong> that was a part of the hydrocarbon will now be a part of <em>water</em>.</p>
<p>So, you know that <strong>one mole</strong> of propene contains </p>
<blockquote>
<ul>
<li><em><strong>three</strong> moles of carbon</em></li>
<li><em><strong>six</strong> moles of hydrogen</em></li>
</ul>
</blockquote>
<p>This tells you that the carbon dioxide produced by the reaction <strong>must contain</strong> three moles of carbon and the water <strong>must contain</strong> six moles of hydrogen. </p>
<p>Since <strong>one mole</strong> of carbon dioxide contains <strong>one mole</strong> of carbon, you can conclude that you need to have <strong>three moles</strong> of <mathjax>#"CO"_2#</mathjax> in the products. </p>
<p>Likewise, since <strong>one mole</strong> of water contains <strong>two moles</strong> of hydrogen, you can say that you need to have <strong>three moles</strong> of water in the products. </p>
<p>Therefore, you can write</p>
<blockquote>
<p><mathjax>#"C"_3"H"_text(6(g]) + "O"_text(2(g]) -> 3"CO"_text(2(g]) + 3"H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>Now, notice that the oxygen atoms are not balanced. More specifically, you have <mathjax>#2#</mathjax> oxygen atoms on the reactants' side and a total of <mathjax>##9#</mathjax> oxygen atoms on the products' side. </p>
<p>Here is where a <em>fractional coefficient</em> comes in handy. Multiply the oxygen molecule by <mathjax>#9/2#</mathjax> to get</p>
<blockquote>
<p><mathjax>#"C"_3"H"_text(6(g]) + 9/2"O"_text(2(g]) -> 3"CO"_text(2(g]) + 3"H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>Finally, to get rid of this fractional coefficient, multiply <strong>all the compounds</strong> by <mathjax>#2#</mathjax> to get the balanced </p>
<p>chemical equation for the combustion of propene</p>
<blockquote>
<p><mathjax>#2"C"_3"H"_text(6(g]) + 9"O"_text(2(g]) -> 6"CO"_text(2(g]) + 6"H"_2"O"_text((l])#</mathjax></p>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">What is the chemical equation for the combustion of #C_3H_6#?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2015-12-14T23:15:50" itemprop="dateCreated">
Dec 14, 2015
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<div class="markdown"><p><mathjax>#2"C"_3"H"_text(6(g]) + 9"O"_text(2(g]) -> 6"CO"_text(2(g]) + 6"H"_2"O"_text((l])#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Propene</em>, <mathjax>#"C"_3"H"_6#</mathjax>, is a <strong>hydrocarbon</strong>, which is the term used to designate a compound that consists of only carbon and hydrogen. </p>
<p>When a <em>hydrocarbon</em> undergoes <em>complete combustion</em>, only <strong>two products</strong> are formed </p>
<blockquote>
<ul>
<li><em>carbon dioxide</em>, <mathjax>#"CO"_2#</mathjax></li>
<li><em>water</em>, <mathjax>#"H"_2"O"#</mathjax></li>
</ul>
</blockquote>
<p>This tells you that <strong>all the carbon</strong> that was a part of the hydrocarbon will now be a part of the <em>carbon dioxide</em> and <strong>all the hydrogen</strong> that was a part of the hydrocarbon will now be a part of <em>water</em>.</p>
<p>So, you know that <strong>one mole</strong> of propene contains </p>
<blockquote>
<ul>
<li><em><strong>three</strong> moles of carbon</em></li>
<li><em><strong>six</strong> moles of hydrogen</em></li>
</ul>
</blockquote>
<p>This tells you that the carbon dioxide produced by the reaction <strong>must contain</strong> three moles of carbon and the water <strong>must contain</strong> six moles of hydrogen. </p>
<p>Since <strong>one mole</strong> of carbon dioxide contains <strong>one mole</strong> of carbon, you can conclude that you need to have <strong>three moles</strong> of <mathjax>#"CO"_2#</mathjax> in the products. </p>
<p>Likewise, since <strong>one mole</strong> of water contains <strong>two moles</strong> of hydrogen, you can say that you need to have <strong>three moles</strong> of water in the products. </p>
<p>Therefore, you can write</p>
<blockquote>
<p><mathjax>#"C"_3"H"_text(6(g]) + "O"_text(2(g]) -> 3"CO"_text(2(g]) + 3"H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>Now, notice that the oxygen atoms are not balanced. More specifically, you have <mathjax>#2#</mathjax> oxygen atoms on the reactants' side and a total of <mathjax>##9#</mathjax> oxygen atoms on the products' side. </p>
<p>Here is where a <em>fractional coefficient</em> comes in handy. Multiply the oxygen molecule by <mathjax>#9/2#</mathjax> to get</p>
<blockquote>
<p><mathjax>#"C"_3"H"_text(6(g]) + 9/2"O"_text(2(g]) -> 3"CO"_text(2(g]) + 3"H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>Finally, to get rid of this fractional coefficient, multiply <strong>all the compounds</strong> by <mathjax>#2#</mathjax> to get the balanced </p>
<p>chemical equation for the combustion of propene</p>
<blockquote>
<p><mathjax>#2"C"_3"H"_text(6(g]) + 9"O"_text(2(g]) -> 6"CO"_text(2(g]) + 6"H"_2"O"_text((l])#</mathjax></p>
</blockquote></div>
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</article> | What is the chemical equation for the combustion of #C_3H_6#? | null |
563 | a8c92a7f-6ddd-11ea-935f-ccda262736ce | https://socratic.org/questions/how-many-grams-of-formaldehyde-are-needed-to-produce-enough-heat-in-a-combustion | 16.67 grams | start physical_unit 4 4 mass g qc_end physical_unit 20 20 17 18 mass qc_end physical_unit 22 22 24 26 temperature qc_end end | [{"type":"physical unit","value":"Mass [OF] formaldehyde [IN] grams"}] | [{"type":"physical unit","value":"16.67 grams"}] | [{"type":"physical unit","value":"Mass [OF] ice [=] \\pu{10.00 g}"},{"type":"physical unit","value":"Temperature [OF] steam [=] \\pu{100 degrees C}"}] | <h1 class="questionTitle" itemprop="name">How many grams of formaldehyde are needed to produce enough heat in a combustion reaction to change 10.00g of ice to steam at 100 degrees C?</h1> | null | 16.67 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And you need to quote the following data:</p>
<p><mathjax>#i. #</mathjax> <mathjax>#DeltaH_"rxn"#</mathjax></p>
<p><mathjax>#ii.#</mathjax> <mathjax>#"Specific heat of ice"#</mathjax></p>
<p><mathjax>#iii.#</mathjax> <mathjax>#"Latent heat of fusion of water"#</mathjax></p>
<p><mathjax>#iv.#</mathjax> <mathjax>#"Specific heat of water"#</mathjax></p>
<p><mathjax>#v.#</mathjax> <mathjax>#"Latent heat of vaporization of water"#</mathjax></p>
<p>So there are 5 pieces of data that you have not supplied. In their absence we cannot help you. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#H_2C=O(g) + O_2(g) rarr CO_2(g) + H_2O(l)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And you need to quote the following data:</p>
<p><mathjax>#i. #</mathjax> <mathjax>#DeltaH_"rxn"#</mathjax></p>
<p><mathjax>#ii.#</mathjax> <mathjax>#"Specific heat of ice"#</mathjax></p>
<p><mathjax>#iii.#</mathjax> <mathjax>#"Latent heat of fusion of water"#</mathjax></p>
<p><mathjax>#iv.#</mathjax> <mathjax>#"Specific heat of water"#</mathjax></p>
<p><mathjax>#v.#</mathjax> <mathjax>#"Latent heat of vaporization of water"#</mathjax></p>
<p>So there are 5 pieces of data that you have not supplied. In their absence we cannot help you. </p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">How many grams of formaldehyde are needed to produce enough heat in a combustion reaction to change 10.00g of ice to steam at 100 degrees C?</h1>
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<div class="markdown"><p><mathjax>#H_2C=O(g) + O_2(g) rarr CO_2(g) + H_2O(l)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And you need to quote the following data:</p>
<p><mathjax>#i. #</mathjax> <mathjax>#DeltaH_"rxn"#</mathjax></p>
<p><mathjax>#ii.#</mathjax> <mathjax>#"Specific heat of ice"#</mathjax></p>
<p><mathjax>#iii.#</mathjax> <mathjax>#"Latent heat of fusion of water"#</mathjax></p>
<p><mathjax>#iv.#</mathjax> <mathjax>#"Specific heat of water"#</mathjax></p>
<p><mathjax>#v.#</mathjax> <mathjax>#"Latent heat of vaporization of water"#</mathjax></p>
<p>So there are 5 pieces of data that you have not supplied. In their absence we cannot help you. </p></div>
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</article> | How many grams of formaldehyde are needed to produce enough heat in a combustion reaction to change 10.00g of ice to steam at 100 degrees C? | null |
564 | ac453a68-6ddd-11ea-8436-ccda262736ce | https://socratic.org/questions/what-would-be-the-formula-between-aluminum-al-3-and-sulfate-so-4-2 | Al2(SO4)3 | start chemical_formula qc_end chemical_equation 7 7 qc_end chemical_equation 10 10 qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] default"}] | [{"type":"chemical equation","value":"Al2(SO4)3"}] | [{"type":"chemical equation","value":"Al^3+"},{"type":"chemical equation","value":"SO4^2-"}] | <h1 class="questionTitle" itemprop="name">What would be the formula between Aluminum #Al^(+3)# and Sulfate #SO_4 ^(-2)#?</h1> | null | Al2(SO4)3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Al^(+3)+SO_4^-2#</mathjax> → <mathjax>#Al_2(SO_4)_3#</mathjax></p>
<p>To get this you would have to balance the charges between the anion and metal cation. </p>
<p>Do this by finding the least common multiple between the parts. In this case it turns out to be 6. Therefore we need 2 moles of <mathjax>#Al^(+3)#</mathjax> and 3 moles of <mathjax>#SO_4^-2#</mathjax> to balance the charges. </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#Al_2(SO_4)_3#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Al^(+3)+SO_4^-2#</mathjax> → <mathjax>#Al_2(SO_4)_3#</mathjax></p>
<p>To get this you would have to balance the charges between the anion and metal cation. </p>
<p>Do this by finding the least common multiple between the parts. In this case it turns out to be 6. Therefore we need 2 moles of <mathjax>#Al^(+3)#</mathjax> and 3 moles of <mathjax>#SO_4^-2#</mathjax> to balance the charges. </p></div>
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<h1 class="questionTitle" itemprop="name">What would be the formula between Aluminum #Al^(+3)# and Sulfate #SO_4 ^(-2)#?</h1>
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Zach
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Dec 24, 2015
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<div class="markdown"><p><mathjax>#Al_2(SO_4)_3#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Al^(+3)+SO_4^-2#</mathjax> → <mathjax>#Al_2(SO_4)_3#</mathjax></p>
<p>To get this you would have to balance the charges between the anion and metal cation. </p>
<p>Do this by finding the least common multiple between the parts. In this case it turns out to be 6. Therefore we need 2 moles of <mathjax>#Al^(+3)#</mathjax> and 3 moles of <mathjax>#SO_4^-2#</mathjax> to balance the charges. </p></div>
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</article> | What would be the formula between Aluminum #Al^(+3)# and Sulfate #SO_4 ^(-2)#? | null |
565 | ac73c180-6ddd-11ea-906b-ccda262736ce | https://socratic.org/questions/if-the-heat-of-combustion-for-a-specific-compound-is-1160-0-kj-mol-and-its-molar-1 | 31.10 grams | start physical_unit 23 24 mass g qc_end physical_unit 7 8 10 11 enthalpy_of_combustion qc_end physical_unit 7 8 17 18 molar_mass qc_end physical_unit 7 8 30 31 heat_energy qc_end end | [{"type":"physical unit","value":"Mass [OF] this compound [IN] grams"}] | [{"type":"physical unit","value":"31.10 grams"}] | [{"type":"physical unit","value":"Heat of combustion [OF] the specific compound [=] \\pu{-1160.0 kJ/mol}"},{"type":"physical unit","value":"Molar mass [OF] the specific compound [=] \\pu{84.45 g/mol}"},{"type":"physical unit","value":"Released heat [OF] the specific compound [=] \\pu{427.20 kJ}"}] | <h1 class="questionTitle" itemprop="name">If the heat of combustion for a specific compound is -1160.0 kJ/mol and its molar mass is 84.45 g/mol, how many grams of this compound must you burn to release 427.20 kJ of heat?</h1> | null | 31.10 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here is to use the compound's <em>molar heat of combustion</em> and its molar mass to determine the <a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of combustion <strong>per gram</strong>. </p>
<p>So, you know that burning <strong>one mole</strong> of this compound will <strong>releases</strong> <mathjax>#"1160.0 kJ"#</mathjax> of heat. This will be equivalent to </p>
<blockquote>
<p><mathjax>#1160.0"kJ"/color(red)(cancel(color(black)("mol"))) * (1color(red)(cancel(color(black)("mole"))))/"84.45 g" = "13.736 kJ/g"#</mathjax></p>
</blockquote>
<p>Now, if your compound gives of <mathjax>#"13.736 kJ"#</mathjax> <strong>per gram</strong> upon combustion, it follows that <mathjax>#"427.20 kJ"#</mathjax> of heat will be given off by the combustion of </p>
<blockquote>
<p><mathjax>#427.20color(red)(cancel(color(black)("kJ"))) * "1 gram"/(13.736color(red)(cancel(color(black)("kJ")))) = color(green)("31.10 g")#</mathjax></p>
</blockquote>
<p>Therefore, burning <mathjax>#"31.10 g"#</mathjax> of this compound will result in <mathjax>#"427.20 kJ"#</mathjax> of heat being <strong>released</strong>. This is of course equivalent to saying that the <em>enthalpy change of combustion</em> when <mathjax>#"31.10 g"#</mathjax> of the compound undergo combustion is </p>
<blockquote>
<p><mathjax>#DeltaH_"comb" = -"427.20 kJ"#</mathjax></p>
</blockquote>
<p>Here the minus sign is used to designate <strong>heat given off</strong>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"31.10 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here is to use the compound's <em>molar heat of combustion</em> and its molar mass to determine the <a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of combustion <strong>per gram</strong>. </p>
<p>So, you know that burning <strong>one mole</strong> of this compound will <strong>releases</strong> <mathjax>#"1160.0 kJ"#</mathjax> of heat. This will be equivalent to </p>
<blockquote>
<p><mathjax>#1160.0"kJ"/color(red)(cancel(color(black)("mol"))) * (1color(red)(cancel(color(black)("mole"))))/"84.45 g" = "13.736 kJ/g"#</mathjax></p>
</blockquote>
<p>Now, if your compound gives of <mathjax>#"13.736 kJ"#</mathjax> <strong>per gram</strong> upon combustion, it follows that <mathjax>#"427.20 kJ"#</mathjax> of heat will be given off by the combustion of </p>
<blockquote>
<p><mathjax>#427.20color(red)(cancel(color(black)("kJ"))) * "1 gram"/(13.736color(red)(cancel(color(black)("kJ")))) = color(green)("31.10 g")#</mathjax></p>
</blockquote>
<p>Therefore, burning <mathjax>#"31.10 g"#</mathjax> of this compound will result in <mathjax>#"427.20 kJ"#</mathjax> of heat being <strong>released</strong>. This is of course equivalent to saying that the <em>enthalpy change of combustion</em> when <mathjax>#"31.10 g"#</mathjax> of the compound undergo combustion is </p>
<blockquote>
<p><mathjax>#DeltaH_"comb" = -"427.20 kJ"#</mathjax></p>
</blockquote>
<p>Here the minus sign is used to designate <strong>heat given off</strong>. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If the heat of combustion for a specific compound is -1160.0 kJ/mol and its molar mass is 84.45 g/mol, how many grams of this compound must you burn to release 427.20 kJ of heat?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"31.10 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here is to use the compound's <em>molar heat of combustion</em> and its molar mass to determine the <a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of combustion <strong>per gram</strong>. </p>
<p>So, you know that burning <strong>one mole</strong> of this compound will <strong>releases</strong> <mathjax>#"1160.0 kJ"#</mathjax> of heat. This will be equivalent to </p>
<blockquote>
<p><mathjax>#1160.0"kJ"/color(red)(cancel(color(black)("mol"))) * (1color(red)(cancel(color(black)("mole"))))/"84.45 g" = "13.736 kJ/g"#</mathjax></p>
</blockquote>
<p>Now, if your compound gives of <mathjax>#"13.736 kJ"#</mathjax> <strong>per gram</strong> upon combustion, it follows that <mathjax>#"427.20 kJ"#</mathjax> of heat will be given off by the combustion of </p>
<blockquote>
<p><mathjax>#427.20color(red)(cancel(color(black)("kJ"))) * "1 gram"/(13.736color(red)(cancel(color(black)("kJ")))) = color(green)("31.10 g")#</mathjax></p>
</blockquote>
<p>Therefore, burning <mathjax>#"31.10 g"#</mathjax> of this compound will result in <mathjax>#"427.20 kJ"#</mathjax> of heat being <strong>released</strong>. This is of course equivalent to saying that the <em>enthalpy change of combustion</em> when <mathjax>#"31.10 g"#</mathjax> of the compound undergo combustion is </p>
<blockquote>
<p><mathjax>#DeltaH_"comb" = -"427.20 kJ"#</mathjax></p>
</blockquote>
<p>Here the minus sign is used to designate <strong>heat given off</strong>. </p></div>
</div>
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</article> | If the heat of combustion for a specific compound is -1160.0 kJ/mol and its molar mass is 84.45 g/mol, how many grams of this compound must you burn to release 427.20 kJ of heat? | null |
566 | a85cb0c3-6ddd-11ea-b5a6-ccda262736ce | https://socratic.org/questions/58fa439511ef6b2bf898ce55 | MgO(s) + H2(g) ->[\Delta] Mg(s) + H2O(g) | start chemical_equation qc_end substance 7 8 qc_end substance 10 11 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"MgO(s) + H2(g) ->[\\Delta] Mg(s) + H2O(g)"}] | [{"type":"substance name","value":"Magnesium oxide"},{"type":"substance name","value":"Dihydrogen gas"}] | <h1 class="questionTitle" itemprop="name">How would we represent the reaction of magnesium oxide with dihydrogen gas?</h1> | null | MgO(s) + H2(g) ->[\Delta] Mg(s) + H2O(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I don't think they use this process to refine magensium ore. You would need FIERCE heat and high dihydrogen pressures, but the reaction is potentially doable.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#MgO(s) + H_2(g) + Delta rarr Mg(s) + H_2O(g)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I don't think they use this process to refine magensium ore. You would need FIERCE heat and high dihydrogen pressures, but the reaction is potentially doable.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How would we represent the reaction of magnesium oxide with dihydrogen gas?</h1>
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<div class="markdown"><p><mathjax>#MgO(s) + H_2(g) + Delta rarr Mg(s) + H_2O(g)#</mathjax></p></div>
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<div class="markdown"><p>I don't think they use this process to refine magensium ore. You would need FIERCE heat and high dihydrogen pressures, but the reaction is potentially doable.</p></div>
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</article> | How would we represent the reaction of magnesium oxide with dihydrogen gas? | null |
567 | abf3afd4-6ddd-11ea-b140-ccda262736ce | https://socratic.org/questions/the-value-of-ksp-for-ni-oh-2-aq-is-6-0x10-16-what-is-the-molar-concentration-of- | 6.0 × 10^(-2) M | start physical_unit 16 17 molarity mol/l qc_end c_other OTHER qc_end physical_unit 5 5 7 9 equilibrium_constant_k qc_end physical_unit 25 25 23 23 ph qc_end end | [{"type":"physical unit","value":"Molar concentration [OF] nickel ions [IN] M"}] | [{"type":"physical unit","value":"6.0 × 10^(-2) M"}] | [{"type":"other","value":"Assume the buffer is sufficiently strong that the pH does not change during equilibration."},{"type":"physical unit","value":"Ksp [OF] Ni(OH)2(aq) [=] \\pu{6.0 × 10^(-16)}"},{"type":"physical unit","value":"pH [OF] saturated buffer solution [=] \\pu{7.0}"}] | <h1 class="questionTitle" itemprop="name">The value of Ksp for Ni(OH)2(aq) is 6.0x10^-16. What is the molar concentration of nickel ions in a saturated pH=7.0 buffer solution? Assume the buffer is sufficiently strong that the pH does not change during equilibration.</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The value of Ksp for Ni(OH)2(aq) is 6.0x10^-16. What is the molar concentration of nickel ions in a saturated pH=7.0 buffer solution? Assume the buffer is sufficiently strong that the pH does not change during equilibration.</p></div>
</h2>
</div>
</div> | 6.0 × 10^(-2) M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There is a big hint in the name: <a href="https://socratic.org/chemistry/chemical-equilibrium/ksp">Ksp</a> is the equilibrium <strong>solubility product</strong> constant. In other words, you are going to take the solubility product, that is, you are going to multiply two solubilities, to get the equilibrium equation! </p>
<p>The equilibrium equation is: </p>
<blockquote>
<p><mathjax>#Ni(OH)_2 rightleftharpoons Ni^(2+)+ 2OH^-#</mathjax></p>
</blockquote>
<p>In this case we are going to multiply the solubility of the <mathjax>#Ni^(2+)#</mathjax> ion by the concentration of the hydroxide ion:</p>
<blockquote>
<p><mathjax>#K_(sp) = [Ni^(2+)]*[OH^-]^2#</mathjax> </p>
</blockquote>
<p>Remember that, as with any equilibrium equation, we always raise the concentration of the chemical to its coefficient in the balanced chemical equation. </p>
<p>So that means that </p>
<blockquote>
<p><mathjax>#6.0 xx 10^(-16) = [Ni^(2+)]*[OH^-]^2#</mathjax></p>
</blockquote>
<p>Now at <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> 7.0 the </p>
<blockquote>
<p><mathjax>#[OH^-] = 1 xx 10^(-7)#</mathjax></p>
</blockquote>
<p>So... </p>
<blockquote>
<p><mathjax>#6.0 xx 10^(-16) = [Ni^(2+)]*[1 xx 10^(-7)]^2#</mathjax></p>
</blockquote>
<p>Dividing both sides by <mathjax>#[1xx10^(-7)]^2#</mathjax></p>
<p>We get </p>
<blockquote>
<p><mathjax>#[Ni^(2+)] = (6.0 xx 10^(-16)) / (1 xx 10^(-7))^2 = 6.0 xx10^(-2)#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>It is <mathjax>#6.0 xx 10^-2#</mathjax>. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There is a big hint in the name: <a href="https://socratic.org/chemistry/chemical-equilibrium/ksp">Ksp</a> is the equilibrium <strong>solubility product</strong> constant. In other words, you are going to take the solubility product, that is, you are going to multiply two solubilities, to get the equilibrium equation! </p>
<p>The equilibrium equation is: </p>
<blockquote>
<p><mathjax>#Ni(OH)_2 rightleftharpoons Ni^(2+)+ 2OH^-#</mathjax></p>
</blockquote>
<p>In this case we are going to multiply the solubility of the <mathjax>#Ni^(2+)#</mathjax> ion by the concentration of the hydroxide ion:</p>
<blockquote>
<p><mathjax>#K_(sp) = [Ni^(2+)]*[OH^-]^2#</mathjax> </p>
</blockquote>
<p>Remember that, as with any equilibrium equation, we always raise the concentration of the chemical to its coefficient in the balanced chemical equation. </p>
<p>So that means that </p>
<blockquote>
<p><mathjax>#6.0 xx 10^(-16) = [Ni^(2+)]*[OH^-]^2#</mathjax></p>
</blockquote>
<p>Now at <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> 7.0 the </p>
<blockquote>
<p><mathjax>#[OH^-] = 1 xx 10^(-7)#</mathjax></p>
</blockquote>
<p>So... </p>
<blockquote>
<p><mathjax>#6.0 xx 10^(-16) = [Ni^(2+)]*[1 xx 10^(-7)]^2#</mathjax></p>
</blockquote>
<p>Dividing both sides by <mathjax>#[1xx10^(-7)]^2#</mathjax></p>
<p>We get </p>
<blockquote>
<p><mathjax>#[Ni^(2+)] = (6.0 xx 10^(-16)) / (1 xx 10^(-7))^2 = 6.0 xx10^(-2)#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">The value of Ksp for Ni(OH)2(aq) is 6.0x10^-16. What is the molar concentration of nickel ions in a saturated pH=7.0 buffer solution? Assume the buffer is sufficiently strong that the pH does not change during equilibration.</h1>
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<div class="markdown"><p>The value of Ksp for Ni(OH)2(aq) is 6.0x10^-16. What is the molar concentration of nickel ions in a saturated pH=7.0 buffer solution? Assume the buffer is sufficiently strong that the pH does not change during equilibration.</p></div>
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<div class="markdown"><p>It is <mathjax>#6.0 xx 10^-2#</mathjax>. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There is a big hint in the name: <a href="https://socratic.org/chemistry/chemical-equilibrium/ksp">Ksp</a> is the equilibrium <strong>solubility product</strong> constant. In other words, you are going to take the solubility product, that is, you are going to multiply two solubilities, to get the equilibrium equation! </p>
<p>The equilibrium equation is: </p>
<blockquote>
<p><mathjax>#Ni(OH)_2 rightleftharpoons Ni^(2+)+ 2OH^-#</mathjax></p>
</blockquote>
<p>In this case we are going to multiply the solubility of the <mathjax>#Ni^(2+)#</mathjax> ion by the concentration of the hydroxide ion:</p>
<blockquote>
<p><mathjax>#K_(sp) = [Ni^(2+)]*[OH^-]^2#</mathjax> </p>
</blockquote>
<p>Remember that, as with any equilibrium equation, we always raise the concentration of the chemical to its coefficient in the balanced chemical equation. </p>
<p>So that means that </p>
<blockquote>
<p><mathjax>#6.0 xx 10^(-16) = [Ni^(2+)]*[OH^-]^2#</mathjax></p>
</blockquote>
<p>Now at <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> 7.0 the </p>
<blockquote>
<p><mathjax>#[OH^-] = 1 xx 10^(-7)#</mathjax></p>
</blockquote>
<p>So... </p>
<blockquote>
<p><mathjax>#6.0 xx 10^(-16) = [Ni^(2+)]*[1 xx 10^(-7)]^2#</mathjax></p>
</blockquote>
<p>Dividing both sides by <mathjax>#[1xx10^(-7)]^2#</mathjax></p>
<p>We get </p>
<blockquote>
<p><mathjax>#[Ni^(2+)] = (6.0 xx 10^(-16)) / (1 xx 10^(-7))^2 = 6.0 xx10^(-2)#</mathjax></p>
</blockquote></div>
</div>
</div>
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</article> | The value of Ksp for Ni(OH)2(aq) is 6.0x10^-16. What is the molar concentration of nickel ions in a saturated pH=7.0 buffer solution? Assume the buffer is sufficiently strong that the pH does not change during equilibration. |
The value of Ksp for Ni(OH)2(aq) is 6.0x10^-16. What is the molar concentration of nickel ions in a saturated pH=7.0 buffer solution? Assume the buffer is sufficiently strong that the pH does not change during equilibration.
|
568 | a96ce676-6ddd-11ea-aa16-ccda262736ce | https://socratic.org/questions/how-many-grams-of-copper-ll-carbonate-are-present-in-4-3-10-24-moles | 882 grams | start physical_unit 4 6 mass g qc_end end | [{"type":"physical unit","value":"mass [OF] copper (ll) carbonate [IN] grams"}] | [{"type":"physical unit","value":"882 grams"}] | [{"type":"physical unit","value":"number [OF] copper (ll) carbonate formula units [=] \\pu{4.3 × 10^24}"}] | <h1 class="questionTitle" itemprop="name">How many grams of copper (ll) carbonate are present in #4.3*10^24# formula units?</h1> | null | 882 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by dividing the number of formula units by Avogadro's number:</p>
<p>#4.3 xx 10^(24)/6.02xx10^(23) = 7.143 moles</p>
<p>Next, since it is <mathjax>#CuCO_3#</mathjax> we are considering, multiply this number of moles by the molar mass of <mathjax>#CuCO_3#</mathjax>, which is <mathjax>#63.5 + 12.0 + 48.0 = 123.5 g/(mol)#</mathjax>:</p>
<p>We get</p>
<p><mathjax>#7.143 mol xx 123.5 g/(mol) = 882g#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>882 g</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by dividing the number of formula units by Avogadro's number:</p>
<p>#4.3 xx 10^(24)/6.02xx10^(23) = 7.143 moles</p>
<p>Next, since it is <mathjax>#CuCO_3#</mathjax> we are considering, multiply this number of moles by the molar mass of <mathjax>#CuCO_3#</mathjax>, which is <mathjax>#63.5 + 12.0 + 48.0 = 123.5 g/(mol)#</mathjax>:</p>
<p>We get</p>
<p><mathjax>#7.143 mol xx 123.5 g/(mol) = 882g#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many grams of copper (ll) carbonate are present in #4.3*10^24# formula units?</h1>
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Dec 22, 2016
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<div class="markdown"><p>882 g</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by dividing the number of formula units by Avogadro's number:</p>
<p>#4.3 xx 10^(24)/6.02xx10^(23) = 7.143 moles</p>
<p>Next, since it is <mathjax>#CuCO_3#</mathjax> we are considering, multiply this number of moles by the molar mass of <mathjax>#CuCO_3#</mathjax>, which is <mathjax>#63.5 + 12.0 + 48.0 = 123.5 g/(mol)#</mathjax>:</p>
<p>We get</p>
<p><mathjax>#7.143 mol xx 123.5 g/(mol) = 882g#</mathjax></p></div>
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</article> | How many grams of copper (ll) carbonate are present in #4.3*10^24# formula units? | null |
569 | aa88f094-6ddd-11ea-a527-ccda262736ce | https://socratic.org/questions/mom-made-a-5-bleach-solution-with-50-ml-of-bleach-and-950-ml-water-how-much-blea | 117.65 mL | start physical_unit 4 4 volume ml qc_end physical_unit 4 4 3 3 percent qc_end physical_unit 4 4 7 8 volume qc_end physical_unit 14 14 12 13 volume qc_end physical_unit 4 4 28 28 percent qc_end end | [{"type":"physical unit","value":"Added volume [OF] bleach [IN] mL"}] | [{"type":"physical unit","value":"117.65 mL"}] | [{"type":"physical unit","value":"Percentage1 [OF] bleach in solution [=] \\pu{5%}"},{"type":"physical unit","value":"Volume [OF] bleach [=] \\pu{50 mL}"},{"type":"physical unit","value":"Volume [OF] water [=] \\pu{950 mL}"},{"type":"physical unit","value":"Percentage2 [OF] bleach in solution [=] \\pu{15%}"}] | <h1 class="questionTitle" itemprop="name">Mom made a 5% bleach solution with 50 ml of bleach and 950 ml water. How much bleach should be added to this in order to obtain a 15% bleach solution?</h1> | null | 117.65 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The current solution has <mathjax>#50#</mathjax> ml of bleach and a total volume of <mathjax>#50+950 = 1000#</mathjax> ml. If we add a certain quantity <mathjax>#b#</mathjax> of bleach, we will have more bleach, but also more solution in general: the ratio bleach-to-total volume will change as follows:</p>
<p><mathjax>#\frac{50+b}{1000+b}#</mathjax></p>
<p>we want this ratio to equal <mathjax>#15% = 0.15#</mathjax>, so we must ask</p>
<p><mathjax>#\frac{50+b}{1000+b} = 0.15#</mathjax></p>
<p>Solve for <mathjax>#b#</mathjax>: if we multiply both sides by <mathjax>#1000+b#</mathjax> we have</p>
<p><mathjax>#50+b = 0.15(1000+b) = 150+0.15b#</mathjax></p>
<p>Subtract <mathjax>#0.15b#</mathjax> from both sides:</p>
<p><mathjax>#50+b-0.15b = 150#</mathjax></p>
<p>Subtract <mathjax>#50#</mathjax> from both sides:</p>
<p><mathjax>#b-0.15b = 150 - 50#</mathjax></p>
<p>Simplify both sides:</p>
<p><mathjax>#0.85b = 100#</mathjax></p>
<p>Divide both sides by <mathjax>#0.85#</mathjax></p>
<p><mathjax>#b = \frac{100}{0.85} \approx 117.6#</mathjax></p>
<p>Let's check our answer: the new solution will have <mathjax>#50+117.6 = 167.6#</mathjax> ml of bleach, for a total volume of <mathjax>#1000+117.6 = 1117.6#</mathjax> ml. The ratio is</p>
<p><mathjax>#\frac{167.6}{1117.6} \approx 0.15#</mathjax></p>
<p>so we're good!</p></div>
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<div class="markdown"><p><mathjax>#117.6#</mathjax> ml</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The current solution has <mathjax>#50#</mathjax> ml of bleach and a total volume of <mathjax>#50+950 = 1000#</mathjax> ml. If we add a certain quantity <mathjax>#b#</mathjax> of bleach, we will have more bleach, but also more solution in general: the ratio bleach-to-total volume will change as follows:</p>
<p><mathjax>#\frac{50+b}{1000+b}#</mathjax></p>
<p>we want this ratio to equal <mathjax>#15% = 0.15#</mathjax>, so we must ask</p>
<p><mathjax>#\frac{50+b}{1000+b} = 0.15#</mathjax></p>
<p>Solve for <mathjax>#b#</mathjax>: if we multiply both sides by <mathjax>#1000+b#</mathjax> we have</p>
<p><mathjax>#50+b = 0.15(1000+b) = 150+0.15b#</mathjax></p>
<p>Subtract <mathjax>#0.15b#</mathjax> from both sides:</p>
<p><mathjax>#50+b-0.15b = 150#</mathjax></p>
<p>Subtract <mathjax>#50#</mathjax> from both sides:</p>
<p><mathjax>#b-0.15b = 150 - 50#</mathjax></p>
<p>Simplify both sides:</p>
<p><mathjax>#0.85b = 100#</mathjax></p>
<p>Divide both sides by <mathjax>#0.85#</mathjax></p>
<p><mathjax>#b = \frac{100}{0.85} \approx 117.6#</mathjax></p>
<p>Let's check our answer: the new solution will have <mathjax>#50+117.6 = 167.6#</mathjax> ml of bleach, for a total volume of <mathjax>#1000+117.6 = 1117.6#</mathjax> ml. The ratio is</p>
<p><mathjax>#\frac{167.6}{1117.6} \approx 0.15#</mathjax></p>
<p>so we're good!</p></div>
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<h1 class="questionTitle" itemprop="name">Mom made a 5% bleach solution with 50 ml of bleach and 950 ml water. How much bleach should be added to this in order to obtain a 15% bleach solution?</h1>
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<div class="markdown"><p><mathjax>#117.6#</mathjax> ml</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The current solution has <mathjax>#50#</mathjax> ml of bleach and a total volume of <mathjax>#50+950 = 1000#</mathjax> ml. If we add a certain quantity <mathjax>#b#</mathjax> of bleach, we will have more bleach, but also more solution in general: the ratio bleach-to-total volume will change as follows:</p>
<p><mathjax>#\frac{50+b}{1000+b}#</mathjax></p>
<p>we want this ratio to equal <mathjax>#15% = 0.15#</mathjax>, so we must ask</p>
<p><mathjax>#\frac{50+b}{1000+b} = 0.15#</mathjax></p>
<p>Solve for <mathjax>#b#</mathjax>: if we multiply both sides by <mathjax>#1000+b#</mathjax> we have</p>
<p><mathjax>#50+b = 0.15(1000+b) = 150+0.15b#</mathjax></p>
<p>Subtract <mathjax>#0.15b#</mathjax> from both sides:</p>
<p><mathjax>#50+b-0.15b = 150#</mathjax></p>
<p>Subtract <mathjax>#50#</mathjax> from both sides:</p>
<p><mathjax>#b-0.15b = 150 - 50#</mathjax></p>
<p>Simplify both sides:</p>
<p><mathjax>#0.85b = 100#</mathjax></p>
<p>Divide both sides by <mathjax>#0.85#</mathjax></p>
<p><mathjax>#b = \frac{100}{0.85} \approx 117.6#</mathjax></p>
<p>Let's check our answer: the new solution will have <mathjax>#50+117.6 = 167.6#</mathjax> ml of bleach, for a total volume of <mathjax>#1000+117.6 = 1117.6#</mathjax> ml. The ratio is</p>
<p><mathjax>#\frac{167.6}{1117.6} \approx 0.15#</mathjax></p>
<p>so we're good!</p></div>
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</article> | Mom made a 5% bleach solution with 50 ml of bleach and 950 ml water. How much bleach should be added to this in order to obtain a 15% bleach solution? | null |
570 | ab12a663-6ddd-11ea-9a36-ccda262736ce | https://socratic.org/questions/59593d6cb72cff186905e875 | 9.45 M | start physical_unit 5 6 concentration mol/l qc_end substance 5 6 qc_end end | [{"type":"physical unit","value":"Concentration [OF] pure toluene [IN] M"}] | [{"type":"physical unit","value":"9.45 M"}] | [{"type":"substance name","value":"Pure toluene"}] | <h1 class="questionTitle" itemprop="name">What is the concentration of pure toluene?</h1> | null | 9.45 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Now <mathjax>#rho_"toluene"=0.871*g*mL^-1#</mathjax>...and thus...</p>
<p><mathjax>#"Concentration"="Moles of toluene"/"Volume of toluene"#</mathjax></p>
<p>And we work from a <mathjax>#1*mL#</mathjax> volume of toluene...</p>
<p><mathjax>#((0.871*g)/(92.14*g*mol^-1))/(1*mLxx10^-3*L*mL^-1)=9.45*mol*L^-1#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Molarity"="Moles of solute"/"Volume of solution"=9.45*mol*L^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Now <mathjax>#rho_"toluene"=0.871*g*mL^-1#</mathjax>...and thus...</p>
<p><mathjax>#"Concentration"="Moles of toluene"/"Volume of toluene"#</mathjax></p>
<p>And we work from a <mathjax>#1*mL#</mathjax> volume of toluene...</p>
<p><mathjax>#((0.871*g)/(92.14*g*mol^-1))/(1*mLxx10^-3*L*mL^-1)=9.45*mol*L^-1#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the concentration of pure toluene?</h1>
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<div class="markdown"><p><mathjax>#"Molarity"="Moles of solute"/"Volume of solution"=9.45*mol*L^-1#</mathjax></p></div>
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<div class="markdown"><p>Now <mathjax>#rho_"toluene"=0.871*g*mL^-1#</mathjax>...and thus...</p>
<p><mathjax>#"Concentration"="Moles of toluene"/"Volume of toluene"#</mathjax></p>
<p>And we work from a <mathjax>#1*mL#</mathjax> volume of toluene...</p>
<p><mathjax>#((0.871*g)/(92.14*g*mol^-1))/(1*mLxx10^-3*L*mL^-1)=9.45*mol*L^-1#</mathjax></p></div>
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</article> | What is the concentration of pure toluene? | null |
571 | ac24d994-6ddd-11ea-b1fc-ccda262736ce | https://socratic.org/questions/how-many-moles-of-nickel-are-in-5-12-10-25-atoms-of-nickel | 85.02 moles | start physical_unit 4 4 mole mol qc_end end | [{"type":"physical unit","value":"Mole [OF] nickel [IN] moles"}] | [{"type":"physical unit","value":"85.02 moles"}] | [{"type":"physical unit","value":"Number [OF] nickel atoms [=] \\pu{5.12 × 10^25}"}] | <h1 class="questionTitle" itemprop="name">How many moles of nickel are in #5.12 * 10^25# atoms of nickel?</h1> | null | 85.02 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key here is <strong>Avogadro's constant</strong>, which tells you the number of representative particles present in <mathjax>#1#</mathjax> <strong>mole</strong> of a substance. </p>
<p>By definition, and this is what Avogadro's constant tells you, <mathjax>#1#</mathjax> <strong>mole</strong> of nickel must contain <mathjax>#6.022 * 10^(23)#</mathjax> <strong>atoms</strong> of nickel. In other words, if you don't have <mathjax>#6.022 * 10^(23)#</mathjax> <strong>atoms</strong> of nickel, you don't have <mathjax>#1#</mathjax> <strong>mole</strong> of nickel. </p>
<p>Now, your sample contains <mathjax>#5.12 * 10^(25)#</mathjax> <strong>atoms</strong> of nickel, so you know for a fact that the sample contains <em>more</em> than <mathjax>#1#</mathjax> mole of nickel.</p>
<p>To find the exact number of moles, set up a conversion factor using Avogadro's constant.</p>
<blockquote>
<p><mathjax>#5.12 * 10^(25) color(red)(cancel(color(black)("atoms Ni"))) * "1 mole Ni"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms Ni")))) = color(darkgreen)(ul(color(black)("85.0 moles Ni")))#</mathjax></p>
</blockquote>
<p>The answer must be 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 the number of atoms present in the sample. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#85.0#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key here is <strong>Avogadro's constant</strong>, which tells you the number of representative particles present in <mathjax>#1#</mathjax> <strong>mole</strong> of a substance. </p>
<p>By definition, and this is what Avogadro's constant tells you, <mathjax>#1#</mathjax> <strong>mole</strong> of nickel must contain <mathjax>#6.022 * 10^(23)#</mathjax> <strong>atoms</strong> of nickel. In other words, if you don't have <mathjax>#6.022 * 10^(23)#</mathjax> <strong>atoms</strong> of nickel, you don't have <mathjax>#1#</mathjax> <strong>mole</strong> of nickel. </p>
<p>Now, your sample contains <mathjax>#5.12 * 10^(25)#</mathjax> <strong>atoms</strong> of nickel, so you know for a fact that the sample contains <em>more</em> than <mathjax>#1#</mathjax> mole of nickel.</p>
<p>To find the exact number of moles, set up a conversion factor using Avogadro's constant.</p>
<blockquote>
<p><mathjax>#5.12 * 10^(25) color(red)(cancel(color(black)("atoms Ni"))) * "1 mole Ni"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms Ni")))) = color(darkgreen)(ul(color(black)("85.0 moles Ni")))#</mathjax></p>
</blockquote>
<p>The answer must be 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 the number of atoms present in the sample. </p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of nickel are in #5.12 * 10^25# atoms of nickel?</h1>
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Feb 9, 2018
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<div class="markdown"><p><mathjax>#85.0#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key here is <strong>Avogadro's constant</strong>, which tells you the number of representative particles present in <mathjax>#1#</mathjax> <strong>mole</strong> of a substance. </p>
<p>By definition, and this is what Avogadro's constant tells you, <mathjax>#1#</mathjax> <strong>mole</strong> of nickel must contain <mathjax>#6.022 * 10^(23)#</mathjax> <strong>atoms</strong> of nickel. In other words, if you don't have <mathjax>#6.022 * 10^(23)#</mathjax> <strong>atoms</strong> of nickel, you don't have <mathjax>#1#</mathjax> <strong>mole</strong> of nickel. </p>
<p>Now, your sample contains <mathjax>#5.12 * 10^(25)#</mathjax> <strong>atoms</strong> of nickel, so you know for a fact that the sample contains <em>more</em> than <mathjax>#1#</mathjax> mole of nickel.</p>
<p>To find the exact number of moles, set up a conversion factor using Avogadro's constant.</p>
<blockquote>
<p><mathjax>#5.12 * 10^(25) color(red)(cancel(color(black)("atoms Ni"))) * "1 mole Ni"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms Ni")))) = color(darkgreen)(ul(color(black)("85.0 moles Ni")))#</mathjax></p>
</blockquote>
<p>The answer must be 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 the number of atoms present in the sample. </p></div>
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</article> | How many moles of nickel are in #5.12 * 10^25# atoms of nickel? | null |
572 | a9194e75-6ddd-11ea-8bc7-ccda262736ce | https://socratic.org/questions/a-solution-contains-180-g-of-glucose-and-162-g-of-water-what-is-the-mole-fractio | 0.10 | start physical_unit 6 6 mole_fraction none qc_end physical_unit 6 6 3 4 mass qc_end physical_unit 11 11 8 9 mass qc_end end | [{"type":"physical unit","value":"Mole fraction [OF] glucose"}] | [{"type":"physical unit","value":"0.10"}] | [{"type":"physical unit","value":"Mass [OF] glucose [=] \\pu{180 g}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{162 g}"}] | <h1 class="questionTitle" itemprop="name">A solution contains 180 g of glucose and 162 g of water. What is the mole fraction of glucose? </h1> | null | 0.10 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find the <strong>mole fraction</strong> of glucose in this solution, you need to know </p>
<blockquote>
<ul>
<li><em>the <strong>number of moles</strong> of glucose</em></li>
<li><em>the <strong>total number of moles</strong> present in the solution</em></li>
</ul>
</blockquote>
<p>To find the number of moles of glucose present in <mathjax>#"180 g"#</mathjax>, use the compound's <strong>molar mass</strong>, which is equal to <mathjax>#"180.16 g mol"^(-1)#</mathjax>.</p>
<p>Your sample of glucose will thus contain </p>
<blockquote>
<p><mathjax>#180 color(red)(cancel(color(black)("g"))) * overbrace("1 mole glucose"/(180.16color(red)(cancel(color(black)("g")))))^(color(blue)("molar mass of glucose")) = "0.9991 moles glucose"#</mathjax></p>
</blockquote>
<p>Now do the same for water, using the fact that its <strong>molar mass</strong> is equal to <mathjax>#"18.015 g mol"^(-1)#</mathjax></p>
<blockquote>
<p><mathjax>#162 color(red)(cancel(color(black)("g"))) * overbrace("1 mole water"/(18.015color(red)(cancel(color(black)("g")))))^(color(purple)("molar mass of water")) = "8.9925 moles water"#</mathjax></p>
</blockquote>
<p>The <strong>total number of moles</strong> present in this solution will be</p>
<blockquote>
<p><mathjax>#n_"total" = "0.9991 moles" + "8.9925 moles" = "9.9916 moles"#</mathjax></p>
</blockquote>
<p>Now, <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> fraction of glucose is equal to the number of moles of glucose divided by the <em>total number of moles</em> present in solution. </p>
<p>In your case, this is equal to </p>
<blockquote>
<p><mathjax>#chi_"glucose" = (0.9991 color(red)(cancel(color(black)("moles"))))/(9.9916color(red)(cancel(color(black)("moles")))) = 0.10#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the mass of glucose. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#chi_"glucose" = 0.10#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find the <strong>mole fraction</strong> of glucose in this solution, you need to know </p>
<blockquote>
<ul>
<li><em>the <strong>number of moles</strong> of glucose</em></li>
<li><em>the <strong>total number of moles</strong> present in the solution</em></li>
</ul>
</blockquote>
<p>To find the number of moles of glucose present in <mathjax>#"180 g"#</mathjax>, use the compound's <strong>molar mass</strong>, which is equal to <mathjax>#"180.16 g mol"^(-1)#</mathjax>.</p>
<p>Your sample of glucose will thus contain </p>
<blockquote>
<p><mathjax>#180 color(red)(cancel(color(black)("g"))) * overbrace("1 mole glucose"/(180.16color(red)(cancel(color(black)("g")))))^(color(blue)("molar mass of glucose")) = "0.9991 moles glucose"#</mathjax></p>
</blockquote>
<p>Now do the same for water, using the fact that its <strong>molar mass</strong> is equal to <mathjax>#"18.015 g mol"^(-1)#</mathjax></p>
<blockquote>
<p><mathjax>#162 color(red)(cancel(color(black)("g"))) * overbrace("1 mole water"/(18.015color(red)(cancel(color(black)("g")))))^(color(purple)("molar mass of water")) = "8.9925 moles water"#</mathjax></p>
</blockquote>
<p>The <strong>total number of moles</strong> present in this solution will be</p>
<blockquote>
<p><mathjax>#n_"total" = "0.9991 moles" + "8.9925 moles" = "9.9916 moles"#</mathjax></p>
</blockquote>
<p>Now, <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> fraction of glucose is equal to the number of moles of glucose divided by the <em>total number of moles</em> present in solution. </p>
<p>In your case, this is equal to </p>
<blockquote>
<p><mathjax>#chi_"glucose" = (0.9991 color(red)(cancel(color(black)("moles"))))/(9.9916color(red)(cancel(color(black)("moles")))) = 0.10#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the mass of glucose. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A solution contains 180 g of glucose and 162 g of water. What is the mole fraction of glucose? </h1>
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<div class="markdown"><p><mathjax>#chi_"glucose" = 0.10#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find the <strong>mole fraction</strong> of glucose in this solution, you need to know </p>
<blockquote>
<ul>
<li><em>the <strong>number of moles</strong> of glucose</em></li>
<li><em>the <strong>total number of moles</strong> present in the solution</em></li>
</ul>
</blockquote>
<p>To find the number of moles of glucose present in <mathjax>#"180 g"#</mathjax>, use the compound's <strong>molar mass</strong>, which is equal to <mathjax>#"180.16 g mol"^(-1)#</mathjax>.</p>
<p>Your sample of glucose will thus contain </p>
<blockquote>
<p><mathjax>#180 color(red)(cancel(color(black)("g"))) * overbrace("1 mole glucose"/(180.16color(red)(cancel(color(black)("g")))))^(color(blue)("molar mass of glucose")) = "0.9991 moles glucose"#</mathjax></p>
</blockquote>
<p>Now do the same for water, using the fact that its <strong>molar mass</strong> is equal to <mathjax>#"18.015 g mol"^(-1)#</mathjax></p>
<blockquote>
<p><mathjax>#162 color(red)(cancel(color(black)("g"))) * overbrace("1 mole water"/(18.015color(red)(cancel(color(black)("g")))))^(color(purple)("molar mass of water")) = "8.9925 moles water"#</mathjax></p>
</blockquote>
<p>The <strong>total number of moles</strong> present in this solution will be</p>
<blockquote>
<p><mathjax>#n_"total" = "0.9991 moles" + "8.9925 moles" = "9.9916 moles"#</mathjax></p>
</blockquote>
<p>Now, <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> fraction of glucose is equal to the number of moles of glucose divided by the <em>total number of moles</em> present in solution. </p>
<p>In your case, this is equal to </p>
<blockquote>
<p><mathjax>#chi_"glucose" = (0.9991 color(red)(cancel(color(black)("moles"))))/(9.9916color(red)(cancel(color(black)("moles")))) = 0.10#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the mass of glucose. </p></div>
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</article> | A solution contains 180 g of glucose and 162 g of water. What is the mole fraction of glucose? | null |
573 | ab1f6575-6ddd-11ea-82dd-ccda262736ce | https://socratic.org/questions/how-many-grams-of-potassium-chloride-are-required-to-many-prepare-500-0-ml-of-a- | 3.73 grams | start physical_unit 4 5 mass g qc_end physical_unit 4 5 15 16 molarity qc_end physical_unit 17 17 11 12 volume qc_end end | [{"type":"physical unit","value":"Mass [OF] potassium chloride [IN] grams"}] | [{"type":"physical unit","value":"3.73 grams"}] | [{"type":"physical unit","value":"Molarity [OF] potassium chloride solution [=] \\pu{0.100 M}"},{"type":"physical unit","value":"Volume [OF] potassium chloride solution [=] \\pu{500.0 mL}"}] | <h1 class="questionTitle" itemprop="name">How many grams of potassium chloride are required to many prepare 500.0 mL of a 0.100 M solution?</h1> | null | 3.73 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In <mathjax>#0.5000*L#</mathjax> of a <mathjax>#0.100*mol*L^-1#</mathjax> solution there are <mathjax>#0.5000*Lxx0.100*mol*L^-1=0.0500*mol#</mathjax></p>
<p>Thus we require <mathjax>#0.0500*molxx74.55*g·mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??g" potassium chloride"#</mathjax></p></div>
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<div class="markdown"><p>A bit under <mathjax>#4#</mathjax> <mathjax>#g#</mathjax> potassium chloride salt.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>In <mathjax>#0.5000*L#</mathjax> of a <mathjax>#0.100*mol*L^-1#</mathjax> solution there are <mathjax>#0.5000*Lxx0.100*mol*L^-1=0.0500*mol#</mathjax></p>
<p>Thus we require <mathjax>#0.0500*molxx74.55*g·mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??g" potassium chloride"#</mathjax></p></div>
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<div class="markdown"><p>A bit under <mathjax>#4#</mathjax> <mathjax>#g#</mathjax> potassium chloride salt.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>In <mathjax>#0.5000*L#</mathjax> of a <mathjax>#0.100*mol*L^-1#</mathjax> solution there are <mathjax>#0.5000*Lxx0.100*mol*L^-1=0.0500*mol#</mathjax></p>
<p>Thus we require <mathjax>#0.0500*molxx74.55*g·mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??g" potassium chloride"#</mathjax></p></div>
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</article> | How many grams of potassium chloride are required to many prepare 500.0 mL of a 0.100 M solution? | null |
574 | a837f71a-6ddd-11ea-bf85-ccda262736ce | https://socratic.org/questions/how-do-you-balance-h-3po-4-hci-pcl-5-h-2o | H3PO4 + 5 HCl -> PCl5 + 4 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":"H3PO4 + 5 HCl -> PCl5 + 4 H2O"}] | [{"type":"chemical equation","value":"H3PO4 + HCl -> PCl5 + H2O"}] | <h1 class="questionTitle" itemprop="name">How do you balance #H_3PO_4 + HCI -> PCl_5 + H_2O#?</h1> | null | H3PO4 + 5 HCl -> PCl5 + 4 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Step 1:</strong> Tally all the atoms based on the subscripts.</p>
<p><mathjax>#H_3PO_4#</mathjax> + <mathjax>#HCl#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#PCl_5#</mathjax> + <mathjax>#H_2O#</mathjax> (unbalanced)</p>
<p><em>left side: </em><br/>
<mathjax>#H#</mathjax> = 3 + 1 (do not add this up yet)<br/>
<mathjax>#P#</mathjax> = 1<br/>
<mathjax>#O#</mathjax> = 4<br/>
<mathjax>#Cl#</mathjax> = 1</p>
<p><em>right side: </em><br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#P#</mathjax> = 1<br/>
<mathjax>#O#</mathjax> = 1<br/>
<mathjax>#Cl#</mathjax> = 5</p>
<p><strong>Step 2:</strong> Find the easiest atom to balance. Remember, in balancing you CANNOT change the subscripts - what you CAN do is multiply the subscripts by an appropriate coefficient. Also remember that with bonded substances, you have to multiply the coefficient, not only to the atom that you intend to balance, but also to whatever other atoms that it is bonded with.</p>
<p><mathjax>#H_3PO_4#</mathjax> + <mathjax>#color (red) 5HCl#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#PCl_5#</mathjax> + <mathjax>#H_2O#</mathjax></p>
<p><em>left side: </em><br/>
<mathjax>#H#</mathjax> = 3 + (1 x <mathjax>#color (red) 5#</mathjax>) = 8<br/>
<mathjax>#P#</mathjax> = 1<br/>
<mathjax>#O#</mathjax> = 4<br/>
<mathjax>#Cl#</mathjax> = (1 x <mathjax>#color (red) 5#</mathjax>) = <strong>5</strong></p>
<p><em>right side: </em><br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#P#</mathjax> = 1<br/>
<mathjax>#O#</mathjax> = 1<br/>
<mathjax>#Cl#</mathjax> = <strong>5</strong></p>
<p>But since <mathjax>#HCl#</mathjax> is a substance, the coefficient 5 should also be applied to the <mathjax>#H#</mathjax> atom.</p>
<p><strong>Step 3:</strong> Continue balancing atoms.</p>
<p><mathjax>#H_3PO_4#</mathjax> + <mathjax>#5HCl#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#PCl_5#</mathjax> + <mathjax>#color (blue) 4H_2O#</mathjax></p>
<p><em>left side: </em><br/>
<mathjax>#H#</mathjax> = 3 + (1 x 5) = <strong>8</strong><br/>
<mathjax>#P#</mathjax> = 1<br/>
<mathjax>#O#</mathjax> = <strong>4</strong><br/>
<mathjax>#Cl#</mathjax> = (1 x 5) = <strong>5</strong></p>
<p><em>right side: </em><br/>
<mathjax>#H#</mathjax> = (2 x <mathjax>#color (blue) 4#</mathjax>) = <strong>8</strong><br/>
<mathjax>#P#</mathjax> = 1<br/>
<mathjax>#O#</mathjax> = (1 x <mathjax>#color (blue) 4#</mathjax>) = <strong>4</strong><br/>
<mathjax>#Cl#</mathjax> = <strong>5</strong></p>
<p><strong>Step 3:</strong> Check if the total numbers per atom on the left is the same with the total numbers per atom on the side of the equation.</p>
<p><mathjax>#H_3PO_4#</mathjax> + <mathjax>#5HCl#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#PCl_5#</mathjax> + <mathjax>#4H_2O#</mathjax></p>
<p>The equation is now balance.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#H_3PO_4#</mathjax> + <mathjax>#5HCl#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#PCl_5#</mathjax> + <mathjax>#4H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Step 1:</strong> Tally all the atoms based on the subscripts.</p>
<p><mathjax>#H_3PO_4#</mathjax> + <mathjax>#HCl#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#PCl_5#</mathjax> + <mathjax>#H_2O#</mathjax> (unbalanced)</p>
<p><em>left side: </em><br/>
<mathjax>#H#</mathjax> = 3 + 1 (do not add this up yet)<br/>
<mathjax>#P#</mathjax> = 1<br/>
<mathjax>#O#</mathjax> = 4<br/>
<mathjax>#Cl#</mathjax> = 1</p>
<p><em>right side: </em><br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#P#</mathjax> = 1<br/>
<mathjax>#O#</mathjax> = 1<br/>
<mathjax>#Cl#</mathjax> = 5</p>
<p><strong>Step 2:</strong> Find the easiest atom to balance. Remember, in balancing you CANNOT change the subscripts - what you CAN do is multiply the subscripts by an appropriate coefficient. Also remember that with bonded substances, you have to multiply the coefficient, not only to the atom that you intend to balance, but also to whatever other atoms that it is bonded with.</p>
<p><mathjax>#H_3PO_4#</mathjax> + <mathjax>#color (red) 5HCl#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#PCl_5#</mathjax> + <mathjax>#H_2O#</mathjax></p>
<p><em>left side: </em><br/>
<mathjax>#H#</mathjax> = 3 + (1 x <mathjax>#color (red) 5#</mathjax>) = 8<br/>
<mathjax>#P#</mathjax> = 1<br/>
<mathjax>#O#</mathjax> = 4<br/>
<mathjax>#Cl#</mathjax> = (1 x <mathjax>#color (red) 5#</mathjax>) = <strong>5</strong></p>
<p><em>right side: </em><br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#P#</mathjax> = 1<br/>
<mathjax>#O#</mathjax> = 1<br/>
<mathjax>#Cl#</mathjax> = <strong>5</strong></p>
<p>But since <mathjax>#HCl#</mathjax> is a substance, the coefficient 5 should also be applied to the <mathjax>#H#</mathjax> atom.</p>
<p><strong>Step 3:</strong> Continue balancing atoms.</p>
<p><mathjax>#H_3PO_4#</mathjax> + <mathjax>#5HCl#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#PCl_5#</mathjax> + <mathjax>#color (blue) 4H_2O#</mathjax></p>
<p><em>left side: </em><br/>
<mathjax>#H#</mathjax> = 3 + (1 x 5) = <strong>8</strong><br/>
<mathjax>#P#</mathjax> = 1<br/>
<mathjax>#O#</mathjax> = <strong>4</strong><br/>
<mathjax>#Cl#</mathjax> = (1 x 5) = <strong>5</strong></p>
<p><em>right side: </em><br/>
<mathjax>#H#</mathjax> = (2 x <mathjax>#color (blue) 4#</mathjax>) = <strong>8</strong><br/>
<mathjax>#P#</mathjax> = 1<br/>
<mathjax>#O#</mathjax> = (1 x <mathjax>#color (blue) 4#</mathjax>) = <strong>4</strong><br/>
<mathjax>#Cl#</mathjax> = <strong>5</strong></p>
<p><strong>Step 3:</strong> Check if the total numbers per atom on the left is the same with the total numbers per atom on the side of the equation.</p>
<p><mathjax>#H_3PO_4#</mathjax> + <mathjax>#5HCl#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#PCl_5#</mathjax> + <mathjax>#4H_2O#</mathjax></p>
<p>The equation is now balance.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you balance #H_3PO_4 + HCI -> PCl_5 + H_2O#?</h1>
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<div class="markdown"><p><mathjax>#H_3PO_4#</mathjax> + <mathjax>#5HCl#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#PCl_5#</mathjax> + <mathjax>#4H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Step 1:</strong> Tally all the atoms based on the subscripts.</p>
<p><mathjax>#H_3PO_4#</mathjax> + <mathjax>#HCl#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#PCl_5#</mathjax> + <mathjax>#H_2O#</mathjax> (unbalanced)</p>
<p><em>left side: </em><br/>
<mathjax>#H#</mathjax> = 3 + 1 (do not add this up yet)<br/>
<mathjax>#P#</mathjax> = 1<br/>
<mathjax>#O#</mathjax> = 4<br/>
<mathjax>#Cl#</mathjax> = 1</p>
<p><em>right side: </em><br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#P#</mathjax> = 1<br/>
<mathjax>#O#</mathjax> = 1<br/>
<mathjax>#Cl#</mathjax> = 5</p>
<p><strong>Step 2:</strong> Find the easiest atom to balance. Remember, in balancing you CANNOT change the subscripts - what you CAN do is multiply the subscripts by an appropriate coefficient. Also remember that with bonded substances, you have to multiply the coefficient, not only to the atom that you intend to balance, but also to whatever other atoms that it is bonded with.</p>
<p><mathjax>#H_3PO_4#</mathjax> + <mathjax>#color (red) 5HCl#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#PCl_5#</mathjax> + <mathjax>#H_2O#</mathjax></p>
<p><em>left side: </em><br/>
<mathjax>#H#</mathjax> = 3 + (1 x <mathjax>#color (red) 5#</mathjax>) = 8<br/>
<mathjax>#P#</mathjax> = 1<br/>
<mathjax>#O#</mathjax> = 4<br/>
<mathjax>#Cl#</mathjax> = (1 x <mathjax>#color (red) 5#</mathjax>) = <strong>5</strong></p>
<p><em>right side: </em><br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#P#</mathjax> = 1<br/>
<mathjax>#O#</mathjax> = 1<br/>
<mathjax>#Cl#</mathjax> = <strong>5</strong></p>
<p>But since <mathjax>#HCl#</mathjax> is a substance, the coefficient 5 should also be applied to the <mathjax>#H#</mathjax> atom.</p>
<p><strong>Step 3:</strong> Continue balancing atoms.</p>
<p><mathjax>#H_3PO_4#</mathjax> + <mathjax>#5HCl#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#PCl_5#</mathjax> + <mathjax>#color (blue) 4H_2O#</mathjax></p>
<p><em>left side: </em><br/>
<mathjax>#H#</mathjax> = 3 + (1 x 5) = <strong>8</strong><br/>
<mathjax>#P#</mathjax> = 1<br/>
<mathjax>#O#</mathjax> = <strong>4</strong><br/>
<mathjax>#Cl#</mathjax> = (1 x 5) = <strong>5</strong></p>
<p><em>right side: </em><br/>
<mathjax>#H#</mathjax> = (2 x <mathjax>#color (blue) 4#</mathjax>) = <strong>8</strong><br/>
<mathjax>#P#</mathjax> = 1<br/>
<mathjax>#O#</mathjax> = (1 x <mathjax>#color (blue) 4#</mathjax>) = <strong>4</strong><br/>
<mathjax>#Cl#</mathjax> = <strong>5</strong></p>
<p><strong>Step 3:</strong> Check if the total numbers per atom on the left is the same with the total numbers per atom on the side of the equation.</p>
<p><mathjax>#H_3PO_4#</mathjax> + <mathjax>#5HCl#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#PCl_5#</mathjax> + <mathjax>#4H_2O#</mathjax></p>
<p>The equation is now balance.</p></div>
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</article> | How do you balance #H_3PO_4 + HCI -> PCl_5 + H_2O#? | null |
575 | aad5b0a4-6ddd-11ea-8cc6-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-6-02-10-23-atoms-of-calcium | 40.08 g | start physical_unit 10 10 mass g qc_end end | [{"type":"physical unit","value":"Mass [OF] calcium [IN] g"}] | [{"type":"physical unit","value":"40.08 g"}] | [{"type":"physical unit","value":"Number [OF] calcium atoms [=] \\pu{6.02 × 10^23}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of #6.02 * 10^23# atoms of calcium?</h1> | null | 40.08 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>How does this help you? Well, it happens that <mathjax>#N_A#</mathjax>, <mathjax>#"Avogadro's Number"#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#6.022xx10^23*mol^-1#</mathjax>.</p>
<p>And it is stated on your trusty Periodic Table that the molar mass of <mathjax>#Ca#</mathjax> is <mathjax>#40.08*g*mol^-1#</mathjax>. So, if we specify a molar quantity, we specify a precise number of atoms and molecules that can be approximated by mass. </p></div>
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<div class="markdown"><p>In <mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> of <mathjax>#Ca#</mathjax> <mathjax>#"atoms"#</mathjax>, there are <mathjax>#N_A#</mathjax> individual calcium atoms.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>How does this help you? Well, it happens that <mathjax>#N_A#</mathjax>, <mathjax>#"Avogadro's Number"#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#6.022xx10^23*mol^-1#</mathjax>.</p>
<p>And it is stated on your trusty Periodic Table that the molar mass of <mathjax>#Ca#</mathjax> is <mathjax>#40.08*g*mol^-1#</mathjax>. So, if we specify a molar quantity, we specify a precise number of atoms and molecules that can be approximated by mass. </p></div>
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<div class="markdown"><p>In <mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> of <mathjax>#Ca#</mathjax> <mathjax>#"atoms"#</mathjax>, there are <mathjax>#N_A#</mathjax> individual calcium atoms.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>How does this help you? Well, it happens that <mathjax>#N_A#</mathjax>, <mathjax>#"Avogadro's Number"#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#6.022xx10^23*mol^-1#</mathjax>.</p>
<p>And it is stated on your trusty Periodic Table that the molar mass of <mathjax>#Ca#</mathjax> is <mathjax>#40.08*g*mol^-1#</mathjax>. So, if we specify a molar quantity, we specify a precise number of atoms and molecules that can be approximated by mass. </p></div>
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</article> | What is the mass of #6.02 * 10^23# atoms of calcium? | null |
576 | a8db5342-6ddd-11ea-8e7e-ccda262736ce | https://socratic.org/questions/597e44e5b72cff1139246c0f | 0.4 g | start physical_unit 5 6 mass g qc_end physical_unit 5 6 1 2 volume qc_end physical_unit 15 15 11 12 volume qc_end physical_unit 15 15 19 20 concentration qc_end end | [{"type":"physical unit","value":"Mass [OF] oxalic acid [IN] g"}] | [{"type":"physical unit","value":"0.4 g"}] | [{"type":"physical unit","value":"Volume [OF] oxalic acid [=] \\pu{10 mL}"},{"type":"physical unit","value":"Volume [OF] NaOH(aq) [=] \\pu{34.40 mL}"},{"type":"physical unit","value":"Concentration [OF] NaOH(aq) [=] \\pu{0.250 mol/L}"}] | <h1 class="questionTitle" itemprop="name">A #10*mL# volume of oxalic acid was titrated with a #34.40*mL# volume of #NaOH(aq)# whose concentration was #0.250*mol*L^-1#. What was the mass of oxalic acid present in the initial #10*mL# volume?</h1> | null | 0.4 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need (i) a stoichiometric equation.....</p>
<p><mathjax>#"HO(O=)CC(=O)OH + 2NaOH"rarrC_2O_4^(2-)Na_2^+ + 2H_2O#</mathjax></p>
<p>And (ii) equivalent quantities of sodium hydroxide......</p>
<p><mathjax>#"Moles of NaOH"=34.40*mLxx10^-3*L*mL^-1xx0.250*mol*L^-1=8.60xx10^-3*mol#</mathjax>.</p>
<p>And HALF this molar quantity was equivalent to the oxalic acid solution contained in <mathjax>#10.00*mL#</mathjax> solution.</p>
<p><mathjax>#"Concentration of oxalic acid"=(1/2xx8.60xx10^-3*mol)/(10.00xx10^-3*L^-1)#</mathjax></p>
<p><mathjax>#=0.430*mol*L^-1#</mathjax></p>
<p>And thus in a <mathjax>#10*mL#</mathjax> volume, there were <mathjax>#10xx10^-3*Lxx0.430*mol*L^-1xx90.03*g*mol^-1=#</mathjax></p>
<p><mathjax>#~=0.4*g#</mathjax> <mathjax>#""#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#[C_2O_4H_2]~=0.5*mol*L^-1#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need (i) a stoichiometric equation.....</p>
<p><mathjax>#"HO(O=)CC(=O)OH + 2NaOH"rarrC_2O_4^(2-)Na_2^+ + 2H_2O#</mathjax></p>
<p>And (ii) equivalent quantities of sodium hydroxide......</p>
<p><mathjax>#"Moles of NaOH"=34.40*mLxx10^-3*L*mL^-1xx0.250*mol*L^-1=8.60xx10^-3*mol#</mathjax>.</p>
<p>And HALF this molar quantity was equivalent to the oxalic acid solution contained in <mathjax>#10.00*mL#</mathjax> solution.</p>
<p><mathjax>#"Concentration of oxalic acid"=(1/2xx8.60xx10^-3*mol)/(10.00xx10^-3*L^-1)#</mathjax></p>
<p><mathjax>#=0.430*mol*L^-1#</mathjax></p>
<p>And thus in a <mathjax>#10*mL#</mathjax> volume, there were <mathjax>#10xx10^-3*Lxx0.430*mol*L^-1xx90.03*g*mol^-1=#</mathjax></p>
<p><mathjax>#~=0.4*g#</mathjax> <mathjax>#""#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A #10*mL# volume of oxalic acid was titrated with a #34.40*mL# volume of #NaOH(aq)# whose concentration was #0.250*mol*L^-1#. What was the mass of oxalic acid present in the initial #10*mL# volume?</h1>
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<div class="markdown"><p><mathjax>#[C_2O_4H_2]~=0.5*mol*L^-1#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We need (i) a stoichiometric equation.....</p>
<p><mathjax>#"HO(O=)CC(=O)OH + 2NaOH"rarrC_2O_4^(2-)Na_2^+ + 2H_2O#</mathjax></p>
<p>And (ii) equivalent quantities of sodium hydroxide......</p>
<p><mathjax>#"Moles of NaOH"=34.40*mLxx10^-3*L*mL^-1xx0.250*mol*L^-1=8.60xx10^-3*mol#</mathjax>.</p>
<p>And HALF this molar quantity was equivalent to the oxalic acid solution contained in <mathjax>#10.00*mL#</mathjax> solution.</p>
<p><mathjax>#"Concentration of oxalic acid"=(1/2xx8.60xx10^-3*mol)/(10.00xx10^-3*L^-1)#</mathjax></p>
<p><mathjax>#=0.430*mol*L^-1#</mathjax></p>
<p>And thus in a <mathjax>#10*mL#</mathjax> volume, there were <mathjax>#10xx10^-3*Lxx0.430*mol*L^-1xx90.03*g*mol^-1=#</mathjax></p>
<p><mathjax>#~=0.4*g#</mathjax> <mathjax>#""#</mathjax></p></div>
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</article> | A #10*mL# volume of oxalic acid was titrated with a #34.40*mL# volume of #NaOH(aq)# whose concentration was #0.250*mol*L^-1#. What was the mass of oxalic acid present in the initial #10*mL# volume? | null |
577 | ac8b18ca-6ddd-11ea-bf77-ccda262736ce | https://socratic.org/questions/iron-is-obtained-commercially-by-the-reaction-of-hematite-fe-2o3-with-carbon-mon | 1117.00 grams | start physical_unit 0 0 mass g qc_end physical_unit 8 8 21 22 mole qc_end physical_unit 11 12 27 28 mole qc_end chemical_equation 35 44 qc_end end | [{"type":"physical unit","value":"Mass [OF] iron [IN] grams"}] | [{"type":"physical unit","value":"1117.00 grams"}] | [{"type":"physical unit","value":"Mole [OF] hematite [=] \\pu{25.0 mol}"},{"type":"physical unit","value":"Mole [OF] carbon monoxide [=] \\pu{30.0 mol}"},{"type":"chemical equation","value":"Fe2O3(s) + 3 CO(g) -> 2 Fe(s) + 3 CO2(g)"}] | <h1 class="questionTitle" itemprop="name">Iron is obtained commercially by the reaction of hematite #Fe_2O_3# with carbon monoxide. How many grams of iron is produced when 25.0 mol of hematite reacts with 30.0 mol of carbon monoxide?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The reaction is <mathjax>#Fe_2O_3(s) + 3CO(g) -> 2Fe(s) + 3CO_2(g)#</mathjax>.</p></div>
</h2>
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</div> | 1117.00 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As can be seen from the chemical equation, each mole of <mathjax>#Fe_2CO_3#</mathjax> needs 3 moles of <mathjax>#CO#</mathjax>. If 30 moles of <mathjax>#CO#</mathjax> are used, only 10 moles of <mathjax>#Fe_2CO_3#</mathjax> can be reduced, generating 20 moles of <mathjax>#Fe#</mathjax>, corresponding to 1117 grams of iron.</p></div>
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<div class="markdown"><p>1117 grams of iron</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As can be seen from the chemical equation, each mole of <mathjax>#Fe_2CO_3#</mathjax> needs 3 moles of <mathjax>#CO#</mathjax>. If 30 moles of <mathjax>#CO#</mathjax> are used, only 10 moles of <mathjax>#Fe_2CO_3#</mathjax> can be reduced, generating 20 moles of <mathjax>#Fe#</mathjax>, corresponding to 1117 grams of iron.</p></div>
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<h1 class="questionTitle" itemprop="name">Iron is obtained commercially by the reaction of hematite #Fe_2O_3# with carbon monoxide. How many grams of iron is produced when 25.0 mol of hematite reacts with 30.0 mol of carbon monoxide?</h1>
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<div class="markdown"><p>The reaction is <mathjax>#Fe_2O_3(s) + 3CO(g) -> 2Fe(s) + 3CO_2(g)#</mathjax>.</p></div>
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il maestro
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<div class="markdown"><p>1117 grams of iron</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>As can be seen from the chemical equation, each mole of <mathjax>#Fe_2CO_3#</mathjax> needs 3 moles of <mathjax>#CO#</mathjax>. If 30 moles of <mathjax>#CO#</mathjax> are used, only 10 moles of <mathjax>#Fe_2CO_3#</mathjax> can be reduced, generating 20 moles of <mathjax>#Fe#</mathjax>, corresponding to 1117 grams of iron.</p></div>
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</article> | Iron is obtained commercially by the reaction of hematite #Fe_2O_3# with carbon monoxide. How many grams of iron is produced when 25.0 mol of hematite reacts with 30.0 mol of carbon monoxide? |
The reaction is #Fe_2O_3(s) + 3CO(g) -> 2Fe(s) + 3CO_2(g)#.
|
578 | ab5625f8-6ddd-11ea-a3aa-ccda262736ce | https://socratic.org/questions/lactic-acid-hc3h5o3-has-one-acidic-hydrogen-a-0-10-m-solution-of-lactic-acid-has | 1.4 × 10^(-4) | start physical_unit 0 1 ka none qc_end c_other OTHER qc_end physical_unit 0 1 8 9 concentration qc_end physical_unit 18 19 21 22 concentration qc_end end | [{"type":"physical unit","value":"Ka [OF] lactic acid"}] | [{"type":"physical unit","value":"1.4 × 10^(-4)"}] | [{"type":"other","value":"Lactic acid HC3H5O3 has one acidic hydrogen."},{"type":"physical unit","value":"Concentration [OF] lactic acid solution [=] \\pu{0.10 M}"},{"type":"physical unit","value":"Concentration [OF] hydronium ion [=] \\pu{0.00363 M}"}] | <h1 class="questionTitle" itemprop="name">Lactic acid HC3H5O3 has one acidic hydrogen. A 0.10 M solution of lactic acid has the concentration of hydronium ion of 0.00363 M. Calculate Ka for lactic acid?</h1> | null | 1.4 × 10^(-4) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you have to do in order to find the value of the <em>acid dissociation constant</em>, <mathjax>#K_a#</mathjax>, is to write a balanced chemical equation for the <strong>equilibrium</strong> that is established when <em>lactic acid</em>, <mathjax>#"HC"_3"H"_5"O"_3#</mathjax>, is added to water. </p>
<p>Being a <strong>weak acid</strong>, lactic acid will <strong>not</strong> dissociate completely to form <em>lactate</em>, <mathjax>#"C"_3"H"_5"O"_3^(-)#</mathjax>, its conjugate base, and <em>hydronium ions</em>, <mathjax>#"H"_3"O"^(+)#</mathjax>, as you can see from the difference between the concentration of the acid and that of the hydronium ions. </p>
<p><img alt="http://ceaccp.oxfordjournals.org/content/6/3/128/F2.expansion" src="https://useruploads.socratic.org/ceg1q92TpeTj56qn1gBd_F2.large.jpg"/> </p>
<p>Use an <strong>ICE table</strong> to help you find the equilibrium concentration of the lactic acid</p>
<blockquote>
<p><mathjax>#" ""HC"_3"H"_5"O"_text(3(aq]) + "H"_2"O"_text((l]) rightleftharpoons "C"_3"H"_5"O"_text(3(aq])^(-) + "H"_3"O"_text((aq])^(+)#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " " " "0.10" " " " " " " " " " " " " " " " " "0" " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" " " "(-x)" " " " " " " " " " " " " " " "(+x)" " " "(+x)#</mathjax><br/>
<mathjax>#color(purple)("E")" " " "0.10-x" " " " " " " " " " " " " " " "x" " " " " " " "x#</mathjax></p>
<p>By definition, the acid dissociation constant will be equal to </p>
<blockquote>
<p><mathjax>#K_a = ( ["C"_3"H"_5"O"_3^(-)] * ["H"_3"O"^(+)])/(["HC"_3"H"_5"O"_3])#</mathjax></p>
</blockquote>
<p>You know that the equilibrium concentration of hydronium ions is</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = x = "0.00363 M"#</mathjax></p>
</blockquote>
<p>This means that the equilibrium concentration of the lactic acid will be </p>
<blockquote>
<p><mathjax>#["HC"_3"H"_5"O"_3] = 0.10 - x#</mathjax> </p>
<p><mathjax>#["HC"_3"H"_5"O"_3] = "0.10 M" - "0.00363 M" = "0.09637 M"#</mathjax></p>
</blockquote>
<p>This means that the value of the acid dissociation constant will be </p>
<blockquote>
<p><mathjax>#K_a = ( 0.00363 * 0.00363)/(0.09637) = 1.367 * 10^(-4)#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the lactic acid solution, the answer will be </p>
<blockquote>
<p><mathjax>#K_a = color(green)(1.4 * 10^(-4))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#1.4 * 10^(-4)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you have to do in order to find the value of the <em>acid dissociation constant</em>, <mathjax>#K_a#</mathjax>, is to write a balanced chemical equation for the <strong>equilibrium</strong> that is established when <em>lactic acid</em>, <mathjax>#"HC"_3"H"_5"O"_3#</mathjax>, is added to water. </p>
<p>Being a <strong>weak acid</strong>, lactic acid will <strong>not</strong> dissociate completely to form <em>lactate</em>, <mathjax>#"C"_3"H"_5"O"_3^(-)#</mathjax>, its conjugate base, and <em>hydronium ions</em>, <mathjax>#"H"_3"O"^(+)#</mathjax>, as you can see from the difference between the concentration of the acid and that of the hydronium ions. </p>
<p><img alt="http://ceaccp.oxfordjournals.org/content/6/3/128/F2.expansion" src="https://useruploads.socratic.org/ceg1q92TpeTj56qn1gBd_F2.large.jpg"/> </p>
<p>Use an <strong>ICE table</strong> to help you find the equilibrium concentration of the lactic acid</p>
<blockquote>
<p><mathjax>#" ""HC"_3"H"_5"O"_text(3(aq]) + "H"_2"O"_text((l]) rightleftharpoons "C"_3"H"_5"O"_text(3(aq])^(-) + "H"_3"O"_text((aq])^(+)#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " " " "0.10" " " " " " " " " " " " " " " " " "0" " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" " " "(-x)" " " " " " " " " " " " " " " "(+x)" " " "(+x)#</mathjax><br/>
<mathjax>#color(purple)("E")" " " "0.10-x" " " " " " " " " " " " " " " "x" " " " " " " "x#</mathjax></p>
<p>By definition, the acid dissociation constant will be equal to </p>
<blockquote>
<p><mathjax>#K_a = ( ["C"_3"H"_5"O"_3^(-)] * ["H"_3"O"^(+)])/(["HC"_3"H"_5"O"_3])#</mathjax></p>
</blockquote>
<p>You know that the equilibrium concentration of hydronium ions is</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = x = "0.00363 M"#</mathjax></p>
</blockquote>
<p>This means that the equilibrium concentration of the lactic acid will be </p>
<blockquote>
<p><mathjax>#["HC"_3"H"_5"O"_3] = 0.10 - x#</mathjax> </p>
<p><mathjax>#["HC"_3"H"_5"O"_3] = "0.10 M" - "0.00363 M" = "0.09637 M"#</mathjax></p>
</blockquote>
<p>This means that the value of the acid dissociation constant will be </p>
<blockquote>
<p><mathjax>#K_a = ( 0.00363 * 0.00363)/(0.09637) = 1.367 * 10^(-4)#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the lactic acid solution, the answer will be </p>
<blockquote>
<p><mathjax>#K_a = color(green)(1.4 * 10^(-4))#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Lactic acid HC3H5O3 has one acidic hydrogen. A 0.10 M solution of lactic acid has the concentration of hydronium ion of 0.00363 M. Calculate Ka for lactic acid?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2015-12-08T01:43:23" itemprop="dateCreated">
Dec 8, 2015
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<div class="markdown"><p><mathjax>#1.4 * 10^(-4)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you have to do in order to find the value of the <em>acid dissociation constant</em>, <mathjax>#K_a#</mathjax>, is to write a balanced chemical equation for the <strong>equilibrium</strong> that is established when <em>lactic acid</em>, <mathjax>#"HC"_3"H"_5"O"_3#</mathjax>, is added to water. </p>
<p>Being a <strong>weak acid</strong>, lactic acid will <strong>not</strong> dissociate completely to form <em>lactate</em>, <mathjax>#"C"_3"H"_5"O"_3^(-)#</mathjax>, its conjugate base, and <em>hydronium ions</em>, <mathjax>#"H"_3"O"^(+)#</mathjax>, as you can see from the difference between the concentration of the acid and that of the hydronium ions. </p>
<p><img alt="http://ceaccp.oxfordjournals.org/content/6/3/128/F2.expansion" src="https://useruploads.socratic.org/ceg1q92TpeTj56qn1gBd_F2.large.jpg"/> </p>
<p>Use an <strong>ICE table</strong> to help you find the equilibrium concentration of the lactic acid</p>
<blockquote>
<p><mathjax>#" ""HC"_3"H"_5"O"_text(3(aq]) + "H"_2"O"_text((l]) rightleftharpoons "C"_3"H"_5"O"_text(3(aq])^(-) + "H"_3"O"_text((aq])^(+)#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " " " "0.10" " " " " " " " " " " " " " " " " "0" " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" " " "(-x)" " " " " " " " " " " " " " " "(+x)" " " "(+x)#</mathjax><br/>
<mathjax>#color(purple)("E")" " " "0.10-x" " " " " " " " " " " " " " " "x" " " " " " " "x#</mathjax></p>
<p>By definition, the acid dissociation constant will be equal to </p>
<blockquote>
<p><mathjax>#K_a = ( ["C"_3"H"_5"O"_3^(-)] * ["H"_3"O"^(+)])/(["HC"_3"H"_5"O"_3])#</mathjax></p>
</blockquote>
<p>You know that the equilibrium concentration of hydronium ions is</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = x = "0.00363 M"#</mathjax></p>
</blockquote>
<p>This means that the equilibrium concentration of the lactic acid will be </p>
<blockquote>
<p><mathjax>#["HC"_3"H"_5"O"_3] = 0.10 - x#</mathjax> </p>
<p><mathjax>#["HC"_3"H"_5"O"_3] = "0.10 M" - "0.00363 M" = "0.09637 M"#</mathjax></p>
</blockquote>
<p>This means that the value of the acid dissociation constant will be </p>
<blockquote>
<p><mathjax>#K_a = ( 0.00363 * 0.00363)/(0.09637) = 1.367 * 10^(-4)#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the lactic acid solution, the answer will be </p>
<blockquote>
<p><mathjax>#K_a = color(green)(1.4 * 10^(-4))#</mathjax></p>
</blockquote></div>
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</article> | Lactic acid HC3H5O3 has one acidic hydrogen. A 0.10 M solution of lactic acid has the concentration of hydronium ion of 0.00363 M. Calculate Ka for lactic acid? | null |
579 | acd557c0-6ddd-11ea-9ffd-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-stock-solution-if-10-ml-is-diluted-to-400-ml-with-a-co | 20.00 M | start physical_unit 6 7 molarity mol/l qc_end physical_unit 6 7 9 10 volume qc_end physical_unit 6 7 14 15 volume qc_end physical_unit 6 7 20 21 concentration qc_end end | [{"type":"physical unit","value":"Molarity1 [OF] stock solution [IN] M"}] | [{"type":"physical unit","value":"20.00 M"}] | [{"type":"physical unit","value":"Volume1 [OF] stock solution [=] \\pu{10 mL}"},{"type":"physical unit","value":"Volume2 [OF] stock solution [=] \\pu{400 mL}"},{"type":"physical unit","value":"Concentration2 [OF] stock solution [=] \\pu{0.5 M}"}] | <h1 class="questionTitle" itemprop="name"> What is the molarity of a stock solution if 10 mL is diluted to 400 mL with a concentration of 0.5M?</h1> | null | 20.00 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You can solve this problem by calculating the <strong><a href="https://socratic.org/questions/how-do-you-calculate-dilution-factor">dilution factor</a></strong> associated with your dilution. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"D.F." = V_"diluted"/V_"concentrated"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#V_"diluted"#</mathjax> - the volume of the <strong>diluted solution</strong><br/>
<mathjax>#V_"concentrated"#</mathjax> - the volume of the <strong>concentrated solution</strong></p>
<p>In essence, the <em>dilution factor</em> will tell you how <strong>concentrated</strong> the stock solution was compared with the dilute solution. </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c_"stock" = "D.F." xx c_"diluted")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you'll have </p>
<blockquote>
<p><mathjax>#"D.F." = (400 color(red)(cancel(color(black)("mL"))))/(10color(red)(cancel(color(black)("mL")))) = 40#</mathjax></p>
</blockquote>
<p>This means that the stock solution is <mathjax>#40#</mathjax> <strong>times more concentrated</strong> than the diluted solution. Since the diluted solution is said to have a <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of <mathjax>#"0.5 M"#</mathjax>, you can say that the stock solution had a molarity of </p>
<blockquote>
<p><mathjax>#c_"stock" = 40 * "0.5 M" = color(green)(|bar(ul(color(white)(a/a)"20 M"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The dilution factor method works because it's based on the <em>underlying principle</em> of a dilution, i.e. the concentration of the solution is <strong>decreased</strong> by <strong>Increasing</strong> the volume of the solution while keeping the <em>number of moles</em> of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>constant</strong>. </p>
<p>The equation for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a> looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(overbrace(c_1 xx V_1)^(color(brown)("moles of solute in concentrated solution")) = overbrace(c_2 xx V_2)^(color(brown)("moles of solute in diluted solution"))#</mathjax></p>
</blockquote>
<p>Here </p>
<p><mathjax>#c_1#</mathjax>, <mathjax>#V_1#</mathjax> - the molarity and volume of the concentrated solution<br/>
<mathjax>#c_2#</mathjax>, <mathjax>#V_2#</mathjax> - the molarity and volume of the diluted solution</p>
<p>This equation can be rearranged as </p>
<blockquote>
<p><mathjax>#c_1/c_2 = V_2/V_1#</mathjax></p>
</blockquote>
<p>This is where the dilution factor comes into play</p>
<blockquote>
<p><mathjax>#"D.F." = V_2/V_1#</mathjax></p>
</blockquote>
<p>Plug this into the above equation to get</p>
<blockquote>
<p><mathjax>#"D.F." = c_1/c_2 implies color(purple)(|bar(ul(color(white)(a/a)color(black)(c_1 = "D.F." xx c_2)color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"20 M"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You can solve this problem by calculating the <strong><a href="https://socratic.org/questions/how-do-you-calculate-dilution-factor">dilution factor</a></strong> associated with your dilution. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"D.F." = V_"diluted"/V_"concentrated"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#V_"diluted"#</mathjax> - the volume of the <strong>diluted solution</strong><br/>
<mathjax>#V_"concentrated"#</mathjax> - the volume of the <strong>concentrated solution</strong></p>
<p>In essence, the <em>dilution factor</em> will tell you how <strong>concentrated</strong> the stock solution was compared with the dilute solution. </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c_"stock" = "D.F." xx c_"diluted")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you'll have </p>
<blockquote>
<p><mathjax>#"D.F." = (400 color(red)(cancel(color(black)("mL"))))/(10color(red)(cancel(color(black)("mL")))) = 40#</mathjax></p>
</blockquote>
<p>This means that the stock solution is <mathjax>#40#</mathjax> <strong>times more concentrated</strong> than the diluted solution. Since the diluted solution is said to have a <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of <mathjax>#"0.5 M"#</mathjax>, you can say that the stock solution had a molarity of </p>
<blockquote>
<p><mathjax>#c_"stock" = 40 * "0.5 M" = color(green)(|bar(ul(color(white)(a/a)"20 M"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The dilution factor method works because it's based on the <em>underlying principle</em> of a dilution, i.e. the concentration of the solution is <strong>decreased</strong> by <strong>Increasing</strong> the volume of the solution while keeping the <em>number of moles</em> of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>constant</strong>. </p>
<p>The equation for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a> looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(overbrace(c_1 xx V_1)^(color(brown)("moles of solute in concentrated solution")) = overbrace(c_2 xx V_2)^(color(brown)("moles of solute in diluted solution"))#</mathjax></p>
</blockquote>
<p>Here </p>
<p><mathjax>#c_1#</mathjax>, <mathjax>#V_1#</mathjax> - the molarity and volume of the concentrated solution<br/>
<mathjax>#c_2#</mathjax>, <mathjax>#V_2#</mathjax> - the molarity and volume of the diluted solution</p>
<p>This equation can be rearranged as </p>
<blockquote>
<p><mathjax>#c_1/c_2 = V_2/V_1#</mathjax></p>
</blockquote>
<p>This is where the dilution factor comes into play</p>
<blockquote>
<p><mathjax>#"D.F." = V_2/V_1#</mathjax></p>
</blockquote>
<p>Plug this into the above equation to get</p>
<blockquote>
<p><mathjax>#"D.F." = c_1/c_2 implies color(purple)(|bar(ul(color(white)(a/a)color(black)(c_1 = "D.F." xx c_2)color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name"> What is the molarity of a stock solution if 10 mL is diluted to 400 mL with a concentration of 0.5M?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-04-15T20:08:46" itemprop="dateCreated">
Apr 15, 2016
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<div class="markdown"><p><mathjax>#"20 M"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You can solve this problem by calculating the <strong><a href="https://socratic.org/questions/how-do-you-calculate-dilution-factor">dilution factor</a></strong> associated with your dilution. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"D.F." = V_"diluted"/V_"concentrated"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#V_"diluted"#</mathjax> - the volume of the <strong>diluted solution</strong><br/>
<mathjax>#V_"concentrated"#</mathjax> - the volume of the <strong>concentrated solution</strong></p>
<p>In essence, the <em>dilution factor</em> will tell you how <strong>concentrated</strong> the stock solution was compared with the dilute solution. </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c_"stock" = "D.F." xx c_"diluted")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you'll have </p>
<blockquote>
<p><mathjax>#"D.F." = (400 color(red)(cancel(color(black)("mL"))))/(10color(red)(cancel(color(black)("mL")))) = 40#</mathjax></p>
</blockquote>
<p>This means that the stock solution is <mathjax>#40#</mathjax> <strong>times more concentrated</strong> than the diluted solution. Since the diluted solution is said to have a <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of <mathjax>#"0.5 M"#</mathjax>, you can say that the stock solution had a molarity of </p>
<blockquote>
<p><mathjax>#c_"stock" = 40 * "0.5 M" = color(green)(|bar(ul(color(white)(a/a)"20 M"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The dilution factor method works because it's based on the <em>underlying principle</em> of a dilution, i.e. the concentration of the solution is <strong>decreased</strong> by <strong>Increasing</strong> the volume of the solution while keeping the <em>number of moles</em> of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>constant</strong>. </p>
<p>The equation for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a> looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(overbrace(c_1 xx V_1)^(color(brown)("moles of solute in concentrated solution")) = overbrace(c_2 xx V_2)^(color(brown)("moles of solute in diluted solution"))#</mathjax></p>
</blockquote>
<p>Here </p>
<p><mathjax>#c_1#</mathjax>, <mathjax>#V_1#</mathjax> - the molarity and volume of the concentrated solution<br/>
<mathjax>#c_2#</mathjax>, <mathjax>#V_2#</mathjax> - the molarity and volume of the diluted solution</p>
<p>This equation can be rearranged as </p>
<blockquote>
<p><mathjax>#c_1/c_2 = V_2/V_1#</mathjax></p>
</blockquote>
<p>This is where the dilution factor comes into play</p>
<blockquote>
<p><mathjax>#"D.F." = V_2/V_1#</mathjax></p>
</blockquote>
<p>Plug this into the above equation to get</p>
<blockquote>
<p><mathjax>#"D.F." = c_1/c_2 implies color(purple)(|bar(ul(color(white)(a/a)color(black)(c_1 = "D.F." xx c_2)color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
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</article> | What is the molarity of a stock solution if 10 mL is diluted to 400 mL with a concentration of 0.5M? | null |
580 | a8de9f9d-6ddd-11ea-b062-ccda262736ce | https://socratic.org/questions/if-20-ml-of-1-0-m-hcl-is-used-completely-to-neutralize-40-ml-of-an-naoh-solution | 0.33 M | start physical_unit 16 17 molarity mol/l qc_end physical_unit 6 6 4 5 molarity qc_end physical_unit 17 17 1 2 volume qc_end physical_unit 16 17 12 13 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] NaOH solution [IN] M"}] | [{"type":"physical unit","value":"0.33 M"}] | [{"type":"physical unit","value":"Molarity [OF] HCl solution [=] \\pu{1.0 M}"},{"type":"physical unit","value":"Volume [OF] HCl solution [=] \\pu{20 mL}"},{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{40 mL}"}] | <h1 class="questionTitle" itemprop="name">If 20 mL of 1.0 M HCl is used completely to neutralize 40 mL of an NaOH solution, what is the molarity of the NaOH solution? </h1> | null | 0.33 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><img alt="" src="https://useruploads.socratic.org/WqZf2InKSzuNfBgLBSZk_XCV.png"/> <br/>
<img alt="My own work (Mert METİN)" src="https://useruploads.socratic.org/X0Lq2ZYySIKUtExpwJe2_photo5857484331545635247.jpg"/> </p></div>
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<div class="markdown"><p>0.33 M</p></div>
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<div class="markdown"><p><img alt="" src="https://useruploads.socratic.org/WqZf2InKSzuNfBgLBSZk_XCV.png"/> <br/>
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<h1 class="questionTitle" itemprop="name">If 20 mL of 1.0 M HCl is used completely to neutralize 40 mL of an NaOH solution, what is the molarity of the NaOH solution? </h1>
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<div class="markdown"><p>0.33 M</p></div>
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<div class="markdown"><p><img alt="" src="https://useruploads.socratic.org/WqZf2InKSzuNfBgLBSZk_XCV.png"/> <br/>
<img alt="My own work (Mert METİN)" src="https://useruploads.socratic.org/X0Lq2ZYySIKUtExpwJe2_photo5857484331545635247.jpg"/> </p></div>
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</article> | If 20 mL of 1.0 M HCl is used completely to neutralize 40 mL of an NaOH solution, what is the molarity of the NaOH solution? | null |
581 | aad81539-6ddd-11ea-8e73-ccda262736ce | https://socratic.org/questions/if-you-mixed-three-gases-8-g-of-helium-2-g-of-nitrogen-and-16-g-of-oxygen-and-th | 6.22 atm | start physical_unit 8 8 partial_pressure atm qc_end physical_unit 8 8 5 6 mass qc_end physical_unit 12 12 9 10 mass qc_end physical_unit 17 17 14 15 mass qc_end physical_unit 3 4 23 24 total_pressure qc_end end | [{"type":"physical unit","value":"Partial pressure [OF] helium [IN] atm"}] | [{"type":"physical unit","value":"6.22 atm"}] | [{"type":"physical unit","value":"Mass [OF] helium [=] \\pu{8 g}"},{"type":"physical unit","value":"Mass [OF] nitrogen [=] \\pu{2 g}"},{"type":"physical unit","value":"Mass [OF] oxygen [=] \\pu{16 g}"},{"type":"physical unit","value":"Total pressure [OF] three gases [=] \\pu{10.0 atm}"}] | <h1 class="questionTitle" itemprop="name">If you mixed three gases: 8 g of helium, 2 g of nitrogen, and 16 g of oxygen and the total pressure was 10.0 atm, what is the partial pressure of helium?</h1> | null | 6.22 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that according to <strong>Dalton's Law of Partial Pressures</strong>, the <a href="http://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of a gas that's part of a gaseous mixture depends on two things </p>
<blockquote>
<ul>
<li><em>the <strong>mole fraction</strong> of said gas in the mixture</em></li>
<li><em>the <strong>total pressure</strong> of the mixture</em></li>
</ul>
</blockquote>
<p>Mathematically, this is written as </p>
<blockquote>
<p><mathjax>#color(blue)(P_A = chi_A xx P_"total")" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_A#</mathjax> - the <a href="http://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of gas <mathjax>#A#</mathjax><br/>
<mathjax>#chi_A#</mathjax> - the <em>mole fraction</em> of gas <mathjax>#A#</mathjax> in the mixture<br/>
<mathjax>#P_"total"#</mathjax> - the total pressure of the mixture</p>
<p>Simply put, the partial pressures of the gases that are part of a <em>gaseous mixture</em> are <strong>proportional</strong> to their number of moles. The more moles a specific gas will have, the <em>higher</em> its partial pressure will be. </p>
<p>So, in essence, all you have to do here is figure out how many moles of each gas you have present in the mixture, since the total pressure is said to be equal to <mathjax>#"10.0 atm"#</mathjax>. </p>
<p>So, use the <strong>molar masses</strong> of the three gases to determine how many moles of each are present. Keep in mind that nitrogen gas and oxygen gas are <strong>diatomic molecules</strong>, <mathjax>#"N"_2#</mathjax> and <mathjax>#"O"_2#</mathjax>, respectively. </p>
<p>You will thus have</p>
<blockquote>
<p><mathjax>#8 color(red)(cancel(color(black)("g"))) * "1 mole He"/(4.003color(red)(cancel(color(black)("g")))) = "2.00 moles He"#</mathjax></p>
<p><mathjax>#2 color(red)(cancel(color(black)("g"))) * "1 mole N"_2/(14.007color(red)(cancel(color(black)("g")))) = "0.0714 moles N"_2#</mathjax></p>
<p><mathjax>#16 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))) = "0.500 moles O"_2#</mathjax></p>
</blockquote>
<p>The <strong>total number of moles</strong> present in the mixture will be </p>
<blockquote>
<p><mathjax>#n_"total" = n_(He) + n_(N_2) + n_(O_2)#</mathjax></p>
<p><mathjax>#n_"total" = 2.00 + 0.714 + 0.500 = "3.214 moles"#</mathjax></p>
</blockquote>
<p>The <em>mole fraction</em> of helium, which is defined as the number of moles of helium divided by the <strong>total number of moles</strong> present in the mixture, will be equal to </p>
<blockquote>
<p><mathjax>#chi_(He) = (2.00 color(red)(cancel(color(black)("moles"))))/(3.214color(red)(cancel(color(black)("moles")))) = 0.6223#</mathjax></p>
</blockquote>
<p>Therefore, the partial pressure of helium will be </p>
<blockquote>
<p><mathjax>#P_(He) = 0.6223 xx "10.0 atm"#</mathjax></p>
<p><mathjax>#P_(He) = "6.223 atm"#</mathjax></p>
</blockquote>
<p>Rounded to one <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a>, the number of sig figs you have for the masses of helium and nitrogen gas, the answer will be </p>
<blockquote>
<p><mathjax>#P_(He) = color(green)("6 atm")#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#P_(He) = "6 atm"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that according to <strong>Dalton's Law of Partial Pressures</strong>, the <a href="http://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of a gas that's part of a gaseous mixture depends on two things </p>
<blockquote>
<ul>
<li><em>the <strong>mole fraction</strong> of said gas in the mixture</em></li>
<li><em>the <strong>total pressure</strong> of the mixture</em></li>
</ul>
</blockquote>
<p>Mathematically, this is written as </p>
<blockquote>
<p><mathjax>#color(blue)(P_A = chi_A xx P_"total")" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_A#</mathjax> - the <a href="http://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of gas <mathjax>#A#</mathjax><br/>
<mathjax>#chi_A#</mathjax> - the <em>mole fraction</em> of gas <mathjax>#A#</mathjax> in the mixture<br/>
<mathjax>#P_"total"#</mathjax> - the total pressure of the mixture</p>
<p>Simply put, the partial pressures of the gases that are part of a <em>gaseous mixture</em> are <strong>proportional</strong> to their number of moles. The more moles a specific gas will have, the <em>higher</em> its partial pressure will be. </p>
<p>So, in essence, all you have to do here is figure out how many moles of each gas you have present in the mixture, since the total pressure is said to be equal to <mathjax>#"10.0 atm"#</mathjax>. </p>
<p>So, use the <strong>molar masses</strong> of the three gases to determine how many moles of each are present. Keep in mind that nitrogen gas and oxygen gas are <strong>diatomic molecules</strong>, <mathjax>#"N"_2#</mathjax> and <mathjax>#"O"_2#</mathjax>, respectively. </p>
<p>You will thus have</p>
<blockquote>
<p><mathjax>#8 color(red)(cancel(color(black)("g"))) * "1 mole He"/(4.003color(red)(cancel(color(black)("g")))) = "2.00 moles He"#</mathjax></p>
<p><mathjax>#2 color(red)(cancel(color(black)("g"))) * "1 mole N"_2/(14.007color(red)(cancel(color(black)("g")))) = "0.0714 moles N"_2#</mathjax></p>
<p><mathjax>#16 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))) = "0.500 moles O"_2#</mathjax></p>
</blockquote>
<p>The <strong>total number of moles</strong> present in the mixture will be </p>
<blockquote>
<p><mathjax>#n_"total" = n_(He) + n_(N_2) + n_(O_2)#</mathjax></p>
<p><mathjax>#n_"total" = 2.00 + 0.714 + 0.500 = "3.214 moles"#</mathjax></p>
</blockquote>
<p>The <em>mole fraction</em> of helium, which is defined as the number of moles of helium divided by the <strong>total number of moles</strong> present in the mixture, will be equal to </p>
<blockquote>
<p><mathjax>#chi_(He) = (2.00 color(red)(cancel(color(black)("moles"))))/(3.214color(red)(cancel(color(black)("moles")))) = 0.6223#</mathjax></p>
</blockquote>
<p>Therefore, the partial pressure of helium will be </p>
<blockquote>
<p><mathjax>#P_(He) = 0.6223 xx "10.0 atm"#</mathjax></p>
<p><mathjax>#P_(He) = "6.223 atm"#</mathjax></p>
</blockquote>
<p>Rounded to one <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a>, the number of sig figs you have for the masses of helium and nitrogen gas, the answer will be </p>
<blockquote>
<p><mathjax>#P_(He) = color(green)("6 atm")#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If you mixed three gases: 8 g of helium, 2 g of nitrogen, and 16 g of oxygen and the total pressure was 10.0 atm, what is the partial pressure of helium?</h1>
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Stefan V.
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Jan 10, 2016
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<div class="markdown"><p><mathjax>#P_(He) = "6 atm"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that according to <strong>Dalton's Law of Partial Pressures</strong>, the <a href="http://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of a gas that's part of a gaseous mixture depends on two things </p>
<blockquote>
<ul>
<li><em>the <strong>mole fraction</strong> of said gas in the mixture</em></li>
<li><em>the <strong>total pressure</strong> of the mixture</em></li>
</ul>
</blockquote>
<p>Mathematically, this is written as </p>
<blockquote>
<p><mathjax>#color(blue)(P_A = chi_A xx P_"total")" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_A#</mathjax> - the <a href="http://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of gas <mathjax>#A#</mathjax><br/>
<mathjax>#chi_A#</mathjax> - the <em>mole fraction</em> of gas <mathjax>#A#</mathjax> in the mixture<br/>
<mathjax>#P_"total"#</mathjax> - the total pressure of the mixture</p>
<p>Simply put, the partial pressures of the gases that are part of a <em>gaseous mixture</em> are <strong>proportional</strong> to their number of moles. The more moles a specific gas will have, the <em>higher</em> its partial pressure will be. </p>
<p>So, in essence, all you have to do here is figure out how many moles of each gas you have present in the mixture, since the total pressure is said to be equal to <mathjax>#"10.0 atm"#</mathjax>. </p>
<p>So, use the <strong>molar masses</strong> of the three gases to determine how many moles of each are present. Keep in mind that nitrogen gas and oxygen gas are <strong>diatomic molecules</strong>, <mathjax>#"N"_2#</mathjax> and <mathjax>#"O"_2#</mathjax>, respectively. </p>
<p>You will thus have</p>
<blockquote>
<p><mathjax>#8 color(red)(cancel(color(black)("g"))) * "1 mole He"/(4.003color(red)(cancel(color(black)("g")))) = "2.00 moles He"#</mathjax></p>
<p><mathjax>#2 color(red)(cancel(color(black)("g"))) * "1 mole N"_2/(14.007color(red)(cancel(color(black)("g")))) = "0.0714 moles N"_2#</mathjax></p>
<p><mathjax>#16 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))) = "0.500 moles O"_2#</mathjax></p>
</blockquote>
<p>The <strong>total number of moles</strong> present in the mixture will be </p>
<blockquote>
<p><mathjax>#n_"total" = n_(He) + n_(N_2) + n_(O_2)#</mathjax></p>
<p><mathjax>#n_"total" = 2.00 + 0.714 + 0.500 = "3.214 moles"#</mathjax></p>
</blockquote>
<p>The <em>mole fraction</em> of helium, which is defined as the number of moles of helium divided by the <strong>total number of moles</strong> present in the mixture, will be equal to </p>
<blockquote>
<p><mathjax>#chi_(He) = (2.00 color(red)(cancel(color(black)("moles"))))/(3.214color(red)(cancel(color(black)("moles")))) = 0.6223#</mathjax></p>
</blockquote>
<p>Therefore, the partial pressure of helium will be </p>
<blockquote>
<p><mathjax>#P_(He) = 0.6223 xx "10.0 atm"#</mathjax></p>
<p><mathjax>#P_(He) = "6.223 atm"#</mathjax></p>
</blockquote>
<p>Rounded to one <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a>, the number of sig figs you have for the masses of helium and nitrogen gas, the answer will be </p>
<blockquote>
<p><mathjax>#P_(He) = color(green)("6 atm")#</mathjax></p>
</blockquote></div>
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</article> | If you mixed three gases: 8 g of helium, 2 g of nitrogen, and 16 g of oxygen and the total pressure was 10.0 atm, what is the partial pressure of helium? | null |
582 | a9130cc0-6ddd-11ea-a6dc-ccda262736ce | https://socratic.org/questions/the-average-mass-of-a-dime-coin-is-2-28-g-and-the-mass-of-an-automobile-is-2-0-x | 6.9 × 10^17 | start physical_unit 26 26 number none qc_end physical_unit 4 6 8 9 mass qc_end physical_unit 14 15 17 20 mass qc_end physical_unit 37 37 34 35 mole qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Number [OF] automobiles"}] | [{"type":"physical unit","value":"6.9 × 10^17"}] | [{"type":"physical unit","value":"Average mass [OF] a dime coin [=] \\pu{2.28 g}"},{"type":"physical unit","value":"Mass [OF] an automobile [=] \\pu{2.0 × 10^3 kg}"},{"type":"physical unit","value":"Mole [OF] dimes [=] \\pu{1.0 mole}"},{"type":"other","value":"Automobiles' total mass is the same as dimes."}] | <h1 class="questionTitle" itemprop="name">The average mass of a dime coin is #"2.28 g"# and the mass of an automobile is #2.0 * 10^3# #"kg"#. What is the number of automobiles whose total mass is the same as #1.0# mole of dimes?</h1> | null | 6.9 × 10^17 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For starters, you should know that in order to have <mathjax>#1#</mathjax> <strong>mole</strong> of dime coins, you need to have <mathjax>#6.022 * 10^(23)#</mathjax> <strong>dime coins</strong>.</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("1 mole of dime coins" = 6.022 * 10^(23)color(white)(.)"dime coins"))) ->#</mathjax> <em><strong>Avogadro's constant</strong></em></p>
</blockquote>
<p>Now, you know that the mass of single dime coin is equal to </p>
<blockquote>
<p><mathjax>#2.28 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = 2.28 * 10^(-3)color(white)(.)"kg"#</mathjax></p>
</blockquote>
<p>Use the mass of a single dime coin to calculate the mass of <mathjax>#1.0#</mathjax> <strong>mole</strong> of dime coins</p>
<blockquote>
<p><mathjax>#6.022 * 10^(23)color(red)(cancel(color(black)("dime coins"))) * (2.28 * 10^(-3)color(white)(.)"kg")/(1color(red)(cancel(color(black)("dime coin")))) = 1.373 * 10^(19)color(white)(.)"kg"#</mathjax></p>
</blockquote>
<p>The mass of an automobile is equal to <mathjax>#2.0 * 10^3#</mathjax> <mathjax>#"kg"#</mathjax>, so you can say that the number of automobiles that will be <strong>equal</strong> to the mass of <mathjax>#1.0#</mathjax> <strong>mole</strong> of dime coins will be</p>
<blockquote>
<p><mathjax>#1.373 * 10^(19) color(red)(cancel(color(black)("kg"))) * "1 automobile"/(2.0 * 10^3 color(red)(cancel(color(black)("kg")))) = color(darkgreen)(ul(color(black)(6.9 * 10^(17)color(white)(.)"automobiles")))#</mathjax> </p>
</blockquote>
<p>The answer is rounded to two <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>#6.9 * 10^(17)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For starters, you should know that in order to have <mathjax>#1#</mathjax> <strong>mole</strong> of dime coins, you need to have <mathjax>#6.022 * 10^(23)#</mathjax> <strong>dime coins</strong>.</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("1 mole of dime coins" = 6.022 * 10^(23)color(white)(.)"dime coins"))) ->#</mathjax> <em><strong>Avogadro's constant</strong></em></p>
</blockquote>
<p>Now, you know that the mass of single dime coin is equal to </p>
<blockquote>
<p><mathjax>#2.28 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = 2.28 * 10^(-3)color(white)(.)"kg"#</mathjax></p>
</blockquote>
<p>Use the mass of a single dime coin to calculate the mass of <mathjax>#1.0#</mathjax> <strong>mole</strong> of dime coins</p>
<blockquote>
<p><mathjax>#6.022 * 10^(23)color(red)(cancel(color(black)("dime coins"))) * (2.28 * 10^(-3)color(white)(.)"kg")/(1color(red)(cancel(color(black)("dime coin")))) = 1.373 * 10^(19)color(white)(.)"kg"#</mathjax></p>
</blockquote>
<p>The mass of an automobile is equal to <mathjax>#2.0 * 10^3#</mathjax> <mathjax>#"kg"#</mathjax>, so you can say that the number of automobiles that will be <strong>equal</strong> to the mass of <mathjax>#1.0#</mathjax> <strong>mole</strong> of dime coins will be</p>
<blockquote>
<p><mathjax>#1.373 * 10^(19) color(red)(cancel(color(black)("kg"))) * "1 automobile"/(2.0 * 10^3 color(red)(cancel(color(black)("kg")))) = color(darkgreen)(ul(color(black)(6.9 * 10^(17)color(white)(.)"automobiles")))#</mathjax> </p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">The average mass of a dime coin is #"2.28 g"# and the mass of an automobile is #2.0 * 10^3# #"kg"#. What is the number of automobiles whose total mass is the same as #1.0# mole of dimes?</h1>
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Stefan V.
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Aug 20, 2017
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<div class="markdown"><p><mathjax>#6.9 * 10^(17)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For starters, you should know that in order to have <mathjax>#1#</mathjax> <strong>mole</strong> of dime coins, you need to have <mathjax>#6.022 * 10^(23)#</mathjax> <strong>dime coins</strong>.</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("1 mole of dime coins" = 6.022 * 10^(23)color(white)(.)"dime coins"))) ->#</mathjax> <em><strong>Avogadro's constant</strong></em></p>
</blockquote>
<p>Now, you know that the mass of single dime coin is equal to </p>
<blockquote>
<p><mathjax>#2.28 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = 2.28 * 10^(-3)color(white)(.)"kg"#</mathjax></p>
</blockquote>
<p>Use the mass of a single dime coin to calculate the mass of <mathjax>#1.0#</mathjax> <strong>mole</strong> of dime coins</p>
<blockquote>
<p><mathjax>#6.022 * 10^(23)color(red)(cancel(color(black)("dime coins"))) * (2.28 * 10^(-3)color(white)(.)"kg")/(1color(red)(cancel(color(black)("dime coin")))) = 1.373 * 10^(19)color(white)(.)"kg"#</mathjax></p>
</blockquote>
<p>The mass of an automobile is equal to <mathjax>#2.0 * 10^3#</mathjax> <mathjax>#"kg"#</mathjax>, so you can say that the number of automobiles that will be <strong>equal</strong> to the mass of <mathjax>#1.0#</mathjax> <strong>mole</strong> of dime coins will be</p>
<blockquote>
<p><mathjax>#1.373 * 10^(19) color(red)(cancel(color(black)("kg"))) * "1 automobile"/(2.0 * 10^3 color(red)(cancel(color(black)("kg")))) = color(darkgreen)(ul(color(black)(6.9 * 10^(17)color(white)(.)"automobiles")))#</mathjax> </p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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SCooke
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<div class="markdown"><p><mathjax># 6.86 xx 10^17#</mathjax> cars</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#2.28 (grams)/(unit) * 6.022 x 10^23#</mathjax> units/mole<mathjax># = 13.7 xx 10^23#</mathjax> grams/mole<br/>
<mathjax>#2.0 xx 10^3 kg * 1000g/"kg" = 2.0 xx 10^6 g/(car)#</mathjax><br/>
<mathjax>#(13.7 xx 10^23)/(2.0 xx 10^6) = 6.86 xx 10^17#</mathjax> cars</p></div>
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</article> | The average mass of a dime coin is #"2.28 g"# and the mass of an automobile is #2.0 * 10^3# #"kg"#. What is the number of automobiles whose total mass is the same as #1.0# mole of dimes? | null |
583 | aa20b3f6-6ddd-11ea-94f1-ccda262736ce | https://socratic.org/questions/at-22-c-a-sample-of-nitrogen-gas-occupies-8-0-l-what-volume-will-the-nitrogen-oc | 14.18 L | start physical_unit 4 7 volume l qc_end physical_unit 4 7 9 10 volume qc_end physical_unit 4 7 1 2 temperature qc_end physical_unit 4 7 18 19 temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] nitrogen gas sample [IN] L"}] | [{"type":"physical unit","value":"14.18 L"}] | [{"type":"physical unit","value":"Volume1 [OF] nitrogen gas sample [=] \\pu{8.0 L}"},{"type":"physical unit","value":"Temperature1 [OF] nitrogen gas sample [=] \\pu{22 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] nitrogen gas sample [=] \\pu{250 ℃}"}] | <h1 class="questionTitle" itemprop="name">At 22°C, a sample of nitrogen gas occupies 8.0 L. What volume will the nitrogen occupy at 250°C? </h1> | null | 14.18 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assume that the gas is enclosed in a piston, expanding against a constant pressure.</p>
<p><mathjax>#V_2=V_1/T_1xxT_2#</mathjax>, where <mathjax>#T#</mathjax> is specified to be <mathjax>#"temperatura assoluta"#</mathjax></p>
<p><mathjax>#=8.0*Lxx(523*K)/(295*K)~=14*L#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#V_1/T_1=V_2/T_2...........#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assume that the gas is enclosed in a piston, expanding against a constant pressure.</p>
<p><mathjax>#V_2=V_1/T_1xxT_2#</mathjax>, where <mathjax>#T#</mathjax> is specified to be <mathjax>#"temperatura assoluta"#</mathjax></p>
<p><mathjax>#=8.0*Lxx(523*K)/(295*K)~=14*L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">At 22°C, a sample of nitrogen gas occupies 8.0 L. What volume will the nitrogen occupy at 250°C? </h1>
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<div class="markdown"><p>We assume that the gas is enclosed in a piston, expanding against a constant pressure.</p>
<p><mathjax>#V_2=V_1/T_1xxT_2#</mathjax>, where <mathjax>#T#</mathjax> is specified to be <mathjax>#"temperatura assoluta"#</mathjax></p>
<p><mathjax>#=8.0*Lxx(523*K)/(295*K)~=14*L#</mathjax></p></div>
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</article> | At 22°C, a sample of nitrogen gas occupies 8.0 L. What volume will the nitrogen occupy at 250°C? | null |
584 | ab02f546-6ddd-11ea-8d78-ccda262736ce | https://socratic.org/questions/30000-tonnes-of-a-copper-ore-containing-6-5-chalcopyrite-was-mined-calculate-the | 675 tonnes | start physical_unit 4 4 theoretical_yield t qc_end physical_unit 4 5 0 1 mass qc_end end | [{"type":"physical unit","value":"Theoretical yield [OF] copper [IN] tonnes"}] | [{"type":"physical unit","value":"675 tonnes"}] | [{"type":"physical unit","value":"Mass [OF] copper ore [=] \\pu{30000 tonnes}"},{"type":"physical unit","value":"Percent [OF] chalcopyrite in copper ore [=] \\pu{6.5%}"}] | <h1 class="questionTitle" itemprop="name">30000 tonnes of a copper ore containing 6.5% chalcopyrite was mined. Calculate the theoretical yield of copper that could be extracted from this ore (in tonnes)?</h1> | null | 675 tonnes | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus percentage mass of copper in chalcopyrite is:</p>
<p><mathjax>#(63.55*g*mol^-1)/((63.55+55.8+2xx32.06)*g*mol^-1)xx100%#</mathjax></p>
<p><mathjax>#34.6%#</mathjax> <mathjax>#"by mass."#</mathjax></p>
<p>But, the ore contains ONLY <mathjax>#6.5%#</mathjax> by mass of copper in this ore......</p>
<p>And thus, the mass of copper is given by............</p>
<p><mathjax>#(6.5)/100xx(34.6)/100xx30000*"tonnes"=675*"tonnes"#</mathjax>.</p>
<p>And of course the rest of the starting mass will either be released to the atmosphere, or end up in a mountainous pile of tailings. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Chalcopyrite"#</mathjax> has a formula of <mathjax>#"CuFeS"_2#</mathjax> (you should have included this formula in the question). </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus percentage mass of copper in chalcopyrite is:</p>
<p><mathjax>#(63.55*g*mol^-1)/((63.55+55.8+2xx32.06)*g*mol^-1)xx100%#</mathjax></p>
<p><mathjax>#34.6%#</mathjax> <mathjax>#"by mass."#</mathjax></p>
<p>But, the ore contains ONLY <mathjax>#6.5%#</mathjax> by mass of copper in this ore......</p>
<p>And thus, the mass of copper is given by............</p>
<p><mathjax>#(6.5)/100xx(34.6)/100xx30000*"tonnes"=675*"tonnes"#</mathjax>.</p>
<p>And of course the rest of the starting mass will either be released to the atmosphere, or end up in a mountainous pile of tailings. </p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">30000 tonnes of a copper ore containing 6.5% chalcopyrite was mined. Calculate the theoretical yield of copper that could be extracted from this ore (in tonnes)?</h1>
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anor277
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Apr 21, 2017
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<div class="markdown"><p><mathjax>#"Chalcopyrite"#</mathjax> has a formula of <mathjax>#"CuFeS"_2#</mathjax> (you should have included this formula in the question). </p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus percentage mass of copper in chalcopyrite is:</p>
<p><mathjax>#(63.55*g*mol^-1)/((63.55+55.8+2xx32.06)*g*mol^-1)xx100%#</mathjax></p>
<p><mathjax>#34.6%#</mathjax> <mathjax>#"by mass."#</mathjax></p>
<p>But, the ore contains ONLY <mathjax>#6.5%#</mathjax> by mass of copper in this ore......</p>
<p>And thus, the mass of copper is given by............</p>
<p><mathjax>#(6.5)/100xx(34.6)/100xx30000*"tonnes"=675*"tonnes"#</mathjax>.</p>
<p>And of course the rest of the starting mass will either be released to the atmosphere, or end up in a mountainous pile of tailings. </p></div>
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</article> | 30000 tonnes of a copper ore containing 6.5% chalcopyrite was mined. Calculate the theoretical yield of copper that could be extracted from this ore (in tonnes)? | null |
585 | ab18fccd-6ddd-11ea-9ac3-ccda262736ce | https://socratic.org/questions/how-would-you-balance-the-following-equation-hi-g-h2-g-i2-g | 2 HI(g) -> H2(g) + I2(g) | start chemical_equation qc_end chemical_equation 7 11 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the following equation"}] | [{"type":"chemical equation","value":"2 HI(g) -> H2(g) + I2(g)"}] | [{"type":"chemical equation","value":"HI(g) -> H2(g) + I2(g)"}] | <h1 class="questionTitle" itemprop="name">How would you balance the following equation: HI(g) → H2(g) + I2(g)?</h1> | null | 2 HI(g) -> H2(g) + I2(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There are 2 Hydrogen atoms and 2 Iodine atoms on the right so the have to be 2 of each on the left as well, so there are <mathjax>#2#</mathjax> <mathjax>#HI#</mathjax> molecules on the left.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#2HI(g) -> H2(g) + I2(g)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There are 2 Hydrogen atoms and 2 Iodine atoms on the right so the have to be 2 of each on the left as well, so there are <mathjax>#2#</mathjax> <mathjax>#HI#</mathjax> molecules on the left.</p></div>
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<h1 class="questionTitle" itemprop="name">How would you balance the following equation: HI(g) → H2(g) + I2(g)?</h1>
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Cameron G.
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<div class="markdown"><p><mathjax>#2HI(g) -> H2(g) + I2(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>There are 2 Hydrogen atoms and 2 Iodine atoms on the right so the have to be 2 of each on the left as well, so there are <mathjax>#2#</mathjax> <mathjax>#HI#</mathjax> molecules on the left.</p></div>
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</article> | How would you balance the following equation: HI(g) → H2(g) + I2(g)? | null |
586 | a94347d5-6ddd-11ea-9f7a-ccda262736ce | https://socratic.org/questions/56077fdd581e2a2b21078d98 | 78.98 grams | start physical_unit 4 5 mass g qc_end physical_unit 15 17 9 10 volume qc_end physical_unit 15 17 13 14 molarity qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] silver chloride [IN] grams"}] | [{"type":"physical unit","value":"78.98 grams"}] | [{"type":"physical unit","value":"Volume [OF] silver nitrate solution [=] \\pu{1.92 L}"},{"type":"physical unit","value":"Molarity [OF] silver nitrate solution [=] \\pu{0.287 M}"},{"type":"other","value":"Enough magnesium chloride."}] | <h1 class="questionTitle" itemprop="name">How many grams of silver chloride are produced when #"1.92 L"# of a #"0.287-M"# silver nitrate solution reacts with enough magnesium chloride?</h1> | null | 78.98 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, you know that you're dealing with a <a href="http://socratic.org/chemistry/chemical-reactions/double-replacement-reactions">double replacement reaction</a> between a solution of <em>silver nitrate</em>, <mathjax>#"AgNO"_3#</mathjax>, and solution of <em>magnesium chloride</em>, <mathjax>#"MgCl"""_2#</mathjax></p>
<blockquote>
<p><mathjax>#2"AgNO"_text(3(aq]) + "MgCl"_text(2(aq]) -> 2"AgCl"_text((s]) darr + "Mg"("NO"_3)_text(2(aq])#</mathjax></p>
</blockquote>
<p>Notice that you have a <mathjax>#1:1#</mathjax> (<mathjax>#2:2#</mathjax>) <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between silver nitrate and silver chloride. This tells you that the reaction will rpoduce <strong>the number of moles</strong> of silver chloride as the number of moles of silver nitrate that reacted. </p>
<p>In your case, you know the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and the volume of the silver nitrate solution, which means that you an find the number of moles that react</p>
<blockquote>
<p><mathjax>#C = n/V implies n = C * V#</mathjax></p>
<p><mathjax>#n_(AgNO_3) = "0.287 M" * "1.92 L" = "0.55104 moles"#</mathjax></p>
</blockquote>
<p>The aforementioned mole ratio tells you that you get</p>
<blockquote>
<p><mathjax>#0.55104color(red)(cancel(color(black)("moles AgNO"""_3))) * "1 mole AgCl"/(1color(red)(cancel(color(black)("mole AgNO"""_3)))) = "0.55104 moles AgCl"#</mathjax></p>
</blockquote>
<p>Use silver chloride's molar mass to get the number of grams that would contain this many moles</p>
<blockquote>
<p><mathjax>#0.55104color(red)(cancel(color(black)("moles AgCl"))) * "143.32 g"/(color(red)(cancel(color(black)("moleAgCl")))) = "78.975 g"#</mathjax></p>
</blockquote>
<p>Rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be </p>
<blockquote>
<p><mathjax>#m_(AgCl) = color(green)("79.0 g")#</mathjax></p>
</blockquote></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"79.0 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, you know that you're dealing with a <a href="http://socratic.org/chemistry/chemical-reactions/double-replacement-reactions">double replacement reaction</a> between a solution of <em>silver nitrate</em>, <mathjax>#"AgNO"_3#</mathjax>, and solution of <em>magnesium chloride</em>, <mathjax>#"MgCl"""_2#</mathjax></p>
<blockquote>
<p><mathjax>#2"AgNO"_text(3(aq]) + "MgCl"_text(2(aq]) -> 2"AgCl"_text((s]) darr + "Mg"("NO"_3)_text(2(aq])#</mathjax></p>
</blockquote>
<p>Notice that you have a <mathjax>#1:1#</mathjax> (<mathjax>#2:2#</mathjax>) <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between silver nitrate and silver chloride. This tells you that the reaction will rpoduce <strong>the number of moles</strong> of silver chloride as the number of moles of silver nitrate that reacted. </p>
<p>In your case, you know the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and the volume of the silver nitrate solution, which means that you an find the number of moles that react</p>
<blockquote>
<p><mathjax>#C = n/V implies n = C * V#</mathjax></p>
<p><mathjax>#n_(AgNO_3) = "0.287 M" * "1.92 L" = "0.55104 moles"#</mathjax></p>
</blockquote>
<p>The aforementioned mole ratio tells you that you get</p>
<blockquote>
<p><mathjax>#0.55104color(red)(cancel(color(black)("moles AgNO"""_3))) * "1 mole AgCl"/(1color(red)(cancel(color(black)("mole AgNO"""_3)))) = "0.55104 moles AgCl"#</mathjax></p>
</blockquote>
<p>Use silver chloride's molar mass to get the number of grams that would contain this many moles</p>
<blockquote>
<p><mathjax>#0.55104color(red)(cancel(color(black)("moles AgCl"))) * "143.32 g"/(color(red)(cancel(color(black)("moleAgCl")))) = "78.975 g"#</mathjax></p>
</blockquote>
<p>Rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be </p>
<blockquote>
<p><mathjax>#m_(AgCl) = color(green)("79.0 g")#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many grams of silver chloride are produced when #"1.92 L"# of a #"0.287-M"# silver nitrate solution reacts with enough magnesium chloride?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"79.0 g"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, you know that you're dealing with a <a href="http://socratic.org/chemistry/chemical-reactions/double-replacement-reactions">double replacement reaction</a> between a solution of <em>silver nitrate</em>, <mathjax>#"AgNO"_3#</mathjax>, and solution of <em>magnesium chloride</em>, <mathjax>#"MgCl"""_2#</mathjax></p>
<blockquote>
<p><mathjax>#2"AgNO"_text(3(aq]) + "MgCl"_text(2(aq]) -> 2"AgCl"_text((s]) darr + "Mg"("NO"_3)_text(2(aq])#</mathjax></p>
</blockquote>
<p>Notice that you have a <mathjax>#1:1#</mathjax> (<mathjax>#2:2#</mathjax>) <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between silver nitrate and silver chloride. This tells you that the reaction will rpoduce <strong>the number of moles</strong> of silver chloride as the number of moles of silver nitrate that reacted. </p>
<p>In your case, you know the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and the volume of the silver nitrate solution, which means that you an find the number of moles that react</p>
<blockquote>
<p><mathjax>#C = n/V implies n = C * V#</mathjax></p>
<p><mathjax>#n_(AgNO_3) = "0.287 M" * "1.92 L" = "0.55104 moles"#</mathjax></p>
</blockquote>
<p>The aforementioned mole ratio tells you that you get</p>
<blockquote>
<p><mathjax>#0.55104color(red)(cancel(color(black)("moles AgNO"""_3))) * "1 mole AgCl"/(1color(red)(cancel(color(black)("mole AgNO"""_3)))) = "0.55104 moles AgCl"#</mathjax></p>
</blockquote>
<p>Use silver chloride's molar mass to get the number of grams that would contain this many moles</p>
<blockquote>
<p><mathjax>#0.55104color(red)(cancel(color(black)("moles AgCl"))) * "143.32 g"/(color(red)(cancel(color(black)("moleAgCl")))) = "78.975 g"#</mathjax></p>
</blockquote>
<p>Rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be </p>
<blockquote>
<p><mathjax>#m_(AgCl) = color(green)("79.0 g")#</mathjax></p>
</blockquote></div>
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</article> | How many grams of silver chloride are produced when #"1.92 L"# of a #"0.287-M"# silver nitrate solution reacts with enough magnesium chloride? | null |
587 | acacfc25-6ddd-11ea-bd33-ccda262736ce | https://socratic.org/questions/how-do-you-calculate-the-empirical-formula-of-phosgenite | Pb2CO3Cl2 | start chemical_formula qc_end substance 8 8 qc_end end | [{"type":"other","value":"Chemical Formula [OF] phosgenite [IN] empirical"}] | [{"type":"chemical equation","value":"Pb2CO3Cl2"}] | [{"type":"substance name","value":"Phosgenite"}] | <h1 class="questionTitle" itemprop="name">How do you calculate the empirical formula of phosgenite?</h1> | null | Pb2CO3Cl2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Phosgenite is a mineral. The empirical formula (which in this case is also the molecular formula) is shown. I don't know what you meant by "calculate" - you can look up the formula for the compound: <a href="http://www.geologypage.com/2014/06/phosgenite.html" rel="nofollow" target="_blank">http://www.geologypage.com/2014/06/phosgenite.html</a></p>
<p>Experimentally the formula would be determined by different reactions to observe the amounts of lead, carbonate and chloride in the molecule. These would then be combined with the known oxidation states and likely ions of the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> to derive the empirical formula. A comparison of the empirical formula to the molecular weight will derive the molecular weight.</p></div>
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<div class="markdown"><p><mathjax>#Pb_2CO_3Cl_2#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Phosgenite is a mineral. The empirical formula (which in this case is also the molecular formula) is shown. I don't know what you meant by "calculate" - you can look up the formula for the compound: <a href="http://www.geologypage.com/2014/06/phosgenite.html" rel="nofollow" target="_blank">http://www.geologypage.com/2014/06/phosgenite.html</a></p>
<p>Experimentally the formula would be determined by different reactions to observe the amounts of lead, carbonate and chloride in the molecule. These would then be combined with the known oxidation states and likely ions of the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> to derive the empirical formula. A comparison of the empirical formula to the molecular weight will derive the molecular weight.</p></div>
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<h1 class="questionTitle" itemprop="name">How do you calculate the empirical formula of phosgenite?</h1>
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<div class="markdown"><p><mathjax>#Pb_2CO_3Cl_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Phosgenite is a mineral. The empirical formula (which in this case is also the molecular formula) is shown. I don't know what you meant by "calculate" - you can look up the formula for the compound: <a href="http://www.geologypage.com/2014/06/phosgenite.html" rel="nofollow" target="_blank">http://www.geologypage.com/2014/06/phosgenite.html</a></p>
<p>Experimentally the formula would be determined by different reactions to observe the amounts of lead, carbonate and chloride in the molecule. These would then be combined with the known oxidation states and likely ions of the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> to derive the empirical formula. A comparison of the empirical formula to the molecular weight will derive the molecular weight.</p></div>
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</article> | How do you calculate the empirical formula of phosgenite? | null |
588 | a8d7d09d-6ddd-11ea-8843-ccda262736ce | https://socratic.org/questions/5908c5f9b72cff4cc31b7526 | 1.16 grams | start physical_unit 11 12 percent_yield g qc_end physical_unit 11 12 5 5 percent_yield qc_end physical_unit 21 22 16 17 volume qc_end physical_unit 21 22 19 20 molarity qc_end physical_unit 29 30 24 25 volume qc_end physical_unit 29 30 27 28 molarity qc_end end | [{"type":"physical unit","value":"Yield [OF] silver chloride [IN] grams"}] | [{"type":"physical unit","value":"1.16 grams"}] | [{"type":"physical unit","value":"Percentage yield [OF] silver chloride [=] \\pu{77.3%}"},{"type":"physical unit","value":"Volume [OF] silver nitrate [=] \\pu{16.7 mL}"},{"type":"physical unit","value":"Molarity [OF] silver nitrate [=] \\pu{0.7 mol/L}"},{"type":"physical unit","value":"Volume [OF] sodium chloride [=] \\pu{1.7 mL}"},{"type":"physical unit","value":"Molarity [OF] sodium chloride [=] \\pu{0.668 mol/L}"}] | <h1 class="questionTitle" itemprop="name">If the percentage yield is 77.3 %, what is the yield of silver chloride when you
mix 16.7 mL of 0.7 mol/L silver nitrate with 1.7 mL of 0.668 mol/L sodium chloride?</h1> | null | 1.16 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>This is a limiting reactant problem.</p>
<p>We know that we will need a balanced equation, masses, and moles, etc., so let's gather all the information in one place.</p>
<p><mathjax>#M_text(r):color(white)(mmmmmmmmmmmmmmm)143.32#</mathjax><br/>
<mathjax>#color(white)(mmmmmmm)"NaCl" + color(white)(ll)"AgNO"_3 → "AgCl" + "NaNO"_3#</mathjax><br/>
<mathjax>#"Moles:"color(white)(mmml)"0.0012"color(white)(mll)"0.010 49"#</mathjax><br/>
<mathjax>#"Divide by:"color(white)(mmm)1color(white)(mmmmll)1#</mathjax><br/>
<mathjax>#"Moles rxn:"color(white)(ml)0.0012color(white)(mll)"0.010 49"#</mathjax></p>
<p><mathjax>#"Moles of NaCl" = 0.0165 color(red)(cancel(color(black)("L NaCl"))) × "0.7 mol NaCl"/(1 color(red)(cancel(color(black)("L NaCl")))) = "0.0012 mol NaCl"#</mathjax></p>
<p><mathjax>#"Moles of AgNO"_3 = 0.0157 color(red)(cancel(color(black)("L AgNO"_3))) × ("0.668 mol AgNO"_3)/(1 color(red)(cancel(color(black)("L AgNO"_3)))) = "0.010 49 mol AgNO"_3#</mathjax></p>
<blockquote></blockquote>
<p><strong>Identify the limiting reactant</strong></p>
<p>An easy way to identify the limiting reactant is to calculate the "moles of reaction" each will produce.</p>
<p>You simply divide the number of moles by the corresponding coefficient in the balanced equation.</p>
<p>I did that for you in the chart above.</p>
<p><mathjax>#"AgNO"_3"#</mathjax> is the limiting reactant because it gives the fewest "moles of reaction".</p>
<blockquote></blockquote>
<p><strong>Calculate the theoretical yield</strong></p>
<p><mathjax>#"Theor. yield" = "0.010 49" color(red)(cancel(color(black)("mol AgCl"))) × (1 color(red)(cancel(color(black)("mol AgCl"))))/(1 color(red)(cancel(color(black)("mol AgNO"_3)))) × ("143.32 g AgCl")/(1 color(red)(cancel(color(black)("mol AgCl")))) = "1.5031 g AgCl"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Calculate the experimental yield</strong></p>
<p><mathjax>#"Experimental yield" = 0.773 × "1.5031 g" = "1.16 g"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The experimental yield is 1.16 g of <mathjax>#"AgCl"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>This is a limiting reactant problem.</p>
<p>We know that we will need a balanced equation, masses, and moles, etc., so let's gather all the information in one place.</p>
<p><mathjax>#M_text(r):color(white)(mmmmmmmmmmmmmmm)143.32#</mathjax><br/>
<mathjax>#color(white)(mmmmmmm)"NaCl" + color(white)(ll)"AgNO"_3 → "AgCl" + "NaNO"_3#</mathjax><br/>
<mathjax>#"Moles:"color(white)(mmml)"0.0012"color(white)(mll)"0.010 49"#</mathjax><br/>
<mathjax>#"Divide by:"color(white)(mmm)1color(white)(mmmmll)1#</mathjax><br/>
<mathjax>#"Moles rxn:"color(white)(ml)0.0012color(white)(mll)"0.010 49"#</mathjax></p>
<p><mathjax>#"Moles of NaCl" = 0.0165 color(red)(cancel(color(black)("L NaCl"))) × "0.7 mol NaCl"/(1 color(red)(cancel(color(black)("L NaCl")))) = "0.0012 mol NaCl"#</mathjax></p>
<p><mathjax>#"Moles of AgNO"_3 = 0.0157 color(red)(cancel(color(black)("L AgNO"_3))) × ("0.668 mol AgNO"_3)/(1 color(red)(cancel(color(black)("L AgNO"_3)))) = "0.010 49 mol AgNO"_3#</mathjax></p>
<blockquote></blockquote>
<p><strong>Identify the limiting reactant</strong></p>
<p>An easy way to identify the limiting reactant is to calculate the "moles of reaction" each will produce.</p>
<p>You simply divide the number of moles by the corresponding coefficient in the balanced equation.</p>
<p>I did that for you in the chart above.</p>
<p><mathjax>#"AgNO"_3"#</mathjax> is the limiting reactant because it gives the fewest "moles of reaction".</p>
<blockquote></blockquote>
<p><strong>Calculate the theoretical yield</strong></p>
<p><mathjax>#"Theor. yield" = "0.010 49" color(red)(cancel(color(black)("mol AgCl"))) × (1 color(red)(cancel(color(black)("mol AgCl"))))/(1 color(red)(cancel(color(black)("mol AgNO"_3)))) × ("143.32 g AgCl")/(1 color(red)(cancel(color(black)("mol AgCl")))) = "1.5031 g AgCl"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Calculate the experimental yield</strong></p>
<p><mathjax>#"Experimental yield" = 0.773 × "1.5031 g" = "1.16 g"#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">If the percentage yield is 77.3 %, what is the yield of silver chloride when you
mix 16.7 mL of 0.7 mol/L silver nitrate with 1.7 mL of 0.668 mol/L sodium chloride?</h1>
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<div class="markdown"><p>The experimental yield is 1.16 g of <mathjax>#"AgCl"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>This is a limiting reactant problem.</p>
<p>We know that we will need a balanced equation, masses, and moles, etc., so let's gather all the information in one place.</p>
<p><mathjax>#M_text(r):color(white)(mmmmmmmmmmmmmmm)143.32#</mathjax><br/>
<mathjax>#color(white)(mmmmmmm)"NaCl" + color(white)(ll)"AgNO"_3 → "AgCl" + "NaNO"_3#</mathjax><br/>
<mathjax>#"Moles:"color(white)(mmml)"0.0012"color(white)(mll)"0.010 49"#</mathjax><br/>
<mathjax>#"Divide by:"color(white)(mmm)1color(white)(mmmmll)1#</mathjax><br/>
<mathjax>#"Moles rxn:"color(white)(ml)0.0012color(white)(mll)"0.010 49"#</mathjax></p>
<p><mathjax>#"Moles of NaCl" = 0.0165 color(red)(cancel(color(black)("L NaCl"))) × "0.7 mol NaCl"/(1 color(red)(cancel(color(black)("L NaCl")))) = "0.0012 mol NaCl"#</mathjax></p>
<p><mathjax>#"Moles of AgNO"_3 = 0.0157 color(red)(cancel(color(black)("L AgNO"_3))) × ("0.668 mol AgNO"_3)/(1 color(red)(cancel(color(black)("L AgNO"_3)))) = "0.010 49 mol AgNO"_3#</mathjax></p>
<blockquote></blockquote>
<p><strong>Identify the limiting reactant</strong></p>
<p>An easy way to identify the limiting reactant is to calculate the "moles of reaction" each will produce.</p>
<p>You simply divide the number of moles by the corresponding coefficient in the balanced equation.</p>
<p>I did that for you in the chart above.</p>
<p><mathjax>#"AgNO"_3"#</mathjax> is the limiting reactant because it gives the fewest "moles of reaction".</p>
<blockquote></blockquote>
<p><strong>Calculate the theoretical yield</strong></p>
<p><mathjax>#"Theor. yield" = "0.010 49" color(red)(cancel(color(black)("mol AgCl"))) × (1 color(red)(cancel(color(black)("mol AgCl"))))/(1 color(red)(cancel(color(black)("mol AgNO"_3)))) × ("143.32 g AgCl")/(1 color(red)(cancel(color(black)("mol AgCl")))) = "1.5031 g AgCl"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Calculate the experimental yield</strong></p>
<p><mathjax>#"Experimental yield" = 0.773 × "1.5031 g" = "1.16 g"#</mathjax></p></div>
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</article> | If the percentage yield is 77.3 %, what is the yield of silver chloride when you
mix 16.7 mL of 0.7 mol/L silver nitrate with 1.7 mL of 0.668 mol/L sodium chloride? | null |
589 | aaac5176-6ddd-11ea-982e-ccda262736ce | https://socratic.org/questions/how-to-complete-this-reaction-hno3-h2o | HNO3 + H2O -> HNO3(aq) | start chemical_equation qc_end chemical_equation 5 5 qc_end chemical_equation 7 7 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reaction"}] | [{"type":"chemical equation","value":"HNO3 + H2O -> HNO3(aq) "}] | [{"type":"chemical equation","value":"HNO3"},{"type":"chemical equation","value":"H2O"}] | <h1 class="questionTitle" itemprop="name">How to complete this reaction?
HNO3 + H2O ?</h1> | null | HNO3 + H2O -> HNO3(aq) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In English: nitric acid and water form a solution, it then solvates into its ions in the solution since <mathjax>#HNO_3#</mathjax> is soluble.</p></div>
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<div class="markdown"><p><mathjax>#HNO_3 + H_2O to HNO_3(aq) to H^+ +NO^(3-)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In English: nitric acid and water form a solution, it then solvates into its ions in the solution since <mathjax>#HNO_3#</mathjax> is soluble.</p></div>
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<h1 class="questionTitle" itemprop="name">How to complete this reaction?
HNO3 + H2O ?</h1>
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<div class="markdown"><p><mathjax>#HNO_3 + H_2O to HNO_3(aq) to H^+ +NO^(3-)#</mathjax></p></div>
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<div class="markdown"><p>In English: nitric acid and water form a solution, it then solvates into its ions in the solution since <mathjax>#HNO_3#</mathjax> is soluble.</p></div>
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</article> | How to complete this reaction?
HNO3 + H2O ? | null |
590 | aa588339-6ddd-11ea-ab21-ccda262736ce | https://socratic.org/questions/582cc55611ef6b078d47eea6 | 15 mol/L | start physical_unit 5 7 concentration mol/l qc_end substance 5 7 qc_end end | [{"type":"physical unit","value":"Concentration [OF] concentrated ammonia solution [IN] mol/L"}] | [{"type":"physical unit","value":"15 mol/L"}] | [{"type":"substance name","value":"Concentrated ammonia solution"}] | <h1 class="questionTitle" itemprop="name">What is the concentration of concentrated ammonia solution?</h1> | null | 15 mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/questions/ammonia-in-the-form-of-bubbled-gas-is-combined-with-water-to-form-aqueous-ammoni">See this older answer.</a> </p>
<p>I think Fisher still sells this as ammonium hydroxide, which is a bit of a misnomer, because the actual reagent is <mathjax>#NH_3*H_2O#</mathjax>, an aqueous solution of ammonia. If you get a whiff of these ammonia <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a>, it will bring literal tears to your eyes. </p></div>
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<div class="markdown"><p>Conc. ammonia is <mathjax>#15*mol*L^-1#</mathjax>.</p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/questions/ammonia-in-the-form-of-bubbled-gas-is-combined-with-water-to-form-aqueous-ammoni">See this older answer.</a> </p>
<p>I think Fisher still sells this as ammonium hydroxide, which is a bit of a misnomer, because the actual reagent is <mathjax>#NH_3*H_2O#</mathjax>, an aqueous solution of ammonia. If you get a whiff of these ammonia <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a>, it will bring literal tears to your eyes. </p></div>
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<div class="markdown"><p>Conc. ammonia is <mathjax>#15*mol*L^-1#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/questions/ammonia-in-the-form-of-bubbled-gas-is-combined-with-water-to-form-aqueous-ammoni">See this older answer.</a> </p>
<p>I think Fisher still sells this as ammonium hydroxide, which is a bit of a misnomer, because the actual reagent is <mathjax>#NH_3*H_2O#</mathjax>, an aqueous solution of ammonia. If you get a whiff of these ammonia <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a>, it will bring literal tears to your eyes. </p></div>
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</article> | What is the concentration of concentrated ammonia solution? | null |
591 | abaa428a-6ddd-11ea-a46d-ccda262736ce | https://socratic.org/questions/how-many-moles-are-in-4-81-10-24-formula-units-of-licl | 7.99 moles | start physical_unit 11 11 mole mol qc_end end | [{"type":"physical unit","value":"Mole [OF] LiCl [IN] moles"}] | [{"type":"physical unit","value":"7.99 moles"}] | [{"type":"physical unit","value":"Number [OF] LiCl formula units [=] \\pu{4.81 × 10^24}"}] | <h1 class="questionTitle" itemprop="name">How many moles are in #4.81 * 10^24# formula units of #LiCl#?</h1> | null | 7.99 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Thus with regard to your problem there are:</p>
<p><mathjax>#(4.81xx10^24" formula units of LiCl")/(6.022xx10^23" formula units of LiCl"*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#mol#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>One mole of stuff consists of <mathjax>#6.022140857(74)xx10^23#</mathjax> individual items of that stuff.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Thus with regard to your problem there are:</p>
<p><mathjax>#(4.81xx10^24" formula units of LiCl")/(6.022xx10^23" formula units of LiCl"*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#mol#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many moles are in #4.81 * 10^24# formula units of #LiCl#?</h1>
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anor277
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<span class="dateCreated" datetime="2016-04-29T19:03:04" itemprop="dateCreated">
Apr 29, 2016
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<div class="markdown"><p>One mole of stuff consists of <mathjax>#6.022140857(74)xx10^23#</mathjax> individual items of that stuff.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Thus with regard to your problem there are:</p>
<p><mathjax>#(4.81xx10^24" formula units of LiCl")/(6.022xx10^23" formula units of LiCl"*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#mol#</mathjax></p></div>
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</article> | How many moles are in #4.81 * 10^24# formula units of #LiCl#? | null |
592 | ab7d23e8-6ddd-11ea-a150-ccda262736ce | https://socratic.org/questions/if-2-4-10-5-l-of-gas-is-at-180-mmhg-what-is-the-pressure-when-the-gas-is-compres | 2.4 × 10^4 mmHg | start physical_unit 16 17 pressure mmhg qc_end physical_unit 16 17 1 4 volume qc_end physical_unit 16 17 9 10 pressure qc_end physical_unit 16 17 21 24 volume qc_end c_other constant_temperature qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the gas [IN] mmHg"}] | [{"type":"physical unit","value":"2.4 × 10^4 mmHg"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{2.4 × 10^5 L}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{180 mmHg}"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{1.8 × 10^3 L}"},{"type":"other","value":"ConstantTemperature"}] | <h1 class="questionTitle" itemprop="name">If #2.4*10^5# #L# of gas is at #180# #mmHg#, what is the pressure when the gas is compressed to #1.8*10^3# #L# at constant temperature?</h1> | null | 2.4 × 10^4 mmHg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's start off with identifying our known and unknown variables. <br/>
The first volume we have is <mathjax>#2.4xx10^(5)#</mathjax> L, the first pressure is 180 mmHg, and the second volume is <mathjax>#1.8xx10^(3)#</mathjax>. Our only unknown is the second pressure.</p>
<p>We can obtain the answer using <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a> which shows that there is an inverse relationship between pressure and volume as long as the temperature and number of moles remain constant.</p>
<p>The equation we use is <mathjax>#P_1V_1=P_2V_2#</mathjax><br/>
where the numbers 1 and 2 represent the first and second conditions. All we have to do is rearrange the equation to solve for the pressure. </p>
<p>We do this by dividing both sides by <mathjax>#V_2#</mathjax> in order to get <mathjax>#P_2#</mathjax> by itself like so:<br/>
<mathjax>#P_2=(P_1xxV_1)/V_2#</mathjax></p>
<p>Now all we do is plug and chug!<br/>
<mathjax>#P_2=(180\mmHg xx 2.4xx10^(5)\ cancel"L")/(1.8xx10^(3)\cancel"L")#</mathjax> = <mathjax>#2.4xx10^(4)#</mathjax> <mathjax>#mmHg#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p>The new pressure is <mathjax>#2.4xx10^(4)#</mathjax> <mathjax>#mmHg#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's start off with identifying our known and unknown variables. <br/>
The first volume we have is <mathjax>#2.4xx10^(5)#</mathjax> L, the first pressure is 180 mmHg, and the second volume is <mathjax>#1.8xx10^(3)#</mathjax>. Our only unknown is the second pressure.</p>
<p>We can obtain the answer using <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a> which shows that there is an inverse relationship between pressure and volume as long as the temperature and number of moles remain constant.</p>
<p>The equation we use is <mathjax>#P_1V_1=P_2V_2#</mathjax><br/>
where the numbers 1 and 2 represent the first and second conditions. All we have to do is rearrange the equation to solve for the pressure. </p>
<p>We do this by dividing both sides by <mathjax>#V_2#</mathjax> in order to get <mathjax>#P_2#</mathjax> by itself like so:<br/>
<mathjax>#P_2=(P_1xxV_1)/V_2#</mathjax></p>
<p>Now all we do is plug and chug!<br/>
<mathjax>#P_2=(180\mmHg xx 2.4xx10^(5)\ cancel"L")/(1.8xx10^(3)\cancel"L")#</mathjax> = <mathjax>#2.4xx10^(4)#</mathjax> <mathjax>#mmHg#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If #2.4*10^5# #L# of gas is at #180# #mmHg#, what is the pressure when the gas is compressed to #1.8*10^3# #L# at constant temperature?</h1>
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Jun 8, 2016
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<div class="markdown"><p>The new pressure is <mathjax>#2.4xx10^(4)#</mathjax> <mathjax>#mmHg#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's start off with identifying our known and unknown variables. <br/>
The first volume we have is <mathjax>#2.4xx10^(5)#</mathjax> L, the first pressure is 180 mmHg, and the second volume is <mathjax>#1.8xx10^(3)#</mathjax>. Our only unknown is the second pressure.</p>
<p>We can obtain the answer using <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a> which shows that there is an inverse relationship between pressure and volume as long as the temperature and number of moles remain constant.</p>
<p>The equation we use is <mathjax>#P_1V_1=P_2V_2#</mathjax><br/>
where the numbers 1 and 2 represent the first and second conditions. All we have to do is rearrange the equation to solve for the pressure. </p>
<p>We do this by dividing both sides by <mathjax>#V_2#</mathjax> in order to get <mathjax>#P_2#</mathjax> by itself like so:<br/>
<mathjax>#P_2=(P_1xxV_1)/V_2#</mathjax></p>
<p>Now all we do is plug and chug!<br/>
<mathjax>#P_2=(180\mmHg xx 2.4xx10^(5)\ cancel"L")/(1.8xx10^(3)\cancel"L")#</mathjax> = <mathjax>#2.4xx10^(4)#</mathjax> <mathjax>#mmHg#</mathjax></p></div>
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</article> | If #2.4*10^5# #L# of gas is at #180# #mmHg#, what is the pressure when the gas is compressed to #1.8*10^3# #L# at constant temperature? | null |
593 | ab3ae919-6ddd-11ea-a7de-ccda262736ce | https://socratic.org/questions/urea-nh-3-2co-is-dissolved-in-100-g-of-water-the-solution-freezes-at-0-085c-how- | 0.27 grams | start physical_unit 0 0 mass g qc_end physical_unit 8 8 5 6 mass qc_end physical_unit 9 10 13 14 freezing_point_temperature qc_end physical_unit 8 8 31 31 kf qc_end physical_unit 1 1 34 34 kb qc_end end | [{"type":"physical unit","value":"Mass [OF] urea [IN] grams"}] | [{"type":"physical unit","value":"0.27 grams"}] | [{"type":"physical unit","value":"Mass [OF] water [=] \\pu{100 g}"},{"type":"physical unit","value":"Freezing point [OF] the solution [=] \\pu{-0.085 ℃}"},{"type":"physical unit","value":"Kf [OF] water [=] \\pu{1.858}"},{"type":"physical unit","value":"Kb [OF] (NH2)2CO [=] \\pu{0.512}"}] | <h1 class="questionTitle" itemprop="name">Urea, #("NH"_2)_2"CO"#, is dissolved in #"100 g"# of water. The solution freezes at #-0.085^@"C"#. How many grams of urea were dissolved to make this solution?
</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>(Given <mathjax>#K_f#</mathjax> of water = 1.858 <mathjax>#K_b#</mathjax> = 0.512)</p></div>
</h2>
</div>
</div> | 0.27 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the equation that allows you to calculate the <strong>freezing point depression</strong> of the solution</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(DeltaT_"f" = i * K_f * b)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#T_"f"#</mathjax> is the <strong>freezing point depression</strong></li>
<li><mathjax>#i#</mathjax> is the <strong>van't Hoff factor</strong></li>
<li><mathjax>#b#</mathjax> is the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of the solution</li>
<li><mathjax>#K_f#</mathjax> is the <strong>cryoscopic constant</strong> of water, equal to<mathjax># 1.858^@"C kg mol"^(-1)#</mathjax></li>
</ul>
</blockquote>
<p>Now, the <strong>freezing point depression</strong> tells you the difference between the freezing point of the pure <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> and the freezing point of the solution.</p>
<p>In other words, the freezing point depression tells you by how many degrees the freezing point of the solution <strong>decreases</strong> compared with the freezing point of the pure solvent. </p>
<p>You know that your solution freezes at <mathjax>#-0.085^@"C"#</mathjax>. Since pure water has a normal freezing point of <mathjax>#0^@"C"#</mathjax>, you can say that the <strong>freezing point depression</strong> will be</p>
<blockquote>
<p><mathjax>#DeltaT_"f" = 0^@"C" - (-0.085^@"C")#</mathjax></p>
<p><mathjax>#DeltaT_"f" = 0.085^@"C"#</mathjax></p>
<blockquote>
<p>This essentially means that the freezing point of the solution is <mathjax>#0.085^@"C"#</mathjax> <strong>lower</strong> than the normal freezing point of the pure solvent. </p>
</blockquote>
</blockquote>
<p>So, rearrange the equation to find the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of the solution.</p>
<blockquote>
<p><mathjax>#b = (DeltaT_"f")/(i * K_f)#</mathjax></p>
</blockquote>
<p>Since urea is a <strong>nonelectrolyte</strong>, which means that it does not dissociate in aqueous solution, its van't Hoff factor will be equal to <mathjax>#1#</mathjax>. </p>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#b = (0.085 color(red)(cancel(color(black)(""^@"C"))))/(1 * 1.858 color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = "0.04575 mol kg"^(-1)#</mathjax></p>
</blockquote>
<p>This tells you that this solution contains <mathjax>#0.04575#</mathjax> <strong>moles</strong> of urea for every <mathjax>#"1 kg" = 10^3 quad "g"#</mathjax> of solvent. Since your sample contains <mathjax>#"100 g"#</mathjax> of water, you can say that it will also contain</p>
<blockquote>
<p><mathjax>#100 color(red)(cancel(color(black)("g water"))) * "0.04575 moles urea"/(10^3color(red)(cancel(color(black)("g water")))) = "0.004575 moles urea"#</mathjax></p>
</blockquote>
<p>Finally, to convert the number of grams to <em>moles</em>, use the <strong>molar mass</strong> of urea</p>
<blockquote>
<p><mathjax>#0.004575 color(red)(cancel(color(black)("moles urea"))) * "60.06 g"/(1color(red)(cancel(color(black)("mole urea")))) = color(darkgreen)(ul(color(black)("0.27 g")))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but don't forget that you have one significant figure for the mass of water. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.27 g"#</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 equation that allows you to calculate the <strong>freezing point depression</strong> of the solution</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(DeltaT_"f" = i * K_f * b)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#T_"f"#</mathjax> is the <strong>freezing point depression</strong></li>
<li><mathjax>#i#</mathjax> is the <strong>van't Hoff factor</strong></li>
<li><mathjax>#b#</mathjax> is the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of the solution</li>
<li><mathjax>#K_f#</mathjax> is the <strong>cryoscopic constant</strong> of water, equal to<mathjax># 1.858^@"C kg mol"^(-1)#</mathjax></li>
</ul>
</blockquote>
<p>Now, the <strong>freezing point depression</strong> tells you the difference between the freezing point of the pure <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> and the freezing point of the solution.</p>
<p>In other words, the freezing point depression tells you by how many degrees the freezing point of the solution <strong>decreases</strong> compared with the freezing point of the pure solvent. </p>
<p>You know that your solution freezes at <mathjax>#-0.085^@"C"#</mathjax>. Since pure water has a normal freezing point of <mathjax>#0^@"C"#</mathjax>, you can say that the <strong>freezing point depression</strong> will be</p>
<blockquote>
<p><mathjax>#DeltaT_"f" = 0^@"C" - (-0.085^@"C")#</mathjax></p>
<p><mathjax>#DeltaT_"f" = 0.085^@"C"#</mathjax></p>
<blockquote>
<p>This essentially means that the freezing point of the solution is <mathjax>#0.085^@"C"#</mathjax> <strong>lower</strong> than the normal freezing point of the pure solvent. </p>
</blockquote>
</blockquote>
<p>So, rearrange the equation to find the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of the solution.</p>
<blockquote>
<p><mathjax>#b = (DeltaT_"f")/(i * K_f)#</mathjax></p>
</blockquote>
<p>Since urea is a <strong>nonelectrolyte</strong>, which means that it does not dissociate in aqueous solution, its van't Hoff factor will be equal to <mathjax>#1#</mathjax>. </p>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#b = (0.085 color(red)(cancel(color(black)(""^@"C"))))/(1 * 1.858 color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = "0.04575 mol kg"^(-1)#</mathjax></p>
</blockquote>
<p>This tells you that this solution contains <mathjax>#0.04575#</mathjax> <strong>moles</strong> of urea for every <mathjax>#"1 kg" = 10^3 quad "g"#</mathjax> of solvent. Since your sample contains <mathjax>#"100 g"#</mathjax> of water, you can say that it will also contain</p>
<blockquote>
<p><mathjax>#100 color(red)(cancel(color(black)("g water"))) * "0.04575 moles urea"/(10^3color(red)(cancel(color(black)("g water")))) = "0.004575 moles urea"#</mathjax></p>
</blockquote>
<p>Finally, to convert the number of grams to <em>moles</em>, use the <strong>molar mass</strong> of urea</p>
<blockquote>
<p><mathjax>#0.004575 color(red)(cancel(color(black)("moles urea"))) * "60.06 g"/(1color(red)(cancel(color(black)("mole urea")))) = color(darkgreen)(ul(color(black)("0.27 g")))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but don't forget that you have one significant figure for the mass of water. </p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">Urea, #("NH"_2)_2"CO"#, is dissolved in #"100 g"# of water. The solution freezes at #-0.085^@"C"#. How many grams of urea were dissolved to make this solution?
</h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>(Given <mathjax>#K_f#</mathjax> of water = 1.858 <mathjax>#K_b#</mathjax> = 0.512)</p></div>
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Stefan V.
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<div class="markdown"><p><mathjax>#"0.27 g"#</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 equation that allows you to calculate the <strong>freezing point depression</strong> of the solution</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(DeltaT_"f" = i * K_f * b)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#T_"f"#</mathjax> is the <strong>freezing point depression</strong></li>
<li><mathjax>#i#</mathjax> is the <strong>van't Hoff factor</strong></li>
<li><mathjax>#b#</mathjax> is the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of the solution</li>
<li><mathjax>#K_f#</mathjax> is the <strong>cryoscopic constant</strong> of water, equal to<mathjax># 1.858^@"C kg mol"^(-1)#</mathjax></li>
</ul>
</blockquote>
<p>Now, the <strong>freezing point depression</strong> tells you the difference between the freezing point of the pure <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> and the freezing point of the solution.</p>
<p>In other words, the freezing point depression tells you by how many degrees the freezing point of the solution <strong>decreases</strong> compared with the freezing point of the pure solvent. </p>
<p>You know that your solution freezes at <mathjax>#-0.085^@"C"#</mathjax>. Since pure water has a normal freezing point of <mathjax>#0^@"C"#</mathjax>, you can say that the <strong>freezing point depression</strong> will be</p>
<blockquote>
<p><mathjax>#DeltaT_"f" = 0^@"C" - (-0.085^@"C")#</mathjax></p>
<p><mathjax>#DeltaT_"f" = 0.085^@"C"#</mathjax></p>
<blockquote>
<p>This essentially means that the freezing point of the solution is <mathjax>#0.085^@"C"#</mathjax> <strong>lower</strong> than the normal freezing point of the pure solvent. </p>
</blockquote>
</blockquote>
<p>So, rearrange the equation to find the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of the solution.</p>
<blockquote>
<p><mathjax>#b = (DeltaT_"f")/(i * K_f)#</mathjax></p>
</blockquote>
<p>Since urea is a <strong>nonelectrolyte</strong>, which means that it does not dissociate in aqueous solution, its van't Hoff factor will be equal to <mathjax>#1#</mathjax>. </p>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#b = (0.085 color(red)(cancel(color(black)(""^@"C"))))/(1 * 1.858 color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = "0.04575 mol kg"^(-1)#</mathjax></p>
</blockquote>
<p>This tells you that this solution contains <mathjax>#0.04575#</mathjax> <strong>moles</strong> of urea for every <mathjax>#"1 kg" = 10^3 quad "g"#</mathjax> of solvent. Since your sample contains <mathjax>#"100 g"#</mathjax> of water, you can say that it will also contain</p>
<blockquote>
<p><mathjax>#100 color(red)(cancel(color(black)("g water"))) * "0.04575 moles urea"/(10^3color(red)(cancel(color(black)("g water")))) = "0.004575 moles urea"#</mathjax></p>
</blockquote>
<p>Finally, to convert the number of grams to <em>moles</em>, use the <strong>molar mass</strong> of urea</p>
<blockquote>
<p><mathjax>#0.004575 color(red)(cancel(color(black)("moles urea"))) * "60.06 g"/(1color(red)(cancel(color(black)("mole urea")))) = color(darkgreen)(ul(color(black)("0.27 g")))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but don't forget that you have one significant figure for the mass of water. </p></div>
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</article> | Urea, #("NH"_2)_2"CO"#, is dissolved in #"100 g"# of water. The solution freezes at #-0.085^@"C"#. How many grams of urea were dissolved to make this solution?
|
(Given #K_f# of water = 1.858 #K_b# = 0.512)
|
594 | a8b248b7-6ddd-11ea-8a10-ccda262736ce | https://socratic.org/questions/what-is-the-concentration-of-oh-in-pure-water | 10^(-7) mol/L | start physical_unit 5 5 concentration mol/l qc_end substance 7 8 qc_end end | [{"type":"physical unit","value":"Concentration [OF] OH- [IN] mol/L"}] | [{"type":"physical unit","value":"10^(-7) mol/L"}] | [{"type":"substance name","value":"Pure water"}] | <h1 class="questionTitle" itemprop="name">What is the concentration of OH- in pure water? </h1> | null | 10^(-7) mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Water undergoes autoprotolysis according to the following equation:</p>
<p><mathjax>#2H_2O rightleftharpoons H_3O^(+) + HO^-#</mathjax></p>
<p>Thru very careful measurement at <mathjax>#298*K#</mathjax> the following value for the ion product has been found to be:</p>
<p><mathjax>#[H_3O^+][HO^-]=10^-14#</mathjax></p>
<p>We could simplify this by taking <mathjax>#log_10#</mathjax> of each side, but clearly if the solution is neutral, then <mathjax>#[HO^-]=[H_3O^+]=10^-7*mol*L^-1#</mathjax>.</p>
<p>If we take logarithms, then we get the useful expression:</p>
<p><mathjax>#pH+pOH=14#</mathjax>.</p>
<p>See this old <a href="https://socratic.org/questions/how-do-you-find-poh-from-ph">answer for further details.</a> </p>
<p>At higher temperatures than <mathjax>#298K#</mathjax>, how do you think the equilibrium would evolve? Remember that this is a bond-breaking reaction. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#[HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^-7*mol*L^-1#</mathjax> at <mathjax>#298*K#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Water undergoes autoprotolysis according to the following equation:</p>
<p><mathjax>#2H_2O rightleftharpoons H_3O^(+) + HO^-#</mathjax></p>
<p>Thru very careful measurement at <mathjax>#298*K#</mathjax> the following value for the ion product has been found to be:</p>
<p><mathjax>#[H_3O^+][HO^-]=10^-14#</mathjax></p>
<p>We could simplify this by taking <mathjax>#log_10#</mathjax> of each side, but clearly if the solution is neutral, then <mathjax>#[HO^-]=[H_3O^+]=10^-7*mol*L^-1#</mathjax>.</p>
<p>If we take logarithms, then we get the useful expression:</p>
<p><mathjax>#pH+pOH=14#</mathjax>.</p>
<p>See this old <a href="https://socratic.org/questions/how-do-you-find-poh-from-ph">answer for further details.</a> </p>
<p>At higher temperatures than <mathjax>#298K#</mathjax>, how do you think the equilibrium would evolve? Remember that this is a bond-breaking reaction. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the concentration of OH- in pure water? </h1>
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anor277
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Feb 10, 2017
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<div class="markdown"><p><mathjax>#[HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^-7*mol*L^-1#</mathjax> at <mathjax>#298*K#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Water undergoes autoprotolysis according to the following equation:</p>
<p><mathjax>#2H_2O rightleftharpoons H_3O^(+) + HO^-#</mathjax></p>
<p>Thru very careful measurement at <mathjax>#298*K#</mathjax> the following value for the ion product has been found to be:</p>
<p><mathjax>#[H_3O^+][HO^-]=10^-14#</mathjax></p>
<p>We could simplify this by taking <mathjax>#log_10#</mathjax> of each side, but clearly if the solution is neutral, then <mathjax>#[HO^-]=[H_3O^+]=10^-7*mol*L^-1#</mathjax>.</p>
<p>If we take logarithms, then we get the useful expression:</p>
<p><mathjax>#pH+pOH=14#</mathjax>.</p>
<p>See this old <a href="https://socratic.org/questions/how-do-you-find-poh-from-ph">answer for further details.</a> </p>
<p>At higher temperatures than <mathjax>#298K#</mathjax>, how do you think the equilibrium would evolve? Remember that this is a bond-breaking reaction. </p></div>
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</article> | What is the concentration of OH- in pure water? | null |
595 | a86cf7b7-6ddd-11ea-b37c-ccda262736ce | https://socratic.org/questions/how-do-you-convert-64-3-g-pbbr-2-to-moles | 0.18 moles | start physical_unit 6 6 mole mol qc_end physical_unit 6 6 4 5 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] PbBr2 [IN] moles"}] | [{"type":"physical unit","value":"0.18 moles"}] | [{"type":"physical unit","value":"Mass [OF] PbBr2 [=] \\pu{64.3 g}"}] | <h1 class="questionTitle" itemprop="name">How do you convert 64.3 g #PbBr_2# to moles?</h1> | null | 0.18 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So the formula we are using is <mathjax>#n=m/M#</mathjax>.<br/>
=> Where <mathjax>#n#</mathjax> is the amount in mols.<br/>
=> Where <mathjax>#m#</mathjax> is the mass in grams.<br/>
=> Where <mathjax>#M#</mathjax> is the molar mass in <mathjax>#g/("mols")#</mathjax>.</p>
<p>First off, we need to find the molar mass of <mathjax>#PbBr_"2"#</mathjax>.</p>
<p>Using a Periodic Table, lead has a molar mass of <mathjax>#207.20 g/("mol")#</mathjax> and bromine has a molar mass of <mathjax>#79.90 g/("mol")#</mathjax>. Including the two atoms of bromine, when we add them all up, we get <mathjax>#367.00 g/("mol")#</mathjax>. </p>
<p>Now, we can plug them into the formula.</p>
<blockquote>
<p><mathjax>#n=m/M#</mathjax></p>
<p><mathjax>#n=64.3/367#</mathjax></p>
<p><mathjax>#n=0.175204359#</mathjax></p>
<p><mathjax>#n=0.175#</mathjax></p>
</blockquote>
<p>Therefore, <mathjax>#64.3#</mathjax> <mathjax>#"g"#</mathjax> of <mathjax>#PbBr_"2"#</mathjax> is <mathjax>#0.175#</mathjax> <mathjax>#"mols"#</mathjax>.</p>
<p>Hope this helps :)</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#64.3#</mathjax> <mathjax>#"g"#</mathjax> of <mathjax>#PbBr_"2"#</mathjax> is <mathjax>#0.175#</mathjax> <mathjax>#"mols"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So the formula we are using is <mathjax>#n=m/M#</mathjax>.<br/>
=> Where <mathjax>#n#</mathjax> is the amount in mols.<br/>
=> Where <mathjax>#m#</mathjax> is the mass in grams.<br/>
=> Where <mathjax>#M#</mathjax> is the molar mass in <mathjax>#g/("mols")#</mathjax>.</p>
<p>First off, we need to find the molar mass of <mathjax>#PbBr_"2"#</mathjax>.</p>
<p>Using a Periodic Table, lead has a molar mass of <mathjax>#207.20 g/("mol")#</mathjax> and bromine has a molar mass of <mathjax>#79.90 g/("mol")#</mathjax>. Including the two atoms of bromine, when we add them all up, we get <mathjax>#367.00 g/("mol")#</mathjax>. </p>
<p>Now, we can plug them into the formula.</p>
<blockquote>
<p><mathjax>#n=m/M#</mathjax></p>
<p><mathjax>#n=64.3/367#</mathjax></p>
<p><mathjax>#n=0.175204359#</mathjax></p>
<p><mathjax>#n=0.175#</mathjax></p>
</blockquote>
<p>Therefore, <mathjax>#64.3#</mathjax> <mathjax>#"g"#</mathjax> of <mathjax>#PbBr_"2"#</mathjax> is <mathjax>#0.175#</mathjax> <mathjax>#"mols"#</mathjax>.</p>
<p>Hope this helps :)</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you convert 64.3 g #PbBr_2# to moles?</h1>
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<div class="markdown"><p><mathjax>#64.3#</mathjax> <mathjax>#"g"#</mathjax> of <mathjax>#PbBr_"2"#</mathjax> is <mathjax>#0.175#</mathjax> <mathjax>#"mols"#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So the formula we are using is <mathjax>#n=m/M#</mathjax>.<br/>
=> Where <mathjax>#n#</mathjax> is the amount in mols.<br/>
=> Where <mathjax>#m#</mathjax> is the mass in grams.<br/>
=> Where <mathjax>#M#</mathjax> is the molar mass in <mathjax>#g/("mols")#</mathjax>.</p>
<p>First off, we need to find the molar mass of <mathjax>#PbBr_"2"#</mathjax>.</p>
<p>Using a Periodic Table, lead has a molar mass of <mathjax>#207.20 g/("mol")#</mathjax> and bromine has a molar mass of <mathjax>#79.90 g/("mol")#</mathjax>. Including the two atoms of bromine, when we add them all up, we get <mathjax>#367.00 g/("mol")#</mathjax>. </p>
<p>Now, we can plug them into the formula.</p>
<blockquote>
<p><mathjax>#n=m/M#</mathjax></p>
<p><mathjax>#n=64.3/367#</mathjax></p>
<p><mathjax>#n=0.175204359#</mathjax></p>
<p><mathjax>#n=0.175#</mathjax></p>
</blockquote>
<p>Therefore, <mathjax>#64.3#</mathjax> <mathjax>#"g"#</mathjax> of <mathjax>#PbBr_"2"#</mathjax> is <mathjax>#0.175#</mathjax> <mathjax>#"mols"#</mathjax>.</p>
<p>Hope this helps :)</p></div>
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</article> | How do you convert 64.3 g #PbBr_2# to moles? | null |
596 | a8e90044-6ddd-11ea-be70-ccda262736ce | https://socratic.org/questions/one-reaction-that-produces-hydrogen-gas-can-represented-by-the-following-unbalan | 182.3 g | start physical_unit 16 16 mass g qc_end chemical_equation 14 20 qc_end physical_unit 34 34 31 32 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] HCl [IN] g"}] | [{"type":"physical unit","value":"182.3 g"}] | [{"type":"chemical equation","value":"Mg(s) + HCl(aq) -> MgCl2(aq) + H2(g)"},{"type":"physical unit","value":"Mole [OF] magnesium [=] \\pu{2.50 moles}"}] | <h1 class="questionTitle" itemprop="name">One reaction that produces hydrogen gas can represented by the following unbalanced chemical equation: #Mg(s) + HCl(aq) -> MgCl_2(aq) + H_2(g)#. What mass of HCl is consumed by the reaction of 2.50 moles of magnesium?</h1> | null | 182.3 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Mg(s) + 2HCl(aq) rarr MgCl_2(aq) + H_2(g)uarr#</mathjax></p>
<p>And, given <mathjax>#2.5*mol#</mathjax> metal (and stoichiometric acid), CLEARLY, there are <mathjax>#5.0*mol#</mathjax> <mathjax>#HCl#</mathjax> consumed.........</p>
<p>And thus mass of <mathjax>#HCl=5.0*molxx36.46*g=182.3*g#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Well we need the stoichiometric equation..............and calculate that <mathjax>#182.3*g#</mathjax> of <mathjax>#HCl#</mathjax> are evolved..............</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Mg(s) + 2HCl(aq) rarr MgCl_2(aq) + H_2(g)uarr#</mathjax></p>
<p>And, given <mathjax>#2.5*mol#</mathjax> metal (and stoichiometric acid), CLEARLY, there are <mathjax>#5.0*mol#</mathjax> <mathjax>#HCl#</mathjax> consumed.........</p>
<p>And thus mass of <mathjax>#HCl=5.0*molxx36.46*g=182.3*g#</mathjax>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">One reaction that produces hydrogen gas can represented by the following unbalanced chemical equation: #Mg(s) + HCl(aq) -> MgCl_2(aq) + H_2(g)#. What mass of HCl is consumed by the reaction of 2.50 moles of magnesium?</h1>
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<div class="markdown"><p>Well we need the stoichiometric equation..............and calculate that <mathjax>#182.3*g#</mathjax> of <mathjax>#HCl#</mathjax> are evolved..............</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Mg(s) + 2HCl(aq) rarr MgCl_2(aq) + H_2(g)uarr#</mathjax></p>
<p>And, given <mathjax>#2.5*mol#</mathjax> metal (and stoichiometric acid), CLEARLY, there are <mathjax>#5.0*mol#</mathjax> <mathjax>#HCl#</mathjax> consumed.........</p>
<p>And thus mass of <mathjax>#HCl=5.0*molxx36.46*g=182.3*g#</mathjax>.</p></div>
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</article> | One reaction that produces hydrogen gas can represented by the following unbalanced chemical equation: #Mg(s) + HCl(aq) -> MgCl_2(aq) + H_2(g)#. What mass of HCl is consumed by the reaction of 2.50 moles of magnesium? | null |
597 | a9572037-6ddd-11ea-9617-ccda262736ce | https://socratic.org/questions/how-many-grams-of-naoh-are-required-to-prepare-200-ml-of-a-0-450-m-solution | 3.60 grams | start physical_unit 4 4 mass g qc_end physical_unit 15 15 9 10 volume qc_end physical_unit 4 4 13 14 molarity qc_end end | [{"type":"physical unit","value":"Mass [OF] NaOH [IN] grams"}] | [{"type":"physical unit","value":"3.60 grams"}] | [{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{200 mL}"},{"type":"physical unit","value":"Molarity [OF] NaOH solution [=] \\pu{0.450 M}"}] | <h1 class="questionTitle" itemprop="name">How many grams of #NaOH# are required to prepare 200 mL of a 0.450 M solution?</h1> | null | 3.60 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's begin with the equation for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a>:<br/>
<img alt="slideplayer.com" src="https://useruploads.socratic.org/4z0vwx7RRla5OizeNQHX_slide_3.jpg"/> <br/>
We are given the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> 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 200mL by 1000mL we will obtain a value of <strong>0.200 L.</strong></p>
<p>Rearrange the equation to solve for moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>:</p>
<p><strong>Moles of solute = Molarity <mathjax>#xx#</mathjax> Liters of solution</strong></p>
<p>Multiply 0.450 M by 0.200:</p>
<p><mathjax>#("0.450 mol")/("1 L") xx "0.200 L" = "0.09 mol"#</mathjax></p>
<p>To obtain the mass of solute, we will need to the molar mass of NaOH, which is 40.00 g/mol:</p>
<p>Finally, multiply the number of moles by 40.00 g/mol</p>
<p><mathjax>#0.09 cancel"mol" xx (40.00g)/(1cancel"mol")#</mathjax> <mathjax>#=3.60g#</mathjax></p>
<p>Boom, you have a mass of:<br/>
<mathjax>#3.60g#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p>3.60g of NaOH is required to prepare that solution</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's begin with the equation for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a>:<br/>
<img alt="slideplayer.com" src="https://useruploads.socratic.org/4z0vwx7RRla5OizeNQHX_slide_3.jpg"/> <br/>
We are given the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> 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 200mL by 1000mL we will obtain a value of <strong>0.200 L.</strong></p>
<p>Rearrange the equation to solve for moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>:</p>
<p><strong>Moles of solute = Molarity <mathjax>#xx#</mathjax> Liters of solution</strong></p>
<p>Multiply 0.450 M by 0.200:</p>
<p><mathjax>#("0.450 mol")/("1 L") xx "0.200 L" = "0.09 mol"#</mathjax></p>
<p>To obtain the mass of solute, we will need to the molar mass of NaOH, which is 40.00 g/mol:</p>
<p>Finally, multiply the number of moles by 40.00 g/mol</p>
<p><mathjax>#0.09 cancel"mol" xx (40.00g)/(1cancel"mol")#</mathjax> <mathjax>#=3.60g#</mathjax></p>
<p>Boom, you have a mass of:<br/>
<mathjax>#3.60g#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many grams of #NaOH# are required to prepare 200 mL of a 0.450 M solution?</h1>
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<div class="markdown"><p>3.60g of NaOH is required to prepare that solution</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's begin with the equation for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a>:<br/>
<img alt="slideplayer.com" src="https://useruploads.socratic.org/4z0vwx7RRla5OizeNQHX_slide_3.jpg"/> <br/>
We are given the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> 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 200mL by 1000mL we will obtain a value of <strong>0.200 L.</strong></p>
<p>Rearrange the equation to solve for moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>:</p>
<p><strong>Moles of solute = Molarity <mathjax>#xx#</mathjax> Liters of solution</strong></p>
<p>Multiply 0.450 M by 0.200:</p>
<p><mathjax>#("0.450 mol")/("1 L") xx "0.200 L" = "0.09 mol"#</mathjax></p>
<p>To obtain the mass of solute, we will need to the molar mass of NaOH, which is 40.00 g/mol:</p>
<p>Finally, multiply the number of moles by 40.00 g/mol</p>
<p><mathjax>#0.09 cancel"mol" xx (40.00g)/(1cancel"mol")#</mathjax> <mathjax>#=3.60g#</mathjax></p>
<p>Boom, you have a mass of:<br/>
<mathjax>#3.60g#</mathjax></p></div>
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</article> | How many grams of #NaOH# are required to prepare 200 mL of a 0.450 M solution? | null |
598 | a919c339-6ddd-11ea-a926-ccda262736ce | https://socratic.org/questions/how-do-you-calculate-the-osmolarity-of-a-6-0-10-2-m-nacl-solution | 0.12 osmolar | start physical_unit 12 13 osmolarity osmol qc_end physical_unit 12 13 8 11 molarity qc_end end | [{"type":"physical unit","value":"Osmolarity [OF] NaCl solution [IN] osmolar"}] | [{"type":"physical unit","value":"0.12 osmolar"}] | [{"type":"physical unit","value":"Molarity [OF] NaCl solution [=] \\pu{6.0 × 10^(−2) M}"}] | <h1 class="questionTitle" itemprop="name">How do you calculate the osmolarity of a #6.0 * 10^-2# M NaCl solution?</h1> | null | 0.12 osmolar | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, <a href="http://socratic.org/chemistry/solutions-and-their-behavior/osmolarity">osmolarity</a> tells you how many <em>particles</em> of a <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> you get per liter of solution. </p>
<p><a href="http://socratic.org/chemistry/solutions-and-their-behavior/osmolarity">Osmolarity</a> is expressed in number of <em>osmoles</em>, or solute particles, per liters of solution. </p>
<p>When dealing with soluble <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a>, you have to keep track of how many ions you get <em>per formula unit</em>. Sodium chloride, <mathjax>#"NaCl"#</mathjax>, will dissociate completely in aqueous solution to produce sodium cations, <mathjax>#"Na"^(+)#</mathjax>, and chloride anions, <mathjax>#"Cl"^(-)#</mathjax></p>
<blockquote>
<p><mathjax>#"NaCl"_text((aq]) -> "Na"_text((aq])^(+) + "Cl"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p>As you can see, each mole of sodium chloride will produce <strong>two osmoles</strong> in solution. </p>
<p>This means that you will have</p>
<blockquote>
<p><mathjax>#6.0 * 10^(-2)color(red)(cancel(color(black)("moles")))/"L" * "2 Osmoles"/(1color(red)(cancel(color(black)("mole")))) = 12 * 10^(-2)"Osmoles/L"#</mathjax></p>
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<div class="markdown"><p><mathjax>#"0.12 osmolar"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, <a href="http://socratic.org/chemistry/solutions-and-their-behavior/osmolarity">osmolarity</a> tells you how many <em>particles</em> of a <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> you get per liter of solution. </p>
<p><a href="http://socratic.org/chemistry/solutions-and-their-behavior/osmolarity">Osmolarity</a> is expressed in number of <em>osmoles</em>, or solute particles, per liters of solution. </p>
<p>When dealing with soluble <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a>, you have to keep track of how many ions you get <em>per formula unit</em>. Sodium chloride, <mathjax>#"NaCl"#</mathjax>, will dissociate completely in aqueous solution to produce sodium cations, <mathjax>#"Na"^(+)#</mathjax>, and chloride anions, <mathjax>#"Cl"^(-)#</mathjax></p>
<blockquote>
<p><mathjax>#"NaCl"_text((aq]) -> "Na"_text((aq])^(+) + "Cl"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p>As you can see, each mole of sodium chloride will produce <strong>two osmoles</strong> in solution. </p>
<p>This means that you will have</p>
<blockquote>
<p><mathjax>#6.0 * 10^(-2)color(red)(cancel(color(black)("moles")))/"L" * "2 Osmoles"/(1color(red)(cancel(color(black)("mole")))) = 12 * 10^(-2)"Osmoles/L"#</mathjax></p>
</blockquote>
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<h1 class="questionTitle" itemprop="name">How do you calculate the osmolarity of a #6.0 * 10^-2# M NaCl solution?</h1>
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<div class="markdown"><p><mathjax>#"0.12 osmolar"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, <a href="http://socratic.org/chemistry/solutions-and-their-behavior/osmolarity">osmolarity</a> tells you how many <em>particles</em> of a <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> you get per liter of solution. </p>
<p><a href="http://socratic.org/chemistry/solutions-and-their-behavior/osmolarity">Osmolarity</a> is expressed in number of <em>osmoles</em>, or solute particles, per liters of solution. </p>
<p>When dealing with soluble <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a>, you have to keep track of how many ions you get <em>per formula unit</em>. Sodium chloride, <mathjax>#"NaCl"#</mathjax>, will dissociate completely in aqueous solution to produce sodium cations, <mathjax>#"Na"^(+)#</mathjax>, and chloride anions, <mathjax>#"Cl"^(-)#</mathjax></p>
<blockquote>
<p><mathjax>#"NaCl"_text((aq]) -> "Na"_text((aq])^(+) + "Cl"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p>As you can see, each mole of sodium chloride will produce <strong>two osmoles</strong> in solution. </p>
<p>This means that you will have</p>
<blockquote>
<p><mathjax>#6.0 * 10^(-2)color(red)(cancel(color(black)("moles")))/"L" * "2 Osmoles"/(1color(red)(cancel(color(black)("mole")))) = 12 * 10^(-2)"Osmoles/L"#</mathjax></p>
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</article> | How do you calculate the osmolarity of a #6.0 * 10^-2# M NaCl solution? | null |
599 | a8df4093-6ddd-11ea-9955-ccda262736ce | https://socratic.org/questions/when-iron-rusts-in-air-iron-iii-oxide-is-produced-how-many-moles-of-oxygen-react | 9.68 moles | start physical_unit 14 14 mole mol qc_end chemical_equation 25 32 qc_end physical_unit 1 1 17 18 mole qc_end end | [{"type":"physical unit","value":"Mole [OF] oxygen [IN] moles"}] | [{"type":"physical unit","value":"9.68 moles"}] | [{"type":"chemical equation","value":"4 Fe(s) + 3 O2(g) -> 2 Fe2O3(s)"},{"type":"physical unit","value":"Mole [OF] iron [=] \\pu{12.9 mol}"}] | <h1 class="questionTitle" itemprop="name">When iron rusts in air, iron (III) oxide is produced. How many moles of oxygen react with 12.9 mol of iron in the rusting reaction? #4Fe(s) + 3O_2(g) -> 2Fe_2O_3(s)#? </h1> | null | 9.68 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We have the balanced equation:</p>
<p><mathjax>#4Fe(s)+3O_2(g)->2Fe_2O_3(s)#</mathjax></p>
<p>And so, <mathjax>#4#</mathjax> moles of iron react with <mathjax>#3#</mathjax> moles of oxygen, and therefore, <mathjax>#12.9 \ "mol"#</mathjax> of iron would react with:</p>
<p><mathjax>#12.9color(red)cancelcolor(black)("mol" \ Fe)*(3 \ "mol" \ O_2)/(4color(red)cancelcolor(black)("mol" \ Fe))=9.675 \ "mol" \ O_2#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#9.675 \ "mol"#</mathjax> of oxygen gas.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We have the balanced equation:</p>
<p><mathjax>#4Fe(s)+3O_2(g)->2Fe_2O_3(s)#</mathjax></p>
<p>And so, <mathjax>#4#</mathjax> moles of iron react with <mathjax>#3#</mathjax> moles of oxygen, and therefore, <mathjax>#12.9 \ "mol"#</mathjax> of iron would react with:</p>
<p><mathjax>#12.9color(red)cancelcolor(black)("mol" \ Fe)*(3 \ "mol" \ O_2)/(4color(red)cancelcolor(black)("mol" \ Fe))=9.675 \ "mol" \ O_2#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">When iron rusts in air, iron (III) oxide is produced. How many moles of oxygen react with 12.9 mol of iron in the rusting reaction? #4Fe(s) + 3O_2(g) -> 2Fe_2O_3(s)#? </h1>
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<div class="markdown"><p><mathjax>#9.675 \ "mol"#</mathjax> of oxygen gas.</p></div>
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<div class="markdown"><p>We have the balanced equation:</p>
<p><mathjax>#4Fe(s)+3O_2(g)->2Fe_2O_3(s)#</mathjax></p>
<p>And so, <mathjax>#4#</mathjax> moles of iron react with <mathjax>#3#</mathjax> moles of oxygen, and therefore, <mathjax>#12.9 \ "mol"#</mathjax> of iron would react with:</p>
<p><mathjax>#12.9color(red)cancelcolor(black)("mol" \ Fe)*(3 \ "mol" \ O_2)/(4color(red)cancelcolor(black)("mol" \ Fe))=9.675 \ "mol" \ O_2#</mathjax></p></div>
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</article> | When iron rusts in air, iron (III) oxide is produced. How many moles of oxygen react with 12.9 mol of iron in the rusting reaction? #4Fe(s) + 3O_2(g) -> 2Fe_2O_3(s)#? | null |
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