Recent Questions - Chemistry Stack Exchange most recent 30 from 2014-04-20T01:26:08Z 0 Is the end of a cigarette on fire when lit? JayBoux 2014-04-20T00:45:39Z 2014-04-20T01:07:24Z <p>Is the end of a cigarette on fire when it is lit? Or is it smoldering? Is there a chemical difference between the two?</p> 0 Calculating the partition function based on molar heat capacity user125598 2014-04-20T00:37:08Z 2014-04-20T00:37:08Z <p>Given the molar heat capacity of a partition function as a function of temperature, how would one determine the partition function?</p> <p>Say the molar heat capacity = $\alpha T^2$ </p> 0 Bond angle of Nitrous Acid Dissenter 2014-04-20T00:34:49Z 2014-04-20T01:22:39Z <p>Why are there two possible bond angles for the nitrous acid molecule? </p> <p>After perusing Wikipedia I found that there are both cis and trans forms of the nitrous acid molecule. I also note that the cis form is sterically hindered. Could this be a reason for the two different bond angles - there are two different geometric isomers and in one case there are larger van der Waals repulsions, which in turn expands the bond angle? </p> 0 What is the relationship between migration and the potential of double layer theory? user5122 2014-04-19T18:00:50Z 2014-04-19T18:00:50Z <p>Double layer theories like that of Gouy-Chapman give the electrostatic potential $\ \phi\ $ in solution. For a symmetrical electrolyte and small potential drops $\ \phi\left(x=0\right)-\phi\left(x\rightarrow\infty\right)=\phi_0-0\ \text V=\phi_0$, the theory gives $$ \phi=\phi_0\exp\left(-\kappa x\right) $$ where $x$ is measured from the electrode/solution interface into the bulk solution. As $\kappa$ is very large $\left(\sim10^{-9}\frac{1}{\text m}\right)$, the electric field $$ E=-\frac{\text{d}\phi}{\text{d}x}=-\kappa\phi_0\exp\left(-\kappa x\right) $$ falls to a constant value (0 V in this example) very quickly.</p> <p>In the equations for the transport of ions in diluted solutions, the <strong>migration</strong> flux of species $j$ is $$ J_{\text {j,migr}}=-z_ju_jFc_j\frac{\text{d}\phi}{\text{d}x} $$</p> <p>As the flux is proportional to $E$ it must be zero everywhere in solution according to Gouy-Chapman. But in every book about electrochemistry migration is said to contribute to the flux $J_j$ of species $j$ except when $c_j$ is much more smaller than the concentrations of other species. But why should there be migration in a macroscopic cell at all? In double layer theories ions gather near the electrode surface and compensate the charge on the metal, so there should be no electric field in solution. Isn't this what the equation for $\phi$ above says?</p> <p>Or are there two different types of "electrostatic potentials" in solution: One associated with the double layer theories and one associated with transport? $$ \phi_{\text{Gouy-Chapman}}\neq\phi_{\text{Migration}} $$</p> <p>If this is the case: How are both potentials related to each other (and why do books always use the same symbol for both)?</p> 0 How much heat (in J) is required to produce 2.0L of $N_2(g)$ at 77.36 K and 1.0 atm? Twilight Pony Inc. 2014-04-19T17:42:41Z 2014-04-19T19:51:37Z <p>The only thing that is given is: </p> <p><strong>The enthalpy of vaporization for $N_2(l)$ is 5.56 kJ/mol.</strong></p> <p>There aren't any examples similar to this problem in the textbook. I only know how to get energy using:</p> <p>$q=mcΔT$ and $w=PΔV$</p> <p>No clue what to do with the enthalpy of vaporization value.</p> <p><strong>UPDATE</strong>: Solved it by first getting number of moles ($n$) via $PV=nRT$ then multiplying moles by 5.56 kJ/mol. Moles cancel out, leaving only kJ, 1.75kJ to be precise.</p> 0 Enthalpy change of a constant pressure process user3551975 2014-04-19T16:31:59Z 2014-04-19T21:37:13Z <p>In the question, we were expected to find <strong>"enthalpy change of a constant pressure process where internal energy of gas is increased $x\ \text{kJ}$ of energy and $y\ \text {kJ}$ of work is done by the system"</strong>.</p> <p>I found the following on <a href="" rel="nofollow">Wikipedia</a>:</p> <blockquote> <p>For quasistatic processes under constant pressure, ΔH is equal to the change in the internal energy of the system, plus the work that the system has done on its surroundings" So the result should be x+y ,however the answer is x-y. </p> </blockquote> <p>I also found this "ΔH of a system is equal to the sum of non-mechanical work done on it and the heat supplied to it."</p> 0 Ways of expressing concentration in activity term of chemical potential formula Vishvajeet Patil 2014-04-19T15:49:47Z 2014-04-19T20:08:42Z <p> <p>We know chemical potential is defined as ${\mu}i={\mu}.i_{standard}+RTln(ai)$. Here $ai$ is the activity of $i^\text{th}$ component of solution.In case of gases instead of $a_i$ it is $f_i$ that is fugacity of $i^\text{th}$ gas out of mixture of gases. Activity is here nothing but activity coefficient multiplied by concentration or in case of fugacity it is product of fugacity coefficient multiplied by partial pressure of gas.</p> <p>But now this is not the only definition of Chemical potential. Chemical potential also means partial molar Gibbs free energy when pressure and temperature of system are constant. We know that Gibbs free energy of a system at constant temperature and pressure also represents Non-PV work done by system. Now how can the term $RTln(ai)$ can be correlated to Non-PV work done by system in order both two definitions will be equivalent.</p> 2 How much hydrolysis will likely occur of the following bonds at pH 1-4 in the time period of 1-3 hours? Brenton Horne 2014-04-19T11:14:06Z 2014-04-19T11:33:42Z <p>The bonds are: (I have given pictures of what I mean in case there's any ambiguity that I leave open in these questions. Also so that if you want to number the R groups you've got a head-start)</p> <ul> <li><strong>Secondary amides</strong>, I suspect I might know the answer, extensive. Example is the one in <strong>folic acid</strong>.</li> </ul> <p><img src="" alt="Secondary amides"></p> <ul> <li><strong>Tertiary amides</strong>. This is one that's a complete mystery to me. Example is the one in <strong>leucovorin</strong>. </li> </ul> <p><img src="" alt="Tertiary amides"></p> <ul> <li><strong>Sulfonamides</strong>. Hydrolysis in this case, should lead a primary amine and a sulfonic acid. Or such is the hope. </li> </ul> <p><img src="" alt="Sulfonamide"></p> <ul> <li><strong>Carboxylic esters</strong>. To give a primary alcohol and a carboxylic acid. Basically the reverse of the <em>Fisher esterification</em>. </li> </ul> <p><img src="" alt="Carboxylic ester"></p> <ul> <li><strong>Sulfonyl ester</strong>. </li> </ul> <p><img src="" alt="Sulfonyl ester"></p> <ul> <li><strong>Secondary/tertiary amines</strong></li> </ul> <p><img src="" alt="Secondary/tertiary amines"></p> <ul> <li><strong>Ether</strong></li> </ul> <p><img src="" alt="Ether"></p> <ul> <li><strong>Sulfyl ether</strong></li> </ul> <p><img src="" alt="Sulfyl ether"></p> <p>Now if you're wondering how exactly is this question relevant? As it might sound pretty random for me to just ask this question but it's incredibly relevant to drugs when it comes to their tendency to be hydrolysed in the stomach. </p> <p><strong>The simple fact is that I'm likely to come up with other questions so if it's possible for you to give me a guide for figuring out this stuff, or estimating it, please do share! If you do this and I can follow it, I will instantly accept your answer as the answer!</strong> <em>Providing me with a <strong>decent ref</strong> will also win you points.</em></p> 2 Draw pyridine as an R Group Daniel Antón García 2014-04-19T10:56:37Z 2014-04-19T14:53:06Z <p>This is a bit of a weird question but here we go. You know how when you carry out a reaction with many compounds, when drawing the reactions you just draw a line at a right angle to the bond with a 'R' at the end (see image)?</p> <p><img src="" alt="R"></p> <p>Then you can specify what group R is in a table (eg: OMe). However, if instead of adding a substituent to a benzene ring, you have pyridine? How can you specify it? Thanks!</p> -3 How to do equilibrium constant [on hold] user4662 2014-04-18T18:10:16Z 2014-04-18T19:39:47Z <p>Consider the reaction: CO(g)+H2O(g)⇌CO2(g)+H2(g) An equilibrium mixture of this reaction at a certain temperature was found to have [CO]= 0.0243M , [H2O]= 0.0145M , [CO2]= 0.200M , and [H2]= 0.0294M . What is the value of the equilibrium constant at this temperature?</p> -2 What is the equilibrium constant? [on hold] user4662 2014-04-18T18:05:02Z 2014-04-18T19:38:45Z <p>Consider the reaction: 2H2S(g)⇌2H2(g)+S2(g) An equilibrium mixture of this reaction at a certain temperature was found to have [H2S]= 0.502M , [H2]= 3.24×10−2M , and [S2]= 7.84×10−3M .What is the value of the equilibrium constant at this temperature?</p> -1 How to do Concentration [on hold] user4662 2014-04-18T17:58:50Z 2014-04-18T19:39:41Z <p>Suppose that the molar concentrations for CO and H2 at equilibrium are [CO] = 0.05 M and [H2] = 0.05 M. Use the formula you found in Part B to calculate the concentration of CH3OH</p> 3 Simple Kinetics Clarification David 2014-04-18T14:45:44Z 2014-04-18T19:39:27Z <p>I just wanted to check my understanding of this. Given the reaction,</p> <p>$$ (1) \; \ce{A-&gt;B} $$</p> <p>the corresponding rate law is $ v=k[A] $</p> <p>Also consider</p> <p>$$ (2) \; \ce{A &lt;=&gt; B} $$</p> <p>with forward rate law: $ v_1=k_1[A] $ and reverse: $ v_2=k_2[B] $. This makes the corresponding rate equation $ v=v_1-v_2=k_1[A]-k_2[B] $.</p> <p>My question is in the difference between the two. I've often seen reactions expressed in both ways. In $ (1) $, is the rate law assuming the reverse reaction not to occur (or at least to be completely negligible? That's why I drew it with $ \ce{-&gt;} $ instead of $ \ce{&lt;=&gt;} $. Am I correct in my thinking? In actuality, is it always more correct to use the rate law of $ (2) $? The rate law for $ (1) $ could also correspond to initial reaction velocity, right? This is because there is no $ B $ present at the start of the reaction.</p> <p>If this is correct, then the integrated rate equations will become substantially more complicated using rate laws of form $ (2) $. I can't use simple equations like $ [A]_t=[A]_0 e^{-kt} $, as this wouldn't take into account the reverse reaction.</p> 1 Which compound can exist as stereoisomers? mastergalen 2014-04-18T11:59:25Z 2014-04-19T23:51:32Z <p>I'm having trouble in understanding how to identify chemical compounds as stereoisomers.</p> <p>In the question below, <strong>D</strong> is the correct answer according to the markscheme. I've drawn out the chemical structure but am unable to see what possible stereoisomers it can form.</p> <p>I also tried visualising the possible isomers on Wolfram Alpha: </p> <p><a href="" rel="nofollow"></a></p> <p>There are two isomers which stand out to me, which I guess are the stereoisomers?</p> <ul> <li>(S)-(+)-2-butanol</li> <li>(R)-2-butanol</li> </ul> <p>However, to me it looks like they are just rotated versions of each other.</p> <p><img src="" alt="enter image description here"></p> 3 If metals want to "lose" electrons, then why will copper ions take electrons from a zinc electrode? Ali Mustafa 2014-04-18T10:48:07Z 2014-04-18T20:01:01Z <p>According to my book, metals want to "lose" electrons to achieve noble gas configuration. But at the same time, copper ions in a galvanic element are going to gain electrons from a zinc electrode. Why is it so? Is it because one of them HAS to gain/lose electrons, and zinc wins out in terms of achieving noble gas configuration?</p> -1 Chromatography Worded Problem - Rf values & length size of paper confused 2014-04-18T08:47:10Z 2014-04-18T09:51:30Z <p><img src="" alt="enter image description here"></p> <p>Can someone correct my working out. I don't know what the correct answer is. </p> <ul> <li>distance travelled by dye x = x</li> <li>distance travelled by dye y = y</li> <li>distance of solvent front = p</li> <li>x/p=0.72 therefore x = p(0.72)</li> <li>y/p=0.8, therfore y = p(0.8)</li> <li>y-x=1, therefore y=1+x</li> <li>new equation: 1+x=p(0.8) where x=0(0.72), if we solve for p we get 12.5</li> </ul> <p>so the piece of paper should be at-least 12.5 cm long. </p> 2 What are the physico-chemical processes at play in ironing of clothes? Michiel 2014-04-18T08:42:25Z 2014-04-18T10:56:36Z <p>The process of ironing of clothes removes (to a large extend) the wrinkles that where introduced in the clothes while drying. Obviously this has to be some physico-chemical process that is initiated/sustained by the heat of the iron and (perhaps) the pressure applied with the iron.</p> <p>My question is: what is the actual process that is going on when ironing? Let's assume we have a cotton shirt, do the polymers only physically rearrange or do they actual undergo a chemical change? If it is the latter, does ironing your clothes often cause more rapid degradation?</p> 8 What is the unit of pH? studen 2014-04-17T09:14:26Z 2014-04-18T14:03:10Z <p>I'm making some graphs and I have to label the axes. I want to be extra careful and put the units in even though the meaning of $\text{pH}$ is well known. But I have a problem (though a simple one): $\text{pH}$ is a minus logarithm (base 10) of concentration of hydrogen ions (or rather their activity). What is the unit then, is it $[-\log(\text{mol}/\text{L})]$? What should I write, could you help me?</p> 1 Law of reciprocal proportion biogirl 2014-04-17T04:26:37Z 2014-04-19T21:36:54Z <p>Is it necessary that the valency of an element must have the same valency?</p> <p>For eg. If 2 elements A and B combine with the same mass of C, is it necessary that they will have the same valency as they would have when combining AB?</p> 1 ESI & MALDI Mass Spectrometry: particle mass vs. measurement error user5209 2014-04-16T18:20:47Z 2014-04-18T10:10:31Z <p>For MALDI or ESI mass spectrometry of large macromolecules (e.g. in the $\approx 100$ kDa to $\approx 10 - 100$ MDa mass range), I've always how measurement error scales with particle mass. Not having any hands on experience with mass spetrometry, and calling anything build in the last 5 - 10 years "modern", are there any published tables of general rules of thumb for how measurement error scales with particle mass (the two must surely be positively correlated)? If we're interested in measuring the mass of large intact nucleic acids or proteins, at what single-molecule particle masses does it make sense to use ESI or MALDI?</p> <p>I'd guess that at higher mass ranges (e.g. in the MDa range) ESI might become preferable since it can produce multiply charged species, allowing for smaller $\frac{m}{z}$ ratios?</p> 0 Aqueous vs. Solid user11629 2014-04-16T16:00:37Z 2014-04-19T12:40:07Z <p>I've been having a conceptual struggle with these two states for a while now, and I'd like to get it cleared up.</p> <p>A compound AB dissolves in water. </p> <p>$$\ce{AB(s) + H2O &lt;-&gt; A+(aq) + B- (aq) + H2O(l)}$$</p> <p>Now, we know that in the case of salts, there can be strong and weak electrolytes based on the degree of dissociation. For example, NaCl is a strong electrolyte because when dissolved in water it dissociates completely. However, in theory, there should be some salts that do not dissociate completely and thus are weak electrolytes, for example, I think, NaCH3CHOO, despite its high solubility.</p> <p>$$\ce{NaCH3COO(s) + H2O(l) &lt;-&gt; NaCH3COO(aq) + Na+(aq) + CH3COO- (aq)}$$</p> <p>Here's an example that confuses me. Assuming that CaO dissociates completely in water:</p> <p>$$\ce{CaO(s) + H2O(l) &lt;-&gt; Ca^{2+}(aq) +O^{2-} (aq) + H2O(l)}$$</p> <p>O(2-)(aq) + H2O(l) --> (irreversibly, I suppose because of the general formula for the combination of anhydride metal oxides with water) 2OH-(aq)</p> <p>Which then finally yields:</p> <p>$$\ce{CaO(s) + H2O(l) &lt;-&gt; Ca^{2+}(aq) + 2OH- (aq)}$$</p> <p>Then, here's the tricky part:</p> <p>$$\ce{Ca^{2+}(aq) + 2OH- (aq) &lt;-&gt; Ca(OH)2(s)}$$</p> <p>Now, I don't know whether this is true. Calcium hydroxide, according to solubility rules, is insoluble in water, but as we know, all compounds are at least a little bit soluble. The KSP may be very low, but it dissolves a bit. My confusion is in how we know if a precipitate forms or not without being given any molar concentration of Ca2+ or the KSP of Ca(OH)2. The latter is necessary to calculate QSP, and this is necessary to determine whether a precipitate actually forms. Without being given any other information in a school setting, is it correct to assume that all salts are strong electrolytes and that all low-KSP compounds form a precipitate in solution?</p> <p>Thanks!</p> 2 Why is the 8-electron rule more important than the 2- or 18-electron rule Hashir Omer 2014-04-16T08:40:46Z 2014-04-19T21:37:57Z <p>Why is the fulfilled electronic configuration of only $p$ orbital is stable. I mean why $II-B $ group with fulfilled $d$ orbital,$II-A$ group with fulfilled $s$ orbital...are not stable. Why makes $p$ orbital so special?</p> 3 Could someone please explain the difference between m/z and m/e in mass spectroscopy confused 2014-04-15T10:42:15Z 2014-04-18T09:52:56Z <p>The definition:</p> <p>The abbreviation m/z is used to denote the dimensionless quantity formed by dividing the mass number of an ion by its charge number. It has long been called the mass-to-charge ratio although m is not the ionic mass nor is z a multiple or the elementary (electronic) charge, e. The abbreviation m/e is, therefore, not recommended. Thus, for example, for the ion C7H72+, m/z equals 45.5.</p> <p>Could someone please explain the definition. I found it a bit difficult to comprehend. </p> <ul> <li>What is the "ionic mass"?</li> <li>what do they mean by "z a multiple or elementary charge"</li> <li>C7H72+, m/z equals 45.5 -> what did they do here</li> </ul> 0 Degenerate Orbitals user2789433 2014-04-14T02:58:11Z 2014-04-19T12:24:15Z <p>How do I know if an atom has degenerate orbitals?</p> <p><strong>My Understanding</strong></p> <p>I understand that degenerate orbitals mean orbitals that have the same energy level for the same n. However, how do I distinguish between an atom that has degenerate orbitals?</p> <p>EDIT: Example of non-degenerate orbitals (<a href="" rel="nofollow">Source</a>)</p> <p><img src="" alt="Source"></p> 1 If two electrons move up to a higher energy level in an atom, is it considered excited? Jonie 2014-03-20T20:27:02Z 2014-04-19T21:58:54Z <p>In Germanium, if two electrons go from 1s go to 4p is it in its excited state?</p> 1 Relating a compound's radii and unit cell dimensions user22323 2013-10-22T18:59:51Z 2014-04-18T20:27:51Z <p>We're asked often to calculate the density of a crystal, and there is a known formula for that related to the unit cell volume. Now, for simple geometric structures such as the bcc structure (and since it is close packed), I know how to relate the different radii of the constituent compounds with the unit cell's (unique in this case) dimension (the length a), but how do I relate it for such other complex structures? </p> <p>In general given a certain crystal geometry (cubic , tetragonal , orthorhombic), how does one relate the unit cell dimension with the radii of the constituent atoms?</p> 1 How did the first amino acids appear on earth? Flux 2013-09-19T20:38:22Z 2014-04-19T22:58:45Z <p>Is there a prove or a theory that the first amino acid synthesis was influenced by a change in heat, gravitation, radiation or something like that?</p> 18 Is a negative pH level physically possible? anorton 2013-07-31T18:58:49Z 2014-04-19T18:04:15Z <p>A friend of mine was looking over the definition of pH, and wondering if it is possible to have a negative pH level. From the equation below, it certainly seems possible--just have a $1.1$ (or something $\gt 1$) molar solution of $H^+$ ions: $$\text{pH} = -\log([H^+])$$ (Where $[X]$ denotes the concentration of $X$ in $\tfrac{\text{mol}}{\text{L}}$.)</p> <p>If $[H^+] = 1.1\tfrac{\text{mol}}{\text{L}}$, then $pH = -\log(1.1) \approx -0.095 $</p> <p>So, it is <em>theoretically</em> possible to create a substance with a negative pH. But, is it physically possible (e.g. can we create a 1.1 molar acid in the lab that actually still behaves consistently with that equation)?</p> 0 Solving carbon black Jonas Stein 2013-05-20T00:09:34Z 2014-04-19T23:57:16Z <p>I want to remove a layer of 2 mm thick carbon black on a copper bowl. I have tried different solvents, but the carbon seems to be very resistive. </p> <p>Which formula solves carbon black without solving copper? The bowl is made from pure copper. It will melt at roughly 1000 degree Celsius.</p> 20 Why is the 2s orbital lower in energy than the 2p orbital when the electrons in 2s are usually farther from the nucleus? Gordon Gustafson 2012-04-28T22:14:02Z 2014-04-19T13:14:44Z <p><img src="" alt="Diagram of P orbital vs. S orbital distance from nucleus"></p> <p>My chemistry book explains that even though electrons in the 2p orbital are closer to the nucleus on average, electrons from the 2s orbital spend a very short time very close to the nucleus (penetration), so it has a lower energy. Why does this tiny amount of time spent close to the nucleus make such a big difference? It seems like it should be the average distance that matters, not the smallest distance achieved at any one point, in determining stability. What makes that momentary drop in energy so important that it is outweighs all the time spent farther away from the nucleus with a higher energy?</p>