text:
How could I cause metal to rust rapidly?
I'm writing a novel set in the mid-nineteenth century. One of my characters needs to cause an iron padlock to rust rapidly, ideally within a matter of hours.
The lock doesn't need to rust away completely - the goal is just to make it noisey and annoying so that it can't be opened in silence.
Is this a realistic goal? If so, what methods could someone use to cause rapid rusting? Bonus points for suggesting period-appropriate materials.
Update: Based on the answer and comments below, it sounds a period appropriate approach would be to mix fine copper filings with saltwater and apply it to the lock.
I've read that bleach is an oxidizing agent which was available in the 19th century. Would adding bleach to this mixture increase the rate of rusting without creating an overly dangerous mixture?
Update: I conducted an experiment in my garage. Iodine worked pretty well.
Experiment Video
A few months later the iodine had eaten through the lid of the jar I tested it in:

I've removed the oxidizing agent thing because I'm afraid they would dissolve the rust...

@user34388 and others - I am not a chemist. What are your thoughts on the risks of trying this experiment? I find the oxidizing agent compelling and it sounds like that might be necessary to reach the goal of rust within hours.

This experiment is not harmful... as long as you don't use oxidizing agents that are too dangerous such as (should I be listing them?) acidified potassium dichromate or acidified potassium permanganate.

@user34388 At worst I'd be using household bleach, however I'm intrigued by your suggestion with beans, that had not occurred to me...

What do you mean by "rust"? Having the whole surface thoroughly cover with a layer of rust is not a problem. Run a piece of (unseasoned) iron through the dishwasher and you'll see it. Having the padlock rust through and through like in the fable of the rope made of ash is a different matter, as the rust on the surface will shield the metal from the stuff you through at it from outside.

@rumtscho - the goal is to make the lock (and associated hinges) on a trapdoor as loud and frustrating as possible so that my protagonist will be able to hear the antagonist open the lock while retaining some deniability - rust can happen naturally, so the antagonist won't immediately assume it's been tampered with.

Picasso's son has reported that there was a period when his father would tell him to "go piss on the goat"  (http://www.moma.org/collection/works/81670) to encourage the desired patina to develop. Salt water...

Just a note to Tyler in regards to the nail in your experiment, most modern nails actually have a vinyl coating on them, I believe this is for rust prevention and as a dry lube also.

The answers and comments should win a prize for obfuscation. Steel /iron rusts when wet, and rusts faster with any salt in the water.

Here are factors that would speed up rusting:

Presence of water: make sure that the iron is wet.
Presence of oxygen: make sure the iron have access to air (the dissolved oxygen in water also works).
Presence of metal below iron in the metal reactivity series (see the picture below): usually copper is used. Tie copper to iron. Make sure that they are in contact. The larger the surface area of contact the better.
Temperature: make the iron as warm as possible.
Presence of salt: explained here. In short, salt increases the conductivity of water which makes the iron rusts faster, since rusting involves electron transfer.
Acidic environment: since rusting involves hydrogen ions ($\ce{H+}$), an acidic environment would increase the concentration of hydrogen ions, making the iron rust faster. However, this would dissolve the rust.
Presence of hydrogen peroxide: hydrogen peroxide ($\ce{H2O2}$) is an oxidizing agent which would not dissolve the rust:
$$\ce{3H2O2 + 2Fe -> 3H2O + Fe2O3}$$

Metal reactivity series:

(source: bbc.co.uk)
Equation of rusting:

$\ce{Fe -> Fe^3+ + 3e-}$
$\ce{O2 + 4H+ + 4e- -> 2H2O}$
$\ce{2Fe^3+ + 3H2O -> Fe2O3 + 6H+}$

More equations can be found here.
Overall equation:
$$\ce{4Fe + 3O2 ->[H+,H2O] 2Fe2O3}$$

Maybe an acidic environment would favour rusting. If you see the electrochemical aspect of rusting, the H+ ions are on the left hand side of the cell equation for rusting

@AmritanshSinghal But wouldn't acid dissolve the rust?

@busukxuan I think the question poster just needs the padlock to go away, so it doesn't matter.

@busukxuan the question only asks methods to speed up the process of rusting. SO,the OP is interested in speeding up the oxidation of iron and for that increasing [H+] would serve the purpose.I do not know if acid would dissolve it or not but  that is asked. However, if you know what shall be the reaction of Fe2O3.xH2O with H+(i.e. dissolution of Fe3+ in acid), do mention in the comment. I would like to know that.

@AmritanshSinghal Not long ago I used HCl to remove rust. Basically it's just the hydroxide reacting with acid to form water, while there acid's conjugated base forms a salt with the iron.

@user34388 the objective is to build up rust to make the lock more difficult to open, so washing away the rust would likely defeat the purpose.

@busukxuan thanks for the explanation.  I was confusing myself with this statement of my chemistry book which discusses rusting only till the generation of Fe2+. It stated that electrons released at the anodic spot move through the metal to another spot to reduce oxygen in presence of H+(from acidic oxides in air). There was also a statement that Fe2+ would further be oxidised to Fe3+ and release some H+ ions (which I did not pay attention to.)

@user34388 your post is now better. I've transformed my vote to and upvote. HOWEVER, note that the actual reactions for rusting are quite complex and proceed in many steps.

@AmritanshSinghal I cannot come up with the equation of $\ce{Fe^2+}$ being oxidized to become $\ce{Fe^3+}$ while giving off $\ce{H+}$...

Nitpick: It's not just salt: $\ce{NaCl}$ is an especially effective salt for inducing corrosion/rust due to the chloride ion. Halides generally (fluoride being the odd duck out, usually) are awesome at inducing corrosion.

When I was 10 or 12 or whatever chemistry-set age is, I remember following some instructions to make a solution that would cause nails to rust within a few minutes. I've no idea whether this was actual rust from the nail or was perhaps simply something deposited on its surface. Any idea what this solution was? (perhaps bleach or ammonia) ¶ This was back in the 1960s, so the range of commonly available chemicals was much greater than it is today (e.g. I remember going to the local pharmacy and buying liquid mercury, potassium permanganate (and glycerine), potassium nitrate (and sulphur), etc.).

Here's a less-known effect:  Iron or steel exposed to iodine vapors will rust within hours.
Also, in a draft-free room, iodine crystals placed well above an iron object will evaporate, creating a descending vapor-plume which rusts everything directly below.   Or just use overkill: affix a sack of black iodine powder over the steel to be rusted.   (Also, iodine has a distinct odor, so an S. Holmes type might sniff the rust and learn much, while normal humans would never think to try.)
Historical:  currently iodine crystals are only sold for treating lifestock, but 170 years ago it was a common photographic chemical (critical in Daguerrotype process,) and also a common antiseptic.  Perhaps your local Victorian druggist would have kilos of raw solid iodine to supply the daguerrotype craze, if not just for mixing up some iodine tincture for treating small cuts.

I set up an experiment to test this. I put approximately 15g of iodine crystals on a plastic mesh about 2" above a steel nail. After about 18 hours I'm seeing a little rust starting to form. It's about 55 defrees Fahrenheit, so the sublimation of the iodine may be slower than yoir response expected.

When I saw this happen, it was above 70F and quite humid as well.   It might be humidity that does it, rather than just heat.   Also, even a slight oil-film (such as with store-bought nails) might interfere  ...which probably removes this idea from the running.   Your door hinge would already have to be grease-free and probably squeaky.

In the book the temperature will be comparable to my test. I expect I'll have the characters use a degreasing solvent (lye?) before applying iodine.

@Tyler  IDEA!   If you have iodine crystals, you can use ammonia to make contact explosive paste.  Ammonium iodide.  Smeared on a hinge or in a keyhole, after drying, it goes POW when disturbed even slightly.   Just use a small stick to thoroughly grind up about a cc of black iodine crystals immersed in a saucer of ammonia cleaner.  Keep adding more ammonia until the violet turns blue-gray.  Let it settle out.  The paste can be made on the spot as needed.  It's dangerous to store it wet in a sealed container, but can be done briefly (just not for ~days.)  Don't make more than 1cc at a time!

@Tyler  Loud crackling floor:  make an entire cc of the black stuff, fling it at the ground hard, so it spatters.  Wait to dry.  Now anyone walking on the floor makes all kinds of crackling noises, even some quite loud pops.   Sometimes it lasts for weeks!  .  .  Also the ammonium iodide makes a small purple cloud of iodine, and if exploding from a keyhole, will stain the hand.    Which perp was messing with a prepared lock?  Check for purple hand!

that's awesome! I may have to try it. I'll comment when i do.

@Tyler But is it Victorian?   Hell yeah.  Discovery announced in 1813.  Studied by Humphy Davy etc.  See:  https://books.google.com/books?id=LmQ2AQAAMAAJ&pg=PA130

excellent. Thanks for the help, I'll set aside a free copy for you when it's done!

I did some more research on this. What you're describing is ammoniated nitrogentriiodide. When it explodes, ammonia triodide is left over.
https://en.wikipedia.org/wiki/Ammonium_iodide

Yep, https://en.wikipedia.org/wiki/Nitrogen_triiodide.   There are LOTS of chemistry demo videos on youtube for this.

I've updated the original post with the results of my garage experiment.

The squeaky performance of the padlock need not come completely from within. If the idea of copper filings is acceptable (they would certainly seem out of place), perhaps a little plaster of Paris could be made into a paste and shoved into the keyhole. After it sets, the operation of the lock would be stiff, perhaps noisy. The plaster could be retarded or not, so as to set hours later or within minutes.
For chemical complexity, you could add iron (not copper!) filings - let the added metal do the rusting. It could look like the padlock was just wearing out. Plaster need not stay white; if the iron filings don't color it enough, there's always carbon black. And NaCl if you want some rusting; and some oxalic acid, or cream of tartar for acidity - or NaHSO4.
This could work even if the padlock is brass, although brass filings would look more innocent. But the plaster should be colored so as to look like dirt.
Perhaps even a little Portland cement, as a paste, maybe a little sand, maybe some iron filings for extra rusting. This would be more adhesive, but would take longer to set, unless you add a little CaCl2 for acceleration.

Science At Play: Instant Rust | Connecticut Science Center describes a children's experiment (including a YouTube video
) for causing iron to rust almost instantly:
In a non-metallic container, add:

eight parts hydrogen peroxide (standard 3% concentration).
one part vinegar (standard 5% acetic acid).

While stirring, slowly add until no more will dissolve:

table salt.

Then:

Place an uncoated non-galvanized nail in the liquid.
Watch it develop rust.
Carefully remove and allow to air dry before touching the rust.

In our experiment, mixing hydrogen peroxide (H2O2) and distilled vinegar together creates a small amount of something called peracetic acid.
Acid is corrosive and can cause things like metal to break down.
Hydrogen peroxide is made of hydrogen and oxygen, but it’s the oxygen that’s key to creating rust on metal.
The molecules of iron on the surface of the nail exchange atoms with the oxygen in the solution and produce a new substance.
You guessed it–rust! (or iron oxide as scientists would call it!)
This whole process is helped along by the salt we added to the solution.
Its job in this whole process is to act as an electrolyte which lowers the electrical resistance in the solution, helping the oxygen and the nail to trade atoms more easily.

This is intended to be kid-safe (with appropriate precautions), but for your situation, a more strongly concentrated solution might be considered.


text:
Why are activities of solids and liquids taken to be unity?
First, I should mention that the question was already asked here: Why is active mass of a pure solid or liquid always taken as unity?
However, the treatment there is qualitative and I do not really understand the reasoning for the disclusion of solids and liquids from the reaction qoutient that is presented there. I have attempted to derive this fact mathematically and failed, which why I ask this question again.
Let us look at an example reaction which involves solids and gases. The treatment for other reactions or reactions with liquids shouldn't differ significantly. A very simple reaction for this matter will be the combustion of graphite
$$\ce{C(s) + O2(g) -> CO2(g)}$$
Reaction quotients are derived from the equation for change in Gibbs free energy at nonstandard pressures, so let's find that. Suppose the reaction is carried out in nonstandard pressure, and $\Delta G^o_r$ is known. Let's (I) compress the system isothermally to standard pressure, (II) carry out the reaction and then (III) return it isothermally to the original pressure.
I. $\Delta G$ for the oxygen here is the well known $RT\ln (1/(P/P_o))$. For the carbon: In any isothermal process $dG=VdP$, let's assume for our ideal solid that the volume is constant (I have also tried this in a model where the volume changes with pressure with bulk moduli, the result does not differ significantly), so $\Delta G=V(P-P_o)$.
II. Here simply $\Delta G=\Delta G^o_r$.
III. Here it's just gas compressing so again $\Delta G=RT\ln (P/P_o)$
Adding the contributions from all three stages we have
$$
\Delta G=RT\ln (1/(P/P_o))+V(P-P_o)+\Delta G^o_r+RT\ln (P/P_o)=\Delta G^o_r+V(P-P_o)
$$
This result is different from what a model of an activity of unity for the carbon would predict. What went wrong? And why are activities of solids and liquids taken to be unity? Any help will be appreciated.

Nothing "went wrong". Hf for CO2 is ~393 kJ/mole. 1 latm = 9.87 kJ, so lets see what a mole of C does Gibbs energy wise going from 1 atm to 11. OK 1 mole is 0.0053 l and multiplying it by 10 atm we get 0.053 latm ~ 0.15 kJ or 0.15% of the heat of formation for the CO2. If you want to get down to this level of accuracy, you gotta throw out the Ideal Gas Law (the simple approximations it is based on aren't that accurate) and all sorts of consequences follow. Then, we have to consider how you actually change pressure "reversibly" (hint:you don't). The point of this level of treatment of thermodynamics is to show you how useful it can be to get an accurate idea of what is going on. It is accurate, unless your idea of "accurate" is errors in the parts per thousand range. (Ideal Gas Law is generally good to between 1 and 3% (at near STP), but it can be off over 10% for some not-so-uncommon gasses (and let's not even consider about gasses near their boiling points, or at cryo-temperatures or at mega-pressures...). A quick search for activity of graphite vs pressure indicates that at 100 bar, its activity coefficient is 1.01. That is, assuming an activity of 1.00 is a great approximation near STP. Liquids (but NOT solutions!) will be similar (but not quite as well 'behaved'). So, pragmatically, the answer to your question is:"Because it gives us an accurate estimate of the system's behavior." Another way to put it is: "Because it is useful." One other factor to consider is what the alternatives are, that is, if we wanted to get more accurate predictions, could we formulate a generally applicable means to do so? The answer is, as far as I know:no. There is no generally applicable (call this a "powerful") method; to get more accurate requires case-by-case considerations: how is heat gained or lost by system, what kind of container walls is it in contact with, how was complete combustion insured, what were the losses due to friction during compression and expansion, etc. etc.

Thank you very much. This is in general what I was suspecting would be the explanation, but I really wasn't sure so thank you.

It is not that case that the activity of solids and liquids is always taken as unity.
Instead, activity is unity for a pure solid or pure liquid at standard pressure.
This is a matter of definition of "activity" and "standard state".
According to the IUPAC definitions:
activity (relative activity), a

Defined by the equation
  $a=e^\frac{\mu-\mu ^0}{RT}$
  where $R$ is the gas constant, $T$ the thermodynamic temperature, $\mu$ the chemical potential
  and $\mu ^0$ the standard chemical potential the definition of which depends on the choice of the standard state.

and standard state

State of a system chosen as standard for reference by convention. Three standard states are recognized: For a gas phase it is the (hypothetical) state of the pure substance in the gaseous phase at the standard pressure p = p °, assuming ideal behaviour. For a pure phase, or a mixture, or a solvent in the liquid or solid state it is the state of the pure substance in the liquid or solid phase at the standard pressure p = p ° ...

So activity is 1 almost by definition when a substance is in its standard state, because $\mu - \mu ^0 = 0$  and $e^0 = 1$

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text:
How many alkanes having a molecular weight of hundred, are chiral?
I successfully found 2, but the answer key says that there are 4 such compounds possible. I found:

3-methyl hexane
2,3-dimethyl pentane

We can't have a cyclic ring as the masses won't match. What else could be made folks?

Are rings allowed?

Yeah, I just forgot about the rings!!

Yeah but then if you make a ring the molecular weight won't match

You are essentially correct.  Of the nine isomers of heptane, only two are chiral.  However, if you count each enantiomer as a separate molecule, then there are "four" chiral isomers:

(R)-3-methylhexane
(S)-3-methylhexane
(R)-2,3-dimethylpentane
(S)-2,3-dimethylpentane

What about a cyclobutyne ring with the triple bond between C1 and C2, and then one H and one ethynyl group -CCH on the third Carbon atom, and a group =C=CH2 on the 4th Carbon atom ? I am not sure this structure does make sense, with such a bent triple bond C-C. But it is C8H4 with molecular weight 100, and it is chiral.

Good find.  But, the question specifies alkanes.

Also Maurice I think cyclooctyne is the first cyclic alkyne that is reasonably stable.  Has anyone every reported a structure anywhere close to a 4-membered annular alkyne?  If not, you could write down infinity more structures that are chiral (e.g. C7H30, C2H100, C8H2, etc. etc.) and argue that maybe they are stable somehow at 100 K and a few terapascals of pressure....who knows after all?

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text:
Why is the value of specific rotation different in different cases?

An unknown compound with a mass of $4.5~\mathrm{g}$ is dissolved in enough carbon tetrachloride to make a total volume of $250~\mathrm{cm^3}$. The observed  rotation of this solution is $+357.75^\circ$ in a $25~\mathrm{cm}$ cell using the sodium D line. But if $4.5~\mathrm{g}$ is dissolved in $125~\mathrm{cm^3}$ we observed  rotation is $+355.50^\circ$. Calculate specific rotation for this compound. (assuming length of polarimeter tube is $1~\mathrm{dm}$)

My approach is starting with the formula stated on Wikipedia.
$$\text{Specific rotation (in first case)} =
\frac{+357.75^\circ}{\left(\frac{4.5}{250}\right)(1)}=19875^\circ=75^\circ$$
$$\text{Specific rotation (in second case)} =
\frac{+355.50^\circ}{\left(\frac{4.5}{125}\right)(1)}=9875^\circ=155^\circ$$
Any idea why the two answers are different? Did I miss out something?

Note that the concentration is doubled in the seconds case. Therefore, the expected rotation should be $2\times357.75^\circ=715.50^\circ$, which is more than a full circle. Substracting one circle $(360^\circ)$ gives the indicated $715.50^\circ-360.00^\circ=355.50^\circ$.

@Loong but the length changes in the second case too!You did'nt take that into account...Initially it was 2.5 dm but finally 1 dm.

Oh.My bad....25 cm is cell length!

Yes, the remarks “in a $25\ \mathrm{cm}$ cell” and “assuming length of polarimeter tube is $1\ \mathrm{dm}$” are confusing. If $2.5\ \mathrm{dm}$ was used in the first case, and $1.0\ \mathrm{dm}$ was used in the second case, then this factor has also to be taken into account.

The "true" rotation may be the experimentally observed rotation or the experimentally observed rotation +/- (n*180°).  You must find at what multiple of 180° the experimental rotations are equal; this is the "true" rotation.

@ron You mean a rotation of $+181^{0}$ should be written as $-179^{0}$ ?

No, an experimentally observed rotation of +181° may actually be +181, +1, +361, -179, etc. That is why the experiment is run at 2 concentrations, then you can solve the equation and determine the "true" rotation.

Wait a sec!I did'nt get you @ron Which equation should I solve? (You may take the question in the post as an example if you wish )

You mean something like $\theta_1 + n\pi=\theta_2+m\pi$ ?But there can be infinite solutions of such an equation...

There is a second condition relating to the change in concentration and path length.  For example, you measure a rotation of 10°, this may really be 10, 190, -170, etc.  Now, repeat the experiment at 1/10 the concentration - is the observed rotation 1, 19, -17, etc.

@ron I'm still not sure if I understood you.Suppose in the first case we get a specific rotation $75^{0}$ and in second case we get a  specific rotation $155^{0}$ what will be the true rotation according to you?

If indeed the path lengths are equal, then rot_1=357.75 + (n180) and, due to concentration 2rot_1 = rot_2.  Solving we find that n=-2, therefore rot_1 = 357.75 +(-2*180) or -2.25.  This "true" observed value may now be converted to the specific rotation using your formula.

@Loong I took the liberty to correct that O.o

If indeed the path lengths are equal, then $rot_1=357.75 + (n*180)$ and, due to concentration $2*rot_1 = rot_2$. Solving we find that $n=-2$, therefore $rot_1 = 357.75 +(-2*180)$ or $-2.25$. This "true" observed value may now be converted to the specific rotation using your formula.
-by @ron in the comments above.

A polarimeter is typically a long cylindrical tube with flat ends. I think the 25 cm was referring to the diameter of the tube. The other wrinkle here is that the angle measurement may be + or -. So 357.75 may in fact be -2.25.
Specific Rotation (in first case)=$$\frac{-2.25^{0}}{(\frac{4.5}{250})(1)}= - 125^{0}$$
Specific Rotation (in second case)=$$\frac{-4.50^{0}}{(\frac{4.5}{125})(1)}=-125^{0}$$

Could you check my calculation in the original question please?Why is the answer not matching if I take the original values with +ve sign?

@ZOZ - How can I do the calculations wrong and get a good answer?

I never said your answer is wrong or your calculation is wrong.I'm asking why is my calculation wrong?Where did I go wrong?See the question...

Your problem was that you assumed that the reading for the solutions were positive when the are in fact negative. Think of a 360 degree compass. If you draw an angle of +357.75 degrees that is the same as drawing an angle of -2.25 degrees.

In other words think of a 360 degree compass from geometry class. They are graduated from 0 to 360. There are no negative markings. A polarimeter is setup the same way. There are no negative angles on the actual dial.

So you mean if the observed angle is >$180^{0}$ then we must take angle as negative in the calculation? I've never read something like that..could you cite a source?

Molecules can rotate light + or -. If a solution rotates light -10 degrees, then half the concentration will rotate the light -5 degrees. But my readings will be 350 degrees for the first solution and 355 degrees for the second solution since a  polarimeter doesn't have - degrees.


text:
Gas densities and the atmosphere
The density of nitrogen is greater than the density of oxygen, hence the denser nitrogen would be present in the lower atmosphere. If so it would be deprived of oxygen as it would be pushed to the higher atmosphere due to its low density. But we live due to the intake of oxygen from the lower atmosphere. How is this possible?

Related: Why ozone being heavier than air does not settle down?, Ozone is above, nitrogen is below, and Chlorofluorocarbons rising

You have it backwards. Oxygen gas is more dense than nitrogen gas.

Gas molecules in the air continuously collide with each, shoot off in different directions etc. This constant motion is like stirring a fluid. You wouldn't observe perfect density separation in a container with oil and water in it if you keep shaking/stirring it. The situation is similar here, the atmosphere is a fluid that is continuously being agitated. I believe this to be the reason, at least off the top of my head.

Although you mixed up the densities of $\ce{N2}$ and $\ce{O2}$, $\ce{O2}$ being more dense than $\ce{N2}$ rather than the other way around, the meaning of you question is still clear and valid.  To paraphrase: "Why  aren't the Earth's gases separated based on density, such that the most dense lie at the bottom of the atmosphere and the less dense gases rise to the top?"
According to this Wikipedia article:

The homosphere and heterosphere are defined by whether the atmospheric
  gases are well mixed. The surface-based homosphere includes the
  troposphere, stratosphere, mesosphere, and the lowest part of the
  thermosphere, where the chemical composition of the atmosphere does
  not depend on molecular weight because the gases are mixed by
  turbulence.

A more thorough discussion of the nature of the homosphere and heterosphere with respect to you question can be found in my answer here.

The density of oxygen is greater than that of nitrogen. At a certain set temperature and pressure, all gases occupy the same volume. Since oxygen has a higher molar mass, it is denser.


text:
Internal pressure of ideal gas
$$\left(\frac{\partial U}{\partial V}\right)_{\!T}$$
I know that this partial derivative is equal to zero for an ideal gas, but how do I determine that? Do I need to use the fundamental thermodynamic relations?

In fact, you need to use two of the fundamental relations.
For an ideal gas, the internal energy $U$ can be considered to be a function of $V$ and $T$. So, the total differential of $U$ would read
$$\mathrm{d}U = \left(\frac{\partial U}{\partial V}\right)_{\!T}\,\mathrm{d}V + \left(\frac{\partial U}{\partial T}\right)_{\!V}\,\mathrm{d}T \tag{1}$$
The first partial derivative is the internal pressure, which we are trying to determine.
The other thing that we know is that (assuming only expansion work and no net reaction)
$$\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V \tag{2}$$
The idea here is that we want to change the term involving $\mathrm{d}S$ into a combination of $\mathrm{d}T$ and $\mathrm{d}V$, so that we can compare coefficients with equation $(1)$. So, we just express $S$ as a function of $T$ and $V$:
$$\mathrm{d}S = \left(\frac{\partial S}{\partial V}\right)_{\!T}\,\mathrm{d}V + \left(\frac{\partial S}{\partial T}\right)_{\!V}\,\mathrm{d}T \tag{3}$$
Substituting $(3)$ into $(2)$
$$\begin{align}
\mathrm{d}U &= T\left[\left(\frac{\partial S}{\partial V}\right)_{\!T}\,\mathrm{d}V + \left(\frac{\partial S}{\partial T}\right)_{\!V}\,\mathrm{d}T\right] - p\,\mathrm{d}V \tag{4} \\
&= \left[T\left(\frac{\partial S}{\partial V}\right)_{\!T} - p\right]\mathrm{d}V + \left[T\left(\frac{\partial S}{\partial T}\right)_{\!V}\right]\mathrm{d}T \tag{5}
\end{align}$$
Comparing the coefficient of the $\mathrm{d}V$ term with $(1)$ we find that1
$$\left(\frac{\partial U}{\partial V}\right)_{\!T} = T\left(\frac{\partial S}{\partial V}\right)_{\!T} - p \tag{6}$$
Equation $(6)$ is entirely general and applies to all substances, not just ideal gases. Now, we turn to the Helmholtz free energy:
$$\mathrm{d}A = -S\,\mathrm{d}T - p\,\mathrm{d}V \tag{7}$$
From this we obtain the Maxwell relation
$$\left(\frac{\partial S}{\partial V}\right)_{\!T} = \left(\frac{\partial p}{\partial T}\right)_{\!V} \tag{8}$$
For an ideal gas $p = nRT/V$, so
$$\left(\frac{\partial p}{\partial T}\right)_{\!V} = \frac{nR}{V} \tag{9}$$
Finally, substituting $(8)$ and $(9)$ into $(6)$ gives
$$\begin{align}
\left(\frac{\partial U}{\partial V}\right)_{\!T} &= T\left(\frac{nR}{V}\right) - p \tag{10} \\
&= \frac{nRT}{V} - p \tag{11} \\
&= p - p \tag{12} \\
&= 0 \tag{13}
\end{align}$$
as desired.

Notes
1 Some texts may write that you can obtain $(6)$ from $(2)$ simply by "dividing" by $\mathrm{d}V$ and imposing conditions of constant temperature:
$$\begin{align}
\mathrm{d}U &= T\,\mathrm{d}S - p\,\mathrm{d}V & & \tag{2} \\
\frac{\mathrm{d}U}{\mathrm{d}V} &= T\frac{\mathrm{d}S}{\mathrm{d}V} - p\frac{\mathrm{d}V}{\mathrm{d}V} & &\text{(dividing by }\mathrm{d}V\text{)} \tag{14} \\
\left(\frac{\partial U}{\partial V}\right)_{\!T} &= T\left(\frac{\partial S}{\partial V}\right)_{\!T} - p\left(\frac{\partial V}{\partial V}\right)_{\!T} & &\text{(constant }T\text{)} \tag{15} \\
&= T\left(\frac{\partial S}{\partial V}\right)_{\!T} - p & &\text{(derivative of }V\text{ with respect to itself is }1\text{)} \tag{6}
\end{align}$$
This gives the correct result, and in general this "shortcut" works, but it is mathematically not rigorous.

Why the shortcut is not rigorous ? is it because $dV$  can be zero ?

That's just how calculus works. dx/dy is not actually a fraction of infinitesimals and you cannot "divide through" by infinitesimal quantities. Sometimes it works, but sometimes it does not. The best example is: $$\left(\frac{\partial x}{\partial y}\right){!z} \left(\frac{\partial y}{\partial z}\right){!x} \left(\frac{\partial z}{\partial x}\right)_{!y} = -1$$ (if you simply cancelled out everything, you would arrive at $+1$ - the wrong answer).

@A---B It doesn't matter whether dV is zero or not. This goes back to the very fundamentals of calculus: although we write it like a fraction, the derivative is not a fraction.

But, I think in non standard analysis we can do that, where derivative is defined as $$y^\prime = \operatorname{st}\left({\Delta y \over \Delta x}\right)$$; $\Delta x$, $\Delta y \in \mathbb{^*R}$ and are +ve infinitesimals and $\operatorname{st}$ is the standard part. I don't know about partial differentiation under non-standard analysis, so I won't comment on that. I have just studied a bit of normal differentiation under non-standard analysis.

Right; you've lost me. I don't know this stuff.

I, myself is lost. I just know the surface of non standard analysis. You can start here https://www.math.wisc.edu/~keisler/calc.html .


text:
Are complex ions thermochromic?
Does raising the temperature of a complex ion (formed from a transition metal ion and ligand(s)) affect its color? For example, if the temperature of $\ce{[Cr(H2O)6]^3+}$ is raised, will its color change?

Most don't change colour by any amount that one can detect. There are some which are thermochromic - I have had the pleasure of working with $\ce{Cu(deen)22}$ - but most aren't.

Indeed, this effect is observed in certain select cases. The effect of spin-spin cross over has been mentioned in this regard. There can be some other effects at play, namely: phase transitions, change in ligand geometry, equilibria between different molecular structure, and change in the number of solvent molecules in the coordination sphere.
Thermochromic behavior can be divided into two main classes: continuous and discontinuous. Some color changes are gradual; as the temperature increases over time, the color slowly changes. This can be due to the breaking or rearrangement of the lattice. This type of rearrangement is also known as continuous thermochromism.
On the other hand, a dramatic change in color can occur at a specific temperature or over a very small temperature range. This is known as discontinuous thermochromism.
I shall attempt to supplement what others have said by providing some illustrative examples.
I. $\ce{[Fe(NH2trz)3]Br2·H2O}$
This compound exhibits a spectacular reversible purple-white thermochromic transition upon heating−cooling with a large hysteresis centered at room temperature. The thermochromic transition is related to the spin crossover (SCO) from the low-spin state (LS, purple, $S = 0$, diamagnetic) to the high-spin state (HS, white, $S = 2$, paramagnetic).

(A) after cooling at $77~\mathrm{K}$ and (B) after heating at $353~\mathrm{K}$. The white form (B) is attracted to the magnet, whereas the purple form (A) is slightly repelled by the magnet.
Spin crossover (SCO) complexes are essentially $\mathrm{d}^4$−$\mathrm{d}^7$ first-row transition-metal octahedral complexes.
The occurrence of ST in coordination compounds of transition-metal ions is governed by the relationship between the strength of the ligand field (the electrostatic field acting at the central metal ion) and the mean spin-pairing energy. Octahedral complexes of $\mathrm{d}^4$–$\mathrm{d}^7$ ions may be either HS or LS, depending on whether the ligand field strength is weaker or stronger, respectively, than the spin pairing energy.
In order for thermally induced ST to occur the difference in energy must be of the order of thermal energy, i.e $k_\mathrm{B}T$. An increase in temperature favours the HS state, while lowering the temperature favours the LS state.
Thermal spin transition occurs nearly exclusively with coordination complexes of $\mathrm{3d}$ metal ions, because for $\mathrm{4d}$ and $\mathrm{5d}$ transition element compounds on the basis of ligand field theory, because the strength of the ligand field increases considerably, relative to analogous $\mathrm{3d}$ compounds and is generally much greater than the spin pairing energy (hence, these mostly show LS behaviour).
II. $\ce{Cu2HgI4}$
$\ce{Cu2HgI4}$, (and the related $\ce{Ag2HgI4}$) undergo a temperature-dependent order-disorder phase change at ca. $67~^\circ\mathrm{C}$, reversibly changing from a bright red solid material at low temperature to a dark brown solid at high temperature, with intermediate red-purple states. the colours are probably due to $\ce{Cu(I)}$-$\ce{Hg(II)}$ charge-transfer complexes.

All the metals involved are $\mathrm{d}^{10}$ so one can rule out $\mathrm{d}$-$\mathrm{d}$ transitions, and spin-spin crossover. An electron transfer of some kind might be occurring, especially since the colours observed are so intense. Based on the pi-donating characteristics of iodide, one can also rule out metal to ligand charge transfer. The colour could be ligand to metal or metal to metal charge transfer, although the electron would have to enter something other than the d orbitals.
As the temperature increases and the lattice is rearranged, the $\ce{Cu(I)}$ and $\ce{Hg(II)}$ ions remain in the same environment relative to the iodide ions, but their environment in relation to the other metal ions is changing, and this supports a metal-metal charge transfer explanation for the colour.
III. Bis(diethylammonium)tetrachlorocuprate(II)
At room temperature, this compound is a bright green colour and changes
discontinuously to bright yellow at the phase transition temperature.
The thermochromic phase transition in this compound is quite complex Structurally, the salt transforms between two monoclinic unit cells, with three independent $\ce{CuCl4^2-}$ anions in the low-temperature phase and two independent $\ce{CuCl4^2-}$ ions in the high-temperature phase. The phase transition is first order due to the lack of any crystallographic
relationship between the two phases.
From reference 5:

No simple structural transformation pathway can be envisioned for converting one phase to the other. This accounts for the stability of the supercooled high-temperature phase as crystals of that phase, grown from the melt, can frequently be kept at room temperature for periods of
several weeks.The crystallographic study of the high-temperature phase shows the presence of disorder for both the cationic and anionic species
in the structure.

At temperatures below ca. $50~^\circ\mathrm{C}$, the four coordinate compound has a square planar geometry even though it is not a $\mathrm{d}^8$ compound, owing to the short hydrogen bonds between the hydrogens of the counter ion, diethylammonium, and the chlorines on the metal compound (this favours the square planar geometry).  When the temperature passes the phase transition temperature, the colour changes to yellow because the hydrogen bonds lengthen to (because of the phase transition between the two crystalline phases) and the geometry changes to a distorted tetrahedral  The change in geometry changes the arrangement of the d orbitals.

Square planar is pictured on the left, and tetrahedral is pictured on the right.
There is no stable intermediate between the square planar and the distorted tetrahedral geometry, so the change is an example of discontinuous thermochromism. Individual clumps of crystals may appear as either green or yellow depending on their exact temperature.
References

http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3596041/

http://pubs.acs.org/doi/abs/10.1021/ed4000487

http://people.umass.edu/mjknapp/Chem242/2004%20Experiments/CuHg2I4Lab.pdf

http://www3.amherst.edu/~thoughts/contents/amberger-thermochromism.html

http://pubs.acs.org/doi/abs/10.1021/ja00230a020

I should add a comment about the complex I mentioned. $\ce{Cu(deen)22}$ (deen = N,N-diethylethylenediamine) is similar to your diethylammonium Cu(II) complex, in that its low-T form is square planar/tetragonal (due to Jahn-Teller effects). Upon heating it moves towards a tetrahedral geometry and as you pointed out, this changes the d-d transition energy and the absorption wavelength. And here is a reference: Inorg. Chem. 1979, 18, 2687.

As correctly commented by @orthocresol, there are instances of thermochromicity for transition metal complexes, but it is rather seldom.
It may occur provided the molecular orbitals of the complex are filled partially only, such that you may shuttle between low- and high-spin state, and if the energy provided by heating is suffient to provide the necessary energy to pass the activation barrier.  This is then called spin switching.  Some of the complexes of iron(II) are examples for this, and this open access article, beside the more senior (behind paywall) here may shed some light on it.

The effect of spin-crossover in some Fe complexes causes thermochromic and photochromic effects. The low spin state, diamagnetic complexes are usually highly coloured while the paramagnetic, high spin ones are weakly coloured. The light induced process is called light induced excited state spin trapping (Liesst). The spin change and hence colour change on cooling & irradiation shows considerable hysteresis. (The theory has been examined by Gutlich in Angew. Chem. Int. Ed. Engl. 1994. vol33, p2024-2054)

Another example of complex compound showing thermochromism is a complex of copper iodide and pyridine known as  copper(I) pyridine iodide cluster complex.

(source)
The complex has a tetrahedral tetrameric structure with a Cu4I4 cube in it’s middle and 4 pyridine ring surrounds this little cage. This crystalline solid shows a bright yellow fluorescence at room temperature and an equally intense blue-violet fluorescence when cooled in liquid nitrogen.

(source)
This behavior is based on the temperature-dependent shift of the same band in the emission spectrum, often due to a change in geometry. This video beautifully explains the thermochromism of copper(I) pyridine iodide cluster complex.
References

http://pubs.acs.org/doi/full/10.1021/ed200736b?src=recsys&
http://labphoto.tumblr.com/post/105201858343/something-really-interesting-fluorescence
https://www.researchgate.net/publication/229968215_Fluorescence_thermochromism_of_pyridine_copper_iodides_and_copper_iodide
https://www.youtube.com/watch?v=8jDFx7yeBvw


text:
Suppose a heated 5.00 g sample of sodium bicarbonate lost a mass of 1.00 g, what percent of impurity is in the sample?
The 5.00 g is a mixture of sodium bicarbonate and an unknown substance, making it impure.
Balanced decomposition equation:
$$\ce{2NaHCO3(s) -> NaCO3(s) + H2O(g) + CO2(g)}$$
There are 0.06 mol of $\ce{NaHCO3}$ in a pure 5.00 g sample of sodium bicarbonate.
Theoretically, 0.0298 moles of $\ce{H2O}$ and $\ce{CO2}$ would be lost durin the decomposition of 5.00 g of sodium bicarbonate.
0.0298 mol converted to grams is:
0.536 g of $\ce{H2O}$
1.31 g of $\ce{CO2}$

First there is a big gotcha. We'll let $x$ be the grams of sodium bicarbonate and $y$ be the grams of the impurity. Since we were given no information about the impurity we must assume that the impurity neither loses mass nor gains mass when heated with the sodium carbonate. So: $$x + y = 5.00$$
Well the balanced equation is really:
$\ce{2NaHCO3(s) -> Na2CO3(s) + H2O(g) + CO2(g)}$
so 2 moles of $\ce{NaHCO3}$ yields 1 mole of $\ce{Na2CO3}$. Thus the mass conversion factor is: $$\dfrac{0.5*\text{MW(}\ce{Na2CO3}\text{)}}{\text{MW(}\ce{NaHCO3}\text{)}} = \dfrac{0.5*106.0}{84.0}= 0.6310$$
So we now also have: $$0.6310x + y = 4.00$$
Subtracting the second equation from the first we get: $$0.3690x=1.000$$ so $$x=2.71g$$ and $$\text{Purity} = \dfrac{2.71}{5} = 54.2\%$$ $$\text{Impurity} = 100\% - 54.2\% = 45.8\%$$


text:
Nomenclature of polyatomic ions
Is there a way to remember polyatomic ions (i.e $\ce{PO4^{3-}}$, $\ce{SO4^{2-}}$ etc.). For example, our professor will ask on an exam what the equation for iron (III) carbonate.  Obviously the (III) means iron has a 3+ oxidation state. However, if one was not too familiar with the number of oyxgens or charges on specific polyatomic ions, how would you know the symbol for carbonate ($\ce{CO3^{2-}}$)? She gives us a periodic table but I dont think that will help if you dont have carbonate memorized?

Pfft, it's crystal clear in comparison with traditional names of compounds.

Initially the idea was that -ate suffix is reserved for a mononuclear acid with highest oxidation state, while -ite suffix was reserved for a mononuclear acid with second highest oxidation state. The per-...-ate construct was reserved for anion with 'extra' oxygen, typically a peroxocompound and hypo-...-ite construct was reserved for acid extra low on oxygen.
However, since for halogens $\ce{XO4^-}$ anions were obtained rather lately, perhalogenates has this formula and simple halognates are $\ce{XO3^-}$. Mn subgroup (Mn, Tc, Re) inherited this in the sense that highest oxidation state corresponds to per- ... -ate construct
Similarly, ortho- prefix was reserved for mononuclear acids with highest number of hydroxogroups possible, while meta- prefix was reserved for an acid with one water molecule less (typically, polymer/oligomer)
Putting it all together using phosphorous as an example

Perphosphate $\ce{PO5^{3-}}$
Ortophosphate - $\ce{PO4^{3-}}$
Metaphosphate - $\ce{(PO3^-)_n}$
Phosphite $\ce{HPO3^{2-}}$ (since $\ce{H3PO3}$ is actually $\ce{H2[HPO3]}$ for some arcane reasons) Note: depending on the chemistry course you are taking (less advanced courses mainly), some will use $\ce{PO3^{3-}}$ as phosphite
Hypophosphite - $\ce{H2PO2^-}$ (since $\ce{H3PO2}$ is actually $\ce{H[H2PO2]}$

And for chlorine (see above about perhalogenates)

Perchlorate $\ce{ClO4-}$
Chlorate $\ce{ClO3-}$
Chlorite $\ce{ClO2-}$
Hypochlorite $\ce{ClO-}$

Not every degree needs to be present, for example only simple carbonates and percarbonates exist, and sometimes extra stages (usually polynuclear) occur, such as dithionites. This system is widespread in areas heavy on traditions, such as medicine and heavy industry.
There is a gentle pressure from IUPAC to move to much less confusing systematic naming, but the success is limited, especially considering widespread chemicals, which are familiar anyway. It is similar to situation with RJ plugs: everyone recognises phone and Ethernet connectors, but very few care about their full formal name.

I tried giving my extra-teaching pupils the following guidance:

Determine the element whose name is central.

Carbonate → carbon
Phosphite → phosphorus
Periodate → iodine
Hypochlorite → chlorine

Initially drop all per- or hypo- prefixes; we’ll get back to them later. Also, initially assume -ate, we’ll determine -ite later.
If this element is in the second period:

for an -ate name, add three oxygens.

carbonate → $\ce{CO3^n-}$

assume the highest oxidation state of the central element.

For carbon, that is $\mathrm{+IV}$ → $\ce{C^{IV}O3^n-}$

If this element is a halide:

for an -ate name, add three oxygens.

iodate → $\ce{IO3^n-}$

assume an oxidation state of $\mathrm{+V}$; not the maximum oxidation state!

iodate → $\ce{I^{V}O3^n-}$

For all other elements (non-halogens, third and higher periods):

for an -ate name add four oxygens.

phosphate → $\ce{PO4^n-}$.

assume the highest oxidation state.

for phosphorus, that is $\mathrm{+V}$ → $\ce{P^{V}O4^n-}$

Calculate the ion’s charge by assuming oxygen’s oxidation state to be $\mathrm{-II}$ (the standard).

Carbonate: $(+4) + 3 \times (-2) = 4 - 6 = -2$
Thus, carbonate is $\ce{CO3^2-}$
Iodate: $(+5) + 3 \times (-2) = 5 - 6 = -1$
Thus, iodate is $\ce{IO3-}$
Phosphate: $(+5) + 4 \times (-2) = 5 - 8 = -3$
Thus, phosphate is $\ce{PO4^3-}$

And because we ignored hypochlorite so far, but we must do the same for chlorate:

Chlorate: $(+5) + 3 \times (-2) = 5 - 6 = -1$

You have successfully determined charge and sum formula of an -ate ion.
If your ion ends in -ite, subtract one oxygen leaving the rest unchanged.

Phosphate is $\ce{PO4^3-}$ thus phosphite is $\ce{PO3^3-}$
Chlorate is $\ce{ClO3-}$ thus chlorite is $\ce{ClO2-}$

If your ion includes a hypo- prefix, subtract another oxygen from the -ite species, leaving the rest unchanged.

Chlorite is $\ce{ClO2-}$ thus hypochlorite is $\ce{ClO-}$

If your ion includes a per- prefix, add another oxygen onto the -ate species leaving the rest unchanged.

Iodate is $\ce{IO3-}$ thus periodate is $\ce{IO4-}$

Any additional hydrogen- prefixes are added as $\ce{H+}$ ions, thus they reduce the charge accordingly.

This method reduces the problem to correctly identifying the element in the periodic table. It may be a lot to remember, but it’s basically the extremes of the main groups (top period and left-most non-noble gaseous group) that have three oxygens while all the rest have four. Remember the halogens’ special case with respect to oxidation states and you’re good.
In time, this will happen automatically, i.e. you will see arsenate and automatically read $\ce{AsO4^3-}$.


text:
Production of hexavalent chromium during electrolysis using a stainless steel as the cathode
I want to do rust removal by electrolysis (sodium carbonate, water solution) Can I use a stainless steel basket as a cathode or will it produce hexavalent chromium ?
I know if I use stainless steel as the anode it does produce hexavalent chromium, but what if it is a stainless cathode ?
I have a solution of sodium carbonate, water and hexavalent chromium. What can I do to eliminate the chromium or make it safe, preferably safe enough to dump in the drain if that is possible.

In anode, things get oxidized. That is why the chromium in anode would become hexavalent chromium (hexavalent chromium is the oxidized form of chromium).
In cathode, things get reduced (opposite of oxidized). That is why you could remove rust, because rust is the oxidized form of iron, so if you reduce rust, you would get iron.
You will not obtain hexavalent chromium, because that is not possible through reduction of chromium; that is only possible through oxidation of chromium, which is impossible at a cathode.

You can reduce hexavalent chromium to trivalent chromium with sulphite in acidic environment:
$$\ce{Cr2O7^2- + 8H+ + 3SO3^2- -> 2Cr^3+ + 3SO4^2- + 4H2O}$$
Then you can precipitate out the trivalent chromium with sodium hydroxide:
$$\ce{Cr^3+ + 3NaOH -> Cr(OH)3(s) + 3Na+}$$

There's no reason a ss basket couldn't work as a cathode, although you need to control current (density) and voltage. If I were to post a question here asking about a solution I made of tri-nitro-toluene (TNT) and how I should "dispose" of it, or of plutonium, I would hope no one would give me the correct technical answer. The practical answer is: if you have to ask, you shouldn't be handling it!! Chromium is an element. It is forever. There are all sorts of LAWS concerning how you must handle waste containing it. Reducing Cr(VI) to Cr(III) does reduce the hazard, but does not eliminate it. You can not legally dump caustic solutions down the drain in the USA, which means using NaOH is a really bad idea. It is against the law. When I was involved in small scale (hundred of gallons) use of Cr(VI), we used ferrous sulfate to reduce it and to precipitate out the Cr(III). There are generally two methods: one uses sulfuric acid to get to a low pH and sodium metabisulfate to reduce it, the other ferrous sulfate and alkaline pH. In either case, you'd have to then neutralize the wastewater back to neutral.  Cr is fairly easy to trace through a sewer system; that means if you dump it, be prepared to spend some time in jail.

The amount is less than 5 gallon and it only contains tiny amounts of chromium, However in an effort to try to be environmentally responsible I ask the question.

"if you have to ask, you shouldn't be handling it" So the people who handle it were just born with the knowledge?


text:
How to find the rate of effusion given rates?
Question
At $25~^\circ\mathrm{C}$, a sample of $\ce{NH}$ effuses at the rate of $0.050$ moles per minute. Under the same conditions, what is the molar mass of a gas that effuses at approximately one half of that rate?
My steps and thoughts
I converted $25~^{\circ}\mathrm{C}$ to $298~\mathrm{K}$ and then I was thinking of using $pV=nRT$ and I got stuck.
I am thinking of using molar mass, but I don't know how to incorporate that.
Please, only give hints not the answer.

HINT:
This question seems to be a simple application of Graham's Law (assuming you know Graham's Law), i.e the rate of effusion of a gas is inversely proportional to the square root of the mass of their particles. This empirical law can be formulated as follows:
$$\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{\text{M}_2}{\text{M}_1}}$$
Solution:

 I don't know what the OP meant by $\ce{NH}$, I assume the gas in question is $\ce{NH3}$ with rate of effusion ($R_1$) = 0.050 $\mathrm{mol \min^{-1}}$.

 $R_2 = \frac{R_1}{2}$ and $M_1 = 17.031$, plugging all of this in

 $M_2 = M_1\cdot \left (\frac{R_1}{R_2} \right )^2 = 4\cdot M_1 = 68.124$


text:
What is the difference between fractional freezing and fractional crystallization?
More simply put, what is the difference between freezing and precipitation when considering a mixture?
Take two scenarios:
First a mineral in water. If the temperature of this solution is lowered, the solubility of the mineral will decrease and eventually precipitate out of solution. This could be done in a cascade type setup so that each time, the solution is cooled further and the liquid phase becomes more and more concentrated in water.
Second, consider a mixture of ethanol and water. Here too, I could lower the temperature until crystals first form. They would be almost pure water, leaving behind a solution higher in concentration in alcohol. I could repeat this procedure until I reach the eutectic mixture.
In both cases, a drop in temperature is allowing a solid phase to form out of an aqueous solution, yet one is referred to as partial crystallization and the other is referred to as partial freezing. I am not sure I see the difference.

@Jan not correct. Many salts are more soluble at higher temperature. A solution just barely saturated at 90 °C will crystallise abundant solids when cooled down to 0 °C. Calcium chloride is one such salt.

@Michael Oh, I see my error. I overlooked the in between mineral and water and thought the question were about mineral water. In mineral water, of course, the concentration of ions is much too low. But for a mineral dissolved in water that’s different.

Surely the key difference is whether you are making crystals of the solvent or the thing dissolved in the solvent. Crystallisation relies on solutes having lower solubility at lower temperature (usually) and this work to purify the dissolved compound. If you want to purify the solvent (or concentrate the remaining components) you partially freeze it removing the purer crystals of solvent.

There is no difference. Strictly speaking, "freezing" does not necessarily include crystalline solids, and may include amorphous materials that precipitate out of solution. But, this does not really matter in the context of these eutectic (or similar) systems.
I suspect the latter is referred to as "partial freezing" because the solid is water, whereas it is more general in the former. We rarely speak of water crystallising to ice, but rather water freezing to ice. This is more of a linguistic issue than scientific.


text:
Why do sweet biscuits melt faster than a salty biscuits?
I live in India and here eating biscuit after dipping it in water is a famous habit. On careful observation my classmate and me found that a sweet biscuit melts faster than a salty one when dipped in water.
I don't know which law or which experimental result is working behind it, it is possible that the result is due to different composition of sweet and salty biscuits or maybe something else. I have got my mind struck there.

I'm voting to close this question as off-topic because...I can't really pinpoint it, but it's just not right for chem.SE.

I removed the clutter about closing or downvoting the question. Such comments are not in the general spirit of the site. The question is too broad, there are way too many biscuits out there to make a definite statement. It might fit at [cooking.se] or not, please check their policies before re-asking it there. || @orthocresol While I agree with your comment, It might be better to not engage in these kinds of situations.

The disintegration of a solid food when it comes in contact with water is due to several factors, one of which might be dissolution but "melting" is unlikely to be occurring (unless the water is significantly above room temperature). I can not answer your question, it has too many culture specific concepts which I (USA) am not familiar with. In the UK (perhaps India?) a biscuit is what we call a cookie. The compositions of most cookies (in the USA) are not "biscuit-like". Even in India, I imagine, there is probably broad differences in the composition of a biscuit from cook to cook and from region to region. There are also probably significant differences in the processing. Wheat flour (only type I have some familiarity with, sorry) can be kneaded so that semi-crystalline polymers become stretched out and increase the materials toughness and strength. Baking temperatures and times also obviously will affect the final consistency (and composition: dehydration is usually a major process during baking, too little and the material has little cohesion (is gooey), too much and the material is more like a brick than a foodstuff. And I am ignoring the effects of yeast and density on the strength and cohesion of the finished biscuit. It should be obvious that a biscuit without large internal voids will be slower to absorb water. Most sugar cooky batter contains little water which leaves a large amount of very soluble sugar behind when it is baked off. Sugar is very soluble and is structurally weak, (unless caramelized). Salt, while also very water soluble and not very strong, need only be present in small quantities to give a cookie a very salty taste. If little sugar and salt is present, then the flour (which is generally not water soluble) will be slow to disintegrate.

i have observed it  in at least 10 kinds of biscuits and have the same results (different time for different biscuit,but sweet biscuit always melt before salty one) that's why i also think that it may be due to the temperature and baking time,which is affecting the composition and hence the melting time.


text:
Workflow for many-molecule/many-conformer calculations in Spartan, GAMESS, AND Gaussian
I wish to investigate the antioxidant properties of several hydrocarbon compounds, which are each mono- or polyhydroxylated, specifically at the phenyl ring. I wish to gain insight into my workflow for deciding which conformers to utilize in these calculations. At the moment, I am most concerned that I am developing the correct algorithm for narrowing down the compounds to be calculated in a fairly-accurate manner. Once I have these compounds, I may then decide how to approach the radical forms (i.e., compound-OH vs. radical form, or compound-O*).
The main subject of this question is choosing which conformers to keep- and thus compute at higher levels of theory- by the use of Boltzmann weights. I am getting these weights from Spartan, automatically, calculated by what I assume is:
$$\frac{N_i}{N_j}=e^{-(\epsilon_i-\epsilon_j)/kT}$$
This formula is what I intend to use for calculating the weights after GAMESS and Gaussian calculations, as I have only reached the last set of Spartan calculations, albeit for only some molecules. Without any further explanation, here is my intended workflow (I will add any edits to the bottom of this original post- thank you for your help):

Spartan

a. Conformer Distribution

i.   MMFF level of theory
ii.  (Monte Carlo algorithm applied)
iii. Max conformers examined=10,000
iv.  Keep 100% of conformers

b.   Energy- Semi-empirical

i.   PM3 Level of theory
ii.  Calculated at both gas and water (PCM)

c.   Energy- Hartree-Fock

i.   3-21+G* level of theory
ii.  Calculated for both gas and water (PCM)

Only calculated for conformers showing SE PM3 Boltzmann weight of ≥.001

d.   Energy- Density Functional

i. ωB97X-V/3-21++G** level of theory
ii. Calculated for both gas and water (PCM)

Only calculated for conformers showing HF 3-21+G*Boltzmann weight of ≥.01

GAMESS

a. Equilibrium Geometry- Density Functional

i.   Only calculated for conformers showing ωB97X-V/3-21++G** Boltzmann weight of ≥.05
ii.  Three separate methods for both gas and water (PCM)

B3LYP/6-31+G* level

ωB97X-D/6-31+G* level

PBE0 (PBE1PBE)/6-31+G* level

b.   Equilibrium Geometry- Density Functional

i.   Only calculated for conformers showing (B3LYP, ωB97X-D, & PBE0)/6-31+G* Boltzmann weight of ≥.05
ii.  Three separate methods for both gas and water (PCM)

B3LYP/6-311++G** level

ωB97X-D/6-311++G** level

PBE0/6-311++G** level

Gaussian

a. Equilibrium Geometry- Density Functional

i.   Only calculated for conformers showing (B3LYP, ωB97X-D, & PBE0)/6-311++G** Boltzmann weight of ≥.10
ii.  One method used in Gaussian, for both gas and water (IEF-PCM), for the corresponding results from the previous GAMESS calculations, or six separate files (i.e, B3LYP H2O and gas, ωB97X-D water and gas, & PBE0 water and gas). To each of these 6 files, a radical form was generated (i.e. RO*), thus 12 separate files are created- 6(ROH) and 6(RO*).

Zero point energy M06-2X/6-311++G(3df,3dp) level

a.   Used to calculate energies of corresponding compounds and their radical counterparts (compound-OH and compound-O*)

edit(19.Sept'16): typos

It would be very useful if you tell us exactly what will you do with them.

Sorry, maybe I'm not understanding you correctly? This is exactly what I will do with the molecules, in trying to understand some of the correlations between bond dissociation energies in homolytic cleavage of the hydroxyl groups. These are mainly biological compounds which will likely never see the light of day, but to understand their properties further is of large interest to my academic pursuits.

In such case, I would only retain one conformer. Differences in bond energies among the conformers will be small enough compeared with the accuracy of DFT.


text:
Conversion of isonitrile (isocyanide) group into cyanide group?
In an isonitrile (isocyanide) group, the carbon atom has a negative charge (I know that this is due to the lone pair of electrons) instead of the nitrogen atom, but in my point of view, this seems a little unstable. So, why doesn't the isocyanide group convert into a cyanide group in which carbon doesn't have a negative charge (carbon valency fully satisfied)?
$$\ce{R-N+\bond{3}C- -> R-C\bond{3}N}$$

isocyanide has two resonance structures of which one has double bond between C and N and also doesn't involve charge separation.

@JM97 That resonance structure is formally a carbene, I would prefer to think of an isonitrile as a stabilised carbene rather than think of a carbene as a stabilised isonitrile.

No dupe, but related: http://chemistry.stackexchange.com/questions/45255/isonitrile-to-nitrile-conversion

Not a dupe, but it does answer the question, I suppose.

Isonitriles will, given the correct conditions, spontaneously transform into nitriles. The reaction mechanism is very interesting; it is basically a cationic [1,2]-migration of the $\ce{R}$ group under retention with a three-membered cyclic transition state. To reach this transition state, the $\ce{C-N}$ bond must be broken and the alkyl residue must form a cation-π interaction with the $\ce{C#N}$ triple bond. This is a non-classical carbonium cation, as shown in scheme 1.

Scheme 1: Mechanism of the [1,2]-migration and sp³-π interaction in the intermediate carbonium ion.
The reaction rates were extensively studied by Meier et al.[1] They note that the typical migration preference of $\text{tertiary} > \text{secondary}>\text{primary} > \text{methyl}$, indicative of cationic carbenium intermediates and observed e.g. in Criegee, Beckmann and Lossen rearrangements, does not apply here. Instead, a migratory aptitude similar to the Curtius or pinacol rearrangements is observed. Another interesting observation is ‘the almost complete lack of a bridgehead effect on the [reaction] rates’, further indicating that a planar carbenium transition state cannot be part of the mechanism and that bond angles and lengths will probably remain unchanged en route to the transition state.[1] When phenylic isonitriles react, next to no substituent effect is observed, indicating that the cation is not delocalised into the phenyl ring.
All of this evidence indicates towards the reaction being ‘properly prototypical for a sigmatropic 1,2-rearrangement via a nonpolarized, tight, hypervalent three-centered transition state.’[1] The authors go on to state that this is in perfect agreement with ab initio calculations, but do not provide an appropriate source.
Isonitriles were known in the 19th century; I am not able to find a definite first description but A. Gautier noted their reactions with acids in 1867.[2] In 1868, he goes on to be the apparant first to describe an isonitrile–nitrile rearrangement.[3] Weith, however, was probably the first to actually observe the reaction as such; he notes:[4]

Zu dem Ende wurde nach der  H o f m a n n ’schen Methode dargestelltes und durch Destillation möglichst von Anilin befreites Cyanphenyl 2—3 Stunden lang im geschlossenen Rohr auf 200–220° erhitzt. Nach dem Erkalten fand sich eine beträchtliche Menge dunkel gefärbter nadlicher Krystalle vor, der Röhreninhalt besass nicht mehr den Geruch des Cyanphenyls, dagegen den des Benzonitrils. [sic! to all apparant errors]
Thus, phenylisonitrile prepared by the Hofmann method and purified from aniline by distillation was heated in a sealed tube to 200–220 °C for 2–3 h. After cooling, a significant amount of dark-coloured, needle-like crystals were found; the tube’s content no longer smelt like phenylisonitril but like benzonitrile.

Therefore, for the rearrangement to take place temperatures above $150~\mathrm{^\circ C}$ appear to be required — which also explains why isonitriles are kinetically inert enough to be isolated under standard conditions and even with 19th century methods. (Gautier used $\ce{AgCN}$ to synthesise them from iodoalkanes.)[3] An activation energy as high as this makes sense, since, as noted above, the $\ce{C-N}$ bond must be broked and a significant rearrangement must take place for the carbonium ion to form.
Some isonitriles, namely those which upon dissociation form very stabilised cations such as benzylisonitrile, can also react according to a different path that is basically an $\mathrm{S_N1}$ rearrangement or can be considered in a chain-like structure. This, however, is a rather special case as Meier et al. note.[1]

References:
[1]: M. Meier, B. Müller, C. Rüchardt, J. Org. Chem. 1987, 52, 648. DOI: 10.1021/jo00380a028.
[2]: A. Gautier, Liebigs Ann. Chem. 1867, 142, 289. DOI: 10.1002/jlac.18671420304.
[3]: A. Gautier, Liebigs Ann. Chem. 1868, 146, 124. DOI: 10.1002/jlac.18681460108.
[4]: W. Weith Ber. Dtsch. Chem. Ges. 1873, 6, 210. DOI: 10.1002/cber.18730060180.


text:
What does oscillator strength have to do with state character?
I was reading S.H.A.R.C. Manual (Surface Hopping in the Adiabatic
Representation Including Arbitrary Couplings) and came across the following paragraph:

"Each state is colored, with one color as contour and another at the core of the line. The contour color represents the total spin expectation value of the state. The core color represents the oscillator strength of the state with the lowest state. See figure 7.3 for the relevant color code. Note that by definition the “oscillator strength” of the lowest state with itself is exactly zero, hence the lowest state is also light grey. This dual coloring allows for a quick recognition of different types of states in the dynamics, e.g. singlets vs. triplets or nπ ∗ vs. ππ ∗ states."

How can one recognize nπ ∗ vs. ππ ∗ from this kind of colour scheme? What does oscillator strength have to do with state character?

I've never seen such a plot, but since the oscillator strength of a transition relates to the transition dipole moment, only such states contribute that have the right symmetry for the dipole operator. By projecting out "self-interaction", the picture is decluttered for states that don't have point inversion symmetry. The plot should show clusters. It'd be nice if you had an example of such a plot. I assume it is 2-D.

This does not seem to make any sense to me either unless I'm missing something very obvious. Suffice it to say that spin changing transitions are generally strongly forbidden, unless paramagnetic electron spin or heavy atoms make them allowed, and that spin 'forbidden-ness' is stronger than symmetry 'forbidden-ness'. Symmetry forbidden transitions can be made allowed by 'intensity borrowing' (Herzberg-Teller) involving a higher electronic state & a vibration of suitable symmetry, as in benzene.  When both spin & symmetry are forbidden transitions are very weak indeed. $n\pi ^*$ transitions are often symmetry forbidden (e.g. in carbonyls) and so weak compared to $\pi \pi ^*$. Other than these general rules you just have to look at the individual molecules to work out what is going on, it might be quicker than trying to understand the diagram.


text:
Why do we integrate square of wave function over N-1 electron coordinates?
Quote form F.Jensen's Introduction to computational chemistry:
"The electron density is the square of the wavefunction, integrated over N-1 electron coordinates..."
Why do we integrate over N-1 instead of N electorn coordinates?
What would happen if we integrated over N coordinates?
I suppose we would get a norm if we integrated over N coordinates, or 1 if wavefunction is normed.
But why N-1?

Well I am not an expert in this but I suppose that if you integrate over all coordinates you get a number. The electron density $\rho$ is a function of $(r,\theta,\phi)$, so if you exclude the integration over one set of coordinates then you get a function of one set of coordinates (which is what you want), just like how if you integrate $\int_0^1\int_0^1 (3x + 4y),\mathrm{d}x\mathrm{d}y$ you get a number, but if you integrate $\int_0^1 (3x + 4y),\mathrm{d}x$ then you get a function of $y$.

$\newcommand{\el}{_\mathrm{e}}
\newcommand{\dif}{\mathrm{d}}
\newcommand{\braket}[2]{\langle{#1}\vert{#2}\rangle}$Consider an $n$-electron system in an arbitrary electronic state represented by the electronic wave function $\psi\el(\vec{q}_{1}, \vec{q}_{1}, \dotsc, \vec{q}_{n})$, where $\vec{q}_{i}$ stands for joint spin-spatial coordinate of $i$-th electron, $\vec{q}_{i} = \{ m_{si}, \vec{r}_{i} \}$.
Integrating the square of the absolute value of the electronic wave function over the spatial coordinates of all the electrons except for electron-one and summing up over the spin coordinates of all of them without any exception gives the probability density of finding electron-one with arbitrary spin at $\vec{r}_{1}$ while having the remaining $n - 1$ electrons with arbitrary spins at arbitrary positions,
\begin{equation*}
    \rho_{\mathrm{e-1}}(\vec{r}_{1})
    =
    \sum_{m_{s1}} \sum_{m_{s2}} \dotsi \sum_{m_{sn}} \iint \dotsi \int
    \left| ψ\el(\vec{q}_{1}, \vec{q}_{2}, \dotsc, \vec{q}_{n}) \right|^{2}
    \dif \vec{r}_{2} \dif \vec{r}_{3} \dotsc \dif \vec{r}_{n} \, .
\end{equation*}
And yes, if we "integrate" over all coordinates in this expression we'll just get an inner product of the wave function with itself,
$$
    \braket{\psi\el}{\psi\el}
    =
    \sum_{m_{s1}} \sum_{m_{s2}} \dotsi \sum_{m_{sn}} \iint \dotsi \int
    \left| ψ\el(\vec{q}_{1}, \vec{q}_{2}, \dotsc, \vec{q}_{n}) \right|^{2}
    \dif \vec{r}_{1} \dif \vec{r}_{2} \dif \vec{r}_{3} \dotsc \dif \vec{r}_{n} \, ,
$$
which is usually equals to 1 since the wave function is usually normalized to unity.
Electrons are, of course, indistinguishable, so that their labeling is arbitrary and can be changed any time.
Thus, the expression above should work equally well for calculating the probability density of finding electron-two, electron-three, etc., or, in other words, it gives the probability density of finding one particular electron, no matter which, with arbitrary spin at $\vec{r}_{1}$ while having the remaining $n - 1$ electrons with arbitrary spins at arbitrary positions.
The electron density, defined as the probability density of finding any of the $n$ electrons with arbitrary spin at $\vec{r}_{1}$, is evidently $n$ times as large as the probability of finding one particular electron,
\begin{equation} \label{eq:electron_density}
    \rho(\vec{r}_{1})
    =
    n \sum_{m_{s1}} \sum_{m_{s2}} \dotsi \sum_{m_{sn}} \iint \dotsi \int
    \left| ψ\el(\vec{q}_{1}, \vec{q}_{2}, \dotsc, \vec{q}_{n}) \right|^{2}
    \dif \vec{r}_{2} \vec{r}_{3} \dotsb \dif \vec{r}_{n} \, .
\end{equation}
Note here that, for a wave function that is normalized to unity, if we "integrate" over all coordinates in this expression we'll just get the number of electrons $n$, which is one of the well-known properties of the electron density,
\begin{equation}
    \int \rho(\vec{r}_{1}) \dif \vec{r}_{1} = n \, .
\end{equation}


text:
Can thionyl bromide produce alkyl bromides?
Thionyl chloride ($\ce{SOCl2}$) is used with alcohols to produce alkyl chlorides. Similarly, can thionyl bromide ($\ce{SOBr2}$) be used with alcohols to produce alkyl bromides?

If you want to learn about the formatting, here's a quick summary: http://meta.chemistry.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here Anyway, Wikipedia has an answer for you.

Would thionyl bromide produce alkyl bromides as easily as thionyl chloride produces alkyl chlorides?

Yes, thionyl bromide can be used with alcohols for synthesis of alkyl bromides.
No, the process is not as efficient.  The reasons for the inefficiency include are primarily (a) thionyl bromide is much less stable than thionyl chloride, and (b) that a bromide-tribromide redox equilibrium is far more relevant than a chloride/trichloride equilibrium, which leads to undesired side pathways instead of the desired reaction.

For more information, see this 1955 paper from Frazer & Gerrard.

On the other hand, pyridinium tribromide is readily formed during the addition of thionyl bromide to base hydrobromide, and its intervention weakens the analogy between the thionyl bromide and the thionyl chloride system.

In the same paper, a yield of 75% is reported for the synthesis of 2-bromobutane from butan-2-ol using pyridinium tribromide (which itself would be synthesized from thionyl bromide and pyridinium bromide).

I think the reaction with thionyl chloride proceeds in the absence of pyridine too. So, what happens with thionyl bromide in the absence of pyridine?

See also http://pubs.rsc.org/en/content/articlepdf/1970/J3/J39700000067 and the other papers that come up when searching "thionyl bromide" "alkyl bromide" on Google Scholar.


text:
Geminal coupling constant of formaldehyde protons in NMR
I read in a book that the protons in formaldehyde have the largest known positive geminal coupling constant (41 Hz). I thought only couplings between non-equivalent protons can be seen in the spectrum. Since formaldehyde has C2v symmetry and (if I remember correctly) coupling constants cannot be calculated from theory accurately, how come this coupling is apparent in the spectrum?

You are right, the normal 1D proton spectrum wouldn't show the coupling. http://u-of-o-nmr-facility.blogspot.com/2008/06/spin-spin-coupling-between-equivalent.html I just can't remember whether there is actually an experimental way to see the coupling.

does this help? https://www.chem.wisc.edu/areas/reich/nmr/05-hmr-04-2j.htm

It's a useful link, but it is still not clear to me whether these are theoretical or experimental. I'd assume this means the values are calculated theoretically but not visible in the spectrum as orthocresol said (as in couplings between equivalent nuclei), but I'm not sure whether that is indeed the case?

Good to see Reich's and Facey's websites referenced as first two comments. They should be prerequisite reading for any [nmr] question here.

This is an extrapolation of an experimentally measured coupling constant. It is possible to determine the coupling constant, $J_{\ce{H-H}}$ based on the $J_{\ce{H-D}}$, where the following relationship is valid:
$$ \left |J_{\ce{H-H}} \right | = (\gamma_\ce{H}/\gamma_\ce{D}) \cdot \left |J_{\ce{H-D}}\right |$$
Shapiro, Kopchick and Ebersole reported the coupling constant for $J_{\ce{H-D}}$ in formaldehyde-d1 in 1963 (The Journal of Chemical Physics 39, 3154 (1963); doi: 10.1063/1.1734163), and calculated the $J_{\ce{H-H}}$ from this. Changes in coupling constants due to the influence of isotopically-modified electron density are apparently very modest for H/D, and so this relationship appears to hold well.
There is no experimental method to measure coupling between magnetically equivalent spins. In order to break the redundancy of the energy transitions, one must remove the equivalence.
You can actually check this isotope relationship very easily with common NMR solvents and the $\ce{C-H}$ and $\ce{C-D}$ coupling if you run carbon experiments ($\gamma_\ce{H}/\gamma_\ce{D}=6.51439$).
$$\begin{array}{ccc} \hline
\text{Solvent} & J_{\ce{C-D}} & J_{\ce{C-H}}\\[1.1em] \hline
\ce{CHCl3} & 32 & 208\\
\ce{DMSO} & 21 & 137 \\ \hline \end{array}$$


text:
How to interpret m quantum number values?

If one electron has the quantum numbers $n = 4,\ l = 3,\ m_l = +2,\ m_s =-1/2$ and another electron has the quantum numbers $n=4,\ l= 3,\ m_l = -2,\ m_s=+1/2$, what are the similarities between the orbitals in which the electron's are found?

My answer is : They have the same number of energy shells. They have the same shape. Their orientations are different and they have different directions of spin.
I checked the answer and it said that the first one lies on the $x$ and $y$ axis and the other one will lie between the $z$ and $y$ axis. But how do they know that? If $m_l= +2$, does that necessarily mean it lies on the $x$ and $y$ axis. And if $m_l = -2$, does that mean it lies between the $x$ and $y$ axis?

OP's answer is technically right while additional details about specific orientations of the real $f$-orbitals mentioned somewhere else are basically wrong. At best such details can be thought of the matter of some convention, but it is better to avoid such conventions.
The thing is that for any value of the azimuthal quantum number $l$ an orbital with a non-zero value of the magnetic quantum number $m$ is complex thanks to $\mathrm{e}^{imf}$ factor. Chemists prefer to work with real orbitals formed by taking linear combinations of complex orbitals. However, such linear combinations are not eigenfunctions of $L_z$, or, in other words, $m$ is no longer a good quantum number (but its absolute value is).
In particular, for both $f_{xyz}$ and $f_{z(x^2−y^2)}$ orbitals $|m|=2$, or $m=\pm2$, but the statement that the former corresponds to $m=-2$ and the later to $m=2$ is technically wrong.

It might be worth looking at McQuarrie & Simon, 'Physical Chemistry' where this is discussed  in chapter 6. Examples of  complex d orbital equations (unfortunately not f) are given in table 6.5 and of the real d orbitals in table 6.6.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Is the carbanion formed in benzyne mechanism non-planar?

I was told by my chemistry teacher that the carbanion formed in the picture above is non-planar. I don't understand why. It looks like benzene with an extra electron pair at one of the carbons so shouldn't it be planar?

I guess that NH3 will form bond perpendicular to the ring. So the carbanion is non-planar.

But I need to know whether the ring will be planar or not

The ring will be planar but the NH3 will come out of the plane due to hindrance because of CH3 group

Assuming that your premise is correct, I would suppose it is just because it wants to allocate the lone pair as much s-character as possible. So, the bond angles change, and the ring loses its planarity. I think that you might get better answers if you try searching for the phenyl anion, because that is what your compound basically is (sans the methyl group). I did a quick search, but couldn't find literature on it.

And it would want to allocate more s-character to the lone pair to keep the electrons as close as possible to the nucleus?

Just five months late for a reply, but yes. (BTW if you want people to receive notifications from comments, you have to ping them by including @orthocresol in your comment.) In organic chemistry terms, more s-character -> closer to nucleus -> lower energy -> more stable lone pair. That was my hypothesis, at least.

I have given one of these molecules a little spin on the computer. With whatever I tried, I was unable to find a non-planar solution. I really tried my best. Below is an animation (sorry for the poor quality) of an attempt which uses a chair like starting geometry. If there was any non-planar geometry lower in energy, this would have found it.

The level of theory is DF-BP86/def2-TZVPP and the program used is Gaussian 09 rev. D.01.
There is really not much more I can add to the discussion.

I was told by my chemistry teacher that the carbanion formed in the picture above is non-planar. I don't understand why. It looks like benzene with an extra electron pair at one of the carbons so shouldn't it be planar?

To be absolutely, brutally honest here: I have no clue why your teacher had this idea. If this wasn't a "naked" carbon, we could talk about it, but it is essentially a deprotonated phenyl ring, and as such, you are absolutely right in assuming the lone pair being in phase with the ring. The π system is hardly different from benzene. To fill this post a little bit more, have those fancy π-orbitals.

(I could not find the completely anti-bonding pi-orbital.)
And for good measure, here is the HOMO, which is the lone pair at the carbon:

I don't know that it would make a difference, but did you include dispersion corrections?

@hBy2Py No. I don't think it'll make a difference. How would they affect the geometry? I think the sigma type orbital would still be that and therefore we have a nice Cs symmetry...

Yeah, I would think it really shouldn't matter, since the only departures from mirror symmetry in the electron density should occur on the ammonium and methyl hydrogens.

I was told the same thing by my chemistry teacher. Thats what brought me here.

I cannot find any published structures — neither computational nor crystal structures — on the actual phenyl anion that is the centre of your question. However, quite a few similar structures are published for the benzyne radical anion and for phenyl lithium and related compounds — the bond between carbon and lithium can generally be considered rather ionic considering the electronegativity difference so it is a reasonable argument to assume phenyl lithium’s structure to be akin to the phenyl anion’s.
Rather unimpressively, Nash and Squires report the calculated structure of the ortho-benzyne radical anion to be $C_\mathrm{2v}$ symmetric.[1] This automatically also means a planar benzyne ring. This may have been an artifact from the calculation, but considering the system’s potentially high symmetry it also makes most sense. The benzyne ring in itself is distorted but planarity is retained.
Likewise, the crystal structure of $\ce{[LiPh(m5dien)]}$ was published by Schümann, Kopf and Weiss (others have also been published but this was my first ‘hit’).[2] Their research also indicates a planar phenyl ring. To quote the paper:

The benzene ring is approximately planar; but the hexagon is distinctly distorted (angle at $\ce{C{1}}$ $113.1$ instead of $120^\circ$, [sic!] while those at $\ce{C{2}}$ and $\ce{C{6}}$ on the other hand, with values of $123.9$ and $125.1^\circ$, [sic!] are significantly enlarged).

Thus, this too is a primarily planar system with most distortions in plane around the anionic carbon to increase the lone pair’s (or $\ce{C-Li}$ bond’s) $\mathrm{s}$ orbital contribution.
In the absence of any further evidence, I would also assume your two structures to be primarily planar; however, featuring significant distortions at the anionic carbon centre. Theoretical calculations (I shall bounty this question in the hope of receiving some) may provide a more conclusive answer.

References:
[1]: J. J. Nash, R. R. Squires, J. Am. Chem. Soc. 1996, 118, 11872–11883. DOI: 10.1021/ja9606642.
[2]: U. Schümann, J. Kopf, E. Weiss, Angew. Chem. Int. Ed. 1985, 24, 215–216. DOI: 10.1002/anie.198502151.


text:
working with ratios and RMS to find variables
question
consider separate 1.0L samples of helium and $UF_6$ both at 1.00atm and containing the same number of moles. what will the ratio of the separate temperature need to be in order to have the same root-mean-square speed?
my steps
So i set up an equation
$\sqrt{\frac{3(8.314T)}{0.004}}$=$\sqrt{\frac{3(8.314T)}{0.352008}}$
but i soon realized that that would be impossible
and that is where i got stuck

Well, obviously the temperatures are not going to be the same. They asked for a ratio of temperatures. So you can't denote both temperatures by the same symbol $T$. If you let the temperature of helium be $T_1$ and the temperature of UF6 be $T_2$, then you need to find $T_1:T_2$.

(1) Setup equation using variables and cancel out all common stuff, solve for the needed variable, then plug in the numbers. (2) How did you get 0.004 for the mass of Helium?!?

ok but i was wondering if you can provide some helpful hints rather than telling me to cancel out the variables, thats kind of obvious. Can you give me a booster step?

@MaxW I think OP is taking mass in kg.

John - Look at orthocresol's comment. The whole point is that if you setup the problem using letters for variables a whole bunch of the variables cancel and you're left with a very simple relationship. By just trying to plug numbers into some formula I don't think you really understand the problem.

As I said solve using variables first and simplify as much as you can, then plug in the numbers. From the Maxwell-Boltzmann Distribution:
$v_p = \sqrt{\frac{2RT}{M}}$
$v_{rms} = \sqrt{\frac{3}{2}}v_p$
where:

$v_p$ most probably velocity
$v_{rms}$ rms velocity
R universal gas constant
T temperature
M molecular mass

so let:

$v_{p,He}$ - most probably velocity for He
$v_{p,U}$ - most probably velocity for $\ce{UF6}$
$v_{rms,He}$ - rms velocity for He
$v_{rms,U}$ - rms probably velocity for $\ce{UF6}$

we're given:
 $v_{rms,He} = v_{rms,U}$
but $v_{rms,He} = \sqrt{\frac{3}{2}}v_{p,He}$
and $v_{rms,U} = \sqrt{\frac{3}{2}}v_{p,U}$
so:
$\sqrt{\frac{3}{2}}v_{p,He} = \sqrt{\frac{3}{2}}v_{p,U}$
which can be simplified to:
$v_{p,He} = v_{p,U}$
Now substituting from the Maxwell-Boltzmann equation for $v_p$ for both He and $\ce{UF6}$ we have:
$\sqrt{\frac{2RT_{He}}{M_{He}} } =  \sqrt{\frac{2RT_U}{M_U}}$
simplying
$\frac{T_{He}}{M_{He}} = \frac{T_{U}}{M_{U}}$
rearranging:
$\frac{T_{U}}{T_{He}} = \frac{M_{U}}{M_{He}} = \frac{352.02}{4.003} = 87.94$
Superfluous information is that $\ce{UF6}$ decomposes about $1800^o \text{ K}$ and is a solid below about $330^o \text{ K}$. Helium liquefies about $4.2^o \text{ K}$. So this only works if Helium is in range $4.2^o \text{ K}$ to $18.4^o \text{ K}$ which would mean that the uranium hexfluoride is between $369^o \text{ K}$ to $1800^o \text{ K}$.


text:
Reversible isothermal expansion and Reversible adiabatic expansion
For a given pressure drop, which of the two- isothermal reversible expansion and adiabatic reversible expansion - will produce a lesser volume increase?

It has been a long time since I took pchem but this doesn't make any sense to me. There seems to be too much left unspecified about the two systems. To me this is like saying I have two variables $x$, and $y$. Which is smaller?

But it is saying that the pressure drop is given, so cant we derive them using the thermodynamic equations?

I was just thinking about PV=nRT. For an isothermal expansion the whole expression nRT is a constant so fine. But for an adiabatic expansion T can can as well as V.

In the reversible adiabatic expansion, the temperature will decrease, while, in the reversible isothermal expansion, the temperature will stay constant.  What does that tell you?

We need additional information.  1. Can we assume the gases are ideal?  2. Can we assume the expansions start and stop at the same pressures?

Let's assume that both expansions are for ideal gases.
Isothermal case
$$ PV = \mathrm{constant} = c_{it}$$
$$ P_1 V_1 = P_2 V_2$$
$$ \frac{V_2}{V_1} = \frac{P_1}{P_2}$$
Adiabatic case
$$ PV^\gamma = \mathrm{constant} = c_{it} = PV^\frac{f}{f+2}$$
Here, $\gamma$ is the ratio of heat capacities, and $f$ is the number of translational degrees of freedom of the molecule (3 for a mono-atomic gas, 5 for a diatomic gas, etc.).  The expression involving $f$ makes clear that $\gamma$ must be $\gt 1$.
$$ P_1 (V_1)^\gamma = P_2 (V_2)^\gamma$$
$$\left ( \frac{V_2}{V_1}\right)^\gamma = \frac{P_1}{P_2}$$
Comparing the two cases
For the same pressure drop, $\frac{P_1}{P_2}$ will be the same in each case.  Say the pressure changes by two-fold, i.e. $\frac{P_1}{P_2}=2$.  Then in the isothermal case, the volume will increase by two-fold.  However in the adiabatic case, the volume will increase by a factor of $2^\gamma$.  Since $\gamma \gt 1$, the volume increase will be more than two-fold.
Thus, the adiabatic case will have the greater volume increase, meaning the isothermal case will have the lesser volume increase.
The reason why
During adiabatic expansion, no energy is removed or supplied to the gas.  However in isothermal expansion, that isn't true.  To keep things isothermal during an expansion (an otherwise cooling process), heat would have to be continually added.  Hotter gases take up more volume than cooler gases (if they are ideal). Thus the process of adding heat counteracts some of the volume change that could have happened if things were adiabatic.  Thus the isothermal case results in a smaller volume change.


text:
Calculating the ionic polarizability from the Clausius-Mossoti relation
The concentration of atoms in a crystal is $N_d = \pu{2.24e28 m-3}$, and the relative dielectric constant is $\varepsilon_{r} = 5.62$. The crystal has a refractive index of $n = 1.5$. Using the Clausius-Mossoti relation,
$$
\frac{\varepsilon_{r} - 1}{\varepsilon_{r} + 2} = \frac{N\alpha}{3\varepsilon_{0}},
$$
I got that
$$
\alpha_\mathrm{i} = \pu{8.26e-40 F m2}.
$$
However, the correct answer is
$$
\alpha_\mathrm{i} = \pu{3.7e-40 F m2}.
$$
Was the refractive index useless info here, or have I missed something?

Yep, you missed quite an important part: Clausius-Mossoti equation in the form you've written is limited  and holds true for the dielectric that consists of individual particles. However, the problem says you are dealing with a crystal, which, by the way, is most likely sodium chloride $\ce{NaCl}$, judging from its dielectric constant and refractive index.
In the first Clausius-Mossoti equation you've written $\alpha$ consists of two main components:

electronic polarizability $\alpha_\mathrm{e}$ caused by the displacement of the electron shell of the atom relative to the nucleus under the action of the field;
ionic polarizability $\alpha_\mathrm{i}$ caused by the displacement of ions with respect to other ions.

For the ionic crystal Clausius-Mossoti equation can be written as
$$\frac{\varepsilon_{r} - 1}{\varepsilon_{r} + 2} = \frac{N}{3\varepsilon_0}(\alpha_\mathrm{e^+} + \alpha_\mathrm{e^-} + \alpha_\mathrm{i})\, ,\label{eq:1}\tag{1}$$
where $\alpha_\mathrm{e^+}$ and $\alpha_\mathrm{e^-}$ are electron polarizabilities of positive and negative ions in crystal lattice. Their sum can be determined using a Lorentz–Lorenz equation, which is derived from the fact that the effect of an external electromagnetic field with relatively high frequencies corresponding to the visible and ultraviolet range of the spectrum results in the displacement of only the electron shells relative to atomic nuclei, while the more massive particles (atoms and ions) do not have time to shift from the occupied positions. Correspondingly, only electronic polarization contributes to the polarization of the medium (e.g. $\alpha_\mathrm{i}\approx0$), and the refractive index is related to the electron polarizability of the particles by the Lorentz-Lorentz formula:
$$\frac{n^2 - 1}{n^2 + 2} = \frac{N}{3\varepsilon_0}(\alpha_\mathrm{e^+} + \alpha_\mathrm{e^-})\label{eq:2}\tag{2}$$
Subtracting \eqref{eq:2} from \eqref{eq:1}, one eliminates the electronic constituent from the equation:
$$\frac{\varepsilon_{r} - 1}{\varepsilon_{r} + 2} - \frac{n^2 - 1}{n^2 + 2} = \frac{N}{3\varepsilon_0}\alpha_\mathrm{i}\tag{3}$$
and ionic polarizability is actually determined as follows:
$$
\begin{align}
\alpha_\mathrm{i} &= \frac{3\varepsilon_0}{N}\left(\frac{\varepsilon_{r} - 1}{\varepsilon_{r} + 2} - \frac{n^2 - 1}{n^2 + 2}\right)\\
&= \frac{3\cdot\pu{8.85e-12 F m-1}}{\pu{2.24e28 m-3}}\left(\frac{5.62 - 1}{5.62 + 2} - \frac{1.5^2 - 1}{1.5^2 + 2}\right)\\
&= \pu{3.7e-40 F m2}\tag{4}
\end{align}
$$