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Pauli exclusion principle and resonance
I've always been a bit uncomfortable with the concept of more than two electrons in a single orbital-like region(probability-wise) which occurs in resonance.
This seems to disobey Pauli's exclusion principle.
Let's take benzene for example. It has a delocalized cloud of electrons. These all had the "original" (as in, if resonance didn't occur) quantum numbers $n=2,l=1,m=\text{(1,0,or -1; but it must be the same for all)},m_s=\pm\frac12$. But, due to delocalization, they're now all in one extra-large orbital-like region (by orbital-like, I mean that the pi cloud is similar to an orbital in the sense that it is also a probability cloud which defines the probabilities of finding one or more electrons at different coordinates).
Now, it's obvious that the original quantum numbers are insufficient here due to Pauli's exclusion principle. They may not even exist for these electrons.
The easiest thing to do would be to "propose" (not exactly propose, since I guess the Dirac equation can prove their existence) an extra quantum number, which can take twelve values. Or a combination of a few extra quantum numbers that have the same effect.
But this raises the question: do the electrons belonging to the pi cloud "retain" their "original" quantum numbers as well?
Basically I want to know how benzene and other resonating compounds are described quantum mechanically, with focus on quantum numbers. (links to papers are fine, probably necessary here--but please explain the gist of the link)
I'd prefer the answer in terms of simpler quantum mechanics, though I'm not averse to the idea of some extra reading.

I would simply says that for the molecule, the original orbitals are no longer adequate… the molecule has a different set of orbitals, which are the solutions to Schrödinger’s equation. These orbitals do not have more than two electrons each in them (or one if you consider spin-orbitals).

I very much like the first answer, especially the superb visuals!
However, I will nonetheless add a simpler approach to answering your question, which is this: Pauli exclusion holds as long as there is something in space or in spin (spin being the pairs of electrons) that firmly distinguishes two "stable" states.
What that means is that your initial intuition is essentially correct if all the electron pairs shared exactly the same molecular orbital.
So the answer to your question in general terms is that electron pairs settle into molecule-sized orbitals (molecular orbitals) that are physically spread out over the entire ring, but which nonetheless have unique geometric features that keep them distinguished in real space. I would add that they are even more unique when represented in an abstract space called momentum space, where the geometric differences translate into actual distances between the electron pairs. These abstract distances in momentum space satisfy the Pauli exclusion principle just as well as physical separation in real space, so that even very similar looking orbitals can remain distinct under Pauli exclusion.
For more details and some easy visual analogies, you can read more in the addendum below.

The same general problem of how to distinguish electron pairs to meet Pauli exclusion exists in metals. In that case, the conduction electrons are free to drift over the entire macroscopic metal. The result is an even more extreme need than in benzene for the pairs to distinguish themselves firmly from each other, since for conduction electrons all of these gazillions [technical term] of pairs must find "unique places" in which they can maintain their unique identity. That's tough!
Now it turns out that these two problems of benzene electrons and metallic conduction electron pairs are related in a rather deep fashion, since a benzene ring can be interpreted (quite correctly) as a special tiny case of a one-dimensional metal. That means that if you understand what happens conceptually in metals, you are well on the way to understanding all forms of hybridization in organic molecules. So, let's take a quick look at how that works.
The situation for electron pairs in metals is most easily understood by imagining the pairs moving back and forth along the $x$ length of a very narrow, highly conductive, and unusually regular metallic wire. Next, borrow the $y$ and $z$ plane to represent the complex plane (complex phase of a wave function). The amplitude for finding an electron from the pair becomes the radius of the complex plain at each point along $x$. For stable electron pair states, these radii and their phases will exhibit a lot of continuity as you move their as along the $x$ length of the wire. They will also rotate at the same speed in the complex plane of $y$ and $z$.
In fact... if you calculate the stable or "stationary" states possible in such a configuration, you will find that they bear a striking resemblance both visually and mathematically to skip rope solutions, that is, to the ways that you can twirl a skip rope so that it maintains stable, unmoving (the "stationary" part) loops.
So let's look at that situation as a good analog solution to the quantum equations. Anyone can make one large loop, and that in fact represents the electron pair state with the lowest possible energy along the length of the wire, or the base state. Incidentally, the volume enclosed by the skip rope is proportional to the square of the radius, which as I mentioned earlier is the "amplitude" of the wave function. So... that means that if you simply look at the volume enclosed by the skip rope as it turns, you pretty much instantly know what the probability of finding an electron from the pair is. In the case of the base state, it means that the electron pair will tend to be found in the center of the wire, and almost never at the ends.
In chemistry such varying probabilities of finding electrons are usually represented as densities in ordinary space, without the benefit of the dynamics of the complex plane. And for good reason, since for anything except wires, trying to show the complex plane would result in five-dimensional spaces. Those are tough to visualize! Instead, the idea of a phase is attached to tell the relative positions of the rotating points in the unseen complex plane.
The skip rope in the one-dimensional wire makes the idea of phase a lot easier to visualized. For example, good skip-ropers can easily create double loop skip ropes. These correspond to the first excited state of the electron pair, one with a bit more energy added (you have to twirl the rope faster) and two distinct loops. Phase just means this for such a double loop: When one loop is towards you, the other is opposite of you. You can label one such loop positive and one negative, but of course it's all arbitrary, and really just a technique for roughly describing those invisible but rope-like complex plane dynamics.
Notice that the volume-equals-probability rule still applies to the double loop. That means that in the first excited state of the electron pair, the odd are good and roughly equal for finding it in the left loop or right loop, but close to zero at the ends and at the center. So, this really is a unique state for the excited electron pair, one that is quite distinct from the first single-loop or base state.
Now there's an interesting bit of mathematics, Fourier transforms, in which any very long sinusoidal shape can always be distinguished very sharply from any other sinusoidal, even if they only differ slightly in wavelength. People use this idea all the time, since in the case of radio waves (which are sinusoidal waves, mostly) it's called tuning a radio!
The uniqueness of long sinusoidal waves is relevant here because it turns out that Pauli exclusion is met equally well if two pairs of electrons are separated from each other in frequency space.
So, another way of saying all of this is that when electrons are "delocalized" either in bonds or in metals -- it really is the same idea -- they can achieve uniqueness by bumping around either in ordinary space or in a more abstract frequency space called momentum space. Both work equally well. The "extra quantum number" you speculated about in your question could be equated to this idea of additional separation in momentum space for example.
So, the benzene orbitals maintain Pauli exclusion uniqueness by finding specific stable solutions that are similar but geometrically distinct, and which are even more unique when interpreted in terms of frequencies or loop counts. Those equate to a quite different-looking, distance-like separations in the abstract space called momentum space.
So to finish up, what about that extreme case I mentioned of metals that have an electron or two for every atom in what may be a very large crystal?
That's a truly fascinating case, a sort of "benzene gone bad" scenario in which the need for Pauli exclusion uniqueness pushes the electron pairs into absolutely bizarre behaviors, ones very important to modern (and ancient) technologies: Some of the electrons are forced to get very hot. Now that's weird!
What happens is easy to see with the above background. The first pair of electrons in a silver wire is very, very cold, near absolute zero, moving so slowly that it forms only a single loop in the center of the wire. To maintain Pauli exclusion -- the momentum space version of it to be precise -- the next pair must look like a skip rope with two loops, and the third three, and so forth. You get the picture.
But each such loop has more energy, even if only by a little bit! So what happens when you have a gazillion such electron pairs all vying for uniqueness?
Someone gets pushed very, very high up in frequency, and thus also in energy, while other lurk in the deeper and colder regions of this pile of moving electrons. The result is almost like a sea in momentum space, and that is in fact what it is called: A Fermi sea, with cold electron pairs at the bottom, and some truly hot ones at the surface. Fermi seas are also referred to by the more common phrase of "conduction bands," meaning the range of energies that the electrons occupy in order to maintain Pauli uniqueness.
You see Fermi surfaces all the time, since it is only the hottest electrons in a metal that interact with photons and bounce them back at you. You are in effect looking directly at a hot Fermi surface every time you look into a silver or aluminum backed mirror.
How hot? Well, believe it or not, X-ray hot. A typical piece of ordinary shiny metal, say a fork or spoon, would kill you from X-ray exposure if all of its hot Fermi surface and near-surface electrons could suddenly fall back into a "ground state" of just one loop. They are tamed and prevented from doing so by all those other cooler electrons that refuse to let them fall to the bottom of the Fermi sea.
And with that I'll close. I realize the parts about metals are a bit remote from the original chemistry of benzene rings. But then again, as you get into longer and longer carbon chains such as the ones that carry off electrons from chlorophyll molecules, you may find that recognizing the "almost metallic" nature of multiple pairs and the way they interact in momentum space can be a good shorthand for understanding how some pretty complicated forms of large-molecule dynamics play out.

I knew it was you behind the long answer! I knew it! :D Excellent explanation, removes all lingering doubts from my mind. I guess my main confusion was that I assumed the electrons to all have the same region in their MOs, which was explained in the above answer--and that "region" includes momentum space.

Manishearth, yep, I am the master of overkill! I didn't realize it was you asking until I got to the very end. If I'd been paying attention to that, I might have kept my answer down to "Pauli exclusion works the same in both ordinary space and momentum space, so clear separation in momentum space allows orbitals to overlap significantly in real space." (To be accurate, it's the unique shapes in all six dimensions that tell the full story.) One the other hand, it really is a delightful topic about which to write, so I appreciate the excuse to talk about ways of visualizing such things.

This answer is mostly nonsensical. The six pi electrons occupy three different molecular orbitals of very different shapes. Momentum space has nothing to do with this. In fact, momentum space representations of molecular orbitals are effectively useless because of the distinct lack of spatial periodicity in molecular systems.

Hmm. Well, sure, the molecular orbitals are unique in shape; that would be why I said "unique geometric features that keep them distinguished in real space" in my, ah, mostly nonsensical answer. I'd be a bit careful about disregarding momentum space even in small molecules, however, since even for the diffuse configurations you get in small molecules, the momentum space view helps explain why orbitals that are unique in shape can have many electrons occupying the same region of ordinary space.

The "shape" of molecular orbitals do not have any meaning in the way you are trying to use them. The pictures that are usually drawn correspond to isosurfaces of a given probability, but an electron always has a nonzero chance to be found outside the isosurface. The properties of orbitals that allow them to obey Pauli exclusion do not happen in real spare, nor in momentum space; they arise from orthogonality properties in Hilbert space. The momentum space picture which works well for crystals works only because of translational symmetries which molecules lack.

Please understand that it is nothing personal, it is simply an incorrect answer based on what appears to be well-intended, but a less than cogent chain of reasoning which happens to not tally with a contemporary understanding of such phenomena.

To be clear: By choice, I mostly answer questions from folks who are unlikely to understand how probability waves and Hilbert spaces work, but who clearly want to learn. If you hit them with barrages of new terms for which they have no mental framework, most will very quickly learn to stop asking. That's bad, since the world needs to get a lot more young people interested science. As for whether a couple of your comments may have rated as more personal than professional, the only answer that counts there is your assessment of yourself. I'm good, and have had far more smiles than frowns today.

I must agree with AcidFlask.  This has elements of truth, but it is far from the simplest answer:  electrons of opposite spin may exist in the same region of space because they have different quantum numbers.  Also, MO

s are orthogonal... (I just noticed AcidFlask's answer so I will stop here)

Far from being repentant, I would simply note that explaining something differently is not the same as explaining it incorrectly. Also, the comment "momentum space has nothing to do with it" is simply incorrect. Or more simply: As a fermion, every electron in the universe must continually maintain a unique seven-number address in that includes space, momentum space, and binary spin. No part of that address is any less important than the other parts, and that includes within orbitals. Look at Peierls condensation for example, and realize that ordinary bonding is a continuum with it.

To understand why the exclusion principle isn't violated in this system, you really need to shift from valence bond theory to molecular orbital theory. The $p_{z}$ orbitals in benzene combine according to the $D_{6h}$ symmetry of the molecule to generate a set of bonding molecular orbitals (MOs) (which are lower in energy than the isolated $p_{z}$ orbitals) and a set of antibonding molecular orbitals (which are higher in energy). Bonding orbitals are generated from sets of $p_{z}$ orbitals which are mostly or totally in phase, whereas the highest energy antibonding orbital of the set is composed of $p_{z}$ orbitals that are totally out of phase with respect to their adjacent neighbors.
Here, for instance, is the MO that is generated from the totally in-phase set (I calculated this at a spin-restricted RI-BP86/6-311G* level of theory in ORCA(1) and visualised the isosurfaces in VMD(2)):

Now, this MO is shared equally across all of the carbons, i.e. it is not localised. However it only contains 2 electrons and thus satisfies the exclusion principle. (F'x mentions spin-orbitals - this is where the orbitals are split on the basis of their spin, which entails a single electron per orbital. Benzene is a closed-shell singlet so we would expect the $\alpha$ and $\beta$ spin orbitals to be spatially indistinguishable.)
Not to sound like a walking advertisment for Housecroft and Sharpe, but H&S has a great visual introduction to MO theory.
P.S.
It was suggested in chat that I expand upon the generated orbitals - here's one of the two highest occupied molecular orbitals, from the same calculation.

As you can see, there's a big node along one of the mirror planes. This orbital is higher in energy than the earlier example because it's composed of two antiphase sets of 3 $p_{z}$ orbitals, so this orbital has antibonding character for two of the carbon-carbon pairs. It should be easy to extrapolate to the highest lying MO of this type - antibonding all the way around.

(1) Neese, F. ORCA – an ab initio, Density Functional and Semiempirical program package, Version 2.6. University of Bonn, 2008
(2) Humphrey, W., Dalke, A. and Schulten, K., 'VMD - Visual Molecular Dynamics', J. Molec. Graphics 1996, 14.1, 33-38.

This is the correct answer to this question. The six pi electrons occupy three different molecular orbitals of very different shapes. These are provided even by the very simple calculations of Hückel theory.

Could you perhaps post the ORCA input you used somewhere? Playing with ORCA is in my to-do list for a long time now and using this example I could just dive in.

What is all the fuss? Pauli exclusion isn't violated, even in MO theory, because the two electrons have different quantum numbers (opposite spin)

The short answer to your question is that no, they don't all occupy the same orbital but yes, they occupy very similar regions in space. What tends to be confusing in these pictures is it sometimes unclear that the pictures are sometimes of the molecular orbitals and sometimes of the electron density, which is an observable quantity and quite another thing altogether.
The six pi electrons occupy three different molecular orbitals of very different shapes, as can be seen here, for example, and in Richard Terrett's answer. However, their corresponding probability densities can be superposed onto similar regions of real space. There is no conflict with Pauli exclusion because you can see (and calculate!) that the orbitals are orthogonal by virtue of having an appropriate number of nodes. While on average the electrons look like they are all in the same region of space, if you measure the positions at any time, no two electrons will ever occupy exactly the same point in space.
Your discussion of orbital and spin angular momenta is completely irrelevant. Angular momentum operators $L_z$ and $L^2$ commute with atomic Hamiltonians because of rotational and Runge-Lenz invariance; however, they do not commute with molecular Hamiltonians because molecules have neither of these symmetries. Therefore, molecular orbitals are not indexable by the same set of quantum numbers as can be done for atomic orbitals. This is not because of Pauli exclusion, but because the Hamiltonian, $L_z$ and $L^2$ no longer form a complete set of commuting observables. (Note that Pauli exclusion must also hold in atoms where this latter statement is true!) Nor is there any need to introduce new fictitious quantum numbers.

(and electrons of opposite spin in the same MO have different quantum numbers, so their Slater determinants don't vanish.)

Great answer! Here is also a more detailed image and explanation of the 3 different molecular orbital shapes that were mentioned in this answer: http://www.chemcomp.com/journal/molorbs.htm


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Why is the benzyne triple bond distorted?
In many places I've seen the "extra" bond in benzyne being labelled as $\ce{sp^2-sp^2}$ overlap or distorted (not parallel) $\ce{p\pi-p\pi}$ overlap. But I've failed to see why we can't have a normal, parallel $\ce{p\pi-p\pi}$ overlap.
Wikipedia says:

In benzyne, however, the p orbitals are distorted to accommodate the triple bond within the ring system, reducing their effective overlap.

That makes no sense to me — why can't a normal, parallel $\ce{p\pi-p\pi}$ overlap be "accommodated" within the system.
So, what is the reason for the distorted $\pi$ bond in benzyne?
I'm fine with answers involving a bit of QM/MO theory.

First, Wikipedia doesn't really say what is the extent of this change. Here are superimposed a benzene and a benzyne molecule (own HF calculation, don't ask for the details):

The carbon atoms involved in the triple bond are displaced by less than $\pu{0.2 Å}$. In benzene, we have with this level of calculation:

$\ce{C–C}$ distance = $\pu{1.39 Å}$
$\ce{C–H}$ distance = $\pu{1.08 Å}$
$\ce{C–C–C}$ angle = $120^\circ$

while in benzyne, we have:

$\ce{C–C}$ triple bond = $\pu{1.22 Å}$
other $\ce{C–C}$ bonds = $\pu{1.38 Å}$ (almost all equal)

So, merely because the triple bond is stronger, the distance will be shorted and the benzyne structure is distorted.
Regarding orbitals, well, the p orbitals of the carbon atoms involved in the triple bond are “naturally” directed towards where the hydrogens would be. Slightly distortion of the system is possible, and indeed evidence above on the structure, but they will not be parallel, as that would disrupt the structure too much and diminish the $\sigma$ bond. So it's a competition between: making the new p orbitals “more parallel” increases overlap, but it weakens the $\sigma$ bonding pattern, which is much stronger. As such, while some distortion happens, you stay mostly close to the benzene-type orbitals.

"disrupt the structure"--that's what I sort of wanted to know.. But I think I got it--basically the rest of the ring would have to distort and that's a Very Bad Thing, right? (I'm going to wait a while before accepting this, let others answer)

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Why does benzene bend in this reaction?

From here
I can see why hapticity changes here, but not why it is necessary for the benzene ring to bend. The resonance energy of benzene is pretty large, and there's no balancing increase in stability elsewhere--unless the decrease of hapticity increases stability, which I doubt (more delocalization=more stability, not vice versa). Also, the concept of a "bent benzene" is a bit strange to me.
I'd like to know what exactly is going on in this reaction and why it happens.

@F'x mentioned that the bent benzene may just be a way of drawing the molecule to clearly show the change of overlap. If this is the case, I'd still like to know how the hapticity changed (because my initial logic behind change of hapticity is invalid if the structure doesn't actually bend).

Do you have any indication that the benzene ring actually distorts that way, or is it merely a convenient way of drawing it because it clearly shows η⁴ hapticity?

@F'x: Is it? Hmmm. If you explain why the hapticity changes in that case (I suspect I know why), then I think it qualifies as an answer(I can edit the question to accommodate that possibility).

I don't know for sure, but I suspect it might be the case…

That diagram is misleading because it’s technically in reference to bis(hexamethylbenzene)ruthenium(0). This is a famous example by the late and great E.O. Fischer. This Ru(0) complex is 18e, three methyl resonances were observed for the bent ring at −10 °C. Crystallography later confirmed this. This structure is surprising to the new student, perhaps, because the classic 18e bis(benzene)chromium is sandwiched between two planar arenes.
This is not an organic chemistry question. It is deep in the realm of organometallic chemistry, with a proper explanation requiring a molecular orbital energy diagram and then one has to invoke a Jahn Teller instability argument that requires lowering symmetry from $D_{6d}$ or $D_{6h}$. Suffice it to say, stability is increased by changing the hapticity.

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Should one pronounce “periodic” the same in “periodic acid” and “periodic table”?
In “periodic table”, the adjective is related to the noun period, and comes from Ancient Greek περίοδος through French périodique. In “periodic acid”, it is formed from the prefix per- and iodic (like peroxide and permanganate).
Wiktionary lists their respective UK pronunciations of as /pɪə(ɹ).iˈɒdɪk/ and /ˌpɜːraɪˈɒdɪk/, markedly different: pɪə (as in piece) vs. pɜː (as in perfume); then i (as in it) vs. aɪ (as in eye).
However, is that distinction really made in practice? Would a native US/UK/Aussie speaker make the difference when talking in the lab?

It only just occured to me that those words are written the same :)

I actually call periodic acid period acid just for fun ;D

Do you "read" books?  I have "read" a few.

I am an Australian English speaker and yes, this distinction is made in practice.
Possibly the only reasonable opportunity to use the pɪə pronunciation in the name of a compound is in the case of the entertaining molecule periodane, which is actually named after the periodic table. This molecule (and later a number of different plausible isomers) was identified computationally by a methodology called 'mindless chemistry' which optimises randomly generated molecular graphs. Periodane is a stable configuration of each atom on the second row of the periodic table, with the exception of neon (although some people are working on that).

I'm not a professional chemist, just a student, but here are some observations from which I can speculate:

Note-- in this post,
/ˌpɜːraɪˈɒdɪk/==per-iodic
/pɪə(ɹ).iˈɒdɪk/==peer-iodic
(since IPA is annoying to read)
Remember, it's written as "per-iodic" acid many times. Which means it must be pronounced as "per-iodic" by whoever who reads/writes those texts. So we have some evidence showing the use of "per-iodic" as a pronunciation. I don't see any such evidence for "peer-iodic", though one of my teachers used it. I myself initially used "peer-iodic", but once I realized what the acid actually was, I switched to "per-iodic".

It is supposed to be pronounced as per-iodic acid schiff but there are many people who dont know that it contains iodine and call it peeriodic acid.


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How are turbostratic graphite phases formed?
Turbostratic graphite is graphite in which there is quenched rotational disalignment between adjacent graphene sheets, i.e. one sheet is rotated with respect to its neighbor. I suppose this could be considered a crystallographic defect of sorts. How do these turbostratic layers form, and can they possess long range order, such as glide or helical symmetry (e.g. similar to a twisted nematic phase)?

"High-temperature phase transformation and low friction behaviour in highly disordered turbostratic graphite," doi:10.1088/0022-3727/46/39/395305
"Phase Transformation Mechanism of Graphite-Turbostratic Graphite in the Course of Mechanical Grinding," CHEM. RES. CHINESE U. 19(2), 216 (2003)
http://www.onxlti.com/product-divisions/contract-manufacturing-products/on-x-pyrolytic-carbon/

Al, this is a good answer to the question, but could you elaborate a bit about what's in the papers?


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What is the difference between melting and dissolving?
What is the difference between melting and dissolving? I am looking some general features.
The answer should be adaptable to the melting/dissolving of ice cube (water) in a class of pure alcohol (ethanol) just below (or at the) melting point of ice, or similar phenomena.
I am now assuming that the ice is dissolving and melting at the same time.
In other words which reaction energy is higher in the following reactions:
$$\begin{align}
\ce{H_2O(s) &\to H_2O(l)\\
H_2O(s) + n EtOH(l) &\to H_2O \centerdot (EtOH)_{n}}
\end{align}$$
Or, are there substances that release more energy in dissolvation than consume in melting?

Hi Juha! Melting and dissolution are two very different physicochemical processes: for one thing, melting is a pure-component equilibrium between the solid phase and liquid phase of the same component, while dissolution involves a solid and a solvent, which is a different chemical compounds. As it currently is, I'm afraid your question is too broad… Could you read up a bit on both melting and dissolution, and edit it to something more specific?

Ice melts, and salt dissolves in water. Ice is made of $H_2 O$ which is the same as water.

What I was actually wondering that is there a substance that exists in two different phases and the first dissolves to the second? or is this always a phase transition (e.g. melting)? In other words, the only difference between melting and dissolving is that the substances are different.

One more question: Can alcohol dissolve ice?

I would also like to ask two yes/no questions: Is melting always endothermic? Is desolvation always exothermic? Do you have (counter)examples?

@Juha regarding melting being endothermic, Wikipedia says “ The enthalpy of fusion is almost always a positive quantity; helium is the only known exception.[1] Helium-3 has a negative enthalpy of fusion at temperatures below 0.3 K. Helium-4 also has a very slightly negative enthalpy of fusion below 0.8 K. This means that, at appropriate constant pressures, these substances freeze with the addition of heat.” I didn't know of this exception, and doubt most chemists would. In short, for a chemist: melting is endothermic.

A question for @F'x. Is melting defined as applying to pure substances only? Is it not correct to say that alloys...solid mixtures...melt? Or am I misunderstanding your terms? (And how do I delete the other comment?)

Juha invited me to write a summary (see comment on my previous answer ) of the differences between melting and dissolving. I’ll try to outline this roughly in the same order as his. I'm giving this as a new answer since my last answer was quite long as it was.
Differences:

Melting and dissolving are completely different processes on the molecular/atomic level that could not be mistaken for each other if you could observe what was happening at that scale.
If you cool the liquid that arose from melting the solid, to a temperature below it’s melting point, you would see the entire sample solidify. If you cool the solution (dissolved solid + solute) to below the melting point of the solid (solute) in it, you would see no change. (Unless you had a saturated or nearly saturated solution.)
Melting requires only a single substance and energy input while dissolving requires a solvent and a solute that are compatible (“like forces”). (This is actually a pretty huge difference and would potentially affect all of the physical and chemical properties.) Dissolving a solid can be either endo- or exo-thermic.
"Phase change" (wording from the previous summary)  In each case you end up with a liquid. Melting caused a phase change (composition of the substance didn’t change) while dissolving a solid in a liquid is not considered a phase change since a change in composition occurred.

Similarities:

In each case, forces between the particles that comprise the solid are disrupted and that takes energy. (Whether it’s chemical bonds or intermolecular forces depends on the process and on the solid and on your definitions. (See this question.) But melting (rare exception noted in previous comments) is endothermic and dissolving can be either endo- or exo-thermic.
In each case you end up with a liquid. Macroscopically, if you walked into a room and saw the liquid on the table, it would be difficult to say whether this liquid came from a solid that had melted or a solid that had dissolved in a solute and made a solution.
But it would be very easy to determine which you had experimentally in a “dozen” different ways.
Both melting and dissolving require interaction among groups of atoms, molecules, or ions.

There are probably more differences that could be given (how to handle thermodynamic calculations, complexity of the system etc.), and possibly more similarities, but that's enough for me on this topic.

Also, melting only happens on an eutectic point - it is defined by the transformation of one liquid to one solid. Non-eutectoid meltings would also be considered a solvation process.

Some additional information relevant to this question.
I'm going to use "forces" and "particles" in this explanation to generalize it. Particles can be ions (ionic compounds held together by strong electrostatic forces), metal atoms (held together by electrostatic forces from attraction of the metal atom nuclei for the "sea of electrons") or molecules (for covalent compounds held together in the solid or liquid by intermolecular forces: London dispersion forces, dipole-dipole attractions, and hydrogen bonding. Note: we are not talking about the intramolecular covalent bonds in this explanation.) So, solids and liquids are held together in that form by forces between the particles.
The reason that melting is always endothermic (except for the exceptions given by F'x...which I also did not know about) is that when you melt a substance, you have to break up some of the forces holding the particles together in the solid so that they can move past each other and form a liquid. This takes energy, the "heat of fusion" (usually given on a per mole basis) mentioned above. The stronger the forces, the more energy it would take to melt the substance. As long as you are melting a substance and holding the temperature at the melting point (i.e. not adding enough energy to raise the temperature) then you will have an equilibrium between the solid and liquid.
I think there is some confusion in the terms you are using. Desolvation means removing the solvent from a solute-solvent mixture. (This can happen in crystallization, for example.) Solvation (which I think you're asking about) is the interaction of solvent particles with solute particles. Dissolving or dissolution is the process in which a solid, liquid, or gas forms a solution with a solvent.
So, as answered by Ashu, dissolution and be either exothermic or endothermic as he's noted above. This is true because in order to dissolve a substance in a solute, it takes energy to break up the solvent-solvent interactions, it takes energy to break up the solute-solute interactions, and you get energy back from new solvent-solute interactions. If you have stronger attractions between the solvent and solute than you had originally, then the process releases energy and is exothermic. The same concepts can be applied to mixing two liquids...which is usually referred to by saying they are "miscible" instead of speaking of one dissolving in the other. Ethanol and water are miscible in all proportions.
Can alcohol dissolve ice ("at or just below the melting point of ice" as phrased in your original question)? There are a couple of ways to answer this. If you put ice in ethanol at 0 degrees (its melting point) it will melt because the melting point of ice in a mixture of water and ethanol is lower than the melting point of pure water. The melting point depends on the proportions of water and ethanol.(That is to say, as some melted, it could not refreeze as you would expect to happen in an equilibrium mixture, so it would continue to melt.) See colligative properties for more information.
You've given two equations,        H2O(s)→H2O(l)
                                   H2O(s)+nEtOH(l)→H2O⋅EtOHn
In either case, you will need to provide the heat of fusion to melt the ice. So the question you asked is really about the energy of solvation of water in ethanol. When you mix water and ethanol, the flask feels warmer, so this is an exothermic process; you get more energy back than you put into it because of good interactions (hydrogen bonding) between water and ethanol particles. That solvation energy would provide some of the energy needed to melt the ice.
For the broader question "can something happen at a given temperature", you're generally asking if it happens spontaneously. To answer this you'd have to calculate the Gibbs Free Energy. This would take into account not only the energy considerations spoken of above, but also entropy considerations (which would matter here). If the Gibbs Free Energy value is negative, then the process or reaction is spontaneous at the temperature given. So, the answer to which would be the preferred process would also depend on entropy considerations.

If I make some summary: In both cases there is an Energy barrier that you have to break. In melting, only one exothermic reaction is known. To me the processes look very similar. If you have two exothermic reactions, one where a solid material X melts in a liquid Y into mixture of liquids X and Y and a second where solid material X dissolves into liquid Y forming a solution of X and Y. How can you tell did it melt or dissolve? Is it just that the reaction components are different (mixture of liquids vs. solution)?

@Juha:
www.thefreedictionary.com/dissolvedis·solve (d -z lv ). v. dis·solved, dis·solv·ing, dis·solves. v.tr. 1. To cause to pass into solution: dissolve salt in water. 2. To reduce (solid matter) to liquid form; melt.

@Juha..So, perhaps the distinction is whether or not there is a solvent present with a bit of "are you above the solids melting point" thrown in. Sugar dissolves in water at room temperature, even if you hold that temperature constant...it doesn't melt. So, if I'm understanding you, your second equation was asking if ice (solid water) would dissolve in ethanol (mp -114) at something like -50 degrees C?

Rereading your comment more carefully: If you were near the melting point of the solid and the two liquids mixed, it would be difficult (impossible?) to tell and could be a mixture of both. (If the liquids were immiscible, then you could watch the solid melt as a separate layer.) If you were at a temperature much lower than the melting point of the solid, and it went into solution, then it dissolved.

So, can I say that melting and dissolving are so similar in some cases that they cannot be separated from each other? And if you would want to demonstrate this, which materials you would use and why? (I can ask this in another thread if this becomes too broad for this)

Could you add some kind of summary to the beginning of your answer? See my answer below what I mean. I really don't want to accept my own answer.

@Juha Melting and dissolving give a similar macroscopic result in some cases (if the two liquids are miscible) but on a microscopic level they are quite different (as F'x noted in his original comment to you). I don't think that most chemists (thinking on a microscopic level) would agree that they are "so similar that they cannot be separated from each other". It would be easy to tell, even if two liquids looked the same, whether you had a mixture (dissolved material) or a single substance that melted, and this difference would affect all of their chemical and physical properties.

@Juha I'll work on summarizing my answer tonight. In your comment above you ask "if you want to demonstrate this, which materials you would use and why?" Can you be a bit more specific about what you're thinking there? A system where your can tell the difference and a system where you can't tell the difference? A system where you know for sure? I'll incorporate that answer into the summary.

Is dissolving always exothermic?

No, it is not always exothermic. For example, dissolving $\ce{NaOH}$ is an exothermic reaction, while dissolving $\ce{NH4NO3}$ is an endothermic reaction (This is from my personal experience)

Is melting always endothermic?

Yes, it always is (except for the exception F'x gave). This is because during melting the body to be melted takes in "latent heat of fusion" from the surroundings, thus bringing down the temperature.  The latent heat of fusion is required to break the intermolecular attraction forces that define a solid. Breaking such bonds requires energy, hence the reaction is endothermic.

Can alcohol dissolve ice?

Lower alcohols like methanol and ethanol may be able to dissolve ice. They can effectively form hydrogen-bonds with water, but higher alcohols may not be able to do that.

To summarize the differences and similarities, here is a list (I will update the list, if new comments arrive):
differences

melting is (almost) always exothermic, dissolving can be endo- or
exothermic
melting is within one substance, dissolvation is between solute and
solvent that are different substances
melting is a collective phenomenon that requires $N_a$ particles, dissolvation can happen one atom at the time

similarities

In both cases, bonds are broken by crossing an energy barrier
phase of matter changes
You can create a situation where it is difficult to say whether the
substance is melting or dissolving.


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What do typical results from inductively coupled plasma mass spectrometry (ICP-MS) look like, and how do I interpret them?
I have a degree in chemical engineering and remember having seen a inductively coupled plasma mass spectrometer (ICP-MS) during the laboratory part of a chemistry course I took while at university, but it is many years ago. Now I would like to know what a typical output from a ICP-MS looks like and how I interpret the results.
There is no picture at the wikipedia page but some frequency domain like plots can be found with a Google search and I understand the Fourier transform and how it is used in number theory.
Are the ICP-MS results anything like the spectrum from the Fourier transform of some signal?

I am not an expert in ICP-MS, but an ICP-MS should be considered an ICP ion source coupled with a mass spectrometer. There are some key issues raised by the coupling between both instruments due to the high temperature of the ICP source as well as due to the fact that one wants to analyse atomic ions and remove any interfering molecular ions (mostly oxides, but other molecules and clusters are possible).
Therefore, to attempt an answer to your question, the standard output of an ICP-MS will be presented as a mass spectrum (intensity versus m/z or intensity versus amu). There are a couple of sample spectra in the brochures published by instrument makers, see for instance the Bruker brochure or a Thermo application note. How this mass spectrum is obtained from the detector will lead to the type of data that is acquired:

For a quadrupole mass filter, as well as for magnetic sector instruments, two modes of operation are possible. Either a single mass is selected, and therefore ont will only get a trace of the abundance of a given mass as a function of time (chromatography or on-line sample analysis). Otherwise the mass filter can be scanned, leading to a mass spectrum.
For time of flight detectors, the mass spectrum is built by the measurement of a flight time for an ion to reach the detector. Therefore the mass spectrum is converted from a time dependent signal to a spectrum.
Finally there have been some attempts to couple an ICP ion source with a Fourier transform mass spectrometer. As far as I know, these are not commercial instruments. In such instruments, the ions oscillate at m/z dependent frequencies, and the recorded image current is converted to a spectrum through the use of Fourier transform.


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Mercury, amalgams and mercury compounds
Mercury is toxic. It forms amalgams and inorganic mercury compounds/salts are available.
My question is on toxicity. I am interested in its toxicity, not the mode of action (as it will be part of biology SE)

Is amalgam toxic? (sort of yes / no)
Mercury salts are toxic? (sort of Yes / no)
If mercury salts / amalgams are non-toxic, what makes them non-toxic. Is it because of the ionic nature / other metal or any other reason?

I suggest to you The Elements of Murder: a History of Poison, by John Emsley… It deals with toxicity of mercury in its various forms, and it is very accessible.

http://en.wikipedia.org/wiki/Dental_amalgam_controversy ; concentration of mercury also matters when assessing toxicity, very small amount of mercury is not considered as toxic.

@test, i think this is debatable, toxic material used in dental problems.

@F'x, I got this link (http://wiki.answers.com/Q/What_is_the_action_of_Mercury_in_ayurvedic_medicine),  which says, mercuric compounds used in ayurvedic preparations are safe. I do understand that this is wiki page. More google search will strenthen the point. Again this is a bit debatable, as part of ayurveda drugs.

OK, I’ve taken time to think and I find that this question is overly broad: first, there are many amalgams (any alloy including mercury is an amalgam) and many mercury salts, so it is effectively a “list question” (which are discouraged on SE sites). Secondly, even toxicity of a single chemical compound is not a yes/no question.

Another source of information is Wikipedia’s page on mercury poisoning…

Toxicities are often greatly influenced by how you are exposed and in what dose.
Mercury is a fascinating metal. You can find it as the pure metal, a salt or in an organic form. In its standard state it is a liquid and is rather volatile. Large amounts are naturally released into the atmosphere, but humans contribute as well, for example: burning fossil fuels, such as coal, will release it into the atmosphere.
The forms of mercury are not set in stone it is quite able to switch forms from the free metal, to the salt, to the organic by processes in nature.
Orally consuming some liquid metal isn't the worst possible route of exposure; mercury was at one time even used as a laxative. The only absorption in the GI tact is thought to be due to mercury vapor. Skin exposure is undesirable, but we know that inhaling the vapor is more toxic because it is soluble in lipids. Upon entering the bloodstream, some makes it way past the blood/brain barrier (hence its neurotoxicity) and some is converted to Hg(II) by catalase.
Mercury's inorganic salts have a long useful history where some were used to treat syphilis, some for lights and some for hat making and so on. "Mad as a hatter" won its name due to chronic exposure to Hg(NO$_3$)$_2$, which was used in hat making. This salt isn't very lipid soluble, but more absorption does occur when swallowed than the free metal. These salts tend to build up in the kidneys after absorption and high exposure will cause renal failure. Not much will pass through the blood/brain barrier, but chronic exposure will eventually allow some across and cause neurotoxicity. They also tend to bind with proteins containing sulfur and this can cause some cell death. As a general rule, Hg(I) salts are less toxic than Hg(II), presumably due to the decreased solubility of Hg(I). This does not mean they cannot cause harm: mercuric chloride was used in a factory and the wastes were discharged into a nearby sea, anaerobic bacteria methylated the mercury, this methylmercury was absorbed by fish that were later eaten and hurt many people.
As an organic, mercury compounds tend to be considerably more lipid soluble and therefore can readily be absorbed or cross the blood/brain barrier, in comparison to the free metal. A terrible example is dimethylmercury, which claimed the life of an inorganic chemist who was accidentally exposed to some she was using in an NMR experiment.
I do not know much about its amalgams aside from gold/silver. I believe most of the toxicity was seen in gold refineries due to the fact that the amalgam is heated to remove the mercury, which leaves behind gold, and results in the release of mercury vapor.


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Resolution in Mass Spectrometry
I have been reading about resolution in mass spectrometry and there are a few things I do not understand.
The resolving power is determined by $m/(m_2-m_1)$ at full width half height of a peak. The higher the value obtained the better. This is where I struggle to form a link between the resolution of the mass spectrometer and the resolving power. In image (a) below the resolution of the mass spectrometer is 1000 and would have a very low mass resolution value, but in image (b) the resolution is 5000 and the nominal peaks masses are separated, therefore providing a high mass resolution value.
Why can a mass spectrometer with a resolution of 5000 resolve the nominal masses and the mass spectrometer with a resolution of 1000 cannot? And how does the mass resolution come into play? Also, what do the values of 1000 and 5000 mean?

I'm not sure I completely understand what exactly your problem is, but:
You want to resolve to signals at approx. m/z = 1060.2 and m/z = 1061.2.
So $\Delta m = 1$ The required resolution is: $\frac{m}{\Delta m} = \frac{1061}{1} \approx 1060$: The spectrometer in (a) has only 1000 which is too low. The spectrometer in (b) has 5000 which is quite more than needed.
Maybe it is easier to think of the reciprocal: $\frac{\Delta m}{m}$ which tells you how far two signal need to be apart so your spectrometer can resolve them.
In (a) this is $\frac{\Delta m}{m} = \frac{1}{1000} = \frac{1.06}{1060}$ two signals can be resolved if they are at least 1.06 m/z apart at m = 1060. I.e. if they were slightly further apart than they are.
The spectrometer in (b) can resolve signals that are at least  $\Delta m/z =  0.21$ apart at  $m/z = 1060$.
FWHM and resolution
The FWHM is basically part of the resolution. The $\Delta m$ is the m/z difference between two signals that are just resolved.
Have a look at the IUPAC's definitions of resolution in mass spec. There are several approaches to specify what "just resolved" means.
The FWHM can be used directly as the $\Delta m$ of the "peak width definition".
If instrument A has $5\times$ the FWHM than instrument B, then resolution of A  $\approx \frac{1}{5}$ resolution of B.
I'm more used to the "10 % valley" which roughly equals the full with at 5 % height: you specify how much signal is allowed at the minimum between the two peaks, e.g. 10 %.
It is important to realize that these two definitions differ by almost a factor 2!
Here are some results of "simulated" peaks at different resolutions and $\Delta m/z$ (all gaussian, which of course doesn't need to be the case in reality). I use the full width at 5% = 10 % valley definitions for the resolution.
The first row is resolution 5000 with the, second row is resolution 1000. You see that the two signals are much better resolved than in your example picture above. Which means that above another resolution definition was used, possibly $\Delta m$ = FWHM. In any case, a resolution of 1000 should be almost sufficient to resolve m/z 1060.2 from 1061.2.
The third row shows two signals of equal height with $\Delta m/z $ = 1 = FWHM. According to the resolution definitions above, this is a resolution of ca. 520. Note that while you can say that there are at least two peaks, you cannot take the two maxima as the m/z of the two underlying signals (grey).
The fourth row is almost the same, but I multiplied the second signal by 56% to account for the frequency of having bradykinin with one $^{13}$C. Looks quite similar to your picture.

Also, have a look at M. P. Balogh: Debating Resolution and Mass Accuracy, LC•GC Europe, 17(3), 152–159 (2004).

Firstly thank you for the answer it has cleared up some issues. But in the second paragraph you looked at the distance between the two peaks, when does the FWHM/FWHH come into this? Also, in the fourth paragraph where were the values 1.06 and 1060 obtained?

@Harpal: the 1060 is approximately the m/z you're looking at. FWHM: I add this to answer.

I didn't think properly when I wrote the first answer - of course $\Delta m/z = 1$. I got confused because I'm more familiar with the 10%-valley definition of the resolution, and the R = 1000 example definitively didn't look like "almost" resolved according to that... It is corrected now and I added a discussion of the two resolution concepts.

Resolution has more than one definition - take a look at Harris' textbook, it has been of great help to me. Especially the subsection "Oh Mass Spectrum, Speak to me!"

@CHM: don't have the Harris here, and I didn't knowt the web site, thx.

@CHM: I guess the huge differences between the definintions are why the IUPAC says you should always state the definition you use. What other definitions besides the "valley" and FWHM are you familiar with for MS? (I'm also quite used to the confusion when the resolution is determined by the data point spacing (full resolution of spectrometer is not needed). In that situation some people refer to the data point spacing as resolution while others use 2x data point spacing ("you need a minimum between two maxima"). And of course there's Rayleigh for gratings.

Is there a physical reason for the definition as $\Delta m / m$ (or the reciprocal)? I wonder why the normalization to the average mass $m$ makes sense. Is 1 at 100 really same as 2 at 200, does this resolving power scale like that or is that just copied from optics?


text:
Striking examples where Kohn-Sham orbitals clearly have no physical meaning
In Density Functional Theory courses, one is often reminded that Kohn-Sham orbitals are often said to bear no any physical meaning. They only represent a noninteracting reference system which has the same electron density as the real interacting system.
That being said, there are plenty of studies in that field’s literature that given KS orbitals a physical interpretation, often after a disclaimer similar to what I said above. To give only two examples, KS orbitals of H2O[1] and CO2 closely resemble the well-known molecular orbitals.
Thus, I wonder: What good (by virtue of being intuitive, striking or famous) examples can one give as a warning of interpreting the KS orbitals resulting from a DFT calculation?

[1] “What Do the Kohn-Sham Orbitals and Eigenvalues Mean?”, R. Stowasser and R. Hoffmann, J. Am. Chem. Soc. 1999, 121, 3414–3420.

+1,000: The use of the KS orbitals drives me crazy and no one ever really tries to justify it. Conversely, I have not seen a good example of it failing, and everyone seems quite happy with it in practice. It's really frustrating: if it works it should have an explanation and if doesn't work people should stop doing it.

Phil Anderson's latest book has a chapter on the issue of Slater vs. Mott, which is really DFT vs. reality. The KS orbitals would be exact in the absence of interactions. In cases where the excitations can be adiabatically linked to non-interacting electrons the error can only be quantitative and may be renormalised away; I think this is what the LDA+U work done by the Cambridge group is doing --- you use bond lengths to establish the "right" U for the current problem. One would expect that on the insulating side of a Mott transition the correspondence would be qualitatively poor.

Cont.: I'm not a DFT expert (just a general theorist) but I would be surprised if accurate results are available for metal oxides, especially ones with d-orbital or f-orbitals.

@genneth: Isn't that the answer? The mott insulator has localized states, but don't the Kohn Sham orbitals in a periodic potential end up delocalized?

@RonMaimon: I would like to think so --- but I'm really not qualified to make a research-level statement. I'd like someone who really works on these things to say something intelligent.

I'm a bit confused by the part of the question comparing KS orbitals with MO theory. MO uses LCAO, and the atomic orbitals are single electron orbitals. So neither are the True Wavefunction, and one linear combination of single electron orbitals will end up much like another, because they represent the basis set of angular momentum orbitals given the Coulomb interaction potential.

These are not necessarily the fault of KS orbitals, but how some people like to interpret them: 1) KS orbitals are delocalised all over the system even for orbitals which are kind of localised according to spectroscopic measurements, 2) ordering of KS orbitals can be very different depending on the functional you use, yet some people like to use orbital energies directly to discuss spectroscopic properties. These are especially visible in research on transition metal systems.

When people say that Kohn-Sham orbitals bear no physical meaning, they mean it in the sense that nobody has proved mathematically that they mean anything. However, it has been empirically observed that many times, Kohn-Sham orbitals often do look very much like Hartree-Fock orbitals, which do have accepted physical interpretations in molecular orbital theory. In fact, the reference in the OP lends evidence to precisely this latter viewpoint.
To say that orbitals are "good" or "bad" is not really that meaningful in the first place. A basic fact that can be found in any electronic structure textbook is that in theories that use determinantal wavefunctions such as Hartree-Fock theory or Kohn-Sham DFT, the occupied orbitals form an invariant subspace in that any (unitary) rotation can be applied to the collection of occupied orbitals while leaving the overall density matrix unchanged. Since any observable you would care to construct is a functional of the density matrix in SCF theories, this means that individuals orbitals themselves aren't physical observables, and therefore interpretations of any orbitals should always be undertaken with caution.
Even the premise of this question is not quite true. The energies of Kohn-Sham orbitals are known to correspond to ionization energies and electron affinities of the true electronic system due to Janak's theorem, which is the DFT analogue of Koopmans' theorem. It would be exceedingly strange if the eigenvalues were meaningful while their corresponding eigenvectors were completely meaningless.

So, in summary, there is no counter-example known where the KS orbitals clearly lack physical meaning, even though it's not theoretically guaranteed. Interesting…

Well, not to my knowledge anyway.

This is a complex issue, particularly because people often like to think in terms of an indepedent-particle picture (i.e. the aufbau filling up orbitals), even though the exact many-body wavefunction has strong electron-electron correlations. So let me rephrase your question:
What is the relationship between the KS eigenfunctions and the exact many-body wavefunction?
Mathematically, as you say, the KS eigenfunctions strictly speaking have no physical meaning (as far as we know). However, the KS eigenfunctions do give a useful qualitative (and sometimes quantitative) picture. The reason for this is that the KS eigenfunctions are a pretty good approximation to something in many-body perturbation theory called the quasiparticle wavefunction. The quasiparticle wavefunction is a well-defined physical property of a system that essentially tells you if you add (or remove) an electron with a certain amount of energy, where it will go. For example, see Phys. Rev. B 74, 045102 (2006).
Are there examples of when the KS eigenfunctions don't give a good desription of the quasiparticle wavefunctions? Well there are certainly many situations where the approximations we typically use in DFT (such as the local density approximation) lead to serious problems. However, I don't know of any examples where someone has shown that the exact KS eigenfunctions (i.e. those obtained with the true exchange-correlation functional) don't agree at least qualitatively with the quasiparticle wavefunctions.
As an aside, everything I have said above applies equally well to the Hartree-Fock wavefunctions. In fact, there is a solid mathematical basis for interpreting the HF wavefunctions as an approximation to the quasiparticle wavefunctions. See Chapter 4 of Fetter's Quantum Theory of Many-Particle Systems.
What about the KS eigenvalues? Strictly speaking, in general they do not correspond to ionization energies (or any other physically useful quantity). The one exception is the highest occupied eigenvalue, which is exactly equal to the ionization energy of the system. Janak's theorem tells us that the other eigenvalues are related to the derivative of the energy with respect to the occupancy of that eigenfunction:
$$\epsilon_i=\frac{dE}{dn_i}$$
See Phys. Rev. B 18, 7165 (1978) and Phys. Rev. B 56, 16021 (1997). It turns out that empirically these eigenvalues are nonetheless pretty good approximations to the true  energy levels of the system with some caveats. In particular, the band gaps of solids are systematically underestimated.

Informative, thank you. Would Fetter's book be overkill for someone interested in DFT?

Fetter's book is not a good resource to learn about DFT, because it is really about many-body perturbation theory. For someone interested in DFT, I would recommend Sholl and Steckel's introductory text or Richard Martin's book. Sholl is more practical, while Martin is more comprehensive and mathematical.


text:
Can dissolution reactions be endothermic?
Is it possible for a dissolution reaction of the form
$$\ce{X(s) ->[solvent] X (solv)}$$
or
$$\ce{M_mX_x (s) ->[solvent] m M^x+ (solv) + x X^m- (solv)}$$
to be endothermic, i.e. $\Delta_\mathrm r H > 0$?

You will find the answer in probably any textbook on chemical thermodynamics, or in a thermochemistry chapter in most physical chemistry textbooks, or in Wikipedia's enthalpy of dissolution page:

Dissolution can be viewed as occurring in three steps:

Breaking solute-solute attractions (endothermic), see for instance lattice energy in salts.
Breaking solvent-solvent attractions (endothermic), for instance that of hydrogen bonding
Forming solvent-solute attractions (exothermic), in solvation.

The value of the enthalpy of solution is the sum of these individual steps. Dissolving ammonium nitrate in water is endothermic. The energy released by solvation of the ammonium ions and nitrate ions is less that the energy absorbed in breaking up the ammonium nitrate ionic lattice and the attractions between water molecules. Dissolving potassium hydroxide is exothermic, as more energy is released during solvation than is used in breaking up the solute and solvent.

So, it can have any sign. The linked page also gives data for common compounds in water, some of them endothermic, some of them exothermic. Ammonium nitrate, $\ce{NH4NO3}$, is a good example of a salt which dissolves in water in an endothermic reaction: if you dissolve some $\ce{NH4NO3}$ in water, the drop in the temperature will be noticeable.

Regarding why the substance is soluble why its dissolution is endothermic, you have to remember that the reaction takes place if $\Delta_\mathrm r G$ is favourable (i.e. negative), and $\Delta_\mathrm r G = \Delta_\mathrm r H - T \Delta_\mathrm r S$. Overall the free energy must be negative for dissolution to occur (on a thermodynamic basis; kinetics are another issue), not the enthalpy.

How can the reaction happen if it is endothermic?

@Juha: Take energy from the surroundings, of course. Lots of reactions are endothermic. Whether they happen or not is determined by $G$, not $H$

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


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When simulating spectral line broadening, which convolution is preferred?
Many computational chemistry packages permit the calculation of vibrational and electronic spectra. These spectra are obtained as a set of discrete eigenvalues however they are often convolved with some distribution to give a continuous spectrum that is realistic for a finite temperature, rather than a sequence of impulse functions.
The ADF package allows the user to convolve the spectrum with either a Gaussian or Cauchy (Lorentzian) function. My understanding is that the latter affords a more realistic line broadening, however the Gaussian must be there for a reason.
Which spectral broadening scheme is preferred? Why the choice?

As I recall, it depends on the conditions the gas is under, if collisions are causing changes in the spectra or not. I can't remember which was which though.

What are the 'raw' signals the vibrational and electronic spectra are calculated from? Do you first have some time-domain signals that are then transformed as frequency-domain signals?

I'm not familiar with the computational packages -- I'm an experimentalist, not a theoretician.
As an example for why both might be present, we can turn to gaseous infrared spectroscopy:
one will quickly find that the line-widths in vibration-rotation spectra depend on pressure. At lower pressures, say less than a torr, Doppler-broadening is the main mechanism. This is Gaussian in nature. While at higher pressures the broadening is instead due to collisions, which are Lorentzian. You'll find combinations of the two and so forth depending on the specifics of your case.

Aha! That makes great sense. Are liquid and solid-state spectral broadenings also Lorentzian?

There are also useful model lineshapes for intermediate cases, such as the Voigt function: http://scienceworld.wolfram.com/physics/VoigtLineshape.html

Liquid are sometimes modeled using the sum of Lorentzian and Gaussian. There are very few deep studies on these bands in solids, but as a general rule the solid bands are typically more narrow than their solution counterparts. The sum would probably be fine for most purposes too.


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How does one recognized Td/Oh symmetry in molecules?
The identification of point groups of a molecule is usually done following a strict scheme, either manually or algorithmically. In all textbooks I could find, however, the first step of the scheme is actually not explicited: in this example scheme drawn from Housecroft and Sharpe,

you can see that there is a very unhelpful step “Does this molecule have $I_h$, $O_h$ or $T_d$ symmetry?”. It is implied that one would recognize e.g. an icosahedral molecular compound on sight. However, I wonder: how can one strictly identify these “special” point groups? What set of rules is to be followed (again, either manually or algorithmically)?

This has been address before, see here: http://scicomp.stackexchange.com/q/135

@Chris thanks for the link! It contains very useful information (such as a link to an generic algorithm and code implementing it). It doesn't provide the specific solution to this question, however (and it couldn't be a duplicate anyway, because it's on another site).

In fact most people see this symmetries "at once". The main problem was the lack of apropriate sketches in books formerly. A org prof of me used to say: progress of sterechemisty was mostly due to progress in printing technology/cost. Another problem is that some persons lack the ability to see/think 3D more or less.

You can detect $O_{h}$, $I_{h}$ and $T_{d}$ symmetry by checking that a molecule has all of the subgroup symmetries of these point groups.
According to this untitled document, which I presume1 is by W.C. Trogler, the elements are as follows:
$T_{d}$: $E$, $4C_{3}$, $3C_{2}$, $3S_{4}$, $6\sigma{_d}$
$O_{h}$: $E$, $3C_4$, $4C_3$, $6C_2$, $4S_6$, $3S_4$, $i$, $3\sigma{_h}$, $6\sigma{_d}$
$I_{h}$: $E$, $6C_{5}$, $10C_{3}$, $15C_{2}$, $i$, $6S_{10}$, $10S_6$, $15\sigma$
Obviously, you don't need to check for the identity symmetry $E$.
If you're a visual sort of person a symmetry spot check is to mentally superimpose a tetrahedron, cube or dodecahedron over the molecule and see whether the view down the surface normal of each face is identical. Cubes and octahedra are duals of each other, as are dodecahedra and icosahedra. The tetrahedron is self-dual.
Interestingly, H&S does not list chiral forms of these point groups, probably because they are so rarely encountered, however researchers have come up with molecules that obligately satisfy $T$, $I$ and $O$ symmetry2 (I have not yet read the paper).

(1) I would be indebted to anyone who can furnish me with a full citation to this work, and confirm the authorship.
(2) Narasimhan, S. K., Lu, X. and Luk, Y.-Y. (2008), Chiral molecules with polyhedral T, O, or I symmetry: Theoretical solution to a difficult problem in stereochemistry. Chirality, 20: 878–884.

Does one really have to check for all the symmetries? Isn’t there a shorter way? (some invariant that these molecules would satisfy, or something like that)

The syllabus for the course makes it clear that the instructor is Bill Trogler

@F'x - Mentally superimposing the corresponding Pythagorean solid is the best heuristic that I know of. However on reflection (see what I did there?) it will fail to distinguish the chiral and achiral forms of the point group. In the end it's only a heuristic, and if you go and check every face and edge rigorously you will be implicitly checking that each and every symmetry element is satisfied. However, hopefully someone savvier will chime in with the easy method.

Adamantane is point group $T_d$, but you won't easily figure that unless you already have a molecular model in front of you. $C_{60}$ is $I_h$, but you'll only realize that if you know the relationships between the dodecahedron, icosahedron, and truncated icosahedron... huh.

The quick way of doing that first step in the flowchart is to look for too many principal $C_n$ axes. (Since the next step is to look for a principal axis anyway, this is a natural step to take.) In particular, you are looking for more than one axis of order > 2. A threefold axis with three twofold axes perpendicular to it? Just $D_{3*}$ (where $* = d$, $h$, or nothing is yet to be determined, by using the flowsheet). But more than one threefold (or fourfold, or fivefold) axis? Has to be one of these special point groups.
The next question, of course, is which one. Again the key is these "excess" principal axes. Lots of fivefold axes is the giveaway for icosahedral symmetry (to be sure, count six of them); lots of fourfold for octahedral (look for three); and neither of these means tetrahedral (you can look for four threefold axes to be sure, but you need to eliminate the other two since both of these also have threefold axes).
In practice you can assume that icosahedral means $I_h$, octahedral $O_h$, and tetrahedral $T_d$. To be complete, however: for octahedral and icosahedral you need to find a centre of symmetry, otherwise it's just $I$ or $O$. For tetrahedral, if you have a centre of symmetry it's $T_h$; if you have mirror planes but no centre of symmetry it's $T_d$, and with neither it's $T$.
Essentially this is just the same answer as Richard's, but my point is just that there are heuristics you can carry around in your head without having to check all 120 (or however many) symmetry elements. To take J.M.'s examples: adamantane has several (you don't really even have to count four) threefold axes but nothing higher: must be $T_d$. Buckyballs have a few fivefold axes: surely $I_h$.

I'm not sure what you mean by the question being "unhelpful," especially if you thought it was worth answering. I suggested an edit.

Thanks for the edit - I was quoting the OP's referring to a "very unhelpful step" in the flowchart rather than suggesting that the question asked was unhelpful!

I understand now. I did not see the word "unhelpful" in the question, so I was confused.


text:
Why does ice cream make soda fizz?
I've noticed that adding a chunk of ice cream to soda makes the soda fizz slightly near the soda-ice cream interface. I thought it was a physical effect due to the temperature, but adding ice has no effect.
It still can be a physical effect due to solubility, or it may be a chemical effect.
I'd like to know which it is, and the details of the mechanism.

The reason for this is mainly because of a factor called Nucleation. Although it is not exactly obvious what is the source of the site for the nucleation to occur, what is clear is that there are sites present on ice-cream that are not present on ice alone. The carbon dioxide (CO$_2$) in the drink nucleates and forms bubbles (sometimes even a frothy foam) on the drink.
There are several possibilities some of which are discussed in the excellent answer by Ashu.
I list some of them here starting with what I feel is the most likely.

Air in ice-cream
Ice crystals in ice-cream
Other chemicals found in the ice-cream

Could be the fat globules(that are part of milk) coupled with the temperature form nucleation centers :\

Soda contains $\ce{CO2}$. Over time the gaseous $\ce{CO2}$ comes to the surface. The ice cream increases the rate of this  $\ce{CO2}$ from the soda causing increased fizzing. This increased rate is a result of the solvation of ice cream particles in the liquid. The particles serve as nucleation sites for the growth of gas bubbles. The bubbles get large quiet fast thus you observe fizzing.
After a bit of experimentation I found the following:

Effect of adding milk — There is a bit of extra bubbling and a curdy precipitate is formed. (Couldn't dare to drink it.)
Effect of adding salt (yes ice cream does contain salt) — Lots of bubbling around the salt. After the salt settles to the bottom, bubbles collect around it become big and then come to the surface.

Only thing left to try is sugar and vanilla essence but I ran out of soda.
So we come to know that the fizzing effect may be due to milk or salt or a combo of both.

why wouldn't it be mainly due to the lower temperature of the ice cream, thereby decreasing the solubility of carbon dioxide?

@Chris: The solubility of gases in liquids decreases with increasing temperature.

your argument is true at equilibrium conditions. In non-equilibrium conditions the ice cream frozen at 0 F will begin mixing with the warmer solution, pop. Upon mixing, an ice sheet will form in some regions for some amount of time at the ice cream/water interface. CO2 hardly dissolves in solid water. This is why it isn't entirely clear to me. Perhaps milk mixed with pop at the same temperature would be more convincing. Time for me to try an exp at home!

@Chris: Yeah, I think it's the milk. Ice doesn't work on its own.

This still doesn't answer the question--what's the actual reaction going on?

@Manis, there is no reaction...

@pica "reaction" isn't limited to a chemical change in this case. Could be a physical one as well--though since this happens to icecream not ice, my bet is on a chemical one.

@Manis, I guess this is what you are looking for.. http://en.wikipedia.org/wiki/Nucleation
In any case, Ashu addresses this pretty well. The  ice does not provide a site for nucleation as well as the ice-cream. Think about diet coke and mentos only working with some types of mentos.

@pica I sort of suspected that the milk particles were serving as nucleation sites, bit I wasn't sure. Expand that into an answer and I'll accept it :)

It is not the milk particles. (Infact the idea of milk particles makes no sense)

I just wanted to add that it can't be the milk because I was pleased to discover it also happens with sorbet & other dairy free 'icecreams'. In Australia we call it a spider

I would say that it is a chemical reaction due to the fact that it is irreversible. Aside from that, the carbon dioxide in the drink reacts to the fat and the bubble (which make ice cream light and fluffy) and wha-la. There is still not solid research as to how it works, but this is my guess due to the fact that most soft drinks have no fat in them, and ice cream has a fair amount. There is research however that states that the carbon dioxide is able to grow in the small air bubbles in the ice cream and this causes the froth. Personally, I think it's both

Could you give the references to which you refer to?

I too think this is a spurious answer. A simple gas-fired piston works by the expansion of a gas. Once it's happened it doesn't reverse spontaneously but we wouldn't call the process of gas expansion 'a chemical reaction'.


text:
Not your ordinary hard water deposit
We have a well that delivers hard water and we do not have a water softener. (There was a “green sand filter” that utilizes permanganate to oxidize other metals on the system, but it has been disconnected.) Analysis of the well water shows the presence of iron in noticeable amounts. No other metal ions were specifically identified. The system does have an ozonator to oxidize Fe(II) to Fe(III). Over the last couple of years, the porcelain toilet bowls have developed a blue ring at the air/water interface. When you scrub on the blue ring, it doesn’t feel like there is a deposit present (like normal hard water deposits). The ring is unresponsive to ammonia (household), strong acid (6 M HCl) or strong base (Drano), as well as to hydrogen peroxide (30 %).  (We haven’t tried a good reducing agent yet.) So, my question is, does anyone have an idea of what is causing the blue color? Is it possible that porcelain reacted chemically with salts, urea, uric acid, oxygen, etc?  Not an earth-shattering theoretical question, but I’m chemically curious as to what this could be if anyone has any ideas.

I know this may seem silly...but I've spent several hours looking up information on colored salts, colored complexes, and trying to find out some information about the reactivity of porcelain and nothing has been useful. And the site has been asking for more questions.....

are you in NC or Georgia?

I'm actually in California.

The blue color is copper from some copper or brass plumbing, presumably. The "bulk" of the deposit is SiO2 (hydrated), doped by Copper.

@Georg, if this is the case then why is there only a blue ring?

@JaniceDelMar, do you know the pH of your drinking water?

I second the idea that the blue is from copper. What part of California do you live in Janice, the Valley? Very hard water around there.

@Chris The pH of our drinking water is 6.7  pre-ozone and 7.4 post ozone and the ozonator is currently on most of the day. I live in Sonoma county. Copper pipes are supposed to be less prone to corrosion when in hard water, so would there be appreciable copper ions present? And how do they dope the porcelain under such mild conditions? (Do you have any references? I can't find anything relevant.) I've used porcelain crucibles and spot plates with various salts and not had discoloration under much harsher conditions. Thanks for any ideas.

What about some silikate in Your water? My idea is some silikate deposit, not copper in the porcelain enamel (which is not possible). Copper does not come from pipes, but more from taps, esp. perlator mesh at water/air interface.

@Georg If it is a copper (II) silicate deposited on the porcelain, shouldn't it be soluble in strong acid? We've tried 6M HCl on it without effect. Is there a simple test you know of that could confirm your suggestion that it's a copper silicate? Thanks.

Electronics people sometimes use ferric chloride to etch copper off PCB blanks. I hope your freshly oxidized Fe(III) ions aren't slowly etching away your plumbing!

As just a quick tip - it could be some kind of ferrous chelate (cyanide, aminoacidic) ?

Have you tried mixing hydrogen peroxide with sulfuric acid in the bowl? Powerful etchant.

I found these instructions: Copper Stains: Combine one cup of white vinegar with 1 tbsp of salt and boil until the salt dissolves. Then dip a cotton rag or towel in the liquid and place directly on top of the copper stain. I repeated this process twice and got rid of ALL our copper stains.

Toilet bowls are made of ceramics which can contain feldspar. Toilet bowls also often have a feldspar glaze. There is a study that disolved Cr(VI) can be removed by feldspar.
Studies on the removal of Cr(VI) from waste-water by feldspar
http://onlinelibrary.wiley.com/doi/10.1002/jctb.280530204/abstract
The IMA currently recognizes 90 Cr bearing minerals. 16 minerals if you only count those that contain also Si. Feldspar can be either K, Na or Ca-Feldspar. Of those 16 minerals 6 contain Na, 2 K and 5 Ca. Reading those papers might even narrow the choices down to a couple of minerals that can form in standard-toilet-bowl-conditions.
Mineral database that lets you select minerals by their chemistry:
http://rruff.info/ima/
I think that unless you are willing to donate your toilet bowl to science that the answer won't get any more precise than this.
Edit:
Cu would be the favorite choice for a blue color mineral (Azurite, ...). There is also a study about the removal of copper by feldspar:
Experimental studies of the interaction of aqueous metal cations with mineral substrates: Lead, cadmium, and copper with perthitic feldspar, muscovite, and biotite
http://www.sciencedirect.com/science/article/pii/S0016703797001178
Again there are 603 Cu-minerals. Cu+Si=32 minerals. With the common Feldspar cations your left with about 10 each. I don't think the minerals could be recognized visually. The favorite method would be x-ray diffraction of a powder sample of the blue stuff.

Thanks for the information..now I need to take some time to go through it. It would be nice if there were some simple solubility tests to ascertain whether this approach has merit...without sacrificing the toilet bowl to science. (Although maybe when redecorating time rolls around....) An interesting idea, at any rate.

XRD of a glaze will not yield any results as glazes are glasses, not crystalline.

I heard it is a high acid content in the water which is eating away at by he copper pipes.  The copper turns blue when oxidized like the statue of liberty, the copper residue stains the toilet bowel.  This was very rare until fracking started.

Oxidized to what? Are you saying it's copper(II) sulfate pentahydrate?


text:
Why does activated carbon preferentially adsorb anions?
Brands like Brita and Pur (in the U.S.) have made a name for themselves for the ability of their product (essentially a large-pore filter with activated carbon/charcoal) to extract the added chlorine from tap water.  I visualize $\ce{C}^*$ as a vast network of channels in which ions are "trapped".  I had assumed that most cations and anions were trapped effectively, but it seems to favor chlorine and iodine.
According to Wikipedia:

Activated carbon does not bind well to certain chemicals, including alcohols, glycols, strong acids and bases, metals and most inorganics, such as lithium, sodium, iron, lead, arsenic, fluorine, and boric acid.

Why would it not bind effectively to fluorine?  Is it an issue of atomic radius?  Does $\ce{F-}$, with a radius of 0.136 nm sneak by, while $\ce{I-}$ and $\ce{Cl-}$ at 0.181 and 0.216 nm, repectively, get caught up in the matrix?  Why do the cations get passed through?

This answer applies to carbon filtration of water.
According to most of the sources I found, activated carbon binds to most substances through London dispersion forces (from the Wikipedia article). This should mean that it adsorbs larger molecules and non-polar molecules preferentially since they would have larger dispersion forces. As your quote indicates, it does not bind polar molecules like "alcohols, glycols, strong acids and bases" well. This source also lists neutral, non-polar compounds, along with organics and compounds with low water solubility as being effectively removed.
I think that you may have misread or misinterpreted the sentence about binding "chlorine and iodine" as saying that it traps chloride, iodide, and (by extension) other anions well. I can't find data to support this statement. It traps molecular iodine well, and as a large (in terms of electrons), non-polar molecule, that makes sense. But it removes molecular chlorine $\ce{Cl2}$ through a chemical reaction by reducing it to the chloride ion $\ce{Cl-}$ which is then soluble in water and flows through the filter.
I did find a couple of sources (1, 2) that said "activated charcoal" has a slight electropositive charge that helps it to attract negatively charged species, which would fit with your statement that it preferentially attracts anions. (However, they don't list what kind of negative species they're talking about and nitrate, sulfate, and fluoride ions were specifically excluded). The idea that it will remove ionic compounds seems to be contradicted by the statement in at least a couple of sources (one given here) that carbon filters don't remove minerals, salts, and dissolved inorganic compounds

I found a paper (web.anl.gov/PCS/acsfuel/preprint%20archive/Files/41_1_NEW%20ORLEANS_03-96_0461.pdf) looking at the attraction of negative ions to activated charcoal and the negative ions they referenced were benzoate, oxalate and fumarate anions (all organic anions). The pH is an important variable for this. Phosphate (a relatively large and highly charged anion) is also adsorbed to some extent.

Chlorine actually does not get adsorbed by activated carbon. It reacts with it and oxidizes the carbon to CO2 while the chlorine is reduced to chloride.
So while activated carbon media can be regenerated when loaded with organics, it cannot be regenerated after chlorine removal applications because it actually gets depleted.


text:
Why are equations of state for a non-ideal gas so elusive?
The ideal gas equation (daresay "law") is a fascinating combination of the work of dozens of scientists over a long period of time.
I encountered Van der Waals interpretation for non-ideal gases early on, and it was always somewhat in a "closed-form"
$$\left( p + \frac{n^2a}{V^2} \right)(V - nb) = nRT$$
with $a$ being a measure of the charge interactions between the particles and $b$ being a measure of the volume interactions.
Understandably, this equation is only still around for historical purposes, as it is largely inaccurate.
Fast-forwarding to the 1990s, Wikipedia has a listing of one of the more current manifestations (of Elliott, Suresh, and Donohue):
$$\frac{p V_\mathrm{m}}{RT} = Z = 1 + Z^{\mathrm{rep}} + Z^{\mathrm{att}}$$
where the repulsive and attractive forces between the molecules are proportional to a shape number ($c = 1$ for spherical molecules, a quadratic for others) and reduced number density, which is a function of Boltzmann's constant, etc (point being, a lot of "fudge factors" and approximations are getting thrown into the mix).
Rather than seeking an explanation of all of this, I am wondering whether a more "closed form" solution lies at the end of the tunnel, or whether the approximations brought forth in the more modern models will have to suffice?

Even if you restrict yourself to a cubic equation of state, you can certainly do better than van der Waals. Some tests I did a few years back point to the Redlich-Kwong equation being one of the best two-parameter equations of state for a good range of pressures and temperatures. Of course, if you need to do better than that, you use virial...

While handling volumetric and thermal departure functions of most hydrocarbons ok, Redlich-Kwong is a poor choice for vapor-liquid-equilibrium calculations.  Much better without sacrificing the basic cubic equation structure is the Soave-Redlich-Kwong or Peng-Robinson EOSs.

Mathematically we can't even solve the gravitational 3-body problem in general with a closed form solution. And that has simpler forces (well, gravity, with a neat inverse-square law) than those that act between molecules in a gas. So that's not very hopeful.

Simple mathematical expressions reflect the underlying mathematical forms of the physical equations are simple. However, the underlying physical theory, namely quantum theory, is not simple :(  Hence, we would naturally expect the exact solution (presumably it exists), even its accurate approximation, not to be  simple...

At the end of the tunnel, you're still trying to approximate the statistical average of interactions between individual molecules using macroscopic quantities. The refinements add more parameters because you're trying to parametrise the overall effect of those individual interactions for every property that is involved for each molecule.
You're never going to get a unified "parameter-free" solution for those without going down to the scale of the individual molecules (e.g. ab initio molecular dynamics), as far as I can tell.

Of course, on scale of the individual molecules, thermodynamics itself becomes meaningless.

It's been known since 1941 that the answer to your question is in the negative, i.e. that there will never be a closed form equation of state for a nonideal gas.
In 1941 Mayer and Montroll developed what is now known as the cluster expansion for the partition function of a nonideal gas whose particles have pairwise interactions. This cluster expansion provides a 1-1 correspondence between various integrals over the interaction potential and virial coefficients in the Kamerlingh Ohnes equation of state which has in principle an infinite number of terms in it. Therefore, it would not be considered a closed form equation of state. In practice the virial coefficients are known to decay so it is usually safe to truncate the expansion for practical calculations.
One could extend the cluster expansion to three-body and higher-order interaction potentials but this will not change the fundamental argument above.

I'll add to Aesin’s answer that in this case, the burden of proof rests on the side of an analytical (or closed-form) equation of state. Statistical mechanics explicitly guarantees that there is a relationship between $p$, $V$ and $T$, i.e. that they are not independent state variables. However, no further generic statement can be made about it, and only by making appropriate approximations can one actually write an actual equation of state (EOS). One such EOS is the ideal gas law, others are as you have cited, but if one were to prove the existence of a closed-form generic EOS for fluids, that would be a major upheaval of the current understanding of thermodynamics of fluids.
For an overview of the thermodynamics of fluids, you can refer to your favorite statistical mechanics or statistic thermodynamics textbook (my own preference goes to McQuarrie). For a introduction to issues specific to the liquid state (but also valid in general for fluids far from the ideal gas behaviour), I would recommend Theory of Simple Liquids, by Hansen and McDonald.

Using the corresponding states approach (normalizing to the critical pressure and critical temperature) usually gives pretty accurate results for design purposes.

There are different factors contributing to the pVT state, and some of them have (different!) temperature dependencies and are interacting.

volume of particles
electrostatic repulsion of electron shells
London dispersion
dipolar interaction
energy distribution of particles

kinetic energy
rotational excitation (quantised, starts at moderate temperatures)
vibrational excitation (quantised, higher temperatures)
electronic excitation (even higher temp.)

and a few more things, like spin states et cetera

To model this in a single equation, you surely need a lot of parameters.
Benedict, Webb and Rubin use eight, and thats not the end of it.
If it is already impossible to find a closed equation for one gas, think about gas mixtures, or even reaction mixtures.

Some general comments on the current use of the VDW equation in both teaching and research (added as an answer at the OP's suggestion):
The OP writes: "[VDW] is only still around for historical purposes, as it is largely inaccurate." That's not quite right—it's like concluding that, since the ideal gas law is inaccurate, it must only persist for historical reasons. Rather, the ideal gas law persists because it provides the simplest model for gas behavior, and VDW persists because it provides the next-most-simple model. Both serve as powerful pedagogical tools for progressively developing student understanding; VDW, in particular, provides a simple and intuitive way to introduce intermolecular interactions, and their effects.
I'd also argue that VDW serves a useful purpose in research, as an ansatz (a simple paradigm) for understanding gas behavior. Specifically, the fact that certain properties can be predicted by this simple two-parameter model, and others can't, provides insight into the nature of those respective properties (into how they arise).
I.e., it's very useful to know whether a certain behavior can be predicted by a minimal model, vs. requiring a more complex one. Knowing this tells you something about that behavior. It's often desirable to know the simplest model that can predict a certain property.

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text:
What is the difference between an angular and linear skeletal formula?
When drawing a skeletal formula, what is the difference between an angular version and a linear version?
I was asked to draw the Z isomer of Resveratrol:

For which I drew:

However the markscheme states that:

skeletal structure must be correct and angular not linear

I haven't come across the difference between the two before and can't find anything on google to suggest one.  Their drawing of the correct answer is equivalent to mine, however I am concerned that I may have drawn the linear version as my benzene rings are in a line and theirs are not. What am I missing here?
Their version:

Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008)

Those are $sp^2$ carbons and should have a triangular planar 120° geometry. The angles you've drawn are clearly not 120° .
I've not exactly come across the terminology before, but I believe this is the distinctions:

In a "linear" formula, you disregard the actual bond angles and just try to keep everything in a line.
In an "angular" formula, you draw everything according to the bond angles.

Not sure of this, though.

Are you getting the definitions from the question? If so, I would wonder that maybe @Rory's structure was simply counted wrong due to the bond angles, not because it's an example of the linear form of the structure. I've heard of "linear structural formulas" as simply the molecular formula written as the structure is bonded (i.e. writing CH3CH2OH instead of C2H6O). http://library.queensu.ca/research/guide/chemistry-how-search-molecular-formulas uses this phrasing too.

Rory - I would nominate that the person marking is potentially being a bit harsh, however the structure as you drew it does disguise the degree of steric collision that one would anticipate in (Z)-resveratrol and indeed in any (Z)-stilbenoid. There's going to be serious competition between the pressure to planar due to conjugation stabilisation and the fact that the rings would be virtually intersecting if the molecule were planar.

@EmmieMC possibly, however they asked for a skeletal formula.  It's me marking a past paper so I was just keen to know if and where I'd gone wrong xD