text:
Does solubility matter when measuring density with displacement method?
Can the volume of a solid (specifically a granular crystal or powder) be accurately determined using the displacement method with a liquid in which the solid is soluble?
Originally I thought by dissolving in the liquid there would (or could) somehow be a change to the density of the mixture which would affect the displacement. As such, it would be a good idea to ensure the substance is not soluble in the liquid being used to determine the solid's volume. But if the dissolution of the solid is only a physical process (i.e., dissolving sugar or salt in water), would this at all affect one's ability to measure the volume of the amount of substance added?
I guess when dealing with an unknown substance, we might not be able to easily determine if the dissolution took place as a physical or chemical process without further testing, so it would make more sense to conduct the measurement with a liquid in which the substance was insoluble.

Volume is not a conserved extensive property of mixtures in the same way that other quantities like mass, net electric charge, number of particles, etc. are (within closed systems). That is to say, in reality, the volumes of components physically mixed together are not simply additive, with the total volume not necessarily equal to the sum of the individual volumes. I.e., $\sum{V_i} \not= V_{mix}$ generally. This occurs chiefly because intermolecular forces of attraction between particles will vary with the composition of the mixture. Mixtures of ethanol and water, for example, are always reduced in volume by comparison to the summed volumes of ethanol and water, respectively, in isolation.
In fact, the total volume can be calculated given the partial molar volumes of the individual components for a specific composition under given conditions. Specifically, $V_{mix} = \sum{n_i \bar{V_i}}$, where $\bar{V_i}$ and $n_i$ are the partial molar volumes and moles, respectively, of the separate constituents of the mixture.
A few good references I found for additional study: 1, 2. Any decent physical chemistry textbook should also cover partial molar properties.

Thank you for this explanation and your references. I've never heard of partial molar properties before. I also didn't direct my search towards volume, as I should have. Does this remain true when the substance does not dissolve in the liquid? Do we have the potential to measure an inaccurate displacement due to the interaction of the outer atoms of our solid and the atoms of the liquid?

@MrCMedlin, if the substance doesn't dissolve, or the liquid components are largely immiscible, then the two phases remain separated and virtually no interaction occurs between the molecules of the disparate pure substances (except at the surface/phase interface). Consequently, as far as I know, the same phenomenon doesn't occur as it does with homogeneous solutions, and the volumes are simply additive (to within some possible tiny deviation).

Yes, I was mostly referring to the contact between the substances at the surface of the solid. I agree this would be negligible and likely near impossible to measure without highly sensitive equipment and/or extremely large amounts of both substances. But thank you, that does answer my second question!

You have to be careful if you are using a powder that can trap air pockets.


text:
Reasonings on the pH of calcium carbonate in ammonium chloride solution
Consider calcium carbonate saturated in $1.0\:\mathrm{M}$ ammonium chloride solution. I have calculated the $\mathrm{pH}$ of this solution analytically (with mass and charge balance) to be $6.92$.
Now, how would the actual solution $\mathrm{pH}$ compare to the theoretical, calculated $\mathrm{pH}$ - in a general sense?
Would one expect the actual solution $\mathrm{pH}$ to be higher or lower than $6.92$?
I know that there are going to be a host of solute-solute interactions in this system that are more than likely to cause deviations from ideal solute behavior. Here's a list of species in the system:

$\ce{Ca^2+}$ and its respective hydrates
$\ce{CO_3^2-}$ and its respective protonation products (i.e. bicarb and carbonic acid)
$\ce{H_4N+}$ and its respective deprotonation products (i.e. ammonia)
$\ce{Cl-}$ and perhaps (unlikely) a bit of its protonation product - hydrochloric acid.
$\ce{H_3O+}$
$\ce{HO-}$

Now, the question I'm fundamentally getting at is how would all these solutes interact with each other. I understand that the calcium ion and the carbonate ion with their +2 charge magnitudes is likely going to cause this solution to deviate significantly from ideal behavior. A quick leafing through of an analytical text lead me to this hypothesis. However, is it possible to actually predict the direction, so to speak, of $\mathrm{pH}$ deviation? Can we model this system with a computer program?
Also, I plan on actually taking $\mathrm{pH}$ measurements on this system. Here is a general outline of the things I plan to do; is there anything else that I need to take into account?

Use of triple distilled water. Our laboratory's "deionized" water runs strongly basic. Water will be distilled with a high partial pressure of nitrogen gas as to prevent carbon dioxide from dissolving in water.
Should we measure $\mathrm{pH}$ in a sealed vessel? I don't want any carbon dioxide to be dissolving into the system, again. Also, I don't want any ammonia gassing off from the system (since ammonia is a possible product upon saturating ammonium chloride solution with calcium carbonate). How would measuring $\mathrm{pH}$ in a sealed vessel work?
Any other lab practical tips/suggestions?

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text:
How does potassium nitrate work as an oxidiser?
Potassium nitrate is very popular compound in pyrotechnics, used as an oxidiser. It can be mixed with sugar to make a "rocket candy", with coal and sulphur to form black powder. You can stick it on rope to create functional fuse.
The saltpeter releases oxygen in all these reactions. But what are the results of reaction? I learned that $\ce{KNO2}$ is also oxidiser. So does the nitrate turn tu nitrite and then reacts again? Or does it form something else?
I have noticed that some strange bubbles come from my homemade fuse (photo will follow). Since potassium nitrate has melting point at 334°C (according to Google) it's likely the saltpeter melting.
And what about fumes? Does potassium nitrate or nitrite evaporate to turn the air poisonous or irritating?

If we assume potassium carbonate ($\ce{K2CO3}$) as the only solid reaction product, the overall reaction can be described as:
$$\ce{2 KNO3 + C12H22O11 + 11 O2 -> N2 + 11 CO2 + 11 H2O + K2CO3}$$
when using sucrose for candy. With sorbitol, an equation like
$$\ce{2 KNO3 + C6H14O6 + O2 -> N2 + 5 CO2 + 7 H2O + K2CO3}$$
seems fitting. In both cases, it is assumed that:

the nitrate is completely reduced to nitrogen
additional oxygen is needed

The latter is plausible, the rocket candy is supposed to burn layer-wise at the surface in contact with air. This allows for propulsion rather than one instant explosion.
Whether the nitrite is formed as an intermediate (or which organic intermediates are formed) is beyond my knowledge. However, once this partial reduction (nitrate to nitrite) has happened, I rather doubt that the reaction goes back again (to regenerate the nitrate).
As far as the bubbles are concerned, is it conceivable that the fuse wasn't completely dried and thus water was driven out?

The bubbles later turn into solid white/black inhomogenous substance. I believe it's potassium nitrate that has been melted by the burning.

I'm not so sure additional oxygen is what we want. The fumes from the rocket will blow all the oxygen away.

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text:
Is pyrene aromatic despite failing Hückel's rule?
Pyrene doesn't seem to be aromatic. However, sources claim that it is aromatic.

Considerations:

Pyrene is cyclic. ✓
Pyrene is flat (planar). ✓
Pyrene has 16 π electrons.
Every atom in the ring structure of pyrene is $\ce{sp^2}$ hybridized. ✓

The problem is that pyrene fails the $4n+2$ rule. $4n + 2 \neq 16$ where $n$ is an integer.
Something leads me to suspect that Hückel's rule is an oversimplification here. Or is pyrene indeed non-aromatic?

while pyrene is formally aromatic (i.e. has ring current), it reacts mostly like diene with isolated double bonds.

I guess the better defining characteristic of an aromatic compound is its response to the magnetic field which is used in NMR spectroscopy. If a compound produces an aromatic ring current as a result of induction due to the applied magnetic field, then it can be categorised as aromatic.

Pyrene is aromatic. The Hückel $4n+2$ rule works best with monocyclic ring systems. If you look at the following resonance structure for pyrene with a central double bond, the monocyclic periphery has 14 π electrons (ignoring the greyed-out central double bond), but that is a rationalization.

Nonetheless, pyrene undergoes reactions characteristic of aromatic systems and has ring currents expected from aromatic systems.

Is that implying that the conjugated pi system does not involve the central double bond? @ron

Correct, in other words, pyrene becomes 14-annulene.  But as I said above, this is an after the fact rationalization.  The 4n+2 rule was really meant to apply to monocyclic pi systems.  You run into problems when it is extended beyond its originally intended use.

What if you drew resonance structures? The central bond then vanishes.

To clarify @ron's point, the general "cheminformatics" rule for deducing aromatic fused-ring systems is whether there is one path (usually the peripheral one) that satisfies $4n+2$:
So for naphthalene:

The bold bonds give you a ring path of $\mathrm{sp^2}$ carbons with 10 π electrons, so it's aromatic. There are some arguments about whether the central bond is aromatic or not. (I say baloney, but it's fair to say aromaticity generates lots of arguments.)

So yes, the bold path gives you 14 π electrons for a $4n+2$ path. Again, I've heard people say the central two carbons aren't aromatic. (IIRC, the $\ce{^13C}$ NMR disagrees.)
I wrote the code for Open Babel and aside from some tricky nitrogen-heteroaromatic rings, this rule generally matches chemical reality. (Setting aside the question of the interior carbons.)
I think another fairly safe rule is "if a fused aromatic system is made up of individually aromatic rings, aromaticity is retained."

What's the debate about the central bond?

@Dissenter it's outside the mainstream of chemistry, but since including the central bond would make 16 $\pi$ electrons, some cheminformatics people question of that bond is truly "in" the aromatic system. As I said above, I think it's baloney - most experimental or computational definitions of aromaticity would indicate aromatic carbons there. Some people are just too tied to Hückel - but ask them about $\ce{C60}$. Heh.

C60 is aromatic? @GeoffHutchinson

I'd say so, but that's a whole different debate.

I absolutely love the statement you give in the comment and I am herewith encouraging you to include it into your answer. I believe, that there is a pretty common misconceptions, that only Hückel aromats are actually aromatic systems. But you are of course right, aromaticity generates a lot of arguments.

While the previous answers are correct in indicating that pyrene has aromaticity, they are often correct for the wrong reason. The proper way to break down the aromaticity in pyrene is to make the most Clar sextets possible. Clar sextexts are sets of six electrons that are cyclically delocalized. There are two unique ways to draw the Clar sextets in pyrene.

The first drawing shows two Clar sextets, while the second only has one, and thus should be less favored. If we perform NICS computations to determine where the delocalization happens, we see more delocalization in rings A and D than in rings B and C.
This can be further expanded to other systems such as pentacene, which shows a much lower singlet triplet gap than many other organic molecules thanks to the triplet having two stabilizing Clar sextets.

For a more complete description of Clar sextets and aromaticity, I would recommend an article by Portella et al.1 Also, as others have mentioned, when moving to 3D, things can get weird.

Portella, G.; Poater, J.; Solà, M. Assessment of clar’s aromatic π-sextet rule by means of PDI, NICS and HOMA indicators of local aromaticity. J. Phys. Org. Chem. 2005, 18 (8), 785–791. DOI: 10.1002/poc.938.

I realized in my answer that I didn't give my stock comment about "many definitions of aromaticity." I do like Clar's sextets with aromatic hydrocarbons. My one problem is that when I've brought them up in classes, it reinforces the false connection between "benzene" and aromaticity.

Nice, but can Mr. Clar do his sextets in cyclooctateraenediide?

According to Wikipedia:

Hückel's rule is not valid for many compounds containing more than three fused aromatic nuclei in a cyclic fashion. For example, pyrene contains 16 conjugated electrons (8 bonds), and coronene contains 24 conjugated electrons (12 bonds). Both of these polycyclic molecules are aromatic, even though they fail the 4n+2 rule. Indeed, Hückel's rule can only be theoretically justified for monocyclic systems.

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text:
When I dissolve sugar in my cup of tea/coffee, does it become a liquid?
When I dissolve sugar in my cup of tea/coffee, does the sugar go from being a solid to being a liquid?

Related: https://physics.stackexchange.com/questions/5562/what-keeps-the-sugar-suspended-in-the-tea

Does NaCl dissolve apart into individual molecules? If so, how then can it be considered a solid anymore? If it doesn't, why?

@User17670 NaCl is not molecular, it is a crystal. You'd expect it to come apart into its constituent ions (atoms).

@Kyle +1 for interesting statement! So, the Na would break apart from the Cl? Would this only happen above the melting point?

@User17670 You could melt it, but you can also dissolve it in e.g. water well below the melting point (melting point of NaCl is quite high, hundreds of degrees, but it dissolves easily at room temperature).

@dani table salt dissolved in water is most definitely not a solid: it is a solution of table salt in water. Under what definition of solid can it possibly be still considered a solid?

When the sugar dissolves, it is broken up into individual particles; each molecule is surrounded by water molecules. Essentially, you are trying to decide if a single molecule is a solid or liquid when dissolve.
The best fit is a liquid since the molecules are free to move independently and will take on the shape of the container.

I've thought about it a bit more and I think that the 'best' answer is that 'it doesn't make sense to consider only the sugar' - e.g. to decide if the sugar is a liquid or a gas, you have to say that it is compressible or not. On the one hand, you have sugar particles that are well spaced like a gas, but on the other hand, you can't really ask whether the sugar is compressible. So, I'm happy to go with 'it becomes part of the liquid'.

Indeed, usually in chemistry, a dissolved compound is indicated with the (aq) subscript, like $\ce{NaCl(aq)}$ rather than (s), (l), or (g). In essence, we're indicating it's not solid, liquid, or gas really, but something a bit different - a mixture.

(aq) specifically means dissolved in water (aqua). There are plenty of other solvents.

@MSalters yes, it specifically means water, but the OP asked about dissolving sugar in tea (i.e., water solution). Thus my answer.

How is the best fit a liquid? Sugar in water isn't a liquid, it is a solution: a completely different thing.

When I dissolve sugar in my cup of tea/coffee, does it become a liquid?

The premise of your question is wrong. There is no more "it" for the sugar when dissolved in water. You can't just consider the sugar alone. Rather the sugar becomes a constituent of the liquid phase.

When we dissolve sugar in tea or coffee,the sugar particle get dissociated into smaller particle. As you might be knowing there is inter molecular distance in case of solid,liquid and gas as well. In case of liquid the particles are loosely held compared to solid and also inter molecular distance is more as to solid. So finally particles get packed in those interstitial spaces.so the sugar particles do not get transformed to liquid or gas rather it get surrounded by water molecules.
.

Sugar added to tea becomes a solution not a liquid
Sugar is very soluble in water. When you add the solid to the tea the key process is that the solid sugar dissolves in the warm liquid: the solid crystals are broken up into molecules which are every dispersed throughout the existing liquid. When well mixed (because sugar doesn't dissolve instantly) the liquid is homogeneous with the sugar molecules evenly distributed in the single phase bulk liquid. There is no point where the sugar crystals become liquid sugar. And no chemist would describe the result as a sugar liquid. The correct description is a solution of sugar in water.


text:
Why does potential chemical energy decrease with increasing oxidation of carbon?
I saw an interesting spectrum in my biology class the other day. On the left end was methane - the least oxidized form of carbon - and the most energetic form of carbon (at least relative to what else was on the slide). On the right end was carbon dioxide - the most oxidized form of carbon - and the form with the lowest energy.

Now, I do agree with the slide - we can combust methane but we can't combust carbon dioxide - at least I don't think we can.
However, is there an intuitive explanation of why increasing oxidation state correlates with lower potential energy, at least for carbon?
Does increasing oxidation state correlate with lower potential energy for all elements?
Why does it seem that oxygen forms rather strong bonds with other elements in many cases?

From a very basic explanation from general chemistry:

Increasing your oxidation state increases the charge. So that means that you move to the left on the periodic table increasing your atom radius and decreasing your ionization energy and electronegativity.
If your IE decreases and you become more electronegative, your atom is more attractive for chemical reactions so it has a lower potential energy.

There is a less handwaving explanation from inorganic chemistry, but that is  a little harder to understand and I don't want to formulate a MO diagram.

is there an intuitive explanation of why increasing oxidation state
  correlates with lower potential energy, at least for carbon?

The heat of formation of methane, carbon dioxide and carbon tetrafluoride are -74.9, -393.1 and -927.2 kJ/mol respectively (the trend remains unchanged even if you divide by MW and compare per unit mass).  As we shift electron density from electropositive elements to electronegative elements we stabilize electrons, their energy is lowered.

Does increasing oxidation state correlate with lower potential energy
  for all elements?

I haven't checked extensively, but the trend seems to hold up.  I'd expect it to hold up.
$\ce{BeH_2}$, $\ce{BeF_2}$  127.2, -796.7 kJ/mol
$\ce{CaO}$, $\ce{CaF_2~~}$      39.1, -789.1 kJ/mol
$\ce{NaH}$, $\ce{NaF~~}$      126.2, -288.0 kJ/mol

Why does it seem that oxygen forms rather strong bonds with other
  elements in many cases?

I'm not sure what you mean by "strong bonds".  The C-O bond in carbon dioxide is much more reactive than the C-H bond in methane.  Perhaps you're alluding to molecular stability which takes us back to the top, e.g. compounds with polar bonds typically have stabilized electrons, electrons in lower energy orbitals, as discussed above.

"Perhaps you're alluding to molecular stability" - I was - and I'll avoid the use of fluffy terms such as "strong bonds" or "stability" in the future! Insightful answer as usual! @ron

This scenario has a very simple explanation. The reduction of any electron to a more electronegative atom will release energy. To understand this, imagine a highly reduced methane molecule; the C-H bonds are unstable and the electrons in the H atoms is itching to escape the methane molecule to pair with an oxygen. In this example the reduction of oxygen by the hydrogen releases energy thereby lowering the reducing agent's( methane) free energy. Oxidation usually will result in a lower free energy state, due to the energy being released in the reaction.
TL DR;  High energy reducing agent ---- donates high energy electron to oxidizing agent. Reaction releases energy which the reducing agent supplied.
oxidizing substance becomes more stable b/c it lost a high energy electron.
Reducing substance becomes more stable b/c it attained an electron for an electronegative atom and the surplus energy is released.


text:
Why does a chemical have more than one oxidation number?
Like for group I metals, it is understood that all the elements in the group would have an oxidation number of +1.
As far as I know, oxidation number is how much electron the element loses/gains.
Hence my question is, why does an element like Iron, Manganese or even oxygen have more than one oxidation state unlike elements like Magnesium or Potassium.
Edit: If I have stated any terms incorrectly, please let me know because I'd want to know where I went wrong.

As you told oxidation number(oxidation state) is number assigned to atom, ion or compound which represent electron gain or lost by atom, ion or compound.
Some element have more than one oxidation number, because they act differently when they combine with other element. Eg. Oxygen (act differently when combined with peroxide or superoxide.)
Some element have more than one oxidation number, because they can gain as well as loss electrons.So they have negative as well as positive oxidation number.
Almost all the elements have more than one oxidation state including Magnesium and potassium. You can see the list of oxidation state of all the elements here

Because a lot of what you have been taught in school is a lie.
In this particular case, you've probably been taught about the octet rule, a rule which makes it seem as if any given element should have exactly two oxidation states; 'n' and '8-n', where n is a simple function of it's location in the periodic table.
The truth, as always, is a little more complicated:
Compared to the (somewhat) well-behaved non-metals in the upper right corner of the periodic table, most metals have a lot for electrons. At 26, iron is a lightweight, but already there you have 4 (common!) oxidation states, the two most common ones being iron(II) and iron(III).
One reason for this is that while iron atoms have a "the lowest energy state", it's difficult to get at without first coming across several "a (locally) lowest energy state".
Its four oxidation states represent local minima, and moving between them requires an activation energy, making each of the states relatively stable, but possible to dislodge, i.e. metastable.


text:
How does Ostwald-Walker method work?
From searching online I found this:

The apparatus consists of two sets of bulbs. The first set of three bulbs is filled with solution to half of their capacity and second set of another three bulbs is filled with the pure solvent. Each set is separately weighed accurately. Both sets are connected to each other and then with the accurately weighed set of guard tubes filled with anhydrous calcium chloride or some other dehydrating agents like $\displaystyle P_2O_5$, conc. $\displaystyle H_2SO_4$ etc. The bulbs of solution and pure solvent are kept in a thermostat maintained at a constant temperature.
A current of pure dry air is bubbled through. The air gets saturated with the vapours in each set of bulbs. The air takes up an amount of vapours proportional to the vapour pressure of the solution first and then it takes up more amount of vapours from the solvent which is proportional to the difference in the vapour pressure of the solvent and the vapour pressure of solution, i.e. $\displaystyle p_0 – p_s$. The two sets of bulbs are weighed again. The guard tubes are also weighed.
Loss in mass in the solution bulbs $\displaystyle \propto p_s $
Loss in mass in the solvent bulbs $\displaystyle \propto (p_0 – p_s) $
Total loss in both sets of bulbs $\displaystyle \propto [p_s +(p_0 – p_s)] \propto p_0 $
Total loss in mass of both sets of bulbs is equal to gain in mass of guard tubes.
Thus, $\displaystyle \frac{p_0-p_s}{p_0} = \frac{\text{Loss in mass in solvent bulbs}}{\text{Total loss in mass in both sets of bulbs}}
= \frac{\text{Loss in mass in solvents bulbs}}{\text{Gain in mass of guard tubes}}$
Further we know from Raoult’s law
$\displaystyle \frac{p_0-p_s}{p_0} = \frac{\frac{w_A}{m_A}}{\frac{w_A}{m_A} + \frac{w_B}{m_B}}$
The above relationship is used for calculation of molecular masses of non-volatile solutes.
For very dilute solutions, the following relationship can be applied.
$\displaystyle \frac{p_0-p_s}{p_0} = \frac{\text{Loss in mass of solvent bulbs}}{\text{Gain in mass of guard tubes}}= \frac{w_Am_B}{w_Bm_A}$

My questions:

Why "The air takes up an amount of vapours proportional to the vapour pressure of the solution first and then it takes up more amount of vapours from the solvent which is proportional to the difference in the vapour pressure of the solvent and the vapour pressure of solution"?
How can we add $\propto p_s$ and $\propto p_0-p_s$ when we don't know what the proportionality constants are or even if they are equal?
It would be better if you could explain in detail the process because I'm not getting it properly from here.For your information the text after "From Raoult's Law" and about "the setup" is purely clear to me.

How are the vapors transported?
The additional question from the comments:

Why [is] the amt. of vapours taken up proportional to vapour press. Why directly proportional? Why not nothing else?

First things first, let's get a definition (Wikipedia, italics added):

Vapor pressure [...] is defined as the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system.

So basically, before you start the experiment (see below) the vapors have already formed, meaning that thermodynamic equilibrium has been achieved. Then you start bubbling through dry gas, which then mixes with the vapor gas already in the bubbles. If you set the gas flow appropriately, you get a near-equilibrium situation (see the graphic below) and the gas that exits the bulb assemply has exactly the same amount of vapor in it as the vapor "layer" (which can not be thought of as a layer anymore because of the bubbling) above the liquid surface.
We do not have an equilibrium situation, which is why so much care has to be taken that we get as close as possible to the equilibrium situation. (Also, we do not have a closed system, so everything starts to get slightly problematic at this point. But to a first degree of approximation, this will hold up fine.)
This is also why it is directly proportional to the vapor pressure: There is no other mechanism that moves solvent molecules from the condensed phase to the vapor phase. Of course you could add a supersonic vibrating membrane and force more solvent molecules into a vapor phase, but that is not what you want to do, because you want to quantify the relative lowering of the vapor pressure.
Is relative vapor pressure lowering a thing?

Why "The air takes up an amount of vapours proportional to the vapour pressure of the solution first and then it takes up more amount of vapours from the solvent which is proportional to the difference in the vapour pressure of the solvent and the vapour pressure of solution"?

I guess the question you really want to ask is: Why is the vapor pressure of a solution lower than the vapor pressure of a pure substance?
The answer is not simple, but the explanation of the vapor pressure lowering effect can be given by assuming a simple model: Imagine that in the solution, there are two types of molecules; the solvent and the solute. In the pure solvent, there is only one kind. Statistically speaking, that means that there is also a smaller amount of solvent molecules at the boundary between the bulk phase and the vapor. Since solute molecules don't go over into the vapor, there is a smaller area of solvent molecules that can go over into the vapor, which means that the vapor pressure of the solution is smaller when compared to the pure solvent.
What about the proportionality constant?

How can we add $\propto p_s$ and $\propto p_0−p_s$ when we don't know what the proportionality constants are or even if they are equal?

While we don't know the magnitude of the proportionality constants, we know that they are equal. The proportionality constant is most likely dependent on ambient pressure, temperature and flow rate of the gas. These three variables are being controlled during the experiment and kept constant for all glass bulbs. (Well, I don't know about keeping ambient pressure constant. But if you perform the experiment fast enough, that shouldn't pose much of a problem.)

The Ostwald-Walker Process

It would be better if you could explain in detail the process because I'm not getting it properly from here.

Why do we do this?
First, we have to remember: Why are we doing this process; what do we want to find out? The answer is: We want to quantify the relative lowering of the vapor pressure.
How do we do this?
Here's what happens during the process:

Dry gas arrives at the first set of bulbs, filled with a solution.
The dry gas gets saturated with solvent vapor.
The solvent-vapor-saturated gas arrives at the second set of bulbs which contain the pure solvent.
Because of the vapor pressure lowering effect (here now applied in reverse) the vapor pressure of the pure solvent is higher. As such, the gas that enters is not saturated. When it exits the second set of bulbs, it is saturated again.
The pure-solvent-saturated gas enters the desiccator and loses all solvent vapor to it.
Dry gas exits the apparatus.

This might still be hard to understand, so here a graphic that shows you what happens to the solvent vapor partial pressure in the gas:

At time 1 the gas enters the first bulb. It exits the first cluster of bulbs at about time 4. Then it travels to the next cluster of bulbs (containing the pure solvent) and loads up there, up until time 8. At time 9 the gas enters the desiccator where it loses all the solvent vapor and exits dry at time 12.
Since we weigh all components, we know exactly how much mass we lost from solution and solvent, and how much mass the desiccant gained (both those numbers [the total of the mass loss of course] should be equal):
$$ \Delta m_\text{solution} + \Delta m_\text{solvent} = \Delta m_\text{desicc.} $$
The following proportionalities hold:

$ \Delta m_\text{solution} \propto p_s $
$ \Delta m_\text{solvent} \propto (p - p_s) $
$ \Delta m_\text{desicc.} \propto p $

The relative lowering of the vapor pressure can now be calculated, it is simply
$$ \frac{p-p_s}{p} = 1 - \frac{p_s}{p}$$
If you know the vapor pressure of the pure solvent, you can simply plug it in the equation, get the proportionality constant and from that calculate the vapor pressure of the solution.
Conclusion
In conclusion, I can only say that I hope to have clarified things for you. I myself spent quite some time researching this method, as I have never heard of this question before. Thank you for the opportunity!

How is $\Delta m_{solvent} \propto (p - p_s)$?

@Archer I don't know if I am correct cuz I have read about this process yesterday but I can tell you what I understood.. So what actually happens is that the saturated aur from the solution chamber has some vapours of the solvent .. So when it reaches the solvent chamber .. the vapours go in the solvent and them the dry gas comes out taking the vapours which was present above the liquid in the beginning.. so actually the solvent gains some mass proportional to $P_s$ and then loses some mass proportional to$P$ . Thus the net change is $P - P_s$...


text:
Balancing the redox reaction
Question:
Balance the following redox reaction:

$$\ce{FeS2 + O2 -> Fe2O3 + SO2}$$

My Efforts:

I tried balancing with oxidation number method.

First of all i determined the oxidation state as follows:
$$\ce{Fe^{(2)}S2^{(-1)} + O2^{(0)} -> Fe2^{(3)}O3^{(-2)} + S^{(4)}O2^{(-2)} }$$
Here, all the element are either oxidized or reduced, so how to move ahead?

I tried balancing with Half reaction method.

$$\ce{Fe^{(2)} -> Fe2^{(3)}} \text Oxidation$$
$$\ce{S2^{(-1)} -> 2S^{(4)} } \text Oxidation$$
$$\ce{O2^{(0)} -> 2O^{(-2)}} \text Reduction$$
Same here, how to move ahead?

P.S. Oxidation state are mentioned in the brackets. Correct me if i am wrong anywhere.

A couple things... Are you sure it's FeS2 and not FeS? Also, in the oxidation state method, the charges don't balance in FeS2.

@jerepierre Yes i am sure it is FeS2. If you can show it by half reaction method then also it will work, because i have test tomorrow.

FeS2 has an odd structure; the iron atom has a +2 oxidation number and each of the sulfurs has a -1 oxidation number.
This can be balanced by inspection.
4 FeS2 + 11 O2 --> 2 Fe2O3 + 8 SO2
Just to check, using oxidation numbers we get:
Sulfur goes from -1 to +4.  (Total change from sulfur = +40)
Iron goes from +2 to +3.    (Total change from iron = +4)
(I'm not sure where you got 4 for the oxidation number of iron in FeS2.)
Oxygen goes from 0 to -2.   (Total change from oxygen = -44)
ETA:
Half reactions are messy here, since both iron and sulfur are oxidized.  Similarly, using oxidation numbers is problematic (except to check the solution) because there are three substances changing oxidation states.  You could write a system of equations to describe it, but that's a lot more trouble than it's worth.
You know from the structure of iron(II) disulfide that there are twice as many sulfur atoms as iron atoms.  That means that the number of sulfur dioxide molecules must be four times the number of iron(III) oxide molecules.  With that relationship in mind, the smallest ratio of molecules that fits the pattern is the one I wrote above.
If you really want to use oxidation numbers, here's what I've come up with:
Let a equal the number of iron atoms, b equal the number of sulfur atoms, and c equal the number of oxygen atoms.
Iron increases its oxidation state by 1, sulfur by 5, and oxygen decreases by 2. So:
1a + 5b - 2c = 0
We also know that iron and sulfur are in a 1:2 ratio because they come from pyrite.
2 a = b
Substitution yields:
1a + 5(2a) - 2c = 0
11a - 2c = 0      Iron and oxygen are in a 2:11 ratio.
Oxygen must be even since it comes as O2, so its smallest possible value is 22 atoms.  Using the ratios listed above, we get 22 O: 4 Fe: 8 S.
ETA #2: Half Reactions
Fe+2 --> Fe+3 + e-
S-1 --> S+4 + 5 e-
O20 + 4 e- --> 2 O-2
We know from FeS2 that there must be twice as many sulfur atoms as iron, so the second equation has to be multiplied by two.
2 S-1 --> 2 S+4 + 10 e-
If we add together all of the species that are being oxidized (Fe and S) then we get an oxidation half reaction of:
Fe+2 + 2 S-1 --> Fe+3 + 2 S+4 + 11 e-
This must be multiplied by 4 and the reduction equation by 11 to balance the number of electrons.
4 Fe+2 + 8 S-1 --> 4 Fe+3 + 8 S+4 + 44 e-
11 O20 + 44 e- --> 22 O-2
Adding this mess together gives:
4 Fe+2 + 8 S-1 + 11 O20 --> 4 Fe+3 + 8 S+4 + 22 O-2
Since we know the structure of the molecules, putting it all back together is fairly straightforward, and it gives the same result as listed above.

I would be thankful if you can saw me by using oxidation number method or by half reaction method. Thank you for correcting my mistake.

Welcome to Chemistry.SE! Please have a look at this and the documentation for mhchem

Think of $\ce{FeS2}$ as the $\ce{Fe(II)}$ salt of hydrogen disulfide. In nature, you'll find it as Pyrite. Looks nice, unless you're in for the real gold.

Balancing With Half Reaction method

Step 1: Determine oxidation number of each element.

$$\ce{Fe^{(2)}S2^{(-1)} + O2^{(0)} -> Fe2^{(3)}O3^{(-2)} + S^{(4)}O2^{(-2)} }$$

Step 2: Determine total increase and decrease in oxidation number. Also maintain $\ce{Fe}$ and $\ce{S}$ ratio(1:2) on both side.

$$\ce{Fe^{(2)} -> Fe2^{(3)}} \text Oxidation -----------(A)$$
$$\ce{S2^{(-1)} -> 2S^{(4)} } \text Oxidation -----------(A)$$
$$\ce{O2^{(0)} -> 2O^{(-2)}} \text Reduction -----------(B)$$

Step 3: Balance Oxidation number of of reaction (A) and (B)

There are total 11 electron transfer take place in (A) and 4 electron in (B)
$$4\ce{[Fe, S2] ->4[Fe, 2S] }$$
$$\ce{11[O2] -> 11[2O]}$$

Step 5: Finally combine above two reactions

$$\ce{4FeS2 + 11O2 -> 2Fe2O3 + 8SO2}$$


text:
Is there an open Chemical database that includes GHS/CLP classifications?
I'm looking for a Chemical database that

is up to date (via a firm, experts or crowdsourcing)
is open and free to use (best via an API)
contains the CLP/GHS classifications for chemicals

I have not found any database that has these three points (e.g. ChemSpider is actually not really open and does not contain CLP/GHS classifications)
I was really hoping any of you might know of one?

Since you're a Dutch neighbour, the European Chemicals Agency might be an option.

That one looks excellent. Do you know if there's any API to access the chemical proprties + classification?

@WillemMulder I'm afraid I don't know that. IUCLID might be another place to look at.

Hi, any updates with this question? Do you guys know of an API that can provide me with GHS info for a given substance / CAS#?


text:
Do acyl fluorides exhibit resonance donation or inductive withdrawal?
I know that in acyl chlorides, there is partial triple bond character in the $\ce{C=O}$ bond not because the chlorine is electronically donating (due to size mismatch with carbon) but because the chlorine is highly electron withdrawing and this stimulates the oxygen into partially donating one of its lone pairs to the partially positive carbon.
This is supported by IR stretch data for the $\ce{C=O}$ bond in acid chlorides; we're looking at frequencies from 1810-1775 reciprocal centimeters - somewhat higher than a vanilla $\ce{C=O}$ bond stretch frequency.
However, in acid fluorides, in which the fluorine is of comparable size to the central carbon, it seems that resonance donation is at least a possibility. However, resonance must compete with the inductive withdrawal effect. So, which one wins out in acid fluorides, and does the $\ce{C=O}$ bond exhibit more single bond character or triple bond character?
I tried looking up acid fluorides, and I can't find any mention of them. So I guess a good starting point would be: are they stable enough to even exist?

Yes, acid fluorides exist.  You can purchase acetyl fluoride from Sigma-Aldrich.
Look back at these two questions for background on how the increased contribution from the triple-bonded carbonyl resonance structure (resonance structure IV below) will lead to an increase in the carbonyl ir stretching frequency (e.g. the triple-bonded carbonyl resonance structure will lead to a stiffening of the carbonyl bond and consequently shift its absorption to higher energy).

Ring Strain and C=O Stretching Frequency
Triple Bond Character in Acyl Chlorides

Here are the resonance structures we considered in the case of an acid chloride.

Resonance structure IV is what we're calling the "triple-bonded" carbonyl structure.
The carbonyl stretch for a acetyl chloride (I, R=Methyl) occurs at 1818 ${cm^{-1}}$.  In the case of acetyl fluoride (replace Cl with F in the above structures) this stretch occurs at 1869 ${cm^{-1}}$, significantly higher.  This suggests that resonance structure IV is a larger contributor in the case of acid fluorides. The stability of $\ce{F-}$ vs. $\ce{Cl-}$ is also a significant factor in increasing the contribution from resonance structure IV in the case of acid fluorides.

Is there any way to prove that resonance structure 3 also plays a larger role in the acid fluoride case?

Also what do you mean by your last sentence @ron

You could probably compare the C-X ir stretches against each other and some model compounds. 2) I mean the heat of formation of aqueous F- is more negative than that of Cl-

I question the validity of your comparison because vibration frequency is dependent on both the force constant, as well as the reduced mass. The reduced mass for the C-F system (7.35 u) is different from that of C-Cl (8.97 u). Considering that the difference in vibrational frequency is not that large as well, your comparison becomes very dubious.

@TanYongBoon I'm talking about the $\ce{C=O}$ stretch, not the $\ce{C-X}$ (where X=halogen)

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


text:
How to solidify sand
I'm doing a research project where we analyzing the structure of sand after a certain process. To get a better idea of the internal structure of the sand, we use different colors to visualize the movement of the particles. However we can't actually observe that structure since sand is not a solid substance. Therefore we need a way to solidify the sand without changing the structure.
We figured we would need some sort of liquid which we can pour into the sand container (the container is 2x1x0.2 meters) so that it will solidify. In other words, how can we solidify sand?
Using a transparent container is not really an option since we are curious about the structure in the center which we cannot see... unless we use transparent sand ;)
Note: we are using 'regular' beach sand.
EDIT:
I have read through all the answers and they are great! I can't accept any answers just yet since we need to figure out which method we will use, but it looks promising! We started with 0 and now we have some great ideas going with your help!

You mentioned colours to visualize the movement of the particles (which I presume consist of quartz, since you are using beach sand). How could you still measure movement if you solidify the material, e.g. by embedding in matrix?

2 x 1 x 0.5 is 400 L or almost 100 gal. of volume. At those volumes (lets say 20% void space), the cost of any binder will be significant.

Which of your dimensions is the height? if it's the 0.2m its a much easier job than if it's 1m.

@steveverrill sorry for my sloppy dimensioning. It's dimension as lxhxw.

@NickT I forgot to mention that we won't be filling the entire container with sand. That 1 meter will vary during each experiment

Based on your comments, it sounds like you have some sort of colored sand and you want to immobilize it. My suggestion would be some sort of polymer matrix.
There are multiple possibilities, but one of the easiest is polydimethylsiloxane PDMS which is optically clear and pretty easy to work with (i.e., safe). There's the "base" and a curing agent which creates cross linking.
Probably the easiest how-to comes from U. Wisconsin.
If done right (i.e., no air bubbles) you should be able to pour in the PDMS, cure, and have a polymerized matrix around the sand. Then you could cut it up or do whatever needed to see different regions.

Geologists prepare soil profiles for a living.  The task is easy in principle - dig a pit, apply glue to the wall, then transfer the glued layer to a sheet of cloth.  The practice is really difficult, the glue recipe and transfer technique are trade secrets.  See here (warning, German language): http://www.sand-abc.de/sandkorn/lackprofil/lackprofil.htm
Why not remove layer by layer and then take photos of the cross-sections?  That would be easier than messing about with casting resin and vacuum chambers.  Alternatively, enquire in your friendly Department of Geology.

I upvoted the water idea, after all it is possible to build sandcastles over 0.2 m high with just water as a binder. I also think coring as mentioned in that answer is a good idea. I also upvoted the comment about the freezer, although expansion could be an issue.
I’m going to suggest a hot gelatine solution (or even dessert jelly just to try it out.) There are various other food gelling agents (pectin, etc.) If it’s strong enough (and you don’t want it too strong, because you want to cut it open) it would be a good option. Gelatine is cheap, and you ought to be able to recover it by heating your used sand and filtering if cost is really an issue.
I’m thinking you may have a serious problem with air trapped at the bottom, if you are planning to distribute the sand using air then add the liquid afterwards. Chromatography columns have a porous frit at the bottom, which initially lets the air out (you see the colour darken as you add the first solvent) and then obviously the liquid, too. A wire or plastic mesh over a platform grating would probably work in your case. Pressure is usually applied above a chromatography column to push the solvent through, but in your case you may find it easier to suck from below, if a pressure difference turns out to be required. A bottom connection could also be useful for preheating the bed with steam.
You also need to think about how to distribute the liquid without disturbing the sand. A matrix of pipes with sprinkler holes, and/or a plastic mesh on top of the sand is probably the best way.
EDIT 1: some more ideas, which while more expensive, will give you a sharp, predictable freezing point. “Glauber’s salt” (sodium sulphate decahydrate) melts at 32 °C, and sodium thiosulphate pentahydrate melts at 48 °C. Sodium acetate trihydrate melts at 58 °C. These compounds can all be supercooled below their melting point, then made to suddenly freeze by adding a single crystal for nucleation (though I expect with sand about they will nucleate immediately.) Videos of this can be found on Youtube. The compounds have been tried for medium term storage of low grade solar heat for space heating, and/or are used in personal heat packs, as all the heat is released at the same temperature (the melting/freezing point) when they cool.
Sodium carbonate decahydrate “washing soda” apparently melts at 32 °C. I’ve never seen any fun experiments with supercooling melted washing soda, so maybe it does not supercool. The situation is confused by the existence of a heptahydrate too. However it is certainly cheap and readily available. It’s strongly alkaline, so would tend to pick up carbon dioxide from the air to become the bicarbonate, but in the volumes required I doubt that would occur to the extent required to produce any change in characteristics. Alkalies are harder on the skin than acids, because they turn your natural oils to soap.
I also thought about candle wax, but with the quantities involved extreme caution would be required to keep it between its melting point and flash point.
Any contact between molten salts and metals should be checked carefully for corrosion. In particular alkali + $\ce{H2O}$ dissolves the protective oxide layer on aluminium, leading it to corrode extremely rapidly with evolution of hydrogen gas.
EDIT 2: even though the OP has now indicated that the height is 1 m not 0.2 m, I still like the gelatine idea, though it’s probably not strong enough to support a metre column. It would probably mean tipping the tank on its side and removing the front (which will have become the top.) It’s a question of balancing the strength of the matrix against the ease of cutting (which will be extremely arduous for some of the higher strength suggestions mentioned here.)

BTW, I am a chemist turned chemical engineer. Hence my answer is far more about chemical engineering than chemistry. Your research would be of interest to chemical engineers, and they may be able to help you out with the problems in performing it. The relevant maths is applicable in both chemical engineering and earth science / civil engineering. I wonder if there is anything relevant on Earth Science stack exchange.

I don't know what kind of resolution you're looking for here.  If you want a true 3d structure you might be able to create one by taking slices off of the sand bed using adhesive paper.  Adding a glue or epoxy to the sand bed will also disturb the sand while holding it in place.
Do whatever process you've got to do to the sand to get it to its final state, then take a picture of the surface.  Apply an adhesive sheet to the top layer and remove.  Take another picture.  Apply a sheet, remove, picture, repeat.  It's going to take a lot of sheets to empty the bed, but if you want a really fine structure of the sand, it ought to work.
Another option would be a container with walls and bottom that can move independently.  Image, slide the sides down by 1mm, use a wiper to remove the protruding sand, repeat.  It might prove difficult to avoid disturbing the lower layers as the upper sand is wiped, but the consumable cost is virtually nothing with this method.  (This idea is basically the reverse of powder bed 3d printers, if it's not clear.)
ETA: A cheaper version of my first idea - Prepare your sand bed.  Image, spray a heavy layer of spray glue on top.  Apply a sheet of printer paper.  Wait a minute, remove.  Repeat.  I think you would pick up more sand with each pass this way and it shouldn't cost too much.

the container is 2x1x0.2 meters

I guess you want something cheap in this case.
option 1: water. Wet fine sand keeps shape. Not strongly, mind you, but still.
option 2: drop it, push glass tube into sand (reasonably wide with thin walls)
option 3: use diluted silicate glue. Would work, but you will have troubles drying it without oven.
Anything else is too costly for fun personal project.

option 1 + an industrial freezer

Sand particles are held together by covalent bonds.  The only way to get 2 separate sand particles to join is to heat them to a temperature at which covalent bonds form between particles.  This is the process used in making glass.  So, to do what you are proposing requires you to make glass from the sand.

Well we can't really put the container into an oven...

Still sounds like the right direction. You may also want to consider using a low-melting plastic for your color tracer. That avoids the need to melt the sand outright. You might be able to get away with grilling the sand.

How about hairspray ?
While dumping the sand (slowly) using a flat and wide board slowly pour the sand while two others generously spray the hairspray onto the falling sand. This way when the sand lands it's sticky, and you capture almost all the sand particles. They will stick together where they land so you don't disturb the original state of the experiment too much.
When dry, you should have a rock-like blob inside a mold.


text:
Why is proton transfer so fast?
Why is proton transfer always kinetically favored? In other words, why are Brønsted acid-base reactions so quick?
Is it because protons are generally unhindered, sterically? This seems plausible; consider an alcohol. The proton isn't blocked by any bulky groups. It's on the end of a oxygen atom.
What else could contribute to the kinetic favorability of a proton transfer? Reasoning backwards - given that protons transfers are very fast - this implies that the activation energy of a proton transfer reaction is small. And as outlined above - the activation energy is small in part because most of the time, protons are unhindered. What else might contribute to the small activation energy of a proton transfer/kinetic favorability of a proton transfer?

Quantum tunneling.

Due to their low mass, protons can tunnel through barriers, they don't always have to go over the barrier in an activated process.  Proton tunneling becomes particularly important at low temperatures where the available thermal energy may be inadequate for an activated process (e.g. over the barrier) to occur.  A nice, brief and not overly-mathematical introduction to tunneling can be found here.
dashed line - thermally activated process
solid line - tunneling process

A second reason why proton transfer is so facile relates to hydrogen bonding.  If hydrogen bonding is present, then the proton only needs to elongate one bond (the initial bond) while simultaneously shortening the hydrogen bond in order to effect transfer from one molecule to the next.  This energetically compensating process reduces the activation energy required for proton transfer.  In solvents with hydrogen bonded clusters or chains (water would be one such example), this process can be repeated within the cluster thus allowing the proton to cover large distances (from one side of the cluster to the other) on a very short timescale.  In other words, only minor mass movement is required; by simply shortening some bonds a bit, and slightly elongating others, a proton at the left side of a group of molecules (top row in the following figure) can "reappear" on the right side of the chain (bottom row) extremely rapidly. This process is referred to as the Grotthuss mechanism after its discoverer.  Some of the finer points of the mechanism still constitute an active areas of research.

Hydrogen bonding is just an elongated covalent bond?

Yes, an elongated, weaker, polar covalent bond.

What about the Steric factor I mentioned? @ron

Yes, steric considerations also play a role.

Proton transfer in many ways exists in its own unique domain of chemistry since the proton does not have any inner shells beneath its valence electron. Recommended classic book: "The Proton in Chemistry" by Bell.

Is electron pair transfer similarly fast? @ron

@Dissenter  faster

@ron what about electron pair transfer to a proton? I was trying to get at whether there is only a facile line between bronsted/lewis acidities.

At temperatures above the region of absolute zero, electron movement will always be faster than the movement of nuclei.


text:
What is the correct net ionic equation for the neutralization of phosphoric acid?
What is the net ionic equation for the reaction between phosphoric acid and sodium hydroxide?
I came up with
\begin{align}
\tag1 \ce{H+ + H2PO4- + OH- &-> PO4^3- + H2O},\\
\tag2 \ce{H+ + OH- &-> H2O}.
\end{align}
Or is it, since phosphoric acid is a triprotic acid, whose ionization is
$$\ce{H3PO4 <=>3H+ + PO4^3-},$$
therefore the net ionic equation will be
$$\ce{3H+ + 3OH- -> 3H2O}?$$

Phosphoric acid dissociates into hydrogen ions and $\ce{H2PO4^-}$ ions in water.  This then can dissociate further, but $\ce{H^+}$ and $\ce{H2PO4^-}$ are the primary ionic species you'd find in a phosphoric acid solution.
$\ce{NaOH}$ dissociates completely into $\ce{Na^+}$ and $\ce{OH^-}$ ions.
The reaction would then be between $\ce{H^+}$ and $\ce{OH^-}$, your option 2.
The difficulty arises because adding more $\ce{NaOH}$ to the reaction pulls off the second then the third hydrogen from the phosphoric acid.  In all cases it is an $\ce{H^+}$ reacting with an $\ce{OH^-}$ ion with the phosphate ions as spectators, but that really doesn't tell the story of the reaction well.  This is a case where the net ionic equation really isn't your best option.emphasized text


text:
Which gas is easier to compress, the ideal gas or a real gas?
Since the pressure of a real gas is less than that of the ideal gas  and its volume is more than that of ideal gas, I am assuming that the real gas is difficult to compress in comparison to an ideal gas.
Is my logic right? Also I'm looking for a clearer explanation!

Let's use the van Der Waals equation for this. (Real gasses also described elegantly by the Viral equation)

You were false that "pressure of a real gas is less than that of the ideal gas". It is greater. Also the volume is smaller.

Thus if it has a higher pressure and smaller volume, it typically is harder to compress (i.e. the compression factor increases) but it also depends on temperature. Here is a nice plot of compression factors vs temp for some common gasses from the cambridge chem guide.
http://www.chemguide.co.uk/physical/kt/realgases.html

Your assessment of the values of the pressure of real and ideal gases seems backward here.  If Pideal = Preal + an^2/V^2, then Preal must be less than Pideal, since the second term is positive.  That makes sense, since the IMFs are pulling the gas molecules together, causing them to hit the walls of the container less forcefully.  Similar reasoning yields a smaller Videal relative to Vreal, since the ideal volume does not account for the additional space taken up by the gas particles.


text:
What is the wave function?
I understand that the square of the wave function is the probability density of finding an electron. However, what is the wave function itself? I am trying to understand molecular orbital theory but this concept is getting in the way of me completely understanding it. Any help would be great. Thanks.

A very similar question was just asked recently here: http://chemistry.stackexchange.com/questions/16467/what-actually-is-the-wavefunction

Thanks. I am looking for a conceptual understanding of what the wave function is. In that thread it was described as a trajectory but there was no explanation. Could you perhaps clarify - at least on a conceptual level?

@user3764899 quantum mechanics is constructed as derivative of Hamiltonian formulation of classical mechanics by changing operator over trajectory to operator over (time-dependent) field. You can look for more here (warning, hardcore math ahead) http://en.wikipedia.org/wiki/Hamiltonian_mechanics

It is often asked "what is waving when a particle is described as a wave?" and the correct response is "existence itself is waving."
The wavefunction is most elegantly described as a description of reality for your system. You then act on the wavefuntion with a hamiltonian to yeild an eigenvalue to quantify the energy.


text:
Gibbs free energy of chemical compounds
How the Gibbs free energy of a chemical compound is theoretically calculated from scratch?
When saying that the Gibbs free energy of NaCl is -384.1 kJ/mol, this value is related to the chemical bonds of Na-Cl.
We should be able to calculate this value from the sum of energies in chemical bonds of one mole of NaCl? Right? How?

In theory, one can use quantum mechanics based codes to calculate bonding energies from first principles.  However, constructing new potentials usually involves making sure that the potential satisfies basic materials properties, such as getting the melting point right.  So it can be a little circular.  Usually in something like DFT you are interested in defect formation of diffusivity, and tailor a potential to get the bulk thermodynamic properties right.

The value you have reported is the standard Gibbs free energy change of formation, which has a specialized definition. A standard free energy change of formation is the free energy change that occurs when 1 mole of substance is formed by reaction of its constituent elements in their standard states at 1 atmosphere of pressure and 25 degrees Celsius.
Thus, for sodium chloride:
$$\ce{Na(s) +\frac{1}{2} Cl2(g) -> NaCl(s)}\ \ \ \Delta_f G^\circ =-384.1 \text{ kJ/mol}$$
Now, all we need to be able to do is measure $\Delta G$ for this reaction at standard temperature and pressure.
In practice, we will separately detemrine $\Delta_f H^\circ$ and $\Delta_f S^\circ$ and plug them into $\Delta_f G^\circ = \Delta_f H^\circ -T\Delta_f S^\circ$.
At constant pressure (which we are by definition operating at),  $\Delta H$ is equal to the heat transfer  $\Delta H = q = -C_p\Delta T$, which is equal to the heat capacity of the surrounding (in practice the calorimeter, which is reasonably isolated from the universe) multiplied by the change in temperature.
$\Delta S$ is still tricky to measure. At equilibrium, for a reversible process at constant temperature $\Delta S = \frac{q}{T} = \frac{\Delta H}{T}$. If the process is not reversible or you are not at equilibrium, then... then it is challenging.

Keep in mind that the $\Delta G$ for the elements at standard state, is by definition 0. So for this particular reaction, you only need to measure the right side - how much energy goes into the $\ce{NaCl(s)}$.


text:
Specific volume of an air parcel
So I have an unsaturated air parcel in the atmosphere that ascends adiabatically from the $1000\:\mathrm{mbar}$ level to the $700\:\mathrm{mbar}$ level with its initial temperature being $10\:°\mathrm{C}$.
I have to find the initial specific volume in $\mathrm{m^3}\:\mathrm{kg^{-1}}$.
I know that the specific volume formula is $v=1/\rho$ and that the equation of the state of an ideal gas can be written as $p=\rho RT$ where $R$ is $8.3143\:\mathrm{J\:mol^{-1}\:kg^{-1}}$
So I have $1000\:\mathrm{mbar}=\rho(8.3143\:\mathrm{J\:mol^{-1}\:kg^{-1}})(283.15\:\mathrm{K})$ and solve then for $\rho$ and then get $0.42477$ and plug that for $v=1/0.42477$ and get $2.354\:\mathrm{m^3}\:\mathrm{kg^{-1}}$ as the initial specific volume. Am I doing this right?

The ideal gas law states
$$ pV = nRT \Leftrightarrow \frac{n}{V} = \frac{p}{RT}\ .$$
For the density $\rho = m/V$ and the mass $m = M\cdot n$ with the molar mass $M$, we get a new formula for the density:
$$ \rho = M\frac{n}{V} = M\frac{p}{RT}$$
With the definition of the specific volume we get
$$ v = \frac{RT}{Mp}\ . $$
So unless you know the molar mass of your parcel of air, you will not get out a numerical value for the specific volume.

assuming air is roughly 80% $\ce {N2}$ and 20% $\ce {O2}$ you can the molar mass no ?
Also, I think you have to put the pressure value in Pascal, not in millibar

Always use SI units, of course. Also, you can assume a molar mass of the air parcel. It doesn't explicitly state that it's air from our atmosphere at sea level.


text:
Where to find common inorganic/organometallic molecules?
I'm working on a project to create a large open repository of quantum calculations, largely for teaching purposes.
I can get thousands of common organic compounds easily from sources like PubChem or drug databases.
I'm looking for sources for teaching or other uses with VSEPR examples or other types of inorganic and organometallic molecules. Ideally I'd want 100s or 1000s with good coverage of the periodic table.
Any ideas?
To be clear, there are lots of databases of solids. I know most of those, although feel free to suggest some for archival. I really want isolated complex ions, molecules, etc.
I don't think there are many, since I haven't turned anything up besides the Cambridge database, which is restrictive about reuse.
Thanks!

try http://www.crystallography.net/browse.html . It is crystallography database, and X-ray crystallography does not detect hydrogen atom positions, plus, crystall structure may differ from structure in gas phase, but still. However, in case of any serious project I recommend to consider splitting the task and outsourcing the parts.

I appreciate the suggestions but some digging found a few possibilities:

As I mentioned in the question, there's the Cambridge Structural Database, which includes over 700,000 compounds (both organic and inorganic/organometallic). It's decidedly not free, but available at many universities.
There is the "Teaching Subset" of 733 compounds from CCDC which offers a Java web view of interesting compounds, including plenty of VSEPR examples and inorganic and organometallic species. This is free to access, but as far as I can tell, not free to redistribute.
Cool Molecules from St. Olaf college, including 900+ structures from experimental data, and lots of inorganic and organometallic species (most elements are reflected), including lots of "cool" or unusual shapes (e.g., hexagonal bipyramidal).
Crystallography Open Database as mentioned by @permeakra. I was aware of this for solid-state materials, but it seems as if they've merged in data from CrystalEye, a resource which grabbed crystallography data from journal articles, including many molecular species now, and can be browsed by journal (e.g., Organometallics).

The last two resources seem the best, since the data can be freely distributed. They also don't require using Java to view the structures.

Gmelin is supposed to be the largest database of inorganic/organometallic chemical information (as Beilstein is for organics). Both are now part of Elsevier's Reaxys, which is as far from open source as you can get.
PubChem does list inorganic molecules and coordination compounds, but I am not sure how easy it is to systematically mine them.

As far as I can tell, Reaxys and Gmelin do not actually give coordinates. They do have an extensive collection of inorganic molecular species, synthesis, facts, etc.

This might be exactly what you're looking for and it has a nice API to help do whatever you want with it:
https://www.materialsproject.org/

To be clear, I want molecules. There are lots of databases of solids, of this is one.

Ah sorry, I misunderstood. This is probably also not what you're looking for since I believe they're almost entirely organic molecules but maybe take a peek: http://link.springer.com/article/10.1007%2Fs10822-014-9747-x


text:
Material that doesn't melt at 800 deg F, but can be dissolved?
I'm a sculptor and I have a chemistry/plastic question. I thought here would be a good place to ask.
I'm looking for a substance (probably a plastic) that has these characteristics:

Doesn't melt at the annealing temperature of Copper (doesn't melt at 700 degrees Fahrenheit)
Can be dissolved/made non-solid/melted by a process that will not affect copper (a specific acid?)
Is available in a wire form
Is relatively flexible

I want to wind long coils of tiny copper wire around a flexible core (the plastic I'm looking for), form that coil into complex shapes, then get rid of the core so as to be left only with the copper wire showing
I understand these requirements are perhaps impossible to meet...Just wondering if someone with good knowledge of plastics maybe has an idea.

Magnesium or aluminium? Available in wire form, dissolve in hydrochloric acid, melt at 1200+ F.

Won`t hydrochloric acid dissolve copper as well though?

No, copper doesn't react with HCl.  It takes an oxidizing acid, like nitric, to dissolve copper (or silver or gold [actually a mix of nitric and HCl for gold, aqua regia.])

@Luke Technically it should not. At least, aluminium/magnesium will dissolve first (though in case of aluminium reaction may start very slowly but react with unhealthy speed once started). However, if left with easy access of air for prolonged period of time, copper indeed may dissolve, especially if iron ions are present. TL;DR: not if you wash it with water immidiately after dissolving aluminium, but if you leave it for days or weeks, some of it may dissolve.

Aluminium and Magnesium
react vigorously with mineral acids.
A lot of heat is evolved.
Try iron/steel wire, which is cheaper and
available in any diameter.
Georg

Alright, seems like aluminium wire and hydrochloric acid is the way to go!
Thank you so much, it would have taken me days of googling to figure this one out!

Careful, it'll evolve hydrogen gas and can be pretty vigorous, depending on how concentrated it is. Make sure you have good ventilation and don't use too concentrated acid.

Always do these kinds of things under a hood. That way if there is a problem you can just close it up.

+1 for vigorous reaction and hydrogen warning. Especially hydrogen: it forms exploding mixtures with air, so good ventilation/fume hood and absence of open fire/sparks are an absolute requirement.

I know it's been a while, but could you add why you chose this for the experiment?


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Contactless non-ferrous metals hardening?
Is it possible to work-harden non-ferrous metals without direct contact?
I know induction hardening is used on ferrous metals, but would like to know if a technique exists for non-ferrous metals.

Depends on which non-ferrous alloys. Induction hardening (IH) isn't work hardening (WH). IH is the application of a electric field to produce heat which raises it the austenitic temperature which quickly cools to martensite. W H is the application of mechanical work to move the material along its stress-strain curve so that its yield strength is higher.

From this link I could find a few non-ferrous systems which show quench-induced or stress-induced martensite . Generally non-ferrous systems dont show this behaviour and they are hardened/strengthened by Precipitation hardening, work hardening, Solid Solution Strengthening.
Also work hardening is basically working/deforming the material plastically to strengthen it. So you can't work harden a material without contact. Induction hardening is used mainly for Case Hardening which is different from hardening(done to bulk of the material). So as long as the non-ferrous alloy is conductive and resistive , Induction hardening is enough for case-hardening.
And if it is hardening , the above mentioned methods are used for non-ferrous. Only the quenching/martensitic transformation is limited to limited systems.


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Which atoms in a given amino acid are able to form hydrogen bonds with water?
Can anyone help me with this question I have tried everything.
I know hydrogen bonding is with $\ce{F}$, $\ce{O}$, or $\ce{N}$, but every time I select those is says it's wrong.

Which part of the question are you getting wrong? Top, bottom, or both? I'm guessing that if it's the bottom, then you need to include H in your list of H-bonding atoms to choose from.

The positive end of water can hydrogen bond with N, O, or F, as you said, but only if A) the geometry of the atom allows the water and N/O/F to get close to one another and B) the N/O/F has a lone pair of electrons.  This happens at each of the green circled nitrogen and oxygen atoms.  The red circled nitrogen is a decoy, as it is neither geometrically available nor does it have the appropriate configuration of electrons.
The negative end of water can hydrogen bond with hydrogen atoms that are attached to the N/O/F atoms, including the ones that are attached to the N atom that was not correct for the first part.


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How do quantum software packages Work?
If one wants to calculate a moderate size Alkane (with say 10-15 carbons, assuming 100 electrons, with Restricted Hartree Fock based methods) we can simply say that the electron-electron part will be $100^4 = 100$ million integrals. If we remove the non-unique ones it will become $\frac{100M}{8}= 12.5 M$ Integrals. If we do a pre-screening it may become around 6 million.  Even if we go for DFT it will be around $\frac{100^3}{8} =125,000$ .
I have calculated a small system like Methane with just 10 electrons (around 400 unique/pre-screened integrals) with an efficient C++ code  (based on Rys Integration which is among the most efficient methods) and it took around 1 hour. When 400 cycles take 1 hour I can assume that 6 million cycles may take years! But software packages do that just in seconds!
My question is if I just want to do simple math like 2+2 for 6 million times it takes way longer than what these packages do with such complicated integrals!!
Does anyone know what their trick is? Do they have a pre-calculated database or some approximation methods?
UPDATE:
there are many good hints in this question and thanks to all for that. A combination of symmetry, finding negligible integrals, and using optimized methods work fine. The only unanswered part is an algorithm for finding symmetric situations.

I think many of these packages use highly optimized numerical libraries for linear algebra like LAPACK and BLAS. They may also implement some parallel programming techniques (eg, multi-threading or MPI) to take advantage of multi-core processors.

The only unanswered part is an algorithm for finding the symmetric situations  || actually, gamess puts this load onto user. Gaussian somehow can guess symmetry without it explicitly mentioning in input. It is not very hard, though. D2h and friends is easily detectable for linear molecules, and all elements of symmetry must go through geometrical center of the molecule, tranforming one atom into anouther.  Since all symmetry operations are representable in matrix form, it is possible for each to points to concstuct such a matrix and try to see if it is indeed and element of symmetry.

I also suggest to google for article "what every programmer should know about memory". It gives interesting insights to optimizations.

Are you sure you C++ code was that efficient? Non-CS guys in scientific computing are notorious about making totally false claims about their codes.

@Aug You should look into the source code of free software computing packages.  Psi4 is written mostly in C++ and Python.  MPQC is mainly C++. There may be others available as well.  There may also be some helpful information in the Q-Chem manual

@Aug Doing Hartree-Fock for methane should be a few seconds if not less. The code you tested is highly inefficient for one reason or another. By the way, Rys integration is among the most efficient if implemented right. Anyway, for your small system choosing the best is not going to change much.

There are many tricks. Since I'm not a developer, I'm not privy to all of them, but there are some simple things to start.

Some integrals are predictably close to zero: Even in fully first-principals calculations, there are parts of the code that prune out integrals that will clearly be negligible. This cuts out a lot of work. Consider, for example, overlap or Coulomb integrals between atoms that are very far apart.
Gaussian Orbitals: Using Gaussian-type orbitals are incredibly fast because most quantum integrals can be derived as simple recurrence relationships. That is, you almost never need to do the integration. See, for example, Obara and Saika and the recent PRISM algorithm from Gill and Pople.

I believe there are some actual quantum developers on here and perhaps they'd be willing to share other tricks. But these are big - eliminate any integral you don't need, and don't actually take the integrals in the first place. ;-)
Yes, there are numeric tricks (using BLAS, LAPACK, etc.) and the use of multi-core processing. But these are actually minor optimizations.
UPDATE
I realized that I forgot to talk about the convergence of the self-consistent field which is also optimized. There are various convergence strategies, most notably DIIS, which help to extrapolate the final orbital coefficients and electron density.

Yes I absolutely agree with the efficacy of rejection of negligible integrals especially in larger molecules as there will be more zero overlaps. I have used Obara-Saika for Overlap Integrals and Kinetic energy. I found that Rys iteration is way faster for e-e integrals . But I couldn't find a definite formula for selection-rejection of the integrals (I only know "Schwarz inequality" but it is not that effective). Do you know any effective selection-rejection criteria? I think the answer of this question may be such criteria.

I'd add that many times you can exploit molecular point group symmetry. Even just having one symmetry element can reduce the storage by 2 and the work by 4. D2h, which has order eight, can reduce the work by 64!

@jjgoings I think this symmetry sounds really promising but it is so difficult to find symmetric situations ( I mean by coding ) because 4 integrals can come from any different atoms but be symmetric yet. By any chance you may know a reliable algorithm to apply this symmetry?

Yes, there are numeric tricks (using BLAS, LAPACK, etc.) and use of multi-core processing. But these are actually minor optimizations. || I.. wouldn't say so. use of multi-core and parallel runs may reduce job time by orders of magniture, though parallelization involves significant overhead. Also, some packages include GPGPU support.

To further reduce storage requirements, integrals may be calculated on the fly, employing different caching strategies. For some methods storing all the integrals is plainly impossible.

First things first, I would check your code: there is no reason 400 unique integrals should take an hour, even with the most naive implementation. You are computing at a rate of an integral every ten seconds. While I agree with all the comments pointing to increasing parallelization, I think efficient scalar code is more important. Why scale up inefficient code? Considering GPUs is a whole 'nother beast, and from the problems, my coworkers have expressed trying to program them I'd recommend avoiding them for the time being.
The best thing you can do is to exploit molecular symmetry. Heads up though: this won't fix everything, because many interesting molecules lack symmetry. But for methane, as you tried, this should help a ton. Of course, this isn't as simple as just looking up some algorithm, because depending on your program you need to rotate the nuclear orientations to be symmetric, or develop routines to detect point group symmetry, etc. That being said, there are --- as far as I know --- two primary ways of exploiting point group symmetry for integrals. The first is to create symmetry-adapted basis functions, so your basis set is symmetric, which means your wave function/integrals/Fock matrix/whatever will be symmetric as well. Ernie Davidson did this in the seventies. A slightly more modern reference by Helgaker and Taylor is given here. Daniel Crawford's electronic structure tutorial explores this as well (it's really well written --- a great introduction).
The other way you can do it, and I believe this is more popular, is to create symmetry unique integrals and then build a partial Fock matrix only to take care of symmetrization later. If you see something like "generating petite list" in the output of your favorite electronic structure program, this is probably what it's doing. This method allows for integral screening more efficiently, I believe. Dupuis and King did this in the late seventies, based on earlier work by Dacre and Elder. These are the standard references you'll find. There may be some more recent work that I'm unaware of. The problem of point group symmetry had been effectively solved by 1980.
Of course, many interesting molecules don't have symmetry to take advantage of, so symmetry won't fix all your problems. The Schwarz equality is well addressed in Molecular Electronic Structure Theory (actually, most electronic structure theory is well addressed in that lovely pink book). If you haven't checked that out, please do. I think it will answer many of your questions about how modern codes work.
I don't want to repeat the other answers (Geoff gave you the papers I'd have suggested as well), so I'll finish with two lesser-considered references. First, there is an awesome (but tiny!) book called Parallel Computing in Quantum Chemistry that lays out the basics for efficient code including integral evaluation. It's not too hard to read, and I found the work on local MP2 to be pretty interesting. I'm often surprised more people haven't heard of it. The other place I would check out is Ed Valeev's LibInt package. He's a pretty talented programmer, and it's used in many modern packages like Psi and MPQC and ORCA, not to mention free for anyone to download and use. Some of my coworkers have played around with it for their own "homemade" electronic structure code, and speak highly of it. The user's manual is actually quite helpful in understanding integral evaluation. He has also written several exercises and tutorials, which you may find useful.

Thank you for the good suggestions. I wrote some "Homemade codes" to find the symmetry ( it is not very efficient as it can not find signed symmetries/ plane symmetries and also is rather slow) but I cannot say how faster the whole calculation is now ( maybe 10 times faster )!  I got surprised and now I know that the main magic is the symmetry and negligible integral rejection. I'm gonna dig into the suggested sources and will post the results here.

There are two things here:

Quantum chemistry problem
Optimization problem

Quantum chemistry or the Schrodinger equation is simplified using DFT and some other techniques:
a) ECP or effective core potentials and pseudopotentials: Instead of solving for all electrons just the outer shell electrons are solved and core electrons are considered using pre-calculated integrals
b) ri approximation: Resolution of identity, it treats two electron terms as a product of single-particle orbital function thus greatly reducing computational cost
And as mentioned above there are some other good techniques are used to deal with quantum chemistry part.
Optimization problem: In optimization usually second derivative (Hessian matrix) is used to find minima or maxima but in computational chemistry code only the first derivative or jacobian is used, thus it is highly sensitive to initial guess and computational setup.
Calculating Hessian is the most costly operation of computational chemistry code, either approximate hessian is used or no hessian is used in this type of optimization problem. If you want to make sure that you got the right geometry you can calculate Hessian with your optimized geometry and for geometry optimization, all the frequencies should be positive and for the transition state, only one frequency should be negative but you will find that to meet this requirement is highly demanding.
Moreover to reduce computational cost we can apply periodic boundary conditions. It enables us to solve the problem for minimum nonrepetitive unit cell ( Irreducible Kpoints). Finding symmetry is also a crucial part of this type of problem.
Cut-off energy is used to reduce the number of basis sets considered. And also optimization techniques also have a great effect on computational cost, you may use a fast but aggressive algorithm but that may not be able to find the minima or maxima.
You can read VASP or TURBOMOLE manual (available freely online), and there you will find some good techniques for reducing the computational cost. And above all, as it is mentioned above efficient implementation of parallelization can help to reduce computational costs dramatically. One another potential method is to use GPU, but it is still in its primary stage.


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Which atom goes in the middle of a lewis dot structure?
I've read online in multiple sites that the least electronegative atom goes in the middle with the exception of H, which always goes on the outside.
However, in the molecule NaCN, C is in the middle. Could anyone explain to me what is going on with this molecule?

It's not really valid to have a Lewis structure for an ionic compound.  Leave the Na out.

@brinnb So is that why the rule does not generalize to this specific molecule?

ETA: I'm unclear about your question.  Are you asking why nitrogen isn't the central atom or why sodium isn't?  If it's the former, the explanation is below.  If it is the latter, sodium, then it's because this is an ionic compound, and we build the lewis structures of the cation and anion separately.
First and foremost, carbon (EN=2.5) is less electronegative than nitrogen (EN=3.1), so by your rule it should be in the center.
A more complete explanation of why that is:
Your molecule is sodium cyanide, an ionic compound.  The cyanide group, CN-, carries a negative charge.  If we consider where that charge lies in the ion, it is a bit confusing.  The basic structure of the cyanide is  ${\ce :C:::N:^-}$  .
At first glance we might think that the nitrogen would be the more negative end of the ion, since it is the more electronegative atom.  However, if we consider the formal charges of the atoms in the ion, we see that nitrogen has a formal charge of 0 and carbon a formal charge of -1.  Formal charge is not the same as actual charge, but it does tell us that the carbon will be the negative end of the ion, and that is what we see in its bonding.
Here, the sodium ion, Na+, bonds with the carbon in the cyanide ion.  Even if it weren't a sodium ion, but something like hydrogen cyanide (with covalent bonding) the bond would be formed at the carbon end.  This is because that structure would allow carbon to minimize its electron excess and reach a lower energy state than if the hydrogen stuck to the nitrogen.  Again, formal charges predict the HCN structure over the HNC structure.


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Letting Copper sit in NaOH for 5 min (for monolayer assembly)
i made a self-assembled monolayer of dodecanethiol on copper, before that i let the copper in a solution of $\ce{NaOH}$ for 5 min, and i need to know what this step is for, anyone can help me?

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Such steps are usually cleaning steps to remove deposited organic materials. $\ce{NaOH}$ is also known as lye and will dissolve basically all organics. In this case, it should leave the $\ce{Cu}$ and copper oxides alone, leaving a clean surface.
I don't have as much experience with copper monolayers, but similar cleaning steps are used in my lab, where you might treat with ozone or some other step to remove any deposited organic "gunk" from the metal surface.
In this case, I believe you don't want to use an acid, because the thiols react with the copper oxide surface, not strictly the copper itself.


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Which material to use for athletic accessories?
I'm looking for some guidance in selecting the right material for dance shoes heels and women's self-defense accessories.
I'm looking for a material that 1) has a superb strength-to-weight ratio, can withstand punishing and frequent impact, and very light (like plastic).
The part for the shoe will be very, very slender (stiletto heel), but will have to be able to endure hours of jumping, sliding, stomping, spinning, and skidding dancers… is compression strength the right term?
2) The material must also be safe to use as jewelry, so nothing toxic. Also, I am looking for something that tends to snap, rather than shatter, when it fails under pressure.
3) Lastly, it would be great if this material can be 3D printed, although that is not a must.
An inexpensive material would be ideal, of course, but I am open to learning about a fuller range of materials, including the more expensive ones. I am willing to consider a higher priced material if it will offer substantial value to the accessories in terms of strength, quality, and safety.

For plastics, my first instinct for the first two properties would be a polycarbonate plastic. Polycarbonates are known to be very strong plastics (tensile strength up to ~70 MPa) and deform without breaking under many conditions, which is why they're commonly used in things like safety glasses and bullet-resistant glass. It also looks very interesting as it's quite optically clear. PC is 3D printable, I've seen a few people using it in hobby printers (http://www.protoparadigm.com/blog/2011/12/printing-polycarbonate/) and in very high end StrataSys printers, but it's not a super common material so it might take some experimenting.
Another option is a nylon. There are a number of different types available and many have similar properties to polycarbonates (minus the optical clarity). Nylons seem to be a bit more popular for 3D printing and I've heard good things about the Taulman filaments.
Of course, you're not limited to a single material and there are many things that could potentially be used in this kind of application. Often, the combination of multiple materials makes sense (e.g. a metal insert in a plastic part).
Something like carbon fibre-reinforced epoxy construction might be appropriate. It's not 3D printable, but can be constructed in layers around a lightweight (expanded polystyrene or the like) form. This approach is often used for parts that need to be very light while having high tensile strength (think bicycle frames, ski poles, etc.) I could see a carbon fibre-wrapped plastic insert making a very strong and light shoe heel.
Ultimately, these things are balancing games between strength, mass, fabrication cost, aesthetics, etc. but these are a few potential materials that come to mind.

+1 definitely agree with polycarbonate, although I didn't know it could be 3D printed. I also agree that experimentation (particularly with durability of 3D printed parts) would be crucial. You might want to order from a commercial 3D printing service to experiment with different options.