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Is it actually possible to dispose of a body with hydrofluoric acid?
In the TV show "Breaking Bad", Walter White frequently gets rid of people who get in his way by submerging them in a plastic container full of hydrofluoric acid. This, at least in the TV show, completely dissolves the body leaving nothing but a red sludge behind at the end.
Is it actually possible to dispose of a body with hydrofluoric acid?
If hydrofluoric acid wouldn't work, are there any acids corrosive enough to achieve the stated effect from the show?

I can't say if this is possible, but concentrated HF poses a very real threat of delayed, life threatening damage by scavenging calcium from the body, either by splash or inhalation. HF is not a strong acid, but its toxicity makes it substantially more dangerous to work with than many strong acids. I tend to think that dissolving a body would be quite difficult, and then you'd have the scrub the bathtub ring...

Related: http://chemistry.stackexchange.com/questions/3949/dissolving-organic-tissues. I wonder when the FBI will come and shut this site down for answering questions about disposing bodies and FOOF ;-)

I guess he used fluoroantimonic acid. :)

or hydrogen peroxide and conc sulphuric acid .. oops there's another finger gone!

I like people who makes experiments Periodic Video I was not a lot of surprised by the experiment, but it is always good to see what happend in an other place than a movies beacause it may sometimes be false. Now I can come back dissolving my stepmother... :)

Related: article from chemistryworld.com.

Huh, while answers are insisting on it to be a myth to debunk, I'm still convinced that big amount of pure HF would do the job. It would take quite a lot of it, more then in the series, and it's not like they could simply buy it, but it should work.

"asking for a friend"

Hydrofluoric acid is toxic and corrosive, but actually isn't that strong of an acid compared to other hydrohalic acids; the fluorine has a very good orbital overlap with hydrogen and is also not very polarizable, therefore it resists donating its proton, unlike other hydrohalic acids which are good proton donators. It will break down some tissues, but it will take a relatively long time and won't turn the entire body into stuff that can be rinsed down the drain. Hydrochloric acid is a much stronger acid, and as it has several uses from pH-balancing pool water to preparing concrete surfaces, it's available by the gallon from any hardware store. However, it isn't very good at dissolving bodies either; while it will eventually work by breaking down the connective tissues, it will make a huge stink and take several days to dissolve certain types of tissues and bones.
The standard body-dissolving chemical is lye aka sodium hydroxide. The main source is drain clog remover because most drain clogs are formed by hair and other bio-gunk that accumulates naturally when humans shower, exfoliate etc. It works, even though the body's overall chemistry is slightly to the basic side of neutral (about 7.35-7.4) because the hydroxide anion is a strong proton acceptor. That means that it strips hydrogen atoms off of organic molecules to form water (alkaline hydrolysis, aka saponification), and as a result, those organic molecules are turned into simpler molecules with lower melting points (triglycerides are turned into fatty acids, saturated fats are dehydrogenated to form unsaturated fats, alkanes become alcohols, etc). Sodium hydroxide is also a ready source of the sodium ion; sodium salts are always water-soluble (at least I can't think of a single one that isn't). The resulting compounds are thus either liquids or water-soluble alcohols and salts, which flush down the drain. What's left is the brittle, insoluble calcium "shell" of the skeleton; if hydrolyzed by sodium hydroxide, the resulting calcium hydroxide ("slaked lime") won't dissolve completely but is relatively easy to clean up.

I like the wording here: "standard body-dissolving chemical"

I like how you answered the question in the first 6 lines of your response, then graced us with another 12 lines of 'how you should do it' information.

"the fluorine is too electronegative to readily donate its proton." What does that mean? If anything, from EN trends alone, we should expect HF to be a strong acid, since the F, as you said, is very EN and is therefore isolating electron density away from the H and thereby  making the H more partially positive and more reactive. That's not the case, as there are other, greater factors in play here, such as conjugate stability. The F- anion simply isn't as stable as the Cl- anion, for example.

What I mean is exactly what I said. Larger anions have higher-energy orbital shells which spread the negative charge of the electrons over a greater space, meaning that the proton is similarly held at greater distance and with reduced attraction to the anion, so the compound more readily dissociates and donates protons. Fluorine is small, and its electron configuration is the densest available of the halogens. So, the attraction to the proton is increased due to the more concentrated charge, thus requiring more energy to break this ionic bond and form another.

If we were talking about a covalent bond, or even an ionic bond involving a cation with more than one electron to its name, then we could discuss electron delocalization and higher bond dipole moments increasing reactivity, but hydrogen ions are a special case. No matter the anion, a hydrogen ion has zero electron density as a halide through the majority of the table (with only HAt being willing to let the hydrogen bond covalently), and so its partial charge is very high in any case; the reactivity is then determined by how easily it can go find something other than the halide to bond with.

I've just realised that, though this is a good answer about alternatives to HF, it doesn't really answer the question about whether HF works.

$\ce{NaBiO3}$ is insoluble in cold water

Sodium hydroxide alone will not dehydrogenate an alkene like an unsaturated fatty acid; is there some biochemical mechanism you're aware of (involving an additional reactant), or are you just speculating that in such a complex and highly basic mixture, such a redox reaction might occur?

I think the use of Hydrofluoric Acid was script-driven rather than fact driven: it sounds scary rather than being a good choice. Also, it allows for the possibility of the darkly comic bathtub scene where the acid dissolves a ceramic bath because Jessie ignores Walter's instructions (which establishes Walter's expertise and Jessie's lack of it).
There is no good reason why Pinkman and White pharmaceuticals needed to have hydrofluoric acid, therefore using large quantities of it is somewhat implausible.
Moreover, it probably wouldn't work as well as several alternatives. Hydrofluoric acid is very nasty stuff, but it isn't a strong acid. Even when dilute it will etch glass and ceramics, but it won't dissolve or burn flesh. I once saw a demonstration where a lecturer showed this by spilling some dilute hydrofluoric acid on his hand and then onto a glass surface. The surface was frosted, his hand unharmed (he was very careful to wash the acid off quickly and take appropriate precautions and I don't recommend trying this at home!)
Its danger to people is its toxicity, not its ability to burn: it insinuates itself into the body and destroys connective tissue and bone slowly by interfering with anything containing calcium. Its danger is worse because it doesn't cause immediate damage and you may receive a dangerous dose without noticing. So it is scary but not corrosive.
Other alternatives are better. Concentrated alkalis such as Sodium Hydroxide are readily available and are very good at dissolving flesh (which is why they are commonly used as drain cleaners). But alkalis don't do a good job on bone. Concentrated sulfuric acid is even better as it does a good job on flesh and will, eventually, dissolve the bone as well. Murderers have used both methods to try to dispose of evidence. For example, John George Haigh who used sulfuric acid and left little other than gallstones (http://en.wikipedia.org/wiki/John_George_Haigh). Using alkali is often done but tends to leave bone fragments even with sophisticated processes that pressure cook the solution (see http://www.slate.com/articles/news_and_politics/explainer/2009/12/soluble_dilemma.html).
So I think the answer is that HF solutions are not a good choice for body disposal as it probably doesn't work well compared to known alternatives.
Update
A lot of the above is theory but good scientists do experiments. So Periodic Videos decided to test this very idea using chicken legs as a model. They compared what happens when raw chicken legs are suspended in strong solutions of HCl, H2SO4 and HF. The HF was the least impressive for flesh-dissolving characteristics, though it did seem to cause other, more subtle damage, to the components of the flesh.
See the actual results here.

Here in Mexico, a guy was arrested a couple years ago for "dissolving" more than 300 bodies killed by the cartels.
They found 55-gallon drums around town with a sludge in it, usually with a note to warn other.
His recipe?
A 55-gallon drum, several bags of lye, add body, fill with h20, and build fire beneath drum.
It took about 24hrs I believe.
Sorry, I got caught up In the comments. And I'm New to this and all (bulletin boards?)
Like the case in Australia with sealed drums w/ HCL, in a bank vault, it Actually did more to preserve the bodies. I am no chemist, but it is my understanding, that when acids burn the skin the results are the same for both Fire and chemicals burns both rapidly dehydrate the body, destroying it. But is in a sealed container, that water can only go to the lid. Stopping the process.
I bet any acid will destroy a body.
When you're in the bath or pool too long you get those wrinkles, I've heard of people with wet feet in boots for weeks, remove their boots, and literally stripping the flesh off their feet to the bone.
The issue is how long will it take like the method used in here in Mexico and another post, boiling seems to be an important component. And again I'm guessing (and vegetarian), the heat tenderizes the meat, helping the acid along, but also the rolling of the water stirs it. If you were to put a piece of meat in acid that was stagnant, the acid might react, dissolve some meat, as the dissolved meat dilutes the acid depending on the makeup. It could settle or form a layer of meat/acid that sits on top of the meat, protecting it. Think layered shots. Or a unflushed end toilet that has had time to settle.
Hydrofluoric acid will eat thru skin, google it. One of the reasons the burns are so bad is @ 50%, it can take 8 hours to realize you were burned.
So that would be a slow process, and fluoride, as we know, preserves teeth, by extension bones. And for hydrofluoric acid burns, the standard treatment is calcium.
So it would be very slow, very likely stop, possible preserve the bones, making the resistant to rotting and harden them, but as your, we'll my fingers have less meat than most parts of my body, it would get to bone faster, ossicle dissolving some calcium, then render all the acid useless.
Again I am not a chemist but have self-studied chemistry for years, I would be very interested to hear anyone else's opinions on my determinations!

I know this is an old question, but it gets viewed a lot so I thought I would update with the fact that this experiment was actually done on the first Mythbusters Breaking Bad special, episode 206. There is a description of the results of the episode inside that link.

Is it actually possible to dispose of a body with hydrofluoric acid?

To summarize from the Mythbusters episode, they tested out hydrofluoric acid on various small scale materials (to simulate the bathtub) as well as pig parts (the body) and found that $\ce{HF}$ certainly deteriorates some of the materials, but not at anything close to the same speed as in the show. It is even less successful at decomposing the actual flesh than the materials. One way of viewing this is to remind yourself that the ability of $\ce{HF}$ to eat through materials actually has more to do with fluorine's reactivity than with the ability to give a hydrogen atom (it is classified as a weak acid), thus it's reasonable that $\ce{HF}$ might be better at disrupting the structural integrity of metals/polymers than acidically eating through flesh.
See this question for details about its reactivity with glass. Mythbusters notes that a fiberglass bathtub actually would've behaved similarly to what we see in Breaking Bad, but over a much longer timescale.
So it's perhaps feasible to dispose of a body with $\ce{HF}$ given a very large amount of time, but it's not possible in the timescale presented in the show.

If hydrofluoric acid wouldn't work, are there any acids corrosive enough to achieve the stated effect from the show?

The real reason I brought up this episode of Mythbusters is that they have a storied tradition of reproducing the results at all costs. In this episode, at the end, they fill up a bathtub with 140 L of concentrated sulfuric acid, a "secret sauce", and a whole pig just to see what happens...
To borrow the words of the episode description from the link above,

After 5 minutes, Adam and Jamie found only black organic sludge where the pig used to be.

And, a word on the secret sauce. Intuition says that this is something which is going to be good at breaking down organic matter. I looked around and found some people speculating on this reddit post, and they come up with pretty convincing evidence that the "secret sauce" is just hydrogen peroxide.
Hydrogen peroxide is a good radical initiator, and thus should be quite effective at assisting in the decomposition of this pig/person. This mixture actually has a name: piranha solution.
So, to answer, yes, a mixture of $\ce{H2SO4}$ and $\ce{H2O2}$ should do the job.

I know Mythbusters and Reddit aren't exactly the standards of scientific authority, but in this case, I think they're doing pretty well.

Oh my got... i have no idea how I got here. I got a panic attack imagining it.

This subject is touched upon here and explained in detail on this Wikipedia page.
Essentially, it is impractical to dispose of a body using an acid, though there exists a process called alkaline hydrolysis, which, as its name implies, makes use of a base to degrade the materials comprising the human body to a liquid state within a matter of hours.

I believe dilute boiling hydrochloric acid is quite destructive to all flesh and bones. You don't want a concentrated acid as it has annoying oxidation effects which slow the reaction. You don't want hydrofluoric as its hideous dangerous to handle at all temperature (a small amount on your skin will be absorbed and lead to inevitable death if you don't inject calcium glutamate into you) and the formation of insoluble calcium fluoride (and others) slows everything down.

A man here in Australia dissolved his wife's body in Hydrochloric acid in a wheelie bin.
He put her body in the bin and poured 20 liters of Hydrochloric Acid in it.
Then 2 days later went and bought another 40 liters and topped it up until everything was liquid.
He then poured it down a storm drain and rinsed it all out with water.
The only thing they found was her prosthetic teeth.
Basically, once the Acid stops dissolving and becomes weak, you have to keep adding more.
But it will eventually dissolve a whole human body.
Cairns man Klaus Andres, who admitted to dissolving wife's body in acid, guilty of murder, jailed for life
http://www.abc.net.au/news/2013-12-12/cairns-man-guilty-of-murdering-wife/5152488

http://en.wikipedia.org/wiki/Larissa_Schuster  This woman dissolved her husband's body in HCl.


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A theoretical molecule that could match uranium mole-for-mole
XKCD #1162 piqued my curiosity. Obviously, there's nothing we generally use as a chemical fuel that can match the energy density of uranium fission. But I wondered what it would take, hypothetically, to match it mole-for-mole?
Looking at the bond enthalpy charts, I see that the carbon-oxygen triple bond has the highest enthalpy of formation (the energy required to break it and form something else, and by extension the energy produced when this bond is formed, minus the energy required to break the previous ones) at $\pu{1072 kJ mol-1}$. However, the carbon-oxygen single bond has an EoF of $\pu{358 kJ mol-1}$, and other bonds likely to be involved ($\ce{H-C}$: $\pu{413 kJ mol-1}$, $\ce{H-O}$: $\pu{463 kJ mol-1}$, etc.) have higher bond enthalpies, meaning the formation of carbon monoxide by combustion of an organic molecule is very likely a losing proposition.
Nitrogen, on the other hand, has an infamously high bond energy ratio; breaking a single nitrogen bond costs only $\pu{163 kJ mol-1}$, while forming a triple bond releases $\pu{941 kJ mol-1}$. Nitrogen single bonds, especially in high percentages of the total molecular mass, are notoriously explosive; the azide anion, covalently bonded, forms the backbone of the eminent Dr. Lowe's Things I Won't Work With. The nitro group ($\ce{RNO2}$), well-known in explosives, can be made even more potent by bonding it to another nitrogen to form a nitramine, and these are central to the well-known explosive cyclotrimethylenetrinitramine (better known in pure form as RDX, and with added plasticizers as C-4) and its big brother, HMX (cyclotetramethylenetetranitranime).
So, let's consider a theoretical molecule composed of only single-bonded nitrogen (eek). $\ce{N4}$, tetranitrogen, has been synthesized, but in a form theorized to be more like two pairs of double-bonded nitrogens that are joined by two single bonds (kind of hard to tell, since it only exists at very high temperature and for less than a microsecond at a time). Let's instead bond each nitrogen to each other one, forming an isohedron. When that structure reverts to diatomic nitrogen (and it will, oh, you betcha), we can expect a heat of detonation of
$$4\cdot\pu{163 kJ mol-1} - 2\cdot\pu{941 kJ mol-1} = \pu{-1230 kJ mol-1}$$
That is a pretty aggressive exotherm; RDX has a detonation energy of 1118 kJ/mol, and its molecular mass is four times that of our theoretical "isohedral tetranitrogen". RDX is as heavy as it is for stability; high molecular weights sacrifice pure energy density for a structure that won't detonate given the slightest excuse (or even no excuse at all).
However, bring in the champion:
$$\ce{^{235}_{92}U + ^1_0n -> ^{142}_{56}Ba + ^{91}_{36}Kr + 3 ^1_0n}$$
(in the atomic chaos of a nuclear reaction, a lot of other products are possible, but this is one of the most common). The masses of the reactants total $\pu{236.0526 Da}$, and the masses of the products, $\pu{235.8658 Da}$, for a delta of mass of $\pu{0.1868 Da}$ ($\pu{3.102e-28 kg}$). $E = mc^2$, so the energy produced by one uranium atom splitting this way is
$$E = \pu{3.102e-28 kg}\cdot(\pu{3.00e8 m s-1})^2 = \pu{2.79e-12 J}$$
which when multiplied by Avogadro's number and converted to kilojoules, gives us
$$\pu{2.79e-12 J}\cdot\pu{6.022e23 mol-1} = \pu{1.68e8 kJ mol-1}$$
That's 168 million kilojoules per mole available by fissioning uranium-235.
Assuming that double the nitrogen equals double the heat of reaction, in order to have the same heat of decomposition, mole-for-mole, our nitrogen molecule would instead have to be an isometric "sheet" of single-bonded nitrogen, or a geodesic dome (buckyazide?), consisting of $546340$ nitrogen atoms single-bonded to each other (a concept that, if ever seriously proposed, should have any student of chemistry falling out of their chair either laughing or horrified). One mole of this material would weigh $\pu{14.0067 g mol-1}\cdot 546340 = \pu{7652.42 kg mol-1} \approx \pu{8.4 t}$. One mole of U-235 weighs $\pu{235 g}$, just a little more than half a pound.
Long live the king.
The question
(thank you for reading this far)
Does the math look right? I've done it three times and come up with three different answers for the heat of detonation of my hypothetical isohetral tetranitrogen, and thus the size and weight of the resulting molecule, possibly based on the use of different enthalpy figures (I've seen the single bond energy of formation of nitrogen to be as little as $\pu{158 kJ mol-1}$, which produces a heat of detonation of $\pu{1250 kJ mol-1}$, meaning, all other things being equal, a required molecule of only $537600$ nitrogen atoms).
Also, can anyone come up with something that would be more energetic to go up against the champ? There are molecules that have been successfully created in the lab, such as octanitrocubane, that are more powerful than RDX while being usably shock-insensitive, but are prohibitively expensive to manufacture.

You could just take nitroylethylene and polymerise it to ultra-high molecular weight 250,000 monomers and same number of nitrooxy groups.

The most violent "chemical" reactions have been proposed as rocket fuels (these guys don't think that chlorine trifluoride is beyond the pale). But the best in theory they have ever com up with is metallic hydrogen (and we don't even know it is possible). See this youtube video

Getting a large exotherm per mole of molecules is a bit of a bad cheat if you're allowed to make molecules arbitrarily big! Energy per mass is the only honest way to do it (or energy per mole of atoms would work too I guess). However, there really is no chance to chemically match a nuclear exotherm; nuclear energies are simply in a class of their own.
Practically all chemical energy is merely a function of electric* repulsion and attraction on distance scales from around a picometre (being very generous, as that is quite small compared to the dimensions of any valence orbital) to perhaps a few micrometres (to include long-distance ion interactions). However, nuclear energies are based on electric attractions and repulsions on distances scales of a few femtometres or less. This makes quite a difference when you remember that electrostatic potentials go as $1/r^n$! Not only that, but the nucleus also plays a whole other ball game, displaying strong nuclear potentials. It is commonly mentioned that, apples to apples, the strong nuclear force is about three orders of magnitude stronger than the electrostatic force. Therefore, no chemical reaction involving structures stable at STP will ever reach the feet of nuclear energies. There is probably no "reasonable" suggestion like yours that could do better by a factor of $10$, much less a factor of $10^5$.
So what can we do about it? Well, if interaction distances being too large is part of the problem, one way to cheat is to simply add pressure, an unreasonable amount of pressure. This effectively brings atoms closer and closer together, so you climb the $1/r^n$ potential for each atom. Chemistry gets pretty weird at ultrahigh pressures. For one, lots of things (perhaps everything?) seem to like turning into a metal. With all those orbitals crammed up against each other, eventually bonds overlap extensively and you get bands.
Keep going, and the bands start to overlap, until there's no gap for electron excitation, and room temperature thermal energy can freely knock electrons all over the place. You talk of making huge polynitrogen molecules. Well, metallic nitrogen would be the natural limit of that. Apparently even at pressures of the order of a terapascal, metallic nitrogen has $\ce{N-N}$ single bonds around $\pu{110 pm}$, which is only a little smaller than STP $\ce{N-N}$ bonds, at around $\pu{145 pm}$.
I'm sure this would already create quite a bang if pressure were suddenly relieved. But how much pressure would it take to get down to N-N distances of $\pu{10 pm}$? And $\pu{1 pm}$? How much energy would be pent up per gram of whatever that material is? It's a crazy suggestion, but it takes beyond crazy to get electrons to carry energy comparable to nucleons. Keep going and at some point you'll arguably leave even the theoretical domain of chemistry and reach astrophysically relevant electron degenerate matter, the stuff in the core of relatively light dead stars.
Go even further, and eventually electronic potentials finally become comparable to nuclear ones, and so it actually becomes favourable for nuclei to gobble electrons up. This is the matter inside a neutron star. The fact that it takes the mass of a Sun in a sphere a few kilometres wide to press electron energies into the realm of nuclear ones should be a sobering thought.
* Magnetism is rarely, if ever, the dominant source of electromagnetic potential energy in chemical systems. Magnetic charges do not exist (except maybe for monopoles, but those aren't chemically relevant!), so all magnetism is induced by electric charges in motion (well, spin magnetism is a little more complicated). As far as I understand, which isn't much, magnetism induced from electric charges takes a $1/c^2$ hit in the equations of electromagnetism, and that's hard to recover from!

Your first paragraph is basically my whole point and one of the reasons I used nitrogen; the idea of single-bonding half a million nitrogens together should itself be laughable, when a substance that has more than two of them is treated with the utmost care in the lab. Nice addition of what it would take to make electro-repulsion match the intranuclear forces.

@keithS: if you can go arbitrarily big with molecules, you don't need to go to extremes with specific energy. Taking one mole of uranium vs one mole of nitrocellulose, I wouldn't be quite sure where to put my money.  If I can get nitrocellulose with ~10^6 mers per chain, I'm really exceeding uranium energies, mole per mole. Sure mole of uranium will be 238 grams, and mole of nitrocellulose will approach 100 tons, but guess which one will make a bigger boom?

It's just hard because c is large. Waiting until it degrades to some more manageable magnitude is, getting into wild hypotheses, your best bet.

This post elegantly connects chemistry to astrophysics, enhancing my intuitional understanding of both fields. Well done!

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What is the effect of substituting oxygen with sulfur in drug molecules?
What sort of changes in the properties of organic or drug molecules can be anticipated if you substitute some or all of the oxygen atoms with sulfur atoms, and vice versa.
My interest in this question is for chemical graph theory perspective. There are numerous parameters calculated on molecular graphs (in trying to relate to chemical or biological activity) which are indifferent to the above substitution. I would like to make an appropriate adjustment if the effects are really significant.

In asking this question - Why do chalcogens (Group VI) stink so badly? - I received a couple answers and comments along the lines of "it's not quite so simple". The same would generally apply here; there isn't one change in behavior, or even a finite set of behavioral changes, that could be predicted by an oxygen-to-sulfur exchange.
Just as a basic example, compare methanol ($\ce{CH_3(OH)}$) to methyl mercaptan, ($\ce{CH_3(SH)}$). Methanol is a colorless liquid at RT with a sweetish odor. Methyl mercaptan, with a higher molecular mass due to the sulfur switch, is nevertheless a gas at RT and smells like rotting cabbage. Methanol metabolizes into formic acid and formaldehyde which in sufficient doses causes nerve damage, specifically to the optic nerve. Methyl mercaptan is pretty much its own metabolite; while also toxic in sufficient doses, it's readily filtered out by the liver and kidneys, as it's produced by bacterial breakdown of many of the foods we eat (especially highly-fibrous vegetables, legumes etc).
Perhaps the most common occurrence of sulfur in otherwise organic molecules is a sulfhydryl group (R-SH), forming a thiol. The alcohol-to-thiol switch is relatively common in human biochemistry and as such is also of interest in pharmacology. The Wikipedia page on thiols lists the following basic changes between an alcohol and its thiol analog:

More easily oxidized, producing disulfides, esters, ethers, etc
As thiols, more acidic than alcohols
Contrasting this, thiols dehydrogenate easily to form thiolate bases
Thiolates complex strongly to most metals

In general, you can expect the sulfur analog of pretty much any organic functional group (alcohols->thiols, aldehydes->thials, ketones->thioketones, esters->thioesters, ethers->thioethers, etc) to be more reactive. However, exactly how the sulfur switch will affect the actual reactions the compound will undergo is best studied case by case, as there is no one way that a particular sulfur-bearing functional group will behave when introduced into the complex chemical stew of human biology.
One near-constant; sulfur in small doses can smell pleasant, because sulfur is an essential not-so-trace element in humans (oranges and grapefruit, for instance, have some volatile sulfur compounds that contribute to their smell and taste). Sulfur in high concentrations stinks, because for the same reasons it is essential in biochemistry, too much of it is toxic (which is a pretty good illustration of the First Rule of Toxicology - every substance, in sufficient dosage, is toxic).

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Electron density definition of single, double, triple bonds?
I was reading this question and the answer by user EricBrown to it, and this got me thinking about covalent chemical bonds.
The way I was always taught is that a single bond contains 2 paired electrons, a double bond 4 etc (The Lewis dots drawings always show this nicely). In his answer Eric mentions that the whole definition of single, double etc bonds is ambiguous because there haven't been measurements of the electron density. This notion strikes me as odd, because a simple googling gives me articles like this: How similar is a molecule to another? An electron density measure of similarity between two molecular structures Int. J. Quantum Chem. 1980, 17 (6), 1185–1189. But it did make me think about the way the different types of bonds are defined.
I read up a bit on the Theory of atoms in molecules and indeed this theory only specifies 1 bond type: the line with maximum electron density between two nuclei. According to the theory this line is unique for a given pair of nuclei in a given molecule. I could imagine that a molecule which has a triple bond (6 bonding electrons in the 'classical' theoretical sense) would have a relatively high electron density along this maximum line as compared to e.g. a single bond.
My question is: if we would be able to measure the electron density in a molecule (are we?) couldn't we use this to define a relation between the value of the maximum electron density and the notion of a single, double or triple bond? Or perhaps use the spread in electron density between the nuclei for this?

Actually, the electron density of the "second" and "third" bonds has a node at the center line. Pi and delta bonds always do. So a triple bond has no more electron density along the center line that a single bond. It will have multiple local maximae of lesser density, and probably one global maxima; the center line. Nice question though, +1. I'll have to think about this one; though doubt I can answer it beyond "quantum mechanics said so".

Below are plots of the electron density for staggered ethane, ethene, and ethyne, computed from electronic structure theory.  For ethene, which is planar, imagine that the hydogens are jutting forward and backward into the paper, so that the contour plots should contain any putative $\pi$ bonds. It is the orthogonal plane that still contains the two carbon atoms.
A couple of points that I wish to make:

You are correct (in this case for sure) that there is some association between stronger bonds and higher electron density at the C-C bond critical point $(\star)$.  This was noted by Bader and co-workers.
There is no evidence (to my eyes) that there are any side-wise $p$ orbital interactions forming $\pi$ bonds. These look like single $\sigma$ bonds!

My question is, where do we draw the dividing line in the correlation between electron density at the bond critical point, and what constitutes single, double, and triple bonds?

Responding to @michaelm: I might have expected this, if for example, ethene were to have a "double bond" between the carbons:

what program are u using to generate this plot using QTAIM?

Awesome! I'm also wondering about the QTAIM code. I would love to play around with it. I agree that from this picture you could define numbers for single, double and triple bonds, but these would probably be wrong if you start looking at e.g. propanol, propenol and propynol

I wrote this from scratch in Mathematica. It's a bit of an "expert system" at this point, but it's the best way (for me) to get precise control over the computation and presentation layers.

It looks beautiful. What I was wondering, to put the nail in the coffin of my proposal, can you come up with an example of a molecule that has either (i) lower electron density in its single bonded than in its double bonded version or (ii) lower electron density at the double bond than Ethane has in its single bond?

@michielm I think what I am trying to say is, if you look at the electron density, there are no particularly distinguishing characteristics between what we call single, double, and triple! They all look just about the same!

Well yes, but that is if you still try to connect the notion of a double bond to 2 critical points. What I am asking if there is some OTHER characteristic (e.g. maximum value of the electron density) that can be used to distinguish between the classical concepts of single, double and triple bonds

Unfortunately, no. Arguments about multiple bondedness, from the perspective of QTAIM, are in the eye of the beholder. :-)

Too bad! Thanks for all your efforts, learned a lot about QTAIM which has certainly caught my attention

People often use other techniques with QTAIM, though, such as the delocalisation index, which uses an orbital overlap integration over atomic basins to give a bond-order-like measure. Unfortunately by reintroducing orbitals, though, it isn't a pure density-only method.

Double bonds can be detected on electron density using electron location function (ELF) methods.
This is a plot for the f-localization domain in the ethylene molecule. Surfaces are colored according to the basin they belong to: magenta for core basins, light blue for the C-H bonds, and green for the C-C banana bonds.

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


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Drying a distillate liquid
So I'm carrying out the distillation of some essential oil and I have to remove the water content in it. The usual method is using diethyl ether but I'm looking for a greener method.
What can I do?

An easy way to remove water from distilled essential oils would be to add some anhydrous hygroscopic salt such as $\ce{MgSO_4}$ to the distillate. I'm not sure what production scale you're talking about, though.

@Nicolau Saker Neto, My production scale is quite small. I'll be producing no more than a few millilitres of the oil. Ah, I initially thought about that. What is the procedure ? Is it as simple as chucking in some anhydrous CuSO4 for example and leaving it overnight ?

I don't know if $\ce{CuSO_4}$ is an acceptable substitute. For one, there's a slight chance that the transition metal ion may catalyse the decomposition of substances in the oil. When I distilled essential oils, I'd add a little bit of $\ce{MgSO_4}$ to the vial with the distillate, swirl it around for a minute, then observe if the $\ce{MgSO_4}$ was very wet. If it got all clumped up, I'd add a little more $\ce{MgSO_4}$ until I could clearly see fine particles still suspended when shaken. To recover the oil just use a pipette with a small amount of cotton on the tip to filter out the solid.

@NicolauSakerNeto, some combination of your comments should be an answer.

Perhaps this (tiny) article on Erowid (Rhodium) is interesting too for you.

I would suggest adding anhydrous $\ce{MgSO4}$ to your distillate, like Nicolau Saker Neto already mentioned in his comments. Mix the suspension thoroughly (e.g. by stirring) and look how long the salt remains dispersed. When it clumps up and settles quickly due to the formation of hydrates, there is still some water in your mixture, and you need to repeat the procedure until the added $\ce{MgSO4}$ remains dispersed as fine particles for a longer time (it should look similar to a snow storm). After the salt has settled, you can recover the dried oil with a pipette. Because magnesium sulfate is not listed as toxic or environmentally hazardous, it can be considered a "greener" alternative to, for example, copper(II) sulfate or diethyl ether (the latter is additionally highly flammable).


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Wavelength of light
how do you calculate the wavelength of light? For chemsitry lab, we calibrated the scale reading of Hg, for example, to be 4.91 cm which is violet and the given wavelength says 436.  For Hg, it says scale reading of 5.51 cm for green and wavelength of 546. But for overhead fluorescent, I took the scale reading of 4.93 for violet but don't know how to calculated the wavelength. Any help with steps shown would be great!

Close wave-lengths would look the same color. I would guess the point of your lab was to use known Hg spectrum calibrate the scale and then use th calibration to determine the wave-length of the line sin the "overhead fluorescent". At least this is exactly what my students were doing today in the lab :)

Well, the equipment appears to be giving you the wavelength; however, the scale seems to be skewed (off-calibration) so you have to correct for it.
The scale readings you give appear to be off a few orders of magnitude; the wavelengths you list as "given wavelengths" are in nanometers. Wavelengths of 4-5 centimeters as your measurements suggest would place the emission spectra in the "ELF" radio frequency range at the extreme low end of detectable electromagnetic spectra, and well out of the range of visible light. So, just for the sake of argument, let's say you interpreted the measurements incorrectly, and these numbers should be in the hundreds and represent nanometers of wavelength (or, as written, they represent tenths of a micrometer).
Given that assumption, the numbers are still off a bit, indicating the instrument is slightly out of calibration. If the instrument itself isn't adjustable to correct the values,  "calibrating" it boils down to a percentage error calculation that is then applied to further readings: (experimental-theoretical)/theoretical * 100%. For your violet wavelength, you got 491nm but expected 436. That is (491-436)/436 * 100% = 12.6% error. The green wavelength reading was 551 when you expected 546; those two are much closer (0.915% error), which indicates that the equipment you're using has some chromatic aberration. That's understandable in an instrument designed to "fracture" light into a prismatic spectrum; other light-sensing equipment, like cameras, generally try to minimize it so all the wavelengths emanating from one point on the subject pass through the lens to expose one point on the film or photocell. But I digress.
Your violet spectral line reading off of an ordinary fluorescent lamp is 493nm. Given the percentage error measures on this end of the scale, this reading is about 12.6% too high; so, we divide by 112.6% to get the actual measured wavelength of 437.8. That measurement, against the known emission spectra of mercury vapor, has a percentage error of only (437.8-436)/436 * 100% = 0.42%, which for a measurement made with classroom lab equipment and other less-than-ideal conditions is pretty darn close.


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Equilibrium constant expression for heterogeneous equilibrium
Consider this reaction:
$$\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$$
for this, $K_c=\left[\ce{CO2(g)}\right]$
For a given temperature, $K_c$ will remain fixed. That means, the  amount of $\ce{CO2}$ released will be same, irrespective of amount of $\ce{CaCO3}$ taken.
But, that means, at the same temperature, irrespective of  whether we start with 1 g of $\ce{CaCO3}$, or 10 kg of $\ce{CaCO3}$, the amount of $\ce{CO2}$ at equilibrium will be same(!) because, according to the $K_c$ expression, it should be constant.
This is against my intuition, because, it seems that if we start with 10 kg of $\ce{CaCO3}$, we should end up with a much larger amount of $\ce{CO2}$ at equilibrium than if we started with just 1 g of the carbonate.
Am I being mistaken somewhere, or is this is what will actually happen? I am confused.

Try putting it in terms of pressure ($K_p$) instead of concentration since $CO_2$ is a gas. So at equilibrium, the partial pressure of $CO_2$ just above the calcium carbonate would be constant for a given temp like you've mentioned. See http://en.wikipedia.org/wiki/Calcium_carbonate#Calcination_equilibrium for more info on this specific reaction.

@kaliaden thanks for the link. But can you please try answering my question directly? the wiki page doesn't mention anything about my doubt- How can we end up with the same partial pressure of $CO_2$ for 10 kg as well as 1 gm of $CaCO_3$ ? that seems to defy common sense.

Sorry about the late response. I haven't been on the site for a while now. I wasn't so sure about the answer which was why I had put it up as a comment. @perplexity has answered your question quite well below.

nilanjana :Why you didn't get CaO(s) ,CaCO3(s) for the Kc expression?

First, as kaliaden commented, the true equilibrium constant
$$K(T,p) = a(\textrm{CaO,s}) a(\textrm{CO$_2$,g}) / a(\textrm{CaCO$_3$,s}) \approx
p_\textrm{CO$_2$} / p^\circ \equiv K_p(T)$$
where $a$ are the activities, $p$ the partial pressure and $p^\circ$ is one bar.
Thus $K$ basically only depends on temperature (unless you are interested in brutally large pressures, and we safely disregard the sublimation of CaCO$_3$ and the effect of the air). Obviously you can obtain
$$K_c(T) \approx K_p(T)/ RT = [\textrm{CO$_2$}]$$
Now your question is: how is it that this constant does not depend on the amount of matter that we have?
The fact is that as long as there is enough matter to reach the equilibrium (to saturate the air with carbon dioxide vapor) it doesn't matter how much CaCO$_3$ you start with. The reaction will proceed until reaching an equilibrium characterized by $K(T) \approx p_\textrm{CO$_2$} / p^\circ$. Obviously if you don't have enough CaCO$_3$ (for instance, a few micrograms) then everything will react and you will not reach the equilibrium.

If I may, the moral of the story is that thermodynamics is not useless.  One can learn counterintuitive things from it!


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Best way to dissolve and consume cysteine HCl monohydrate powder
Does cysteine break down in filtered water?  Should this made more acidic?
If I mix cysteine powder in juice (something acidic maybe), does it remain at cysteine?  It seems that it can oxidize into cystine and lose some of its qualities.

I see an article titled Cysteine stability in aqueous amino acid solutions.  Looks like that might do the trick.  It goes on to say Optimal conditions for cysteine stabilization in aqueous solutions of amino acids, available for parenteral protein feeding, were found.  It is in Rushan; maybe I can use a translator.  I'm having trouble finding where to download the article.  I see only the abstract.  http://www.ncbi.nlm.nih.gov/pubmed/6431404

did you get the article? I am looking for it but I can't find it.

I wouldn't worry at all about the oxidative dimerization of cysteine to cystine when you consume freshly prepared solutions.
In the lab, you'd use oxidants, such as $\ce{FeCl3}$ (in the presence of air) or hydrogen peroxide to achieve this reaction; both of which you're not likely to add to your amino acid diet!
As far as the article which you mentioned is concerned: The stabilization methods in question will only be relevant to maintain the composition of prefabricated and sterile-packed solutions over a long period of time.


text:
Degeneracy of orbitals?
Why is that in an external magnetic field(uniform) the degeneracy of d,f orbitals is lost but the degeneracy of p orbitals remain intact if the main cause of losing degeneracy is the difference in energy of differently oriented orbitals in the magnetic field?

Do you have a source stating the degeneracy of p orbitals in the presence of an external magnetic field?

My textbook states that the energy of the p orbitals will split in a magnetic field. (Physical Chemistry, McQuarrie, p.210)

On a sidenote, do you recommend that book? I've been looking at it.

I do like this textbook - it's fairly straightforward and not too bad to understand. However, I haven't used other pchem textbooks in depth, so I'm not sure if this is the best out there. Heard the Levine text was good as well.

I used to teach from McQuarrie, it was heavy! but definitely the best out there.

Any energy that belong to orbitals with non-zero angular moment (ie. anything, except s orbitals) will split in magnetic field.

Can you provide more details?

Check the Zeeman-effect (http://en.wikipedia.org/wiki/Zeeman_effect) in your preferred textbook. It describes how magnetic field interacts with atoms (molecules etc). The electronic part of it generally can be factored to an angular and a spin contribution. Sometimes they are strongly couples (e.g. lanthanides) and it doesn't work, but at basic level we don't care about those cases. From that it comes that angular contribution exist always when there is a non-zero angular moment.


text:
Why Inductive effect operates only through single bonds?
I am not able to find any examples of the inductive effect operating through multiple bonds. Why can't electron displacement take place in multiple bonds?

it might be beneficial if you could post an example of two systems that you are comparing, or if you are talking about some "process" or context in particular.

@EricBrown I have just started studying Organic chemistry , I am confused why electron displacement doesn't occur in bonds like C=O .

@VivekKhandewal The physics of "single bonds" and "double bonds" are the same, and so both would experience an "inductive effect" for whatever that term means. It will take me some time to come up with a comprehensive answer to a very good question.

http://en.wikipedia.org/wiki/Inductive_effect

The argument is false.  Inductive effects happen through any covalent bond.
It is true that as far as systems with multiple bonds are concerned, conjugation, hyper conjugation, or resonance effects are often more significant than inductive effects.  These effect may reach a much larger distance and have richer chemistry, and frankly are more interesting. As a result, text books focus mainly on these effects in double bonded systems. Presence of a myriad of strong effects also makes it harder to identify the role of induction in system with multiple bonds.
However, this by no means eliminates the presence of inductive effect and in fact induction often do have significant contribution to reactivity in such systems.
The best example might be fluorobenzene (Ph-F).  Electrophilic substitutions of fluorobenzene is significantly slower than those of benzene, and almost predominantly the para- product is obtained.  This is caused by the induction of F atom.  This can be attributed solely to inductive effects because the conjugation effect of F atom is electron donating and would cause an increase in reactivity instead, and would cause near statistical distribution between para and ortho substitution products(which is the case in chlorobenzene and bromobenzene).
In fact inductive effect also occur through C=O.  You could not find an example only because indicator of electron charge on the oxygen atom of C=O is harder to come by.  The examples do exist.  For example, pKa of cation acid $\textrm{R2C=O-H}^+{\longleftrightarrow}\textrm{R2C=O + H}^+$ is strongly influenced by substitutions on the R groups.

If I'm not mistaken a good example of the inductive effect through multiple bonds is trifluoro-acetic-acid, or $\ce{F3CCOOH}$. The proton is so electron deficient that it is easily abstracted by anything that can act as a base. The not 'fluorinated' equivalent, acetic acid, is not nearly as strong an acid.
The reason for the difference in acidity, is the difference in electro negativity between hydrogen atoms and fluorine. Due to the very large electro-negativity of fluorine atoms the electrons in the acid protons are 'pulled' towards it, leading to a clear example of the inductive effect over multiple bonds.
A comment on one of the comments below the question: Electron displacement does occur in double bonds. A good example is the carbonyl functional group (C=O) in a molecule. The inductive effect results in a slightly more positive charge on the Carbon atom and slightly higher at the O-atom. This is also due to electro-negativity and results in the reactivity of the C=O functional group.
Hope this helped!

The example using F3CCOOH is not pertinent, it is not about inductive effect propagation through a double bond. Well it does on the conj base, but then there is not really a double bond.


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Partial pressure of water in a submerged balloon

A balloon is half filled with water at sea level ($P = \pu{1 atm}$, $T = \pu{298K}$) such that the partial pressure of water is $\pu{24 mmHg}$.  It is then submerged to a point where $P = \pu{2 atm}, T = \pu{298K}$.  What will be the vapor pressure of water at that point?

My first instinct is to find the mole fraction of water in the balloon.  Using Dalton's law,
$P_{total} = P_{\ce{H2O}} + P_{\mathrm{atm}} = 24 + 760 = \pu{784 mmHg}$
$P_{\ce{H2O}} = X_{\ce{H2O}}\,P_{total}$
Solving for $X_{\ce{H2O}}$, I get $24/784$.  Then I use that to find the new partial pressure at the new depth, $(24/784)\cdot 1520$.  But the answer seems to be that the partial pressure of water remains constant. Can someone explain why the partial pressure will be constant?

Excellent question. On the one hand, the vapor pressure of water (or any other volatile liquid) is a function of the temperature alone, but on the other, if you assume that the mole fraction is constant you find that the vapor pressure of the water changes even though the temperature is constant.
The short answer
The water vapor pressure stays constant because as the pressure in the balloon increases, the mole fraction of the water reduces proportionally to maintain the equilibrium it has established.

The longer answer
Let's dig into the former approach I mentioned in the introduction first. We can begin with the Clausius-Clapeyron relationship.
The Clausius-Clapeyron relationship (often is simplified to):
$$\ln \left(P_{vapor}\right) = -\frac{L}{R}\left(\frac{1}{T}\right) + C$$
Where $L$ and $C$ are constants and $R$ is the ideal gas constant.
Using this model, we would expect the vapor pressure of the water to remain constant and therefore the mole fraction of water in the gas phase to decrease.
Dalton's law tells us: $$X_{water}=\frac{n_{water}}{n_{tot}} = \frac{P_{water}}{P_{tot}}$$
Using this expression we can see that, as the total pressure of the gas increases and the vapor pressure of the water stays constant, the mole fraction of water in the gas phase must go down. If we assume, as a first approximation, that the total number of gas molecules in the gas phase is constant, then a reduction in mole fraction can only be accomplished by some of the molecules of water in the gas phase being reabsorbed into the liquid water phase. This consequence is in agreement with the Le Chatelier principle prediction. A decrease in volume available to the water molecules will shift the equilibrium below to the left:
$$H_2O\,(l) \rightleftharpoons H_2O\,(g)$$
Thank you for such a rich question! How interesting that there are so many topics it relates too! Excellent!

Other thoughts

When we calculate the initial pressure of the other gases in the volume, we should use:

$$P_{tot} = P_{water} + P_{all\,other\,gas}$$
$$P_{tot} = P_{external} = 760\,mm\,Hg$$
$$P_{all\,other\,gas} = P_{tot} - P_{water} = 760\,mm\,Hg - 24\,mm\,Hg = 736\,mm \,Hg$$

The dissolved gas in the water in the balloon will also increase at higher pressures. How will this affect our calculations?
We have not considered energy concepts explicitly in this argument about the pressures being equal (in the first bullet above), but we could have considered the elastic energy stored in the bonds of the balloon. How could we include this to get a better estimate of the pressure of the other gases in the balloon?

+1 for this terrific answer! And welcome to chemistry.SE :)

Thanks michielm! I am so excited that this resource is available to the chemistry community! :)


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Consumer Report for Lab Equipment
I'm just wondering if anybody here knows of a Consumer-Report-like publication for lab equipment. Something that would give the specifics, as, for example, the flow rate of a distilling condenser, etc.

This is an interesting question, but the specific example you mention depends more on factors you control than on the condenser: how much heat you give it, what solvent you are distilling, what is the external pressure, are you distilling a compound that is relatively pure or is it part of a complex mixture, etc.

My advisor tells us to always ask the sales rep for references, other people who purchase and used that equipment. You can ask them how it works. If they won't give a reference, don't buy it. This stuff is mostly learned through experience and talking with other scientists.


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Prediction of surface atoms in molecule from its graph
If we are given a molecular graph i.e. al the atoms involved and their connectivity, how can we make reasonable prediction if the atom would lie in the surface or in the inner part. On way might be check it degree (how many atoms it is bonded to), higher the valency higher it surface accessibility.
I will most of the time have small molecules (drug molecules) which have in average 50 atoms.
My broader interest is to find out which atoms directly affect the activity of the molecule more than the other. Probably, atoms lying at surface have more role to play. Any other suggestions aimed at this identification would be helpful.

Just to clarify: do you have an approximate geometry, as well, or not? (You may be able to use some sort of van der Waals radius space-filling method.)

I want to use only graph topology and no 3D geometry.

The Quantum Theory of Atoms in Molecules (QTAIM) provides a fairly unambiguous answer. First, I would like demonstrate the concept of an Atomic Basin. A region of space belongs to a given atom if the steepest ascent path through the electron density terminates at that atom.  Consider a plane that contains the nuclei of the water molecule:

and the atomic basins could also be represented in 3D:

These atomic basins are the QTAIM definitions of ATOMS in MOLECULES. Note that there are gradient paths for each atom that extend to infinity, hence each atom could be considered an "external atom."  (This is a quite trivial case, of course.)
Now consider formamide in this configuration:

The atomic basins of form amide look like:

You can see that the carbon atom is fairly "inaccessible" in the plane of the molecule, but it does have some extent "peeking" out of the plane of the molecule (here, the maroon-colored basin, sorry for change-up in color):

This is a completely general concept, and so can be extended to any scenario. Here is a molecular graph of thiol on a silver cluster (meant to mimic a surface) (note the bond paths unambiguously show which atoms are bonded to what, even for C-Ag, S-Ag, H-Ag interactions where these "cooked-up" rules of hybridized bonding and "valency" are totally out of their league):

Plotting the atomic basins is just an exercise:

The final answer: you can define "interior atoms" as QTAIM atomic basins that do not have steepest ascent trajectories starting at infinity that terminate at that atom, such as an interior silver atom in the above example. If you are conceptually separating a system into a substrate and an adsorbate, then you could say "exterior" atoms of the substrate include the ones that have bond paths connecting the two.
I must caution that as two or more atoms approach each other, their atomic basins change their shape.  (Of course they do!)  And so one must consider the possibility that atomic basins of "exterior" atoms move out of the way and lead to exposure of what was, in isolation, an interior atom.
(PS:  the atomic basins of peripheral atoms actually extend to infinity.  Here, I have applied cutoffs so that one can actually plot the molecule in a finite region of space.]

I appreciate your effort. This is interesting. However, in my case I want simple (low computing time ) method to given general (it need not be accurate) guidelines on which atoms would be inaccessible. Using this qtaim analysis, can we develop some simple method to do that using only graph (i.e not using actual quantum calculation but using idea motivated by them), its ok if that method has some exceptions.

I did these on my laptop. :-)   You might try a HF/3-21G* calculation, which is very easy. Also, I try to confine myself to physics rather than heuristics.  It's been my experience that every time we try to do something "on-the-cheap" it just leads to more problems than the fundamental solution.

Said another way, I think that you would find yourself spending more time developing "rules" than you would performing even a coarse electronic structure theory calculation.

what software do you use to make those plot? Can you please give link to tutorial introduction to this.

I use Mathematica to generate my plots and perform QTAIM.  This gives me a lot of control.  My current setup is "an expert system" because I use it for quite intricate and pedagogical things. You may wish to try AIMALL--it was written by Todd Keith who is one of the founders of QTAIM. (I've never used it, so I can't really endorse it for this purpose.)


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Michaelis Menten derivation for 2 enzyme substrates
We know the Michaelis Menten derivation for the following reaction:
$$\ce{E + S <=> ES -> E + P}$$
However, what if the reaction took place in a different scenario whereby:
$$\ce{E + S <=> ES1 -> ES2 -> E + P}$$
What would the derived Michaelis Menten equation be now?

Welcome to Chemistry.SE. This is a good question, but it seems like it is phrased like a textbook or homework question. It can become a great question if you shared with us what you have tried and where you are stuck. You will get a better answer that way. I am tagging it as [tag:homework], which means that answers will tell you how to solve the problem, but not just give you the answer. See our [faq] for more information.

I will propose an outline for determining an expression for the rate of substrate conversion in the given case:

Set up the reaction with rate constants, assuming $k_{-2}\approx k_{-3}\approx0$: $$\ce{E + S <=>[k_1][k_{-1}] ES_1->[k_2] ES_2 ->[k_3] E + P}$$
Set up the differential equations describing the reaction, i.e. the rate of change for each component with time. The rate of substrate change, for example, will be $\frac{d[\ce{S}]}{dt}=-k_{1}[\ce{E}][\ce{S}] +k_{-1}[\ce{ES_1}]$.
Choose initial conditions and set up two equations for conservation of mass. For example, the initial concentration of enzyme must equal the sum of the concentrations of E, ES1 and ES2.
Make the pseudo-steady-state assumption (PSSA): assume that the concentrations of the intermediate complexes do not change on the time-scale of product formation, i.e. $\frac{d[\ce{ES_1}]}{dt}\approx \frac{d[\ce{ES_2}]}{dt}\approx0$.
Solve for $-r_S$, the negative rate of substrate conversion, obtaining the Michaelis-Menten expression describing the kinetics of the given situtation.


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How To Obtain A Reducing Sugar At Home?
I'm looking for a reducing sugar such as glucose or maltose (any reducing sugar will do) for some home chemistry experiments. Is there a readily available reducing sugar that I might source without contacting a chemical company? A simple "synthesis" of glucose from sucrose would work as well.

Glucose can be found in drug stores and does not require a recipe in some countries.

As permeakra mentions, glucose can be purchased pretty easily. It might be labeled as "dextrose".
Another commonly available is maltose (or malt sugar).
Sucrose (or table sugar) is not a reducing sugar, but one of its constituents (glucose) is. Glucose can be made from sucrose by heating with a little bit of acid to make invert sugar syrup. Invert sugar is approximately a 1:1 ratio of glucose to fructose, and only the glucose in this mixture is reducing. An example procedure can be found here:How does adding lemon juice to sugar make better caramel?. Honey is a naturally occurring invert sugar.


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How do octahedral seed crystals yield polyhedral Cu₂O crystals?
In an article I am trying to understand, the following occurs: (summarized below)

Synthesis of octahedral $\ce{Cu_2O}$ seeds:In a typical synthesis, 2.9946 g of $\ce{Cu(CH_3COO)_2}$ was dissolved in deionized water (20 mL) using a beaker under a constant stirring at 70°C for 2 min. A dark precipitate was produced when a sodium hydroxide solution (9 M, 10 mL) was added dropwise to the above solution. After being stirred for 5 min, D-(+)-glucose powder (0.3 g) was added into the dark precursor with a constant stirring for another 60 min at 70°C,and then was allowed to cool to room temperature naturally. Afterward, the obtained products were centrifuged at 5000 rpm for 1min (XIANYI TG16-WS centrifuge). The precipitates were centrifuged twice more in deionized water and anhydrous ethanol, respectively. And finally were dried at 70°C for 12 hours in a vacuum oven.

My understanding of this paragraph is that copper acetate reacted with sodium hydroxide to form copper hydroxide and sodium acetate. A reducing sugar reduces the copper hydroxide to $\ce{Cu_2O}$. The ethanol dissolves the sodium acetate to leave only $\ce{Cu_2O}$ (octahedral) crystals.
The next morphology of $\ce{Cu_2O}$ crystals, "polyhedral M-pyramid $\ce{Cu_2O}$ architecture," was grown through essentially the same process, except using seed crystals from the first batch of octahedral crystals.
How do octahedral seed crystals give rise polyhedral ones?

It would seem that by very slight changes in conditions they can get the crystals to grow either at 8 corners, giving a cube, or 6 corners, giving an octahedron, or even somewhere in between, giving 14 facets. How they get 50 facets is beyond me. That would be 4x(8)+3x(6)=32+18=50 I guess. Iron pyrites is cubic and has several different growth habits, which can lead to stripy-faced crystals. This article explains that face-controlled and diffusion-controlled growth give different forms, but I'm having problems linking it correctly: www.researchgate.net/...pyrite.../0fcfd50b7c008b64e9000000


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What is octanamide used for?
I read a paper about enzyme-catalysed synthesis of octanamide.
What is octanamide used for?

Well, one thing is as plastics. Nylon is a linear polyamide; it's traditionally produced from alkyl-diamines and a dicarboxylic acid (to donate the acyl group of the amides, as well as a second alkyl chain joined directly to the carboxyl), and the typical product is a poly-hexanamide, but nylon-8,x variants are also possible, as are reactions involving a more direct route of welding the monomers together and arriving at the same place.
Some Google-fu turns up a number of possible end products for which octanamide aka caprylamide would be a useful intermediate. By itself, it's an emulsifier and common solvent in organic synthesis. Other alkyl amides including derivatives of octanamide, often with a few additional alkyl or amino groups welded on, are used in drugs, inks/dyes, fragrances, and skin care items. N,N-dimethyloctanamide and N,N-diethyloctanamide appear to be common ingredients in deodorants and sunblocks, though I'm not sure if they're an actual fragrance or a carrier/stabilizer (Edit: they seem to be emulsifiers and stabilizers for essential oils of the fragrances in water-based products).

Good overview, +1 if you can provide me with the reference for the imidazole synthesis from butanamide.

Grr, I had it when I wrote it. The closest I can find now is that 2-aminobutanamide is an intermediate in the synthesis of Levetiracetam (Keppra, antiepileptic) but that's only a pyrrole compound and not put together from the intermediate the same way I saw the imidazole synthesis.

This is the closest I can find: http://ccc.chem.pitt.edu/wipf/current%20literature/mike_2.pdf - page 12 shows the reaction of an amidine (which exchanges the oxygen of the carboxyl of the amide to an amine group) with a haloether to form imidazoles. Butyramidine is a possible candidate but it's not quite the same as butyramide (or octanamide FTM), and I can't find a procedure for synthesis of amidines directly from amides; most procedures will get you one or the other from the same starting material (often the carboxylic acid) depending on variations in workup.

The first part on polyamides is plain nonsense, there is no polyamide from a monoamide. Second Nylon 6,6 is not produced any longer, all "6" polyamid is Nylon 6, called Perlon correctly. The switch was due to technoeconomic reasons.


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Calculating volumes of acid/base required to make buffer using pH and pKa
How would we use our $\mathrm{p}K_{\mathrm{a}}$ of the acid component of the buffer to calculate $\large \frac{[\ce{A-}]}{[\ce{HA}]}$?
I know $$\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)$$ rearranging gives me $$\log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)=\mathrm{pH} - \mathrm{p}K_{\mathrm{a}}$$
if for example the $\mathrm{pH}=4.5$, $\mathrm{p}K_{\mathrm{a}}=3.74$,
Is it correct to say
$$\dfrac{[\ce{A-}]}{[\ce{HA}]}=\dfrac{19}{250}$$
since
$$
\begin{align}
\log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)&=\mathrm{pH} - \mathrm{p}K_{\mathrm{a}}F\\
\implies\log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)&=4.5- 3.74\\
\implies\frac{[\ce{A-}]}{[\ce{HA}]}&=\frac{\operatorname{e}^{0.76}}{10}
\end{align}
$$
if $\dfrac{[\ce{A-}]}{[\ce{HA}]}=\dfrac{V_\text{b}}{V_\text{a}}= \dfrac{19}{250}$
how would I use this to find the volumes of acid and base needed to make my buffer?

First of all, note that:
$$\log_{10}\left(\dfrac{\ce{[A^{-}]}}{[\ce{HA}]}\right)=4.5- 3.74$$
$$\left(\dfrac{\ce{[A^{-}]}}{[\ce{HA}]}\right)\neq\dfrac{e^{0.76}}{10}$$
but
$$\left(\dfrac{\ce{[A^{-}]}}{[\ce{HA}]}\right)= 10^{0.76}$$
This said, the volume will depend on the concentration of the acid and of the salt (if it's a solution) that you are provided with. This ratio will give you the relative concentrations required to get a buffer of the pH 4.5.


text:
Is perchloride the same as perchlorate ion?
I read about perchloride in a talk and after trying to search for its chemical formula, the search results imply that perchloride is actually named perchlorate. Is this correct?

Short answer: Yes, sometimes "perchlorate" can be called "perchloride" in older literature, but because of the classical definition of a "perchloride", it's so confusing and technically wrong that the $\ce{ClO_4^-}$ anion should be called either "perchlorate" or "chlorate(VII)".
In older naming systems, pre-IUPAC, a "per-ide" is quite simply "any substance containing an unusually high proportion of the named element". So, a "perchloride" is any substance with more chlorine than is "normal" for that mixture of elements. This name was typically given to metal halide salts where the metal has multiple possible oxidation states; the highest oxidation state of the metal forms the metal "per-X-ide". for instance, titanium perchloride aka titanium tetrachloride, $\ce{TiCl_4}$ (IUPAC name titanium(IV) chloride), as differentiated from the trichloride and dichloride salts of titanium, and iron perchloride aka ferric chloride (that name's rarely used as iron only has two stable chlorides, and so the traditional "ferric" and "ferrous" descriptors work fine).
However, the perchlorate anion isn't actually a "perchloride" by that definition; it's more a "peroxide", as there's only one chlorine (which is relatively "normal") that has been "superoxidized" by four oxygens (not normal).
This use of "perchloride" as a synonym for "perchlorate" is more the result of a bad simplification of the traditional naming system, where salts of "-ic" acids are "-ides". While the rule follows for the hydrohalic acids (HF, HCl, HBr, HI) and most oxoacids (which, when they end in "-ous", form "-ites", as in hypofluorous acid, chlorous acid and sulfurous acid), the highly oxygenated oxoacids like chloric, perchloric, sulfuric, nitric, phosphoric and boric acids form "-ates", not "-ides", to mirror the low-oxygen "-ous" acids and their "-ite" anions/salts.
When not talking about the perchlorate anion, you sometimes hear about organic molecules that have had all their hydrogens replaced by halogens. Perchlorocarbons and perfluorocarbons are the chlorine and fluorine derivatives of hydrocarbons (for instance, the dry-cleaning chemical perchloroethylene, "Perc", and the perfluoroethylene monomer of polytetrafluoroethylene, aka Teflon).
"Peroxide" has found a home in the IUPAC system as the official name of the $\ce{(O_2)^{2-}}$ anion, composed of two single-bonded oxygens (often as the related "hydroperoxide", which is an OOH functional group bound to a carbon or to hydrogen). However, because each chlorine anion in a "perchloride" is typically bound to whatever the chlorine is oxidizing, the multiple chlorides are not a single anion and so aren't named as such in IUPAC.

-ide means the lowest oxidation state. For $\ce{Cl}$, if it gains 1 e$^-$, oxidation state is -1 (i.e. $\ce{Cl-}$); in case of $\ce{S}$ (16th group) it's 2 e$^-$ (i.e. $\ce{S^{2-}}$)
per- means higher than usual oxidation state. For $\ce{Cl}$ it's +7 ($\ce{ClO4-}$); for $\ce{S}$ look at compounds with O-O bond (peroxomonosulfate $\ce{SO5^{2-}}$)
So per*ide like your perchloride is an oxymoron.
Compare http://en.wikipedia.org/wiki/Chlorite

per- -ide

Refers, as far as I know, to double anions. For example: $\ce{HgCl2, H2O2}$ are named "Mercury dichloride" and "hydrogen peroxide" respectively. I believe that "per- -ide" has been dropped for most cations (e.g. $\ce{HgCl2}$ was previously called "mercury perchloride"). "Peroxide" is the only one I've still see a lot around.
On the other hand,

per- -ate

Refers, as far as I know, to anions derived from 'per- -ic' acids. Perchloric acid is $\ce{HClO4}$ and perchlorate is $\ce{ClO4-}$
This system of naming isn't so common now. We generally refer to 'chlorate' or 'chloride' and indicate the oxidation number.
E.g. $\ce{NaClO4}$ is now often referred to as "sodium chlorate (VII)" instead of "sodium perchlorate".
You might want to look a bit into the wikipedia article.

"Peroxide" stuck around because IUPAC adopted the name as the official term for the $\ce{O_2^2-}$ anion.

Perchloride is a word that has been used 100 years ago in the expression Iron perchloride to describe the Iron(III) chloride $FeCl_3$, so as to avoid any risk of confusion with the more usual Iron(II) chloride $FeCl_2$ which is obtained in the reaction of Hydrochloric acid with metallic iron. This prefix shows that there is more $Cl$ atom in $FeCl_3$ than in $FeCl_2$.
The prefix per- is used for translating more. Today it is only used in expressions like perchloric acid or perchlorate ion. It points out that the Chlorine atom is more oxidized in these substances than in any other chlorine compounds.
Today nobody uses the prefix per- before chloride any more when speaking of $FeCl_3$, because it is not the $Cl$ atom that is oxidized, but the $Fe$ atom.


text:
Reduction of carboxylic acids by LiAlH4
I've read that when a carboxylic acid reacts with $\ce{LiAlH4}$ the corresponding alcohol is formed:

But when I try to think of the mechanism, I get stuck here:
$\ce{LiAlH4}$ produces $\ce{H-}$. Since $\ce{H-}$ is a strong base it should immediately abstract a proton from the carboxylic acid to give the corresponding carboxylate ion (just like in the reaction of carboxylic acids with Grignard reagents), instead of undergoing nucleophilic addition to give the alcohol.
How do I resolve this?

I am curious to know where you read that reduction of the carboxylate to form two alcohols. Do you have a link? The other oxygen becomes attached to the aluminum species as shown in Jerry's answer. Now, a carboxylic acid anhydride is a different beast altogether. Anhydrides will reduce to a pair of alcohols.

The $\ce{Li^+}$ ion substitutes the $\ce{OH}$ in $\ce{COOH}$ first. The $\ce{C=O}$ is then reduced by $\ce{AlH_3}$ to give an aldehyde. See below:

Another $\ce{LiAlH_4}$ then reduces the aldehyde to alcohol:

I found the pictures here.

$\ce{LiAlH4}$ is such a powerful reducing agent that it will attack carboxylate anions (which are resistant to most other nucleophiles).


text:
The reasoning behind predominance diagrams
Consider a one dimensional predominance diagram for a simple acid-base reaction:
$$\ce{AH <=> A^- + H^+}$$
With equilibrium constant:
$$K_a = \frac{[A^-]_{eq}[H^+]_{eq}}{[AH]_{eq}}$$
If I understand correctly, $[H^+] > K_a$, then $[AH]>[A^-]$. At first I thought I understood why this was so: the $[H^+]$ cancels out on both sides of the inequality, and then you bring $[AH]$ to the other side. Only now I realized that you can't actually cancel $[H^+]$ because on one side of the inequality it's actually $[H^+]_{eq}$, not the current amount of protons but the amount at equilibrium. So now I'm lost. What justifies this rule for determining predominance?

I think you might be making a fundamental mistake when you differentiate between the "current amount of protons" and the "amount at equilibrium." All dissociation constants are based on equilibrium concentrations, and predominance diagrams refer to the concentrations present only after equilibrium has already been established.
Consider the concrete example of the dissociation of ammonium: $\ce{NH4+ <=> NH3 + H+}$. The equilibrium constant expression for this chemical process is:
$$
K_\text{a} = \frac{\ce{[NH3][H+]}}{\ce{[NH4+]}}
$$
The $\mathrm{p}K_\text{a}$ of this process has been experimentally determined to be approximately $9.25$. Therefore, the following equation should hold:
$$
-\log(K_\text{a}) = -\log\frac{\ce{[NH3][H+]}}{\ce{[NH4+]}} = \mathrm{p}K_\text{a} = 9.25
$$
If a given solution has a measured $\mathrm{pH}$ precisely equal to the $\mathrm{p}K_\text{a}$ of the above process, then what we have is the following (keeping in mind that $\ce{[H+]} = 10^{-\mathrm{pH}}$):
$$
\log\frac{[\ce{NH3}]\cdot 10^{-9.25}}{[\ce{NH4+}]} = -9.25
$$
Taking the anti-log of both sides of the equation:
$$
\frac{[\ce{NH3}]\cdot 10^{-9.25}}{[\ce{NH4+}]} = 10^{-9.25}
$$
Then dividing both sides of the equation by $10^{-9.25}$, we find:
$$
\frac{[\ce{NH3}]}{[\ce{NH4+}]} = 1
$$
That is to say, the concentrations of ammonium and ammonia in the solution are perfectly equal when the $\mathrm{pH}$ and $\mathrm{p}K_\text{a}$ are equal. If the $\mathrm{pH}$ is higher (that is to say, more basic), then you'll find that your conjugate base (in this case, $\ce{NH3}$) will predominate by factor of $10^{\mathrm{pH} - \mathrm{p}K_\text{a}}$. Conversely, if the $\mathrm{p}K_\text{a}$ is higher, then the conjugate acid ($\ce{NH4+}$) will predominate by a factor of $10^{\mathrm{p}K_\text{a} - \mathrm{pH}}$.
If you prefer to use an inequality and some generic conjugate acid/base pair $\ce{HA/A- }$, then the logic goes as follows.
If $\mathrm{pH} > \mathrm{p}K_\text{a}$:
$$
\mathrm{pH} > \mathrm{p}K_\text{a}\\
-\log\ce{[H+]} > -\log\frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\
\ce{[H+]} < \frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\
1 < \frac{\ce{[A- ]}}{\ce{[HA]}}\\
\ce{[HA]} < \ce{[A- ]}
$$
If $\mathrm{pH} = \mathrm{p}K_\text{a}$:
$$
\mathrm{pH} = \mathrm{p}K_\text{a}\\
-\log\ce{[H+]} = -\log{\frac{\ce{[H+][A- ]}}{\ce{[HA]}}}\\
\ce{[H+]} = \frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\
1 = \frac{\ce{[A- ]}}{\ce{[HA]}}\\
\ce{[HA]} = \ce{[A- ]}
$$
If $\mathrm{pH} < \mathrm{p}K_\text{a}$:
$$
\mathrm{pH} < \mathrm{p}K_\text{a}\\
-\log\ce{[H+]} < -\log\frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\
\ce{[H+]} > \frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\
1 > \frac{\ce{[A- ]}}{\ce{[HA]}}\\
\ce{[HA]} > \ce{[A- ]}
$$

If you rearrange the equation to $$K_a[HA] = [A^-][H^+]$$ the relationship becomes more obvious. If $[H^+]>K_a$, then $[HA]>[A^-]$ in order to keep the two products equal. otherwise $[H^+]>K_a$ and $[A^-]>[HA]$, then the relationship would be an inequality: $$[A^-][H^+] > K_a[HA].$$


text:
How much ascorbate to neutralize one liter of chlorinated tap water?
In Las Palmas de Gran Canaria, tap water is chlorinated. In an Internet forum, someone   claims that in 2008 the water had a chlorine concentration of 0,5 - 1mg/l. On the official site of Emalsa, the water provider, I couldn't find any information about chlorine concentration. I tried contacting them online, but the contact form is broken.
In another forum, I read that ascorbic acid (vitamin C) can be used to neutralize the chlorine: My taste buds say it works!
How much ascorbate/l do I need to neutralize the above amount of chlorine?

The first result from a Google search yielded the following website: http://www.fs.fed.us/t-d/pubs/html/05231301/05231301.html, which appears to have precisely the information you're searching for.

@GregE. I suggest you make an answer out of this. In the article, they write: "Approximately 2.5 parts of ascorbic acid are required for neutralizing 1 part chlorine."

I'm not sure that's enough meaningful content for a full answer. But if you'd care to answer your own question using data from that website, I'd have no issue with that.

@GregE. I don't feel qualified to answer this question, and of course, I prefer a detailed answer over just a link to an article.

Have you tried calling your water provider about the average chlorine content? Sometimes picking up the phone is much more likely to get you what you want.

@BenNorris Only my Spanish is not sufficient for a phone call like that.

Here you go: https://chemistry.stackexchange.com/a/116242/79678.

@EdV Thanks! Maybe someone can add that answer here too. I'll accept it.

This is just a very short explanatory note on the article suggested by Greg E..
The reaction between ascorbic acid and hypochlorous acid is as follows:
$$ \ce{C5H5O5CH2OH + HOCl → C5H3O5CH2OH + HCl + H2O}$$
so for 1 mole of hypochlorous acid you need 1 mole of ascorbic acid. This can be recalculated into a mass ratio by the use of their molecular masses:
$$M(\ce{C5H5O5CH2OH})=176\;\text{g/mole}$$
$$M(\ce{HOCl})=52.5\;\text{g/mole}$$
From this we can see that if the molar ratio is $1:1$, the mass ratio is $176:52.5$ which is $3.3:1$ somewhat higher than the article claims.
I am not exactly sure of the reason, it could be that there is some side reaction of ascorbic acid that also removes some additional hypochlorous acid although I wouldn't know which. Perhaps the article is talking about the amount of ascorbic acid you need to get a concentration which you don't taste anymore?

In addition to the first oxidation, $\ce{HOCl}$ can oxidize the alcohol side chain of ascorbic acid to an aldehyde and then an acid. These reactions are slower than the first oxidation, and so the 2.5 mass equivalents probably comes from the extent of overoxidation in a reasonable time. $$\ce{C5H3O5CH2OH + HOCl -> C5H3O5CHO + HCl + H2O}$$ and $$\ce{C5H3O5CHO + HOCl -> C5H3O5CO2H + HCl}$$


text:
Entropy, enthalpy, and reversibility
In my introductory chemistry course, we learned that for reversible processes, entropy is defined as $ΔS = q/T$. However, we did not spend that long on this before moving on to other things like Gibbs free energy. My questions are:
1. What exactly is a reversible process?
2. Why must we use things like standard molar entropies to calculate $\Delta S$ instead of just using $\Delta H$ according to this equation?

Your text will best explain reversibility.  The short answer is that a reversible process is one whose direction can be reversed by a microscopic change in external conditions.
For example a gas in a cylinder expanding against a piston will do so reversibly as long as the external pressure and the internal pressure are essentially identical.  Of course for expansion the internal pressure must be infinitesimally greater than the external pressure.
Next, entropy is really only defined for systems at equilibrium.
Next, the equation for the entropy above is not correct except at constant temperature.  In reality it is $dS = dq/T$, which generally must be integrated in order to determine the entropy by this method.
For the second question, there is some confusion here. $\Delta H$ is equal to $q$ only at constant pressure.
In fact there are lots of ways to compute the entropy change in a process, but this space is too small to go into them.  They should be in your textbook.


text:
Chemical reasons for choosing Salt Water or Chlorinated pool?
I am currently researching as too which is better for maintaining the chemical balance in a swimming pool. I have obviously tried the standard Google search for Salt Water or Chlorinated Pools and have come up with lots of results. The only problem, is that most of these results are based more around peoples opinions or psuedo-scientific observations. So, to the question:
What chemical reasons are there for choosing a salt water pool over a chlorinated pool?
Note: Salt water meaning the chlorine is generated and chlorinated meaning that you add the tablets/pucks etc.

A salt water pool still has hypochlorous acid $(\ce{HOCl})$ as the active component so essentially it CAN be the same thing as a chlorinated pool. For a chlorinated pool the $\ce{HOCl}$ is directly added, whereas for a salt water pool the salt water is electrolysed to create $\ce{Cl2}$ which then reacts with water to form $\ce{HOCl}$.
However, there is indeed often a difference and the explanation is partly chemistry and partly chemical engineering. In terms of chemistry  the overabundance of chloramines $(\ce{NH2Cl})$ in chlorinated pools over salt water pools makes the difference. When hypochlorous acid, the active component in both, comes into contact with ammonia (from sweat, saliva etc) you get the formation of these chloramines:
$$\ce{NH3 + HOCl → NH2Cl + H2O} $$
Chloramines are the major reason for irritated eyes and skin.
Now the difference between salt water and chlorinated pools has a chemical engineering reason. Salt water pools have a constant generation of chlorine as opposed to a 'batch' delivery in the case of chlorinated pools. Therefore the control over the concentration of hypochlorous acid in the water is better, which also allows better control over the concentration of chloramines.
So to sum up: salt water pools allow higher control over the concentration of chloramines and therefore might be better, however, if this control is not properly applied, it could also even be worse!


text:
Do halogens bond more commonly with each other or themselves?
To me it seems like diatomic elements like $\ce{F2}$, $\ce{Cl2}$, etc. occur more prevalently than diatomic molecules with both, like $\ce{ClF}$.
Is this the case, and if so, what causes diatomic elements to be favored?

There's a lot of circumstantial evidence that heterodiatomics are more stable than homodiatomics. It is usually ascribed to the observation that atoms of different "electronegativity" tend to have stronger bonds than atoms of the same electronegativity.
In your particular example, consider the reaction:
$$\ce{Cl2 + F2 -> 2 ClF}$$
I don't have the tables handy, but I calculate with electronic structure theory that this reaction is exothermic by about 25 kcal/mol. That's quite a bit! (The equilibrium lies far to the right.)
This would suggest that a mixture of Chlorine and Fluorine, heated up to very high temperature and then cooled down, would tend to reform as ClF.

Thanks! But why don't we find ClF more frequently then? Do they simply not have enough energy to combine?

@DarkLightA Might be that the activation energy is very high? I can't seem to find it though to get an actual value.

Alright, I guess it might be like ionic triple bonding then? Takes a lot of energy to obtain, but once in the bonding, is hard to break?

I have not seen evidence that $\ce{Cl2}$ and $\ce{F2}$ are more prevalent in nature than $\ce{ClF}$.  They may be synthesized for particular purposes more often, or they may be talked about more often--but that's a totally different matter.

Please see my answer, which addresses the activation energy point.  Heat up very hot (thereby overcoming activation barrier) and then cool down (thus forming thermodynamically-stable product).  Again, I don't know if there is some reaction in nature that preferentially forms diatomic halogens--but I highly doubt there is.  Heterodiatomics are usually more stable than homodiatomics. If you have some basis for assuming homodiatomic halogens are more prevalent in nature, then please cite your evidence/context (e.g. observed more in the atmosphere by some spectroscopic method)


text:
Calculate Ethanol content after heating for N minutes
Does there exist a formula I can use to calculate alcohol content of a solution after heating it for some time?
I have the following use-case:
I am making a pecan pie. I have a thick sugary solution. I am not sure how much the makeup of the solution matters. Just assume its more viscous than water. I add 75 ml of 40%v  alcohol and I heat it at a temperature high enough to cause boiling for 20 minutes. How much alcohol remains at the end of 20 mins.
Obviously there are a lot of variables here that would make it impossible to give a precise answer. I would be interested in an answer that solves an "easier" version of this problem, for example: having 1 liter of water, adding to it 100ml of 100%v alcohol (Ethanol), boiled for 100 minutes, how much alcohol is left?
How quickly does alcohol boil off? Is it dependent on is concentration within a solution? is there a formula for that?

For a pure substance, to be able to calculate how much has boiled off after $N$ minutes you will need the enthalpy of vaporization, $L$ (in kJ/kg) and the heat input, $Q$ (in kJ) which will give you:
$$\tag{1} m=\frac{\epsilon \, Q}{L}$$
where $m$ is the mass that has boiled off and $\epsilon$ is the fraction of the heat input that actually goes into the liquid. $1-\epsilon$ would be the fraction of heat lost to the surroundings.
The very crude story
An extremely crude calculation would be to take the mixture enthalpy of vaporization and assume that only ethanol vaporizes such that the fraction of ethanol can be directly related to the total mass loss. Note that this calculation will give you a lower bound for the time needed to get rid of the ethanol
We can calculate the mixture enthalpy of vaporization by averaging the enthalpies of vaporization of water and ethanol:
$$\tag{2} L_{mixture}=x_{\ce{EtOH}} L_{\ce{EtOH}}+x_{\ce{H2O}}L_{\ce{H2O}} $$
Now we need to figure out the heat input. A precise calculation is impossible here, but we can use some reasonable estimates. According to an estimate on this (dutch) website when we are cooking we use about $164$ l/h of methane. Assuming that that is for cooking with 2 pots this translates to a gas flow rate of $\dot{V}=1.4$ l/min for a single pot. With the heat of combustion of methane, $\Delta H_c$ and the density $\rho$ we can then calculate the amount of heat released per minute:
$$\tag{3} \dot{Q}=\rho \dot{V} \Delta H_c $$
plugging in the values for methane we find $\approx 50$kJ/min
The enthalpy of vaporization of the mixture will constantly change, because the composition of the mixture constantly changes over time during the boiling process. Therefore, instead of simply using Eq. 3 in Eq. 1 to find $\dot{m}$ and multiplying with $N$ we will need to do an integration:
$$\tag{4} m_{boil-off}=\int_0^N \dot{m}\; dt $$
where $\dot{m}=\frac{\epsilon \, \dot{Q}}{L(t)}$. Now this is a bit problematic because $L(t)$ depends on $m_{boil-off}$, but it can be solved numerically (I feel like this should also be possible analytically, but can't figure it out right now, anybody?!).
Your example question (with $\epsilon=0.5$) leads to the following curve for the fraction of ethanol in your mixture:

As you can see the ethanol is gone after about 7 minutes. To do this calculation for your pecan pie mixture you will need to know the heats of vaporization of all your ingredients and the mass of each.
The more precise story
The more precise story is that not only the ethanol, but also part of the water will vaporize therefore slowing down the vaporization rate of ethanol. If you are 'lucky' you could assume that the vapor above your pan is at equilibrium with the boiling liquid. In that case you can use the theory available for distillation columns and something called the Ponchon-Savarit Method (the complete diagram for water-ethanol can be found here), which I will not recap here, because it is basically a book chapters worth of undergrad chemical engineering.
Note that the equilibrium vapor assumption is probably not a very good one due to air flows over the pot which keep pushing the system out of equilibrium. For the evaporation of ethanol this is good news by the way, because it will speed up the process a bit. Also note that with a true equilibrium you would run into the azeotrope of ethanol-water mixtures.

""If you are 'lucky' you could assume that the vapor above your pan is at equilibrium with the boiling liquid."" This is not luck, but obvious! If the liqiuid boils, the vapor comes directly from contact with the liquid, and must have the same temperature. The unknown in the question is the reflux conditiones. Look for fractional distillation.

@Georg - The same temperature yes, heat transfer is fast, but not the equilibrium vapor-liquid composition right? (please explain if I'm wrong here). The VLE is only reached in steady-state conditions which I think are hard to achieve in an open pan with (I presume) the exhaust hood running


text:
Why do we neglect the protons from water's autoprotolysis?
Why don't we need to consider the $\ce{H+}$ ions of water when finding the $\ce{pH}$ of $0.01~\mathrm{M}$ of $\ce{HCl}$ solution?
The $\ce{pH}$ value of something is depending on the molarity of $\ce{H+}$ ions. But the teachers says the calculation is just $-\log(0.01)$, don't we need to consider the $\ce{H+}$ ions of water?

You could consider them, but what would be the result?
$0.01 M$ of $\ce{HCl}$ with $1\times10^{-7} M$ of $\ce{H^+}$ from water gives you $0.0100001 M$ of $H^+$. Find the pH of this concentration of $\ce{H^+}$ and compare with the pH you would obtain if you used $0.01 M$ directly.
You shouldn't see that the resulting pH differ significantly.

Do note, however, that the self-ionization of water technically must be taken into account. Though very often justifiable, it is an approximation to leave it out. Ask yourself, what is the pH of a $10^{-8}\  mol.L^{-1}$ solution of $\ce{HCl}$? It doesn't make sense to add a strong acid to pure water, however dilute, and obtain a basic pH! If you don't rely on the approximate formula, you will find that such a solution has a pH of about 6.98.
Here is a good source to get you going. The more proper way of solving for $\ce{[H^+]_{eq}}$ is based on the solution of a polynomial found from the combination of the charge balance equation and chemical entity balance equations along with equilibria constant definitions, and it is detailed here. Though used specifically to analyse the behaviour of a weak acid solution, the same equations can be used for the strong acid solution, with the added approximation that $\ce{[HA]_{eq}} \approx  0$ (i.e. $\ce{c_{HA}} \approx \ce{[A^{-}]}$).


text:
What are the main axis of research in Chemistry?
I would like to know what are the main problems currently studied in Chemistry.
For some reason, it seems that there is far less vulgarisation in chemistry than the other fields, and it's hard to find an overview of the field accessible to a layman (compared to math or physics for instance).
I only found this wikipedia page, but I have no idea how relevant it is.

Welcome to Chemistry.SE. This question sounds like you are asking for a big list of things, which is generally discouraged since there is unlikely to be a single correct answer. Check out the [faq] for more info.

That being said, the question is a fantastically interesting one in the philosophy of chemistry, and I do not believe that Wikipedia page captures the most important unsolved problems in chemistry.

Even though I voted to close this question as not constructive, these are among the most important unsolved problems in chemistry:

Liquid Water - We still know so little about the molecular level interactions of water molecules with themselves and other species. The great extent of hydrogen bonding networks in liquid water make it hard to model. Just how many water molecules are solvating a proton in acid? Why are organic reactions between hydrophobic molecules accelerated in the presence of water?
Molecular origins of life, especially homochirality. The formose reaction is believed to be the origin of carbohydrates. Amino acids, lipids, and nucleic acids have similar hypothetical origins. However, how did these molecules come together to create living organisms? Why are most (not all) biomolecules present as one enantiomer and not the other?
Protein folding. Can we predict the three-dimensional shape of a protein simply from the order of the amino acids? If so, then we could design proteins to do whatever we wanted.
True designer proteins. Once we solve the folding problem, then we can design proteins with active sites of specific geometries and specific amino acids in those active sites to do whatever reaction we need.
High Temperature superconductors. Although room temperature superconductors would be nice, inexpensive superconductors that operate at liquid nitrogen temperature (77 K) instead of liquid helium (5 K) temperatures would make NMR, MRI, and a host of other important techniques much cheaper.
The Energy Crisis. We need a chemical solution to replace fossil fuels as the world's primary energy source. Hydrogen, fuel cells, biomass, and solar power all have serious drawbacks currently.
Replacement of petrochemical feedstocks. The petroleum industry remains the primary source of organic compounds. What will we do when the petroleum runs out?
Bonding theory. Modern bonding theories have developed into bloated mathematical formalisms that have removed any resemblance of intuition from the study of the structure of molecules. Can a unified modern bonding theory be developed that retains only the best features of valence bond and molecular orbital theories?
Improved ideality and efficiency of synthesis. Can the synthesis of compounds be improved so as to remove wasteful steps (like adding and removing protecting groups) or complex purifications to better adhere to the principles of Green Chemistry?
Structure-Property Relationship. All established relations between structure and activity is empirical, especially when the desired activity is biochemical. Can we design the best drug with the fewest side effects on the first try?

I find some of the unsolved problems on the to be too narrow (stability of norbonyl cation - this is a subset of the structure-activity problem) or trivial (origin of rotation barrier in ethane - the problem is important but the answer is useless).

I would like to read about chemists' approaches to problem 8: making a unified modern bonding theory that retains the scope of VB and MO theory but still has intuition.  Where should I start?

Chemistry is the study of the transformation of matter. People are studying every single aspect of how this happens.
You can't get a simple answer for this question.  For example, if you were to ask, "what are the important aspects of chemistry research?"  You might get back drug design, nanotechnology, etc. etc.  But chemical engineers are cranking out new paints and lacquers (notice that almost everything in our human world is covered by some sort of lacquer.)  Food chemists are working in genetics and flavorings.  Petroleum chemists are working on better cracking and new products.  The list goes on.
Importance is in the eye of the beholder.  Quantification of importance is impossible.  If you want to know what is deemed newsworthy by, e.g. the American Chemical Society, pick up the past issue of Chemical & Engineering news and take a look.

"You can't get a simple answer for this question." It's possible in every other field (Hilbert and others). Why not in Chemistry?

@LSA Chemisty is to e.g. "nanotechnology" as Mathematics is to e.g. "Hilbert space"  You are talking about a radically different scope.  Even within a subfield, there are different areas of focus.  I have some familiarity Hilbert (and Banach) spaces, and folks are working on all sorts of different problems even in those subdisciplines.

I was not talking about Hilbert spaces. I was saying that Hilbert was able to describe to describe the mathematics of the 20th century with his famous list (not perfectly of course, but not that far).

Well, I'm pretty sure the person working on the 24th most important problem thought Hilbert was a complete idiot.

Another thing, you may ask a different question: "Is there a luminary in chemistry who has compiled a list of what he/she thinks are the most important problems in chemistry?"