text:
Why aren't there iron bones and eggs in nature?
As far as I know, most of the "hard things" in living creatures composed of calcium (for example, bones plated with calcium over the keratin structure, and eggshells and seashells with $\ce{CaCO3}$). The calcium ion has an oxidation state of +2, and so does iron.
Why don't we find in nature creatures with bones of iron or eggs with iron shells, given that living creatures do use ionic iron for other purposes like in the heme group of hemoglobin?

iron would get oxidized, whereas calcium wont get oxidized as calcium has a maximum oxidation state of 2+. Iron could get oxidized to its 3+ state, ie, it would rust reducing the strength of the bones.

Structures made form calcium carbonate can be really tough (bones are stronger than concrete)  and are much less dense than iron. Plus calcium is more abundant than iron in food.

@vapid "Plus calcium is more redundant than iron in food" is a circular argument: The reason it is not abundant in food is the very same reason why it is not used in nature.

Iron eggs? I guess little birdy would have a hard time trying to break out of an iron egg, just saying ...

http://biology.stackexchange.com/questions/9419/why-are-there-no-organisms-with-metal-body-parts-like-weapons-bones-and-armou

@ketbra "Food" is an oversimplification, you got that right. I meant dietary sources down the food chain. Water is an important source of minerals to animals, and it usually has much more calcium than iron. Carnivores eat herbivores, herbivores eat plants, and plants get their minerals from soil, which again has more calcium than iron.

@vapid I see ; )

yes, but again the same reason that the soil and water contains more calcium is beacuse of the decadence and wear of  seashells years ago.

One disadvantage of iron carbonate shells is they are unlikely to last in the oxidizing environment of Earth. Even in the laboratory, preventing the iron from converting to rust is problematic. Wikipedia:

Care must be taken to exclude oxygen $\ce{O2}$ from the solutions [to prepare iron(II) carbonate], because the $\ce{Fe^{2+}}$ ion is easily oxidized to $\ce{Fe^{3+}}$, especially at pH above 6.0.[1].

"pH above 6.0" would include our oceans, which we must assume are also aerated. Thus proposed iron carbonate shells would rust out.
Cited Reference

Philip C. Singer and Werner Stumm (1970): "The solubility of ferrous iron in carbonate-bearing waters". Journal of the American Water Works Association, volume 62, issue 3, pages 198-202. https://www.jstor.org/stable/41266171

I don't think the original question was assuming the iron would be in iron carbonate: it was why not metallic iron?

The question mentions ionic iron, and as a comment suggested baby birds might find hatching from metallic shells a bit challenging.

Nature works with what it has not with what it could have
Biological systems are ingenious and often find clever ways to create materials useful to survival. Humans are even more ingenious than other creatures and organised groups of humans have found ways to create new materials not known in nature for the benefit of human survival. But human intelligence can do many things nature cannot by way of collective, coordinated action driven by intelligence and tools not just biology.
The modern world is familiar with many human-invented materials that transcend the limitations of natural materials. Like iron and steel. This has led many to wonder whether it might be better to have things like bones or shells made from those materials. Crabs and molluscs have hard shells to protect them from predators, would't they have better protection if they could make their shells of steel? Humans do this by building tanks, after all. Wouldn't bones made of steel be more resistant to breaks?
But the limitation preventing such adaptation is the availability of the materials to biological systems and the lack of mechanisms to exploit them.
It is worth remembering that humans have only been exploiting iron for just over 3000 years. The iron age only started around 1,200 BC despite coordinated groups of humans having been around for at least 10,000 years before than. The benefits of iron were large and transformed civilisation, but didn't appear quickly. Two factors were critical. People had to find concentrated sources of the right raw materials (minable iron ores) and the technology to forge iron products from them (this is hard as getting iron requires, inter alia, far higher temperatures than earlier metals).
For biological systems to have evolved to be able to exploit iron as a metal (I assume that is the question's intent as there don't seem to be any structural benefits of using known iron minerals as structures) then biology would have to solve the same problems of availability and processing.
And there is no easy way to achieve either access to enough iron to make large structures from it or to create metallic iron from known sources via available biological mechanisms.
Small amounts of iron are used in many living creatures (eg in oxygen carrying proteins in blood). But the amounts are tiny compared the the amounts that would be needed to make metallic iron shells or bones. There just isn't enough iron generally available to make large structures. Humans had to discover concentrated iron ores to get enough for their weapons and tools.
But how could living creatures–even if they could get access to iron in sufficient quantities–turn that into metallic structures inside their bodies? And, even if they could, would iron structures be useful? Pure iron is tough but not that hard. It rusts, easily. Many of the iron-based products we are used to seeing are not pure iron, but complex alloys that do a better job of balancing stiffness, corrosion resistance and other features. Most require deployment of other trace elements in their structures increasing the problems of access to the elements in a living environment and compounding the problems of deriving the resulting material from any available biological process. Humans need to use a lot of high-tech processes to create the common structures made from iron and most of those involve high temperatures that biological systems can't use.
Contrast this with the materials living systems actually use. Many shells are complex structures built from carbonate and phosphate minerals (bone is mostly phosphate not carbonate). But Calcium is common as is the carbon dioxide use in carbonates and the phosphorous in phosphates. And there are plenty of biological mechanisms for acquiring and processing those components and building structures with them at biological conditions.
So nature uses what it has access to and has available mechanisms to process under biological conditions. It can't easily leap to use other harder to access materials and processing mechanisms that are–as far as we know–impossible under biological conditions. And it does a good job as many of the structural materials it manages to generate are, weight or volume-wise, better than the common materials produced by human technology.


text:
How this tricky Wagner-Meerwein rearrangement occurs?
In 1979, Karel Frantisek Wiesner performed a tremondous total synthesis in $59$ steps of the 13-desoxydelphonine. In one of those steps there is a very tricky Wagner-Meervien rearrangement. I tried to find something by myself, I also look on the net, I found this and you can see page $8$ they have the same problem than me.
What is the mechanism of this rearrangement ?

Just to be a bit more clear, the acetate which looks to appear magically obviously don't, it comes from the previous step because of the solvant, they just heat several days after so the solvant is still the same as you can see in the picture below.
If someone has an idea, I would be glad! Thank you in advance :)

Something is missing here.  There's an extra acetate in the product. Is glacial acetic acid the solvent?

Also, if you look at the reaction on page of your link. The structure after this rearrangement is the same as the one after reacting the intermediate with CAN and acetic acid. That's probably where the acetate comes from, but it also means that the structure of the final product is wrong. I recommend looking at the original reference for what the actual product is.

@Zhe if you look carefully the structures are not the same, a methyle is transform as aldehyde with CAN and AcOH.

Oh drat... But we're still missing an acetate. I guess the previous reaction is in acetic acid/acetic anhydride mixture. If they're just taking this pot forward, then I guess that's where the acetate comes from...

My literal back of the envelope solution. Notable points:

I invoked a nonclassical carbocation in the first step of the rearrangement. This was the easiest way to explain the stereochemistry of the new acetate group.
This is a reaction from 1959-1960. It's almost certainly racemic. I think the product as drawn is actually the enantiomer of the one that you would expect from the reactant.

My bet would be the following hydride shift, followed by nucleophilic attack of the acetate:

Though I must admit that Zhe's answer is probably correct. Still weird that nonclassical carbocation. Probably possible because of the +M effect of the OMe group.

I actually thought of something like this to start. But I couldn't get the correct stereochemistry on the acetate group, so it was back to the drawing board...

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text:
How is graphite a covalent solid?
I know it is covalent because there are covalent bonds between atoms of carbon in the same plane. But there are Van der Waals interactions between different planes. Isn't this a feature of molecular solids?

We think of graphite as being a "pure" chemical which it really isn't since the graphite sheets can have different size. But most pure chemicals have a consistent 3D structure, which graphite does not. So graphite is odd in that sense too.

You seem to already know what graphite really is. Don't bother with meaningless categorisation such as the one you ask about.

@MaxW: I see.  $    $

"covalent solid" is of course a very useful category if you compare one with ionic or molecular solids.

Graphite is odd in its categorisation and does by no means rigorously follow the conventions of either a covalent solid, or a molecular solid (for example it can conduct electricity). Graphite has properties of both molecular solids (it is soft), and covalent solids (it has a very high melting point. In fact it will break down before it melts.).
For this reason, the categorisation of some materials, such as graphite, can be somewhat complicated, and the idea of graphite being a covalent solid really just comes from its similarity with other covalent solids.

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text:
Why does ammonia "clean" scratches?
I read that ammonia can be used to clean scratches of certain materials, for example from wedding rings. I was thinking about how this would work but I actually don't know. I suspect that the question I might have to ask is "what is a scratch" and that thus the reason for my misunderstanding comes from some ambiguity in what a scratch might be in the English language.
This is how I would phrase my understanding of a scratch: A scratch on a material is a place on which parts of the surface are removed, creating a "cut" in the surface. Thus there are less particles in one place than the surrounding.
Either way, that is pretty much how I imagine it. If this is the case, why does ammonia "remove" the scratch? Does the ammonia-water solution fill the crevice left and attach itself to the surrounding material, for example the gold. If the ammonia-water solution is transparent you would just see the gold underlying the scratch and it would seem to have "filled" the scratch. But for that to happen the ammonia would have to become solid, which does not happen at room temperatures.
Perhaps the reason why a deep cut can not be "cleaned" with the solution, whilst the small cuts can, is that the word "scratch" is used for two different physical phenomena.
I'm hoping to get a deeper understanding of what is actually happening to the material being cleaned by the water-ammonia solution. Or, perhaps, a link to a paper or other resource explaining what happens.
Links:

https://www.leaf.tv/articles/how-to-get-scratches-out-of-a-gold-ring/
http://www.wikihow.com/Remove-Scratches-from-Glass

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This seems more like folklore than a tested method. However, gold is extremely soft (malleable), and it must be alloyed with other metals to make it hard enough to survive wear as jewelry. Simply rubbing a gold item with a cloth, with or without "ammonia" (ammonium hydroxide, $\ce{NH4OH}$), is enough to polish the gold, i.e. relocate some gold particles to fill in the scratch.
Household ammonia does not solidify, nor will it fill scratches. In fact, it simply evaporates into the air. It is an effective cleaner, though, and would help remove anything adhered to the gold, which might appear like a scratch.
For the same reason, ammonium hydroxide is an effective glass cleaner, but don't expect it to remove any scratches. The suggestion at your second reference to use toothpaste, though, is a good idea, particularly for those brands containing silica.
BTW, "milk of magnesia", $\ce{Mg(OH)2}$, is an effective mildly abrasive scratch remover for some plastics, if rubbed very well with a soft cloth.


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Fast and complete ways to remove chloride from solution (sub ppb levels)
Main question: We are trying to reduce the chloride concentration in a solution to sub-ppb levels. I have to dissect a fruit fly in this solution so I am interested in removing any chloride that is released during that process. Is there a method that will reduce chloride below 1ppb in a timescale of minutes? The best I have found is the addition of silver nitrate, which will form silver chloride precipitate, but I am not sure how fast or how thorough this process is.
Background: I have been performing an experiment where I inject current into a cell of a fruit fly for a research project. For various reasons, we believe the cell membrane in one fruit fly line is more permeable to chloride than another fruit fly line and are trying to confirm this. We inject with 20nA of current for about 2 minutes and can't change these parameters much. We hope to completely replace the chloride anion in our bath with another anion, to see if this changes the readings we get from our injection. For that small of a current, I believe I need less than 1ppb chloride concentration for it not to participate much in carrying charge. It is very difficult to find a replacement salt that doesn't have a minor chloride impurity, so we are hoping to find ways to remove the chloride.

Yea, silver chloride is the least soluble common chloride salt I can come up with right off, and it is still soluble to about a ppm at room temperature.  So, when everyone tells you "silver chloride is insoluble", its pretty close, but there is always going to be some solubility, in you case way too much.  Sorry, good luck!


text:
What is the final product of a carbohydrate treated successively with hydrogen cyanide, water, and hydrogen iodide/ phosphorus?

An unknown carbohydrate is successively treated with (1) hydrogen cyanide, (2) water, and (3) hydrogen iodide and phosphorus:
  $$\ce{C6H12O6 ->[HCN] X ->[H2O] Y ->[HI,P]Z}$$
  What is the final product $\ce{Z}$?

I could figure out that in $\ce{Y}$ there is a $\ce{CH(OH)(COOH)}$ group instead of a $\ce{CHO}$ group, but I could not find out $\ce{Z}$.

Your starting material is an unspecified hexose ($\ce{C6H12O6})$. Leaving the stereochemistry aside, it is either an aldose or a ketose.

Addition of $\ce{HCN}$ results in the formation of X, a cyanohydrin (α-hydroxy nitrile) with seven carbon atoms ($\ce{C7H13NO6})$.
Hydrolysis of X yields the α-hydroxy carboxylic acid Y ($\ce{C7H14O8})$.
The reaction of Y with $\ce{P_{red}/HI}$ is likely to result in reductive dehydroxylation. Overall, the oxygen atom of the carbonyl group in X has been replaced with $\ce{H}$ and a $\ce{COOH}$ group.
Z therefore is $\ce{C7H14O7}$.

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text:
What advantage is there in reporting the Lennard-Jones well depth in wavenumber units?
The Lennard-Jones well depth $\epsilon$ is typically given in energy units ($\mathrm{kJ}$ or $\mathrm{kcal}$; sometimes per mole). Jasper and Miller, however, reported $\epsilon$ values in reciprocal centimeters ($\mathrm{cm^{-1}}$).
Why use $\mathrm{cm^{-1}}$?

It seems unnatural using $hc$ as a conversion factor for a classical
description.
Absorption peaks (e.g. in wavenumbers) are not related
to the LJ potential well, which is for non-bonded interactions.

Ahren W. Jasper, James A. Miller, Combust. and Flame 2014, 161 (1), 101–110. (preprint pdf)

Why not? It's just a conversion factor.

@ToddMinehardt it seems unnatural using hc as a conversion factor for a classical description.

I don't think there's any inherent advantage as one can always convert to another unit. It seems natural to me, however, because chemists generally have a good idea of the energy of vibrational and rotatial states in $cm^{-1}$, so being able to compare the depth of the well with some familiar chemical process is quite nice. For instance, if the well-depth were quite small, one could immediately identify that the system may not spend a lot of time there if it is comparable to the zero-point energy or something similar.
It's similar to how physicists report everything in electron-volts ($eV$). They do it because it's familiar and they have many physical processes they can use as a reference. For instance physicists know that the band gap of semiconductors is between $2-5\ eV$, so it's quite natural to report other energies in the same unit.

are absorption peaks (e.g. in wavenumbers) related to LJ potential well? (LJ is for non-bonded interactions)

They are related in some sense. Non-covalent interactions absorb light like covalent bonds do. For instance, a water dimer has intermolecular vibrational normal modes, and these normal modes have a certain energy which corresponds to a frequency of light that might be absorbed when using spectroscopy. These modes can only exist because the water molecules are bound in some potential energy well. If I observed one of the modes were at an energy of $1000\ cm^{-1}$, then I know the well must be deeper than that, and I would express the well-depth in $cm^{-1}$.


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Amber solvatebox: How do I remove water that has traveled away
I used Amber's solvatebox function with a small polypeptide and tip3p water with a distance of 15 angstroms between the molecule and the edge of the box.
After I ran dynamics, I tried to view the output by removing water from view in Jmol, but I had a really hard time rotating the polypeptide. When I viewed it with water, I found that the difficulty in rotating was because the radius of rotation (are those the right words?) was extremely large because some waters floated off in the distance.

I want to remove these stray waters. How would I go about doing that? Would I just crack open the output and remove those waters by deleting lines that contain information for those particular waters?

That looks pretty peculiar. I've had similar things happen when trying to visualize a large system using jmol, and it usually is a problem with going from the MD to jmol and not actually a problem with the simulation itself. It usually has something to do with periodic boundary conditions... I would maybe try using solvateShell instead of solvateBox which creates a shell around the peptide of some specified thickness which will probably avoid problems with the boundaries that solvateBox might create.

If you use solvateShell, you can use a harmonic restraining potential to keep your solvent from flying off, but if I remember correctly the simulation is no longer periodic.

This has nothing to do with box walls (there are no walls!) or restraining the waters. You do not want to do that!
Do not confuse periodic boundary conditions with wrapping to the central cell. Look at the amber tutorials, it is all explained there.
You either need to set iwrap=1 in your amber minor put your trajectory trough cpptraj and use autoimage.
Again, all in the tutorial and manuals.

Sounds like you know what you're talking about here. Can you expand on how one would use the iwrap, cpptraj, and autoimage keywords, perhaps by mocking up an input file containing them? See here for a tutorial on Markdown formatting to aid comprehensibility.  Welcome to Chem.SE -- thanks for posting!

Could you tell me a bit more about what iwrap and autoimage do?


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Are hydrogen bonds stronger than ionic bonds?
What is the order of strongest bonds to weakest? Are hydrogen bonds stronger than ionic bonds? Are hydrogen bonds covalent bonds?

http://chemistry.stackexchange.com/questions/41472/which-are-stronger-intermolecular-forces-ionic-bonds-or-hydrogen-bonds

http://chemistry.stackexchange.com/questions/5503/what-exactly-is-hydrogen-bonding-and-why-cant-it-exist-with-atoms-other-than-f

Hydrogen bonding is very weak, it follows that order of forces:
Ionic bond > Covalent bond > Hydrogen bond > Van der Waals forces. (I have not included electrostatic force for sake of simplicity).
Hydrogen bonds can vary in strength from very weak ($1–2~\mathrm{kJ/mol}$) to extremely strong ($161.5~\mathrm{kJ/mol}$ in the ion $\ce{HF2−}$).
Typical enthalpies in vapor include:
\begin{array}{cl}
\ce{F−H\bond{...}:\!F}& \text{161.5 kJ/mol or 38.6 kcal/mol}\\
\ce{O−H\bond{...}:\!N}& \text{29 kJ/mol or 6.9 kcal/mol}\\
\ce{O−H\bond{...}:\!O}& \text{21 kJ/mol or 5.0 kcal/mol}\\
\ce{N−H\bond{...}:\!N}& \text{13 kJ/mol or 3.1 kcal/mol}\\
\ce{N−H\bond{...}:\!O}& \text{8 kJ/mol or 1.9 kcal/mol}\\
\ce{HO−H\bond{...}:\!OH3+}& \text{18 kJ/mol[15] or 4.3 kcal/mol}\\
\end{array}
{source-Wikipedia}

Please also insert a link to the Wikipedia source pages.


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Why is the manufacture of sulfuric acid known as the contact process?
I was after some detail as to why the contact process was named as such.

According to this site:

The Hot gases( $\ce{SO2}$
) evolved from burning of sulfur ore comes in
contact with catalyst bed,  So the name of this process is called
contact process.

The word "Contact process" has an historical origin.
Before the $18$th century, sulfuric acid was difficult to obtain. It was synthetized by first oxidizing pyrite $\ce{FeS2}$ in wet air, in order to obtain iron sulfate $\ce{Fe2(SO4)3)}$, which had to be later pyrolized according to : $$\ce{Fe2(SO4)3 + 3 H2O -> Fe2O3 + 3 H2SO4}$$ Unfortunately, the initial oxidation of pyrite in humid air was extremely slow and inefficient. A huge advance was made in the middle of the $18$th century, when a new process was invented, the so-called lead chamber process, although it was based on the incorrect phlogisticon theory. First sulfur $\ce{S8}$ or pyrite $\ce{FeS2}$ was burnt in air, producing gaseous $\ce{SO2}$ which was quickly oxidized by droplets of nitric acid $\ce{HNO3}$ falling like a mist or a shower in a chamber filled with the combustion gas $\ce{SO2}$. The reaction is quick and occurs according to : $$\ce{SO2 + 2 HNO3 -> H2SO4 + 2 NO2}$$ The chamber is made of lead, in order to resist to sulfuric acid. The advantage of this process is that the gas getting out of the reaction ($\ce{NO2}$) is not lost. It is pushed into another lead-chamber where droplets of water were injected in order to regenerate the original nitric acid according to the equation : $$\ce{4 NO2 + O2 + 2 H2O -> 4 HNO3}$$ This process was much quicker than the old process using iron sulfate. But it needs controls and maintenance of these two lead chambers, which is not easy.
A new progress happened in $1832$. The gas mixture containing $\ce{SO2}$ + air could be directly transformed into $\ce{SO3}$ without lead chambers, but with simple contact with a solid stuff, namely platinum wires, which had a catalytic effect in the reaction $$\ce{2 SO2 + O2 -> 2 SO3}$$. Then $\ce{SO3}$ is quickly transformed into sulfuric acid, according to $\ce{SO3 + H2O -> H2SO4}$. Lead chambers were replaced by a simple tube equipped with some wires. It was a big improvement. Later on, the platinum was replaced by vanadium oxides, which is not so sensitive to impurities. The process was named "contact process" at this time, when the catalysis was not yet well understood.

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text:
Solving for H+ given dissociation of weak acid, c and Ka Can the quadratic equation yield two nontrivial solutions?
For example, if a question asks for $[\ce{H+}]$ of a $0.100~\mathrm{M}$ solution of picric acid (monoprotic), given that its $K_\mathrm{a}$ at the system temperature is $4.2 \times 10^{-1},$ following my teacher's instructions I would use the ICE table, which looks like this (for a different example).
$$\begin{array}{cccccccc}
 & \ce{CO (g)} & + & \ce{H2O (g)} & \ce{<=>} & \ce{CO2 (g)} & + & \ce{H2 (g)} \\
\text{Initial} & 0.100~\mathrm{M} & & 0.100~\mathrm{M} & & 0 & & 0 \\
\text{Change} & -x~\mathrm{M} & & -x~\mathrm{M} & & x~\mathrm{M} & & x~\mathrm{M} \\
\text{Equilibrium} & (0.100-x)~\mathrm{M} & & (0.100-x)~\mathrm{M} & & x~\mathrm{M} & & x~\mathrm{M} \\
\end{array}$$
I would then solve for $x$ by using the expression for $K_\mathrm{a}$. However, since $0.100/K_\mathrm{a} < 100$ (lowest initial concentration divided by $K_\mathrm{a}$ is less than 100), $x$ is significant and cannot be omitted from the $0.100-x$. Thus, you need to use the quadratic equation, which yields two solutions for my example above: $x=0.0834, 0.503$. The latter is obviously incorrect because it is greater than the initial concentration of the undissociated compound.
However, my question is: Is it possible that both solutions to a given reaction will be nontrivial, that is, is it possible that both answers appear to work at a glance? (No conflicts with the initial concentration, etc.) This question refers not specifically to the example in question, but rather looks at if it is possible at all (is there a reaction and starting concentration for which both answers would seem correct). In addition, if it is possible, how would one determine which answer is nontrivially impossible?
Whenever I get into a new subject I'm not able to "do" it until I work out some small details that stick with me.  It would be best if the answer had some logical explanation, or even a proof as to why the two answers will never be both possible at a glance, or if the answer shows a way to determine without fail which answer is correct. I don't see an obvious reason why both solutions resulting from the quadratic equation will always have one blatantly incorrect result.
Please ask for clarifications if necessary.

This has no real bearing on the question itself, but please pay attention to your units. Prior to my edit, the way you earlier defined $x$ in the ICE table, $x$ needed to have units of concentration. It cannot be a pure number because then $0.100~\mathrm{M} - x$ makes no sense. You either define $x$ as a dimensionless quantity, or define $x$ as a concentration and then tack the units along with it. Either way is fine but just be consistent.

With more complicated systems, you might end up with a cubic or even a higher order equation. Still, the nature guarantees the existence of a unique physically meaningful solution, no matter what.

What you have got here is a strong acid ($\mathrm{p}K_\mathrm{a} \approx 0.38$) that is not diluted, whereof the latter tells you that you can safely ignore the autoprotolysis of water.
The pH approximation to choose is therefor:
$$[\ce{H+}] = \frac{[\ce{HA}]_0 K_\mathrm{a}}{[\ce{H+}] + K_\mathrm{a}}$$
or in a quadratic form:
$$[\ce{H+}]^2 + [\ce{H+}] K_\mathrm{a} - [\ce{HA}]_0 K_\mathrm{a} = 0$$
which has two solutions:
$$[\ce{H+}] = -\frac{K_\mathrm{a}}{2} \pm \sqrt{\left(\frac{K_\mathrm{a}}{2}\right)^2 + [\ce{HA}]_0 K_\mathrm{a}}$$
From those two solutions, only one is chemically meaningful, which is the positive combination which yields one of your solutions ... $[\ce{H+}] = 0.0834$.

Just to make it crystal clear for OP: the other solution is meaningless because it's negative.

The question is asking in general, not specifically referring to the example. I've edited the question to make it more clear.

It's impossible to have two meaningful results for a pH calculation as a solution simply can't have two pH values.


text:
Why do noble gases, although not molecules, use London dispersion forces?
Why do noble gases, although not molecules, use London dispersion forces?

Noble gases typically do not form molecules as these atoms have filled valence shells rendering them chemically very stable and therefore unreactive. Despite that, there are numerous compounds synthesized using noble gas atoms.
The London dispersion force is a temporary attractive force that results when the electrons in two adjacent atoms occupy positions that make the atoms form temporary dipoles. This force is sometimes called an induced-dipole-induced-dipole attraction. Noble gases are able to liquefy because of this force. For example, Helium liquefies at 4K, without this force, its boiling temperature would be lower or Helium would not liquefy at all.

Have also a look here.


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Reactivity of Alkanes
Does the tetrahedral structure of the alkanes contribute to their lower reactivity? I thought that because a tetrahedral structure suggests ${sp^3}$ hybridization, it should contribute to its lower reactivity. Is this right?

The lower reactivity is attributed to the nonpolar C-H bonds that are present in alkanes. They are very stable and typically do not want to react... unless you do some exotic organometallic chemistry. That is a whole another topic, though.

@TOLA3HPPA can you give me example of "exotic" organometallic please? I'm inquisitive :)

Does the tetrahedral structure of the alkanes contribute to their lower reactivity?

Hybridization is at the center of the answer.
Due to poor P-P overlap, it only takes something on the order of 60 kcal/mol to break a pi bond and produce cis-trans isomerization about a double bond involving ${sp^2}$ hybridized carbons, whereas it takes something like 90 kcal/mol to break a carbon-carbon single bond involving ${sp^3}$ hybridized carbon atoms with highly directional orbitals that result in strong bonds. This poorer overlap with the pi bond causes ethene to be higher in energy than ethane. Alternatively one can view ethene as a two-membered ring (ref. 1), no pi bond, just 2 sigma bonds between the 2 carbons forming a two-membered ring. It is easy to imagine that such a ring system would contain a significant amount of strain (ref. 2).  Either way (and by the way these two views of ethene are equivalent and complementary) - using a poorer overlap argument or a higher strain energy argument, we see why alkenes (and even more so alkynes) would generally be more reactive than alkanes.
From this argument we would expect that if we were to somehow increase the strain in an alkane, then we should increase its reactivity. A good example of such behavior can be found in cyclopropane.  Placing 3 "tetrahedral" carbon atoms into a ring dramatically increases the strain energy in the system (it changes the carbon hybridization too, see ref. 3) and the reactivity increases dramatically.  Indeed, cyclopropane adds bromine much like an alkene (ref. 3)
Since someone above commented that ${sp^3}$ $\ce{C-H}$ bonds are "very stable and typically do not want to react". Let me just point out that $\ce{C-H}$ bond energies vary as follows
\begin{array}{|c|c|c|c|} \hline
\ce{C-H} & \text{Bond Strength}  \\  \text{bond type} & \text{(kcal/mol)} \\ \hline
\ce{sp^3 C-H} & 101 \\ \hline
\ce{sp^2 C-H} & 111 \\ \hline
\ce{sp C-H} & 133 \\ \hline
\end{array}
These bond dissociation energies (ref. 4) don't fit the observed reactivity pattern.  They indicate that an ${sp^3}$ bond is actually easier to break than an ${sp^2}$ or ${sp}$ $\ce{C-H}$ bond

The carbon-hydrogen bonds are only very slightly polar; therefore, there are no portions of the molecules that carry any significant amount of positive or negative charge that can attract other molecules or ions.
The carbon-carbon bonds are not polar at all and so these do not contribute to any charges on the molecule at all.
To my knowledge, the relative unreactivity of alkanes compared to other hydrocarbons such as alkenes or alkynes is due to the high amount of saturation across the carbon atoms.
I have heard little of the hybridization of the carbon atoms contributing to the reactivity if the alkanes. I was under the impression that the trends in reactivity were as I have described above.

You might want to amend your post to address the OP's query, especially regarding (a) tetrahedral geometry and; (b) hybridization in such geometry, both in the context of reactivity.

Thank you for the comment, I have updated the post. Was my mistake, I neglected to directly address the question by accident. Many thanks.


text:
Do the products in exothermic reactions have higher kinetic energy than the products? Do they also have lower chemical energy?
I understand that in an exothermic reaction energy is transferred to the surroundings. More energy is produced from making bonds than absorbed to break bonds.
I would imagine that the products have lower chemical energy (bond energy perhaps?) but have higher kinetic energy as measuring the temperature of the surroundings is effectively measuring the kinetic energy of the specific molecules? I only doubt myself because on enthalpy change graphs the products are always lower than the reactants in an exothermic reaction, so increasing KE doesn't seem likely. If someone could clear up my confusion that'd be great!

The energy diagrams that you see solely plot chemical potential energy. Any kinetic energy that the particles possess is not reflected in the diagram.

In line with the comment of orthocresol, these potential energy diagrams only display the change in the chemical potential as a function of the reaction coordinate.
To clarify: Let's consider the theoretical situation of a reaction mixture in complete vacuum that undergoes a reaction towards a product state. If we assume that the reaction is exothermic and there is no loss of energy due to irradiation, then this means that the products are heated up by an amount of energy that is gained from the loss in chemical potential (i.e. the exothermic reaction). In other words, if you would plot the chemical potential + the kinetic energy (the two terms combined) as a function of the reaction coordinate, you would obtain a straight line due to the law of conservation of energy, because chemical energy is converted to kinetic energy (heat).

How then we say that during bond formation (e.g. a reaction between two atoms to form a diatomic molecule) energy of the system goes down if the energy is conserved?

There are four main ways that a molecule can store energy: translation, rotation, vibration, and electronic energy. The first three of these are kinetic energy in one way or another, and the fourth is less commonly important when looking at the products of a reaction, so forget about that.
From the outset we can say yes, of course, the products have more kinetic energy because energy must be conserved. I wanna go into more detail though to hopefully demonstrate some of the intricacies of where the product energy is stored.
velocity flux contour map http://202.141.40.218/wiki/images/Velocityfluxcontourmap.jpg
The picture above is a miracle of experimental chemistry.
This plot shows you all observed trajectories of the $\ce{DF}$ molecule in the reaction,$$\ce{F + D_2 -> DF + D}$$
I can hear the naysayers now, "who would possibly care about such a useless reaction?" Well, there are many interesting things to be learned from very simple systems.
The plot is made by preparing all incoming product molecules with a total kinetic energy of, in this case, $1.82\ \frac{kcal}{mol}$. After the reaction takes place, that incoming kinetic energy, which was all in the form of translational energy (it need not be but here it was) must be stored in either vibrations, rotations, or translation.
Each of the dashed lines represents a vibrational state of the product $\ce{DF}$ molecule. Each of the contours represents a rotational state. All the rest of the energy is then stored in translational kinetic energy.
There are many details that can be gathered from this one picture, but in this case I will highlight a detail that might change how you think about the products of a reaction. In this case, the products are most often in the $v=3$ vibrational state. That is, when this reaction happens and you look at the products, you'll find a molecule meandering along pretty slowly, but vibrating quite vigorously. You can see that $v=3$ is most populated by looking at the density of the contour lines.
Also, notice that the reaction almost never proceeds and ends up in the $v=1$ vibrational state. How strange that the molecules have a preference!!
So, to tie this back to something we're more familiar with such as exothermic reactions taking place in a beaker. When a reaction happens, and there is excess energy to be stored in the products, it will almost always go into some combination of vibration, rotation, and translation. In the experiment above, after the reaction happens, nothing else is around to take that energy so we can see what the true product state looks like. In a beaker, a molecule that is made rotationally excited quickly collides with something else and gives away some of its energy. Eventually, it is quite likely everything will settle down to just plain old translational kinetic energy and some rotations, but initially, it is quite feasible the products start out vibrationally excited. Hopefully that answers your question, but also gives you some other things to think about.

The figure was found online somewhere, but it comes from this paper.
Neumark, D. M., Wodtke, A. M., Robinson, G. N., Hayden, C. C., Shobatake, K., Sparks, R. K., ... & Lee, Y. T. (1985). Molecular beam studies of the F+ D2 and F+ HD reactions. The Journal of chemical physics, 82(7), 3067-3077.


text:
Calculating the pH of buffer solution made of two salts of a polyprotic acid
When we are making a buffer solution by solutions of a weak acid and its salt like $\ce{CH3COOH}$ and $\ce{CH3COONa}$, or by a weak base and its salt like $\ce{NH3}$ and $\ce{NH4Cl}$, we can use the Henderson equation to determine its $\mathrm{pH}$. But, when we are making a buffer solution by two salts of a polyprotic acid like $\ce{Na2HPO4}$ and $\ce{NaH2PO4}$, how can we calculate the resulting $\mathrm{pH}$? Can Henderson's equation still be used? Or, is there any other formula?
For example, the pH of a solution of $\ce{NaHCO3}$ is given by the formula $\frac{1}{2}(\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2})$. But what would be the formula if the salt is from a triprotic acid? Again, what will be the formula if the solution is a mixture of salts?
I need a theoretical calculation of the $\mathrm{pH}$, and this is not for any laboratory type experiment.

You can use Henderson-Hasselbach. The equations will independently hold for both equilibria. You use use the total amount in solution to compute the actual amounts.

A useful internet search for this type of problem is 'systematic calculation of pH'. These calculations can get messy pretty quickly, so take it slow.

@Zhe How can I use the equation?

http://chemistry.stackexchange.com/questions/8254/calculating-approximate-ph-of-polyprotic-acids?rq=1


text:
Why is maltose a reducing sugar but not sucrose, even though they're both disaccharides?
I came across this on Wikipedia:

Many disaccharides, like cellobiose, lactose and maltose, also have a reducing form, as one of the two units may have an open-chain form with an aldehyde group. However, sucrose and trehalose, in which the anomeric carbons of the two units are linked together, are nonreducing disaccharides since neither of the rings is capable of opening.

I took the following Haworth projection of sucrose for comparison-

That's all fine and dandy, but the thing is, I can't seem to find any source that (explicitly) explains why this whole 'ring-opening' business even happens in disaccharides like maltose, but not in sucrose.
The Wikipedia article probably implies that this has to do with the fact that both the monomers of sucrose are bound through their anomeric carbons, which is not the case with maltose. A quick look at the structures of maltose and sucrose confirms this, however; pardon me if I'm being dense here, but I don't see how that piece of information helps.
Can anyone explain this discrepancy?

The difference in stability between maltose and sucrose boils down to the different structural elements their aldehyde/ketone groups have been turned into during the formation of the disaccharide.
In maltose, you combine two glucose units using the 1-hydroxy group of one and the 4-hydroxy group of the other. The 1-hydroxy group has been created in a process called acetal formation by the attack of what used to be the 5-hydroxy group and is now the ring oxygen. This is shown in equation $(1)$, the explicit carbon atom being carbon-1; the aldehyde.
$$\ce{R'-OH + RCHO <=>> R'-OH+-CHRO- <=>> R'-O-CHR-OH}\tag{1}$$
Notice that there are two oxygens bound to that carbon after the reaction; if we wanted, we could label it $\ce{RCH(OR)(OH)}$; this structure is called hemiacetal. If we use that hemiacetalic hydroxy group to bond to a second glucose unit,[1] we have replaced the hydrogen with an alkyl residue. Like the difference between alcohols ($\ce{R-OH}$) and ethers ($\ce{R-OR'}$), this group has a different reactivity and is termed full acetal or O,O-acetal.
Hemiacetals are rather easy to break down as two mechanisms are available: you can either initially protonate the other oxygen or you can use a base to facilitate proton removal of the hydroxy group — ideally perhaps while offering the other oxygen a proton by another base. Full acetals, however, are not as easy to break down; the require protonation, implying a sufficiently strong acid.
Coming back to maltose, remember we have two glucose subunits. We used one to generate a full acetal but the other one (on the right-hand side of your scheme) is still hemiacetalic and can reconvert back to an aldehyde. This is why you preserve the reducing abilities; remember that equation $(1)$ is fully reversible and according to Le Châtelier, removing aldehyde (e.g. by oxidation) will enhance the reverse reaction.
In sucrose, on the other hand, carbon 1 of glucose and carbon 2 of fructose are connected. These are the two carbon atoms that form hemiacetals in isolated glucose/fructose. Because of this, there is no hemiacetal group $\ce{RCH(OR)OH)}$ in sucrose — only full acetals — and no way it can easily revert back to an aldehyde or a ketone. Therefore, it has lost its reducing ability.
This is the key information of your quotation by the way. The anomeric carbon is the one that will be a hemiacetal. If two anomeric carbons are connected, they have formed a full acetal. Therefore, identifying the anomeric carbons and seeing if they still have a free hydroxy group (i.e., if they are hemiacetals) allows quick determination whether a disaccharide is reducing or not.

Note:
[1]: While formally correct; the actual reaction mechanism involves the loss of this hemiacetalic hydroxy group.

Note that this is not about digesting/cleaving disaccharides into monosaccharides first.
Opening has to occur on the stage of the intact disaccharide.
What is requirement for an opening that results in an aldehyde?
The carbohydrate has to have a hemiacetal moiety!


text:
Travelling wave ion mobility spectrometry: determination of size?
I performed an ESI / travelling wave ion mobility spectrometry / MS experiment and observe $\ce{M + H+}$ at $m/z ~701$ and $\ce{M + K+}$ ions with $m/z ~740$. The drift times are 6.75 ms for $\ce{M + H+}$ and 7.0 ms for $\ce{M + K+}$. My first instinct was that this means that the proton adduct is smaller than the potassium adduct. However, if the mass is taken into account as well, i.e.
$$
   7.0 ~\mathrm{ms} ~/ ~740 = 0.0095 ~\mathrm{ms}
$$
for the potassium adduct, and for the proton adduct
$$
   6.75 ~ \mathrm{ms} ~/~ 701 = 0.0096 ~\mathrm{ms}
$$
which would indicate that the two are very similar in size, the proton adduct being slightly larger. For a linear drift tube it would be very easy for me to calculate the relationship (i.e. Mason-Schamp equation), but for travelling wave IMS I am pretty unsure. Which of the two reasonings is correct?

In terms of practical data analysis standpoint, travelling wave-IMS data can generally be treated similarly as drift tube-IMS for which the technical differences is in resolution (better or worse). However, one point of consideration is that

Low gas pressure in TW IMS means strong fields in E/N terms
and thus substantial heating of ions. This may cause fragmentation
or distortion of macromolecular structures, complicating their
detection and characterization. [1]

Your first instinct to assume that the $\ce{[M + H]+}$ adduct is quite rational, however remember that ion conformation is always in flux and can be influenced by collisions with buffer gas. What you may be experiencing is the proton adduct producing a "looser" structure, resulting in a larger collisional cross section. The $\ce{[M + K]+}$ adduct is heavier in mass but may produce a tighter conformation.
A good example of this is in the fragmentation and mobility of polyethylene glycol (PEG) ions $\ce{[PEG + H]+}$ and $\ce{[PEG + Na]+}$. When carrying out fragmentation studies of PEG, it was found that the sodiated ion produced less extensive fragmentation and required higher collision energy.[2] Ion mobility studies have also found that PEG chelates the $\ce{Na+}$ ion similar to that of crown-ether, producing tightly wound "rigid" structures.[3]
These conditions are highly compound/structure dependent however, it should be something to keep in mind especially in the case of analyzing other molecules such as peptides which respond to adduction in "weird and mysterious" ways.
References

Alexandre A. Shvartsburg and Richard D. Smith, Anal. Chem. 2008, 80 (24), 9689–9699.
Jennifer Gidden, Thomas Wyttenbach, Anthony T. Jackson, James H. Scrivens, and Michael T. Bowers, J. Am. Chem. Soc. 2000, 122 (19), 4692–4699.
Rui Chen, Xinlei Yu, and Liang Li, J. Am. Soc. Mass Spectrom. 2002, 13 (7), 888-897.

Especially the second reference looks very promising. Thanks : )


text:
Atomic orbitals: how is bond strength related to stability, energy, bond length?
I have a lot of confusion regarding atomic orbitals and I don't know if what I am having in mind is even correct or not.

The bond dissociation energy of H-H is greater than that of Cl-Cl and as far as I can make out, it is because in H-H, there is an s-s overlap whereas in Cl-Cl, there is a p-p overlap. Because s-s is more compact, therefore more energy is required to break the bond.
Assuming what I wrote above to be correct, this means an s-s bond is stronger than p-p bond. But then, bond strength of p-p is more than that of s-s because of greater extent of overlap.
So there is definitely something/everything wrong in what I am thinking.
In short, how is the bond strength related to other things like stability, energy, bond length (bond length of s-s < s-p < p-p?).
Kindly help because I cannot find any of my queries anywhere else.

I'm not sure that there's a good way to compare between H-H and Cl-Cl solely based on the type of orbital. Your rationalisation doesn't sound convincing at all to me.

There are several factors that affect bond disassociation energies (bond enthalpies) but one quick correlation you can make is bond distance vs bond strength. The interatomic distance of the H2 bond is 74 pm vs 199 pm in Cl2. The H2 bond consists of 1S electrons that are very close to the nucleus, giving a short bond. The valence electrons in Cl2 are 3p (relatively far away). This leads to other properties such as polarizability.
In short your first assumption is incorrect, the trend you are seeing has nothing to do with s or p bonds, but on the bond lengths. Therefore comparing s-s bonds and p-p bonds is meaningless unless you take into account the atomic orbital number and therefore an estimate of bond length.
For your final questions, the stronger the bond, the more stable, the higher energy, and the shorter the bond.
For more info this UF link has good information:http://www.science.uwaterloo.ca/~cchieh/cact/c120/bondel.html

The stronger the bond, the more stable, the higher energy, and the shorter the bond. So in this way, a 2s-2s bond is shorter than a 2p-2p bond, and also stronger? But then the strength of a p-p bond > s-s bond. Can you kindly explain?

@Cheapstrike a 2s-2s bond is practically impossible because you will run into p-orbitals (which are directional) on both atoms.

There is a vague correlation between bond energy and length, but only in that longer bonds involving larger atoms are weaker. The figure shows some data.  The data are for bonds between many different types of atoms and involve single, double and triple bonds.

data from http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html
and references therein.


text:
At equilibrium are the rates of the forward and backward reactions equal to zero?
At equilibrium, the concentrations do not change with time. So, is it true that the rates of the chemical reactions are zero at equilibrium? Wikipedia says that they are not zero. Why is this?

Who say these two r zero??

In equilibrium, the forward and backward rates are equal to each other. The net is zero, but the individual rates are not zero.
Consider something as simple as water — we know that the water molecule can dissociate in to an $\ce{OH-}$ and $\ce{H+}$, and the pH is a direct measure of the amount of $\ce{H+}$. Once equilibrium is achieved, it isn't like $\ce{H2O}$ stops splitting, or $\ce{H+}$ and $\ce{OH-}$ get back together — it is just at a macro level nothing changes.

The magic word here is "dynamic equilibrium". Let's look into that.
Let's say we got a simple reaction like: $$\ce{A <--> B}$$
Now how fast does A react to B? For this case we assume first order in both directions, which means the reaction rate would be:
$$r_1= [\ce{A}] \times k_1$$
and of course for the reaction back it's
$$r_2=[\ce{B}]\times k_2$$
So in both cases the reaction rate is dependend on the concentration of the reactant, that's very important.
Now let's say we start off with only A, in this case $r_1$ will be large and $r_2$ will be 0, since there's no B there. Now A gets less and B is getting more, so $r_1$ is going down and $r_2$ is going up. At some point $r_1$ will be equal to $r_2$, but this means there will be as much of A produced from B, than A will react to B. And the same for B. So over all the reaction is still going in both directions, but it's in a dynamic equilibrium.
Here's an example which describes this pretty well:
Let's imagine you got a backyard and right on the boarder to the next one is an apple tree and a lot of old apples are laying on the ground. You don't like those in your garden, so you go around, pick them up and throw them into your neighbours garden. He doesn't like that and starts throwing them back. Now he is old and is moving much slower than you are, so what is happening?
At first there are a lot of apples on your side and you might need to walk some meters but then you can pick one up and throw them. So you will throw a lot of apples very fast. On the other hand there aren't many apples on the other side and the old man is slow, so he needs much longer to grab one. So he isn't throwing them back to you very rapidely. But here's the problem: the more apples you throw the less apples there are, and you need to move longer distances to grab one. So the rate in which you are throwing apples decreases. On the other side there are more and more apples, so the old man doesn't need to move far to grab one, so his rate in throwing apples increases.
After some time there will be a point where for every apple you throw to your neighbour he will throw one back. And if you count the total number of apples on each side it will pretty much stays the same, even if there are apples flying in both directions the whole time.

So, is it true that the rates of the chemical reactions are zero at equilibrium?

Here you are making a  mistake. Just take a simple look at the definition of equilibrium you will find that:
In a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to change with time. Usually, this state results when the forward reaction proceeds at the same rate as the reverse reaction.
What the last line says is the state of equilibrium results when the forward reaction proceeds at the same rate as the reverse reaction i.e rate if reactions (forward and backward) become equal not each zero.

Correction to above statement which assumes that there must be equal concentrations of products and reactants in an equilibrium reaction.
In equilibrium there is no net change of products or reactants. The forward and backward rates are only equal to eachother if there are equal amounts of products and reactants which is not always the case.
A reaction can be 99% reactants, 1% products and be in equilibrium as long as those concentrations are not changing, and in that case the forward and reverse reaction rates are not equal.
Consider for example the disassociation of a weak acid:
HA <> H+ + A-
since it is a weak acid you already know the equilibrium will lie to the left with most of the acid not disassociated. The forward rate is smaller than the reverse rate. If the [HA], and [H+][A-] are not changing over time, the reaction has reached equilibrium. There is always HA disassociating, and H+ and A- combining, but the overall concentrations do not change.

I don't see anywhere that it says that the concentrations are equal.


text:
What would an ammonium sulfide fire look like?
I am writing a novel and the climax involves killing my protagonist in an explosion. My science knowledge is poor but I believe my scenario is entirely plausible except possibly a couple of details.
My character is in a small, poorly ventilated room. It has an equipment or instrumentation cart with some 1970s era machinery. Within it is a jar of ammonium sulfide from which the gas is pumped via leaky tubes. The leak has gotten worse and the room is very smelly. He bends over to unplug the equipment, which is not grounded. The smell is worse near the floor. Pulling the plug out of the outlet causes a spark and the gas ignites.
I hope this is perfectly logical so far. Now here are a couple specific questions.
What would that fire look like? I currently have a line of fire in the air back to the source by the jar. It then explodes. I then have gases like hydrogen sulfide produced and there is a larger, secondary explosion which kills the injured character.
Would it be better to say that all of the gas burns in a flash. Not much damage is actually done, but the jar of ammonium sulfide is heated, begins to produce hydrogen sulfide and other pyrophoric gases and then there is a spectacular secondary explosion?
I am not necessarily going to explain the science to the reader, which gives me some wiggle room. But I do want these events to be plausible.

It's not clear to me what gas is igniting. If you are imagining the hydrogen sulfide is the gas that ignites your character may have another problem. In that hydrogen sulfide is incredibly dangerous to inhale. As in levels above 100 ppm are regarded as immediately dangerous to life. I guess a ppm level that could be ignited would probably render unconscious kill anyone that walked into that room in minutes. See this terrifying document for more: http://www.vinnueftirlit.is/media/greinar-og-skyrslur/2009_oecd_analysis_of_h2s-incidents.pdf

Thank for that!  I have the ammonium sulfide in the air burn first, then the bottle of the stuff explodes, as it stands now. There is then a secondary explosion Involving the hydrogen sulfide.

Note that pure ammonium sulfide would basically be a source of H2S, which is toxic via inhalation.  I think I've seem quotes where ammonium sulfide at 600 ppm in air caused fatalities.  It also REALLY smells bad.  I'd guess it would be fatal at levels below those necessary to combust.

If your container is supposed to contain solid ammonium sulfide $\ce{(NH4)2S}$, it better be cooled to -20 °C or below.
(Typically, ammonium sulfide is only used in aquous solution to precipitate metal cations of cobalt, nickel and others as their sulfides in the Kationentrennungsgang).
Long before the solid ammonium sulfide reaches room temperature, it will start to decompose and ammonia and hydrogen sulfide are liberated.


text:
Rate Law Equation
We are given the following two equation,
The first reaction is an equilibrium reaction.
$$A_2 \rightarrow A+A \quad (fast)$$
$$A+B\rightarrow P \quad (slow)$$
Where A is the intermediate.
Now I have to find the rate law using the steady state approximation.
I have come to this following equation for the net change of [A],
$$\frac{d[A]}{dt}=k_1[A_2]-k_2[A]^2-k_3[A][B]=0$$
Where I have said that $k_1$ is for the forwards reaction, $k_2$ is the for the backward reaction in the first one. $k_3$ is the final reaction's rate constant. Now what I dont understand is this - How can I solve the above equation and find an expression for $[A]$.
Is there some trick that I can use, or something? I haven't been able to solve it, since it's a quadratic formula and I cannot really solve a quadratic formula with just this given information. I even tried using the quadratic formula, but that was a dead end after some lines of working out.

Why are you trying to find $\ce{[A]}$? The third term is the rate of reaction, almost certainly what you actually want. Combine the first two terms into the equilibrium expression for the fast reaction.

Well but [A] is the intermediate, so that is why i can not use the third term. I need to find an expression in terms of only $[A_2]$ and [B].

No, the second step is fast. There will not be a $\ce{B}$ term in the rate. But the third term is the rate of product formation. I'll try to write this out for you when I get back to my desk. Typing LaTeX on a phone is hard...

Also, I can't read, so the second step isn't fast... :(

I apologize for my comments; I shouldn't be answering Chem.SE questions during my commute without writing stuff out...
I renamed some of your rate constants. In addition, notice that as written, the reverse of the first reaction does not lead directly to a component of $\frac{d\ce{[A]}}{dt}$ because two $\ce{A}$'s are being consumed.
$$\ce{A2 \underset{k_{-1}}{\overset{k_{1}}{<=>}} 2A}$$
$$\ce{A + B \overset{k_{2}}{->} P}$$
$$\mathrm{rate} = \frac{d\ce{[P]}}{dt} = k_{2}\ce{[A][B]}$$
Your steady state equation is almost correct:
$$\frac{d\ce{[A]}}{dt}=2k_{1}\ce{[A2]}−\frac{k_{-1}}{2}\ce{[A]}^{2}−k_{2}\ce{[A][B]}=0$$
I actually ended up solving the quadratic equation. It wasn't that bad:
$$\ce{[A]} = \frac{\sqrt{k_{2}^{2}\ce{[B]}^{2}+4k_{1}k_{-1}\ce{[A2]}}-k_{2}\ce{[B]}}{k_{-1}}$$
so:
$$\frac{d\ce{[P]}}{dt} = \frac{k_{2}\left(\sqrt{k_{2}^{2}\ce{[B]}^{2}+4k_{1}k_{-1}\ce{[A2]}}-k_{2}\ce{[B]}\right)\ce{[B]}}{k_{-1}}$$
Sanity check the two limits:

$k_{2}\ce{[B]} \gg \sqrt{4k_{1}k_{-1}\ce{[A_{2}]}}$
$k_{2}\ce{[B]} \ll \sqrt{4k_{1}k_{-1}\ce{[A_{2}]}}$

Case 1 means that the second reaction is very fast. Case 2 means that the second reaction is very slow.
Case 1:
$$\frac{d\ce{[P]}}{dt} = \frac{k_{2}\left(\sqrt{k_{2}^{2}\ce{[B]}^{2}+4k_{1}k_{-1}\ce{[A2]}}-k_{2}\ce{[B]}\right)\ce{[B]}}{k_{-1}}=\frac{k_{2}\left(k_{2}\ce{[B]}\sqrt{1 + \frac{4k_{1}k_{-1}\ce{[A2]}}{k_{2}^{2}\ce{[B]}^{2}}}-k_{2}\ce{[B]}\right)\ce{[B]}}{k_{-1}}$$
Taylor expand the root: $\sqrt{1+x^{2}} \approx 1+\frac{x^{2}}{2}$
$$\frac{d\ce{[P]}}{dt} = \frac{k_{2}\left(k_{2}\ce{[B]}\left(1 + \frac{2k_{1}k_{-1}\ce{[A2]}}{k_{2}^{2}\ce{[B]}^{2}}\right)-k_{2}\ce{[B]}\right)\ce{[B]}}{k_{-1}}=2k_{1}\ce{[A2]}$$
In other words, the intermediate is consumed in the forward direction at the rate that we can fragment $\ce{A2}$. This seems right.
Case 2:
$$$$
$\sqrt{k_{2}^{2}\ce{[B]}^{2}+4k_{1}k_{-1}\ce{[A2]}}-k_{2}\ce{[B]}$ reduces to $\sqrt{4k_{1}k_{-1}\ce{[A2]}}$.
$$\frac{d\ce{[P]}}{dt} = \frac{k_{2}\sqrt{4k_{1}k_{-1}\ce{[A2]}}\ce{[B]}}{k_{-1}}$$
The thing you want is somewhere between the complicated first rate I wrote and the limit in case 2. This seems reasonable in that it is first order in $\ce{B}$ as you might expect from a transition state that involves a single $\ce{B}$.

The square root term in your last equation is in effect the equilibrium constant for the $\ce{A2 = A + A}$ reaction as would be expected as this is the fastest reaction

If the first reaction is fast, then the third term can be neglected in comparison to the other two terms in the equation $$\frac{d[A]}{dt}=k_1[A_2]-k_2[A]^2-k_3[A][B]=0$$This immediately gives:$$[A]=\sqrt{\frac{k_1[A_2]}{k_2}}$$Therefore, $$\frac{d[P]}{dt}=k_3\sqrt{\frac{k_1}{k_2}}[B]\sqrt{[A_2]}$$


text:
Can rearrangement in carbocations cause inversion?

Can rearrangement reactions in carbocations cause inversion?

If you have a carbocation rearrangement (i.e. a methyl shift or a hydride shift), is it possible for inversion of stereo chemistry to take place?
The carbocation has the free p-lobe, so is it possible for the hydride/methyl group to bond to either lobe, such that one can cause retention of stereochemistry while the other can cause inversion of stereochemistry? Does this logic even make sense?

Firstly, I think you have to clarify some stuff. 1) Which carbon are you looking at the configuration of? The carbon undergoing migration, or something else? 2) By inversion, do you only mean a change in the Cahn-Ingold-Prelog descriptor from R to S (or vice versa), or do you mean an inversion in stereochemistry at the migrating centre (such as that seen in an SN2)? Concerted 1,2-alkyl shifts always proceed with retention of configuration at the migrating center. Whether this leads to a change in the CIP descriptor depends on the substituents present.

[...] The phrase "inversion" in organic chemistry typically has absolutely nothing to do with the CIP descriptor. In that diagram, any and all of the compounds can be either R or S, depending on the identity of the groups A, B, C, X, Y. For example, SN2 reactions always proceed with inversion, but the CIP descriptor may or may not change: see this question.

Hmm thanks for the link. Uhm, so R/S designations just depend on the group acting as the nucleophile and whether it will change the priorities of groups. Thanks.

I edited the question so it is a little bit clearer now that I've remedied some of the misunderstandings that I had...

I think that the question you're asking about inversion of stereochemistry upon migration is a good one, and for that general topic I would point you to this surprisingly good (and picturesque) wikipedia page about sigmatropic migrations. I don't think what you're specifically asking works as well though because if a migration occurs due to the formation of a carbocation, that carbocation has a pure p orbital on it, and thus anything that adds there will add 50/50 to either side. You'd need to migrate into a stereocenter.

Originally this was a comment, but I figure it's good enough for an answer.
As far as I can tell, you seem to be confusing the notion of suprafacial/antarafacial with retention/inversion.
You're asking whether the migrating group can loop around to the other side of the carbocation, right? That's known as an antarafacial process, not an inversion.

Retention, or inversion, refer to the stereochemistry at the migrating centre. If it is a hydride shift, then the migrating centre is a hydrogen, and there is no stereochemistry to speak of. The terms retention or inversion are therefore not applicable to hydride shifts.

In any case, the crux of the matter is that this migration is a concerted process (it occurs in one step), and an antarafacial migration cannot occur due to geometric reasons. The lobe of the p orbital on the other side is way too far away, and the $\ce{C-R}$ $\sigma$ orbital (which contains the electrons that you are trying to migrate) is not going to be able to reach it.

[1,2]-alkyl (or hydride) shifts are always going to be suprafacial!
(As a side note, the suprafacial [1,2] shift has to proceed with retention, as per Woodward-Hoffmann rules.)

"Originally this was a comment, but I figure it's good enough for an answer." More than an year later, do you mind deleting your original comments on the question, if they are obsolete now? Thank you.

Supplement to o-cresol's answer
The Woodward-Hoffman rules are central to this discussion, and succinctly put concern themselves with the conservation of orbital symmetry. Conservation of orbital symmetry requires the transformation of the
molecular orbitals of reactants into those of products to proceed continuously by following a reaction path along which the symmetry of these orbitals remains unchanged.
First, a note about notations. In a sigmatropic rearrangement  are concisely described by an order term [i,j], which is defined as the migration of a $\sigma$-bond adjacent to one or more $\pi$ systems to a new position (i−1) and (j−1) atoms removed from the original location of the $\sigma$-bond. When the sum of i and j is an even number, this is an indication of the involvement of a neutral, all C atom chain.
Now, given below is a visual summary of how a suprafarcial and an antarafarcial process takes place.

Now, let's look at it the Wagner-Meerwein rearrangement i.e a 1,2-migration of an alkyl or aryl residue to a carbocation center. It represents the most simple and smallest system in which a sigmatropic rearrangement can take place.

Now if this reaction is to take place with stereochemical retention, the migrating group has to translate, (without rotation), into the new bonding position. The image given below illustrates this process.

Now turns out conservation of orbital symmetry in the transition state, demands that the process take place this way.
Imagine, if the migrating group were to rotate, we will have inversion (however, we can't form a new bond because the symmetry requirement is violated).


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Caffeine extraction
I'm working on a caffeine extraction experiment, using dichloromethane and instant coffee. I am testing how the amount of caffeine extracted from coffee will be affected when different concentrations of NaOH are added to the coffee. However, I am not getting any results. When the extraction is being done, no caffeine is extracted. Is this correct? What should technically happen when NaOH is added to coffee?

What, with or without NaOH, you do not get any coffeine? How do you do this extraction? And why are you doing this experiment if you don't know what the NaOH should do? Please explain!

http://chemistry.stackexchange.com/questions/1217/can-caffeine-be-separated-from-coffee


text:
Reason for endothermic nature of NaHCO3 + C2H4O2 reaction
Why is the classic baking soda and vinegar reaction endothermic? Here are some possible answers I've come up with:

The dissolution of the sodium acetate absorbs energy,
The release of the carbon dioxide from solution absorbs energy, or
Some combination of the above.

Your "all of the above" answer is correct, with emphasis on number 2.  The phase change from a condensed phase (solid or liquid) to gas is almost always endothermic.  It's the same thing as wetting your hand then blowing on it.  The liquid water undergoes and endothermic phase transition which takes in heat from its surroundings, i.e. it evaporates and your hand gets cold.


text:
Difficulty understanding redox in terms of hydrogen and oxygen transfer?
I have come across 4 ways of defining oxidation:

loss of electrons
increase in oxidation number
gain of oxygen
loss of hydrogen

From these definitions, I have an idea of what I think 'oxidation' and 'reduction' really refer to, but I would appreciate if someone could verify or tell me if I am going wrong.
I think oxidation and reduction simply refer to the degree to which the electrons of an atom 'dissociate' or 'distance' themselves from the nucleus of an atom. We define this extent of distancing to be 0 when an atom is bonded to another atom of the same element.
Thus oxidation really is 'loss of electrons', and reduction is 'gain of electrons', but the electrons do not have to be completely gained or lost. It is more like, oxidation is the valence electron shell moving further from the nucleus and reduction is the valence electron shell coming in closer to the nucleus. I.e. in a chemical reaction when the atoms surrounding a particular atom, say A1, are changed, the terms 'oxidation' and 'reduction' give a measure as to whether the atoms surrounding A1 have an overall effect of being more electronegative than the atoms that were bonded to A1 previously (in which case A1 has been oxidised) or less electronegative (A1 has been reduced).
Then the increase in oxidation number coincides with these definitions.
When it comes to REDOX in terms of hydrogen and oxygen transfer, I think this is only valid for organic chemistry. In general, loss of gain of hydrogen is to a carbon atom or an oxygen atom, or maybe nitrogen and sulfur too. As these are more electronegative than hydrogen, a gain in hydrogen means that these atoms can attract more electrons towards themselves, whereas a loss of hydrogen means that they lose electrons that they could be attracting towards themselves.
In terms of oxygen, in most of these organic systems it is gain of oxygen to carbon, sometimes nitrogen, which are less electronegative than oxygen. So by the same idea as above with hydrogen, a gain in oxygen corresponds to decrease of electron density being held around the atom, i.e. oxidation.

It is always true that an oxidising agent is an electron acceptor, a reducing agent is an electron donor. Have a look at my answer here http://chemistry.stackexchange.com/questions/59748/determine-products-of-redox-reaction-kmno4-by-standard-electrode-potential/59762#59762

You are quite right, those rules of thumb are not always valid.
To be precise, the "adding hydrogen = reduction" rule only works for elements more electronegative than hydrogen, and the "adding oxygen = oxidation" rule only works for elements less electronegative than oxygen. For example, sodium is less electronegative than hydrogen, and as such adding hydrogen to sodium ($\ce{Na -> NaH}$) is in fact an oxidation of sodium from 0 to +1.
However, I wouldn't be so quick to limit their application to "organic chemistry". The elements that are less electronegative than hydrogen are just the metals, and the only element more electronegative than oxygen is fluorine.
Therefore, adding oxygen to chlorine is an oxidation, although the oxides or oxoanions of chlorine are classified under inorganic chemistry. Likewise, nearly the entirety of the p-block obeys these rules (except for Group 13). Think about germanium, arsenic, etc.. those are undoubtedly not "organic chemistry".


text:
Understanding the bond-line representation of glycine
I'm confused because I see now two different way to represent the glycine amino acid. while I understand the first way of the representation the second one I don't understand at all. I believe that both represent the same amino acid but I see that some of the componenets are absent in the second way (no COOH and no carbon). I would like to know the explanation or simple instruction for understand and correct reading of such diagrams.
This is the first and the simplest way:

And this is the second (and non-understandable) way:

https://en.wikipedia.org/wiki/Skeletal_formula

See also Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008). These recommendations mainly describe the second case; however, the first case is also mentioned in Section GR-13 Linear Drawing Style.

Thank you. I think that you considered that one: https://www.iupac.org/publications/pac/pdf/2008/pdf/8002x0277.pdf but it's very long... I hope to find a shorter way to study these chemical diagrams. I always become curious  to understand them.

The second structure is a common way chemists use to draw molecules because it is simpler and faster to draw - by omitting the labeling of "C" for carbon atoms and "H" for hydrogen atoms on carbon. Hydrogens on heteroatoms (Oxygen, Nitrogen, Sulfur, etc...) are explicitly stated
Instead of explicitly labeling the carbon atom, a carbon atom is represented by a "bend" or a "stick" where its valence is fulfilled by remaining hydrogens.

Another example is in the structure of another amino acid, Alanine. You can see that none of the carbons with hydrogens attached are not labeled and each carbon is either at the end of the stick or at a bend. Ignore the big wedge for now, but know that it has significance in spacial orientation.

If you want to learn more, Khan Academy has a cool series on labeling, drawing, and naming these molecules.
Bond Line Structures

Thank you for your answer, it really answered me the question. I'll use this videos of Khan academy as well.

Can you just explain me the meaning of the bold line of alanine? What is the reason that it's bold? is it because of the Carbon over there?

Sure thing, the bold line specifies its isomeric stereochemistry. As a simple definition- isomers have the same elemental composition but different structures.
Notice how the wedge is pointing toward you? This means that the -CH3 group is pointing towards you. The alternative would be -CH3 pointing away from you.
This simple geometric feature causes both molecules to be mirror images of each other (also known as Enantiomers) - yet they are not the same. More detail can be found here at the wikipedia page for Stereochemistry

Thank you for your nice explanation. I found further explanations here: http://chemistry.stackexchange.com/questions/28503/what-are-the-meanings-of-dotted-and-wavy-lines-in-structural-formulas

And also here: https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/intro3.htm

There are different ways to represent chemical structures. The first one is called structural formula, in this case we write down every atom and lines between them for every bond. This is usually the first thing you learn, since it makes understanding the formula very easy.
The second one is called skeletal formula and this is the one commonly used by chemists. In this formula we just draw the "carbon skeleton" of the structure with lines. Each corner (or end of line) represents a carbon atom, unless there's a symbol for another atom. We leave away the hydrogens, because it's clear how many of those are on each carbon if you look at the number of bonds drawn from the carbon.
In addition there are several other representations, like Lewis Dot structures.
So yes, both structures represent the same molecule.
Let's look at the second one starting from the left

we got a NH2 group here just like in Figure one
from that we got one bond to a corner, so there's a carbon
from this carbon we got a total of two bonds (one to the left were we came from, one to the right) and we know carbon usually has 4 bonds, so there are two hydrogens there, just like for the green carbon in the first figure.
now we go further to the right, there's one corner, so another carbon.
this carbon has one bond to the left, two to the oxygen above it (a double bond) and one to the oxygen on the right. So a total of 4 bonds, no hydrogen here.
Then there's the double bonded Oxygen to the top and an OH to the right, this whole part, the carbon + double bonded oxygen + OH is the "COOH" group

Thank you for your answer, but I'm not sure that I understood it. Is there a way that I can comment to you with a picture? (in that way I can upload the 2nd diagram and to put the 'missed' atoms in  the place that they are supposed to be, and then I can be sure that I understood)

You can just edit your post and add this to the bottom of it.

Ok I got it after I saw Patrick's answer. Thank you:)


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Bond angles of dichlorine and dibromine monoxides
Most online sources suggest that the bond angle of $\ce{OBr2}$ is greater than the one of $\ce{OCl2}.$ It is explained by the electronegativity difference and electron repulsion within the VSEPR theory.
However, I think that the extent of back bonding would be more for $\ce{OCl2}$ and hence its bond angle should be greater. I have not been able to find any experimental data from reliable sources to verify my assumption.
Can someone please explain why I may be wrong and provide experimental data on bond angles along with it?

It is indeed larger, but the difference is only 1.1 degree. This is negligible and a lot of different effects could cause it.

Related: http://chemistry.stackexchange.com/q/18697/7475

I found some old data (J. Chem. Soc. A, p658,1968)  on $\ce{Cl2O}$ which gives an experimental bond angle of $111^{\circ}$ and bond length of $171$ pm. There are several sites that quote an angle of $105^{\circ}$ for $\ce{Br2O}$ but don't say whether this is measured.

$\ce{F2O}$

Phase
Method
∠F−O−F
Reference

solid (−223.8 °C)
powder x-ray diffraction
98.0(1.5)°
[1], ICSD-430183

gas
microwave spectroscopy
103.07(5)°
[2]

$\ce{Cl2O}$

Phase
Method
∠Cl−O−Cl
Reference

solid (−183(2) °C)
single-crystal x-ray diffraction
111.27(4)°
[3], ICSD-407768

gas
microwave spectroscopy
110.96(8)°
[4]

$\ce{Br2O}$

Phase
Method
∠Br−O−Br
Reference

solid (−143 °C)
single-crystal x-ray diffraction
114.2(2)°
[5], ICSD-50198

gas
microwave spectroscopy
112.24(20)°
[6]

There are differences for the bonding angles for gaseous and solid phases due to packing effects and intermolecular interactions.
The trend, however, remains: there is an increase of $\ce{X-O-X}$ angle from $\ce{F2O},$ $\ce{Cl2O}$ to $\ce{Br2O}$ as “a result of increasing repulsion forces between the heavier halogen atoms”. [5] Overlay of the three structures obtained via x-ray diffraction is shown below by merging the location of the common oxygen atom and fixing the direction of a single $\ce{X-O}$ bond as well as the plane of each molecule:

References

Marx, R.; Seppelt, K. Structure Investigations on Oxygen Fluorides. Dalton Trans. 2015, 44 (45), 19659–19662. DOI: 10.1039/C5DT02247A.
Pierce, L.; Jackson, R.; DiCianni, N. Microwave Spectrum, Structure, and Dipole Moment of $\ce{F2O}$. J. Chem. Phys. 1961, 35 (6), 2240–2241. DOI: 10.1063/1.1732240.
Minkwitz, R.; Bröchler, R.; Borrmann, H. Tieftemperatur-Kristallstruktur von Dichlormonoxid, $\ce{Cl2O}.$ Zeitschrift für Kristallographie - Crystalline Materials 1998, 213 (4), 237–239. DOI: 10.1524/zkri.1998.213.4.237.
Jackson, R. H.; Millen, D. J. Microwave Spectrum and Nuclear Quadrupole Coupling Coefficients for Chlorine Monoxide. Proc. Chem. Soc. 1959, 10.
Hwang, I.-C.; Kuschel, R.; Seppelt, K. Structures of Bromine Oxygen Compounds. Z. Anorg. Allg. Chem. 1997, 623 (1–6), 379–383. DOI: 10.1002/zaac.19976230160.
Brenschede, W.; Schumacher, H.-J. Über Die Darstellung Und Einige Eigenschaften Eines Bromoxyds von Der Formel Br 2 O. Z. Anorg. Allg. Chem. 1936, 226 (4), 370–384. DOI: 10.1002/zaac.19362260409.

I think steric effects may play a role in that chlorine atoms are smaller than bromine atoms. The bromine atomic radius is about 0.15 Å larger than the chlorine atomic radius. The larger bond angle can be in part explained by the need to accommodate the larger atoms.


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Why are there so many reactions for the discharge of LiCoO2?
I've just managed to understand a bit the meaning of having the notation of 0.5 in a molecular formula, when I found a set of multiple notations for the same discharge process that started to confuse me.
One can find this set of notation for the discharge reaction of a LiCoO$_2$ cathode:

$CoO_2 + Li^{+} + e^{-} \rightarrow LiCoO_2$ (see here)
$LiCoO_2 - Li^{+} - e^{-} \rightarrow Li_{0.5}CoO_2$ (see here)
$Li_{1-x}CoO_2 + xLi^+ + xe^- → LiCoO_2$ (see here)
$2Li_{0.5}CoO_2+Li^{+}+e^-\rightarrow 2LiCoO_2$ (see here)
$Li_{0.5}+0.5Li^+ + 0.5e^-\rightarrow LiCoO_2$ (see  here)

and the list can continue...
Why are there so many notations for the same thing? It's so confusing for someone who is trying to understand this. Aren't they representing the same discharge process? What changes? And why do we have to have so many notations?
Sorry if this is too puerile, but notations were always a source of confusion for people trying to learn something from different sources.


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Does DCl have a higher rotational partition function (qrot) than HCl? And if so why?
I know that DCl has a higher moment of inertia than HCl, but I can't really explain how this actually affects the rotational partition function, I know larger molecules have a higher moment of inertia and this means the energy levels are closer, but I don't know how to explain how this has an effect on the partition function of rotation, can someone help?

Do you know how to obtain an expression for the rotational partition function (specifically, by approximating the sum as an integral and evaluating the subsequent integral)?

The partition function is the sum over states, which means over all the energy levels, thus
$$Z=\sum _i g_i \exp(-E_i/(k_BT))$$
where $g_i$ is the degeneracy of each level with energy $E_i$. The rigid rotor rotational energy levels of a diatomic molecules are $$E_J= \hbar^2J(J+1)/(2I)$$ where I is the moment of inertia $I=\mu r^2$ for reduced mass $\mu=m_1m_2/(m_1+m_2)u $ (in kg) and bond length r. The degeneracy of rotational levels is $g_J=2J+1$.
As the moment of inertial of DCl is greater than that of HCl (due to reduced mass effect) the energy levels are lower in energy in DCl at the same J than in HCl and so at the same temperature more levels are populated and the partition function is greater. This means that the exponential term in the partition function is greater in DCl at a given J and so this increases the partition function. (recall that $e^0 = 1$ and $e^{-\text{big number}} \rightarrow 0$). The energy levels of DCl and HCl are shown in the left figure and the terms in the partition function on the right as $g_J \exp(-E_J/(k_BT))$. (The curves are plotted for clarity assuming that J is  continuous not discrete). The respective partition function is the sum of the values at each J in the right-hand figure.

(To plot the curves I used a mass of Cl = $35$ and H and D, $1$ & $2$ respectively. The unified atomic mass unit $u = 1.660\cdot 10^{-27}$ kg, and assume a temperature of $300$ K. The bond length is $74.15$ pm.)
(Note that the often used integral form of the partition function has a systematic error that does not imporove as J increases.)


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How do amino acids look like at the isoelectric point?
At the isoelectric point amino-acids are neutral to the outside. It is however not known to my knowledge if the molecules are indeed in a zwitterionic form or just in its "normal" totally uncharged form, i.e. the dissociation constant for this reaction:

is not known. Does anyone know how they actually look like? What form is more probable, how realistic is the zwitterionic form?
Edit: I did a test calculation for cysteine with DFT, using a PBE functional and implicit water and found that the zwitterionic form is about 7 kcal/mol more stable, which would rather strongly indicate that the zwitterionic form is indeed dominating a lot and you would not at all find the structure on the left as the bars from wikipedia indicate (supposing that I found low lying minima in both calculations and not only local ones). Are there some other studies proving this?
Edit2: Ok the zwitterionic form is definitely more stable, that one can somehow reason from chemical knowledge or pKa's. But how much more stable, can one somehow get a dissociation constant for the reaction above?
Edit3: Since there is much confusion about my question. I want a dissociation constant for the reaction (COOH)(NH2) -> (COO-)(NH3+). Until now I think only the dissociation constants for (COOH)(NH3+) -> (COO-)(NH3+) and (COO-)(NH3+) -> (COO-)(NH2) are known if I am right and I cannot tell how these would help to calculate the constant I want...
Edit4: To clear up my question even a bit more, here is a diagram:

The red arrow and question mark is what I want to know about. The pks1 and pks2 are known, that was clear to me, but how to get any information about the equilibrium in the middle?

What studies?  -COOH group in amino acids is just like that in other carboxylic acids, and as such, should be expected to have about the same pKa, so in less acidic media than $\rm pH\approx4$ it should be mostly deprotonated. -NH2 group in amino acids is just like that in primary amines, and as such, should be expected to have about the same pKb, so below $\rm pH\approx9$ it should be mostly protonated. Putting these observations together gives...

Ok thanks for the comment, I have reformulated the question a bit

Does this answer your question? Zwitterion (amino acid) equilibrium and the equilibrium constant

While the other answers are undoubtedly correct with respect to water, the better answer is that it depends on your solvent.
The zwitterionic form is the one preferred in solid state and in solutions where the solvent can accomodate charges well, i.e. polar solvents. In less polar solvents, the neutral form can predominate to allow dissolution (rise in entropy) in spite of non-dissociation (not as favourable). However, cyclic hydrogen bonds are to be expected.
But in the end, this complicated discussion boils back down to what the other answers have said again. The only difference when changing solvent is that the $K_\mathrm{a}$ equation becomes different. Water is no longer the co-acting principle, it is a different solvent resulting in different $\mathrm{p}K_\mathrm{a}$ values (which are solvent-dependent to a certain degree).

That's of course true, it will depend on the solvent also and I did not specificy this in the question, although I was only thinking of water. It does however not provide an answer, how to get such equilibrium constant or estimate it...

@Guiste That’s what the other answer say (or maybe: gloss over).

Thanks for the answer anyways, did not think about the entropy factor for less polar solvents.

The other answers however assume that the first pKa of a protonated aminoacid is the same as that of the neutral species we are talking about...

By Henderson relation we have
$$\mathrm{pH=pI=p}K_\mathrm{a}+\log\left(\frac{[\ce{A-}]}{[\ce{AH}]}\right)$$
So if $\mathrm{pH - p}K_\mathrm{a} < 0$ then $[\ce{A-}] < [\ce{AH}]$
And if $\mathrm{pH - p}K_\mathrm{a} > 0$ then $[\ce{A-}] > [\ce{AH}]$
If you want then to know the exact proportion you need to know the concentration of your amino-acid.

Thanks for the comment, but that was not my question. It is totally clear that at the isoelectric point, we have a neutral species to the outside. But how much of this neutral species is the also partially neutral species and how much is the zwitterion (the two species in the chemical reaction I showed above)

@Guiste and if you have the exact proportions you have the equilibrium constant then.

I think you still don't understand the question, please read my Edit 3. Of course knowing the concentrations of both species (COOH)(NH2) and (COO-)(NH3+) would give us the equilibrium constant I want, but is it possible to measure this concentrations? And has someone done this?

@Guiste read my answer correctly. I have not said to know the concentration of both, I just said the total concentration of your amino acid.

Then you'll be able to find the concentration of both species. And then the constant.

You don't understand. The problem is the pKa value, we do not know this!

@Guiste if you don't know everyting ...

You ask to have other theory, I give you a possible theory with values.

@Guiste If you prefer to have only big theorical calculations tell in in your question. Everything is theory, and while experiments are still true I prefer give an answer based on experiments. I guess if you're able to understand DFT you'll be able to understand this point of view. I just gave you a simple way to do it.

The pKa value is measured. I don't understand what you are saying...

The pKa for which reaction is measured? For the reaction (COOH)(NH3+) -> (COO-)(NH3+). But how does that help me? I need the pKa for the reaction (COOH)(NH2) -> (COO-)(NH2) It must not necessarily be theory what I am asking for. Maybe there are new experimental studies on that, spectroscopic or anything, I just wanted to know if anyone have heard about something related.

I modified the question, hope it gets clearer now

This is totally known based on the pKa values of the carboxylic acid and the ammonium. The zwitterion is the predominant form.

Then what would be the pKa of the carboxylic acid for cysteine? I don't think that you can find it, since the ammonium group will be always protonated, so the constant you are mentioning is an artificial equilibrium constant which you might not be able to get...

What I am meaning is, if you want to use a thermodynamic cycle to evaluate an equilibrium constant for the reaction I showed above, you would need to start from the left educt...

Hexacoordinate-C actually has numbers. Upvote his answer. :)

He does not. I want a dissociation constant for the chemical reaction I presented above. To my knowledge this is not known so far! The pKa he presents denote the dissociations (COOH)(NH3)+-> (COO)-(NH3)+, where AH2 is the twicely protonated amino acid. To calculate a dissociation constant for the reaction I presented you would need a constant for the reaction  (COOH)(NH2)-> (COO)-(NH2), which is not the same!

Since you can change the reaction diagram from the one you want to one that branches to both (NH3+)(COOH) and (NH2)(CO2-) then reconvenes to the final zwitterion, we should be able to use the simple relationships between free energy summed over those reactions (a kJ is a kJ after all) to obtain a valid pKa for the net reaction.  I am inclined to believe Zhe is correct in asserting that said math has been done and favors the zwitterion (in fact I think I learned this in AP chem), but even if it not so certain, the method for calculating it is trivial.

Delta G (which determines pKa) for a net reaction is the sum of delta Gs of the reactions in between.  Conservation of energy ensures that it does not matter which path across your diagram you take, the pK between the two endpoints is the same.

@sqykly, please check the diagram I posted above. The path that is marked with a questionmark (red arrow) is unknown and can NOT be constructed by any route along thermodynamic paths since all reactions along the outside arrows are unknown. The only thing known are routes to the equilibrium in the center which is the constant I am interested in.

Maybe the better question we should be asking is why do you care what this value is?

Because I need to do a calculation and compare to experiment and do not know which structure to take or how to weight both


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Why is the reaction enthalpy the same as change in enthalpy for each of the reactants and products?
How come the $\Delta H$ of the reaction is the same $\Delta H$ for each of the reactants and products? For example, in an equation, the $\Delta H$ of the reaction is $244 \,\mathrm{kJ}$. When my professor did the problem, I saw that she used that same delta when converting the moles of a reactant to $244 \,\mathrm{kJ}$. But isn't that $\Delta H$ the sum of all of the products or maybe all of them combined since it says that it is the $\Delta H$ of the REACTION?

Your statement is not fully correct. $\Delta H$ for the reaction is the difference between the sum of $\Delta H_f$ for products minus the sum of $\Delta H_f$ for reactants. This works out because enthalpy is a measure of a potential (it is a state function), so the exact method of conversion (i.e. the path) between reactants and products doesn't matter, only the fact that we have a specific reactant and a specific product.


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How to make a CCl3F Fluorine NMR standard?
I am running some fluorine NMR and would like to make a bulk solution of $\ce{CCl3F}$ in $\ce{CDCl3}$ to aid in assessing the amount of fluorinated product present by having my NMR solvent act as a reference. Can anyone suggest a good ratio of the $\ce{CCl3F}$  standard to $\ce{CDCl3}$ NMR solvent?

The amount of standard should be in the same order of magnitude as the amount of compound in the NMR sample (i.e. not as a v/v% ratio of chloroform-d). This is because unlike in H-NMR where the solvent peak is just residual chloroform-h, the fluorine standard is all fluorine, so if you add too much the reference signal will swamp your actual signal and it'll be difficult to get the peak shape of your compound to come out right (it also screws with the integration but this isn't such an issue with 19F NMR as you probably dont have that much fluorine to worry about)

@NotWoodward -you should expand this into an answer, as it essentially answers the question.

You need the concentration of your reference peak to be the same order of magnitude as your solute. Accuracy of your integration depends on something called dynamic range, which refers to the relative intensities of the peaks being integrated. The greater the dynamic range; the greater the difference in peak intensities, the bigger the error.
You might want consider that for most medium-high field NMR applications, CDCl3 is used with TMS at concentrations of about 0.03%v/v. Low field and bench top applications use TMS at 0.5-1%v/v concentration.

While your answer is certainly correct, the dynamic range of modern spectrometers is large.

On second thought, it is actually not correct. Only when the concentration difference comes close to the dynamic range of the instrument, you suddenly get a big error.

@Karl, this is not DR in terms of limit of detection but rather limit of quantification - very different. Error in measurement is determined by the error of your minor component; the error increases with decreasing S/N. A S/N decrease by an order of magnitude gives an increase in uncertainty by an order of magnitude. For S/N of 1000, uncertainty is ~0.1%; for S/N 100, uncertainty ~1%. So, as per the question, it is very important to get the relative concentrations correct. There are many other things to consider for quantitative measurements also, and depends on level of accuracy needed.

And FIW, selection of correct relaxation delay for 19F is critical, depending on the types of systems you are analysing. 19F have a big range in T1 - something like CCl3F will have a T1 of a few seconds. Many F-containing molecules will have much shorter T1s. This will compound error across multiple scans if not taken into account, when comparing multiple samples.

But that is not a matter of dynamic range, but only of sensitivity. The uncertainty is the sum of the inverses of the S/N ratios of each component to be compared if I'm not very mistaken.

There are three points to take into account, imo:

Dynamic range: If the concentration of the reference peak is very large, your analyte vanishes in the digital noise. Modern spectrometers have a large dynamic range, which ameliorates the problem, but you should adjust the gain.
Signal overlap: A huge reference peak (which has a Poisson lineshape, i.e. broad "feet") distorts the baseline to a great distance. Less of a problem with 19F, because the CS range is large.
Sensitivity: It'd be bad if you had to measure four or eight times as many scans, just to get a proper signal from your reference peak.

So I'd recommend a reference with as many fluorine atoms as your analyte in the smallest expected concentration (per peak, of course!), and perhaps not more than one order of magnitude more.


text:
Reaction between magnesium and hydrochloric acid
I am reacting different masses of Magnesium with hydrochloric acid to find the temperature change.
I need 2 explanations:
First of all why is the reaction exothermic?
Secondly, why does the difference in temperature increase as the mass of magnesium increases?

(1)
The reaction I think you're talking about is regarding magnesium solid and (concentrated?) HCl:
$Mg(s) + 2HCl(l) \rightarrow H_2(g) + MgCl_2(aq)$
The following are approximate bond enthalpies from this site and this site.
The bonds that would be broken are:

H-Cl single bond (x 2), $\Delta{H_{break}} = \text{+431 kJ/mol}$

The bonds that would be made are:

Mg-Cl single bond (x 2), $\Delta{H_{make}} = \text{-406 kJ/mol}$
H-H single bond (x 1), $\Delta{H_{make}} = \text{-436 kJ/mol}$

Approximately, $\Delta{H_{rxn}} = \sum{|\Delta{H_{break}}|} - \sum{|\Delta{H_{make}}|} = \text{-386 kJ/mol}$.
The reaction is exothermic because the energy absorbed in breaking bonds is smaller overall than the energy released in making bonds (and exothermic by definition is a negative enthalpy). Even though these numbers are approximate, they are clearly different enough to make such a conclusion.
(2)
The difference in temperature is related to the enthalpy. In "lab bench conditions", otherwise known as constant pressure conditions, when one performs a (coffee-cup-) calorimeter experiment to determine the change in temperature, the heat flow $q$ is essentially equal to the enthalpy $\Delta{H}$.
Recall that $q$ is often reported in $J$, joules. $q$ is an extensive property, so the more substance (in terms of mass) the process involves, the higher the magnitude of $q$ involved. Since $q$ $\alpha$ $\Delta{T}$, the more heat involved, the greater the change in temperature.
For an exothermic process, heat flows out from the reaction into the calorimeter solution, so $\Delta{T_{sys}} > 0$, i.e. the reaction releases heat into the solution, and the solution gets hotter. Note that $q_{cal} = -q_{rxn}$.
As a result, although $q_{cal} > 0$, $[q_{rxn} = \Delta{H_{rxn}}] < 0$, which has been established from (1) for this exothermic process, and when there is more $Mg(s)$ (assuming excess $HCl$), $q_{rxn}$ is larger, which corresponds to a larger $\Delta{T_{sys}}$.