text:
Storing kinetic energy in bonds
Let's assume a setup with a static linear molecule with three identical atoms connected by bonds and a single atom, identical to the other three, being shot at the molecule. Let's also assume that everything happens in 1D and can therefore be illustrated by this simple scheme:
(projectile) A ->    A-A-A (molecule)
Where the direction of the arrow is the direction of motion of the projectile and the dash signs in the molecule are the bonds between atoms.
[EDIT - must change next paragraph as the description does not fit what would happen: stand-by, no time to do it right now!]
Assuming a low-energy, non-relativistic, perfectly elastic collision, the projectile will stop dead in its track while the molecule will be set in motion at 1/3rd of the projectile's speed, as it has 3 times the projectile's mass (conservation of Momentum). Balancing also the equations for the Kinetic Energy easily reveals that 2/3rds of projectile's kinetic energy must have been locked away in internal vibrations of the molecule.
Now, it's easy for us to calculate these numbers by considering the projectile and the molecule as just two separate systems and conveniently attaching a total mass to each. But in reality the universe doesn't group atoms into molecules, does it? Each bond in the setup doesn't "know" about the existence of the other bond and each atom could be said to be unaware of all but the atoms bonded or colliding with it.
In this context, not "knowing" total masses, how does the universe "calculates" that in this case the molecule must store 2/3rds of the projectile kinetic energy into internal vibrations, no more, no less?

""Assuming a low-energy, non-relativistic, perfectly elastic collision,"" means that the molecule will store nothing at all, but the single atom impactor will recoil carrying away part of the engergy. Before starting questions resting on wrong assumptions, read about elastic collisions.

Georg, I've read about elastic collision but I might still be making mistakes. Are you saying that my description of what happens is not correct? Are you saying that the projectile will not only stop after impact but also bounce backward? I have in mind some experiments we did in the physics lab during high school (20 years ago!) on an air-rail, and I remember little carriages stopping on their track, not bouncing back. But I could be wrong.

Yes of course! The impactor will stop dead only if both masses are equal! If a ball hits the ground, does the earth fly away?

Ah,thanks for reminding me about the case where that happens. That helps: I will reformulate the question in light of it. Thankfully, the specific example is not the focus of the question.

The universe doesn't calculate anything. Energy and it's conservation and everything that follows from it is a human-made concept, a physical model. It describes the things that happen in nature quite well (in the right circumstances), but it is only a model. In how far this lends some kind of "reality" to the concept/model I couldn't say. It's more of a philosophical question. Physical laws only describe how nature works, but not why it works the way it does.
Edit:
Concerning the physical content of the collision you described: I think of it in the following way: The projectile hits the first atom of the molecule, let's call it A1, and transfers its momentum to it. Now, this atom moves in the direction along the bond axis directly towards the next atom, A2. While it does so, it has to work against the spring force of the bond between A1 and A2. Thus, kinetic energy of movement is converted to kinetic energy of vibration. A2 will now be accelerated in the direction of the next atom, A3, and it has to work against the spring force of the bond between A2 and A3 and(!) the restoring force brought about by the bond between A1 and A2. Again kinetic energy is converted to vibrational energy. But now you have a system of coupled oscillators. This process will be repeated for every atom in the molecule until every bond vibrates. So, it is not true that

Each bond in the setup doesn't "know" about the existence of the other
  bond and each atom could be said to be unaware of all but the atoms
  bonded or colliding with it.

They are coupled through the spring forces of the bonds and thus each atom "feels" the mass of every other atom in the molecule as well as the bond strength of every bond in the molecule.

Thanks for your answer Philipp. My suspicion is that there is an answer at a quantum-mechanical level, or perhaps even at a classical level thinking bonds as springs with oscillations partially cancelling each other. But I can see how we are otherwise potentially close to the edge of metaphysics and therefore philosophy. Let's stick to the physical world then, and let's see if anything comes out.

@manu3d I added some more physical thoughts to my answer, which I hope are helpful.

Dear Philipp, thank you, your later edit is much appreciated. Indeed it is a good qualitative description of how things happen. Clearly the molecule's atoms would like to move at the same speed as the projectile, so much so that if no bonds were involved we'd be dealing with a newton's cradle and the last atom would depart the molecule with the speed of the projectile. What is left unclear is how the individual interactions between single atoms and bonds generate that split 2/3rds (vibration) + 1/3rd (net motion) without "counting" the atoms involved prior to the impact.

Concerning your end-of-answer comment, I disagree. Of course all atoms in a molecule "feel" -indirectly- each other, as a movement or impact somewhere in the molecule is likely to be "felt" everywhere else in the molecule. -However-, if you think of a linear molecule 1000 atoms long, unless you invoke some of the strangest (and minimal) quantum effects such as probability waves having non-zero values even at astronomical distances, we can reasonably assert that A1 is highly unlikely to be influenced by A1000 and therefore, for all intents and purposes, doesn't know about it.

@manu3d Concerning your first comment: Unfortunately, I can't give you the exact equations on a microscale for this. Maybe one would get to it by solving the equations of motion for a system of coupled harmonic oscillators. This might be a question for physics.SX.
Concerning you second comment: I see your point. Maybe I have used the word "feel" injudiciously here, since during the first impact the influence of the other atoms on the movement of A1 will be very small. The "feel" part comes later when the bond starts to swing as then A2 is influenced by A3 and so on. That's how I meant it.

Read questions careful: ""perfectly elastic" says all, including no/low knowledge in elementary physics.

@Georg While reading the question I've thought that "perfectly elastic" can't be true. But since I didn't have the time to go deeper into the matter I simply ignored that bit and gave a description of the process that seemed logical to me. If you think it necessary I could add a short note that such a collision where kinetic energy is (partly) transformed into vibrational energy must be inelastic.

Do You think someone dwelling in such basic errors will understand a sophisticated answer?

@Georg I guess it could point him in the right direction to learn more about it. But since I too thought that a complicated answer will be of little use to the OP I held it as intuitive as I did and of course I was too lazy to look into differential equations of coupled harmonic oscillators at the time :).

I just noticed that someone downvoted my answer. Could the one doing this please tell me why, so that I could improve my answer?

Thank you Phillip for providing Georg with some basic motivation on the matter. He doesn't seem to be particularly interested in providing an answer but it is somehow puzzling how much time he's wasting trying to convince me I shouldn't have even asked! As if that's ever going to work! =D

No idea about the downvote. While I cannot just yet accept it as -the- answer, I would gladly upvote it had I enough reputation to do it.


text:
Anti-Markovnikov Effect
We have seen that $\ce{HBr}$ shows an anti-Markovnikov effect when it reacts with alkenes in the presence of peroxides. Why does only $\ce{HBr}$ shows this effect? How about the other hydrogen halides $\ce{HX}~(\ce{X} = \ce{F}, \ce{Cl}, \ce{I})$?

See here for detailed discussion or here for brief overview.

Main reason is behind is the reaction mechanism involved. $\ce{HBr}$ or other halides usually react with alkenes through an electrophilic addition. Due to the weak $\ce{O-O}$ bond, peroxides act as a source of free radicals. Thus in the presence of peroxides, a free radical chain pathway is followed.
The other hydrogen halides are reluctant to this reaction, because the chain reaction can't proceed due to energy requirements for the generation of the corresponding halogen free radicals.

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text:
Why does sulfate have this structure?
Look at the structure for sulfate:

Why does sulfur form so many covalent bonds. Its valency is only $-2$, it only needs two electrons, yet here it's getting $6$.
The same thing happens with phosphate.

Phosphorus has a valency of $-3$, yet, it forms $5$ covalent bonds.

I think the key thing is that Sulphur can have a range of valencies. It just usually has an oxidation number of -2. In this case each Oxygen is -2 and the Sulphur is +6.

Why does sulphur have multiple valencies?

Although it seems that current research favours a 3c,4e bond model with the D orbital contribution being small. I can't read the papers though as they are paywalled.

I think an important point to mention here is that Lewis dot structures, and the octet rule, are simply models that describe experimental observables. For example, the concept of valency works well with organic molecules but was challenged by Alfred Werner in the development of bonding models that adequately described coordination compounds.
The electron-dot model of bonding is so meaningful because it can be adapted to explain various experimental observables.  You can draw two perfectly reasonable electron dot structures for $\ce{SO4^{2-}}$: one with all $\ce{S-O}$ single bonds and one with two $\ce{S=O}$ double bonds (and the additional resonance structures).  The question is, which electron-dot structure best represents the real structure?  We can use the concept of formal charges  to predict that the structure with double bonds is most likely closer to reality.
We then look at the experimental data (which is reported on the sulfate wikipedia page) for S-O single and double bond lengths.  The S-O bond length in $\ce{SO4^{2-}}$ is 149 pm, which is shorter than that observed in sulfuric acid (157 pm) and very close to the gas-phase bond length of sulfur monoxide which is 148 pm.
So, at the end of the day, an electron dot structure is a type of model that describes bonding.  It has some additional features such as expanded octets and formal charge that help broaden its applicability to more situations; however, models such as this one don't tell us what an atom "needs"; rather, it provides an explanation for how we observe atoms "behaving".

Ultimately those structures are wrong.  Most of the community of chemistry educators teaches that formal charge reduction creates hypervalency.  All modern research and models shows this not to be the case.  You can't hybridize a d orbital and even get the proper tetrahedral geometry here.  Reducing formal charges to 0 makes no real world sense, when sulfur has a measured empirical charge of around 1.7 in sulfate.
Sulfate is 4 single bonds.  The bond length difference is due to the fact that the sulfur carries extra charge, as does the oxygens.  This ionic like bond is of course going to be shorter than more neutral forms of S-O.  Further, why would two double and two single hybrid bonds average to be the same length as a double bond?
Essentially any general chemistry solution that invokes d-orbitals as it's solution, outside of transition metal chemistry, is wrong.  Sulfur hexafluoride is not in fact 6 single bonds.
Unfortunately, biochemists are not going to update their picture of phosphate anytime in the next fifty years because no one cares.
The fact that basically every science major in the world learns this wrong is abhorrent.

Why blame "the biochemists"? In biochemistry textbooks, the formula for phosphate is often $\ce{P_i}$, sometimes even with a nice circle around it.

It's a highly debated topic and seems that none has won. I thought it to be fully explainable through MOT. However, while searching I found this link. It cleared few of my misunderstanding.
Have a look. You should be able to find the answer yourself. If not leave a reply, I will update the answer accordingly.
Again thanks to Nick for citing the (3c,4e)-bond model.

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text:
Identifying the gas given ratio of its specific heats
The ratio of the heat capacities $\frac{C_\mathrm p}{C_\mathrm v}$ for one mole of a gas is $1.67$. The gas is:

a) $\ce{He}$
  b) $\ce{H2}$
  c) $\ce{CO2}$
  d) $\ce{CH4}$

I know how to answer this by analysing the degrees of freedom of the gas molecule. Is there any other way to derive the ratio?

This question does not make sense without giving the temperature of the gases.

I'd go about such a question looking at the easiest thing first: ideal gases. So, what do you know about the relations between $C_{p}$ and $C_{v}$ (both being the molar heat capacities at constant pressure and constant volume, respectively) for ideal gases? For one thing, it is well known that
\begin{equation}
  C_{p} - C_{v} = R
\end{equation}
with $R$ is the universal gas constant. You can rearrange this equation for the ratio you are after:
\begin{equation}
  \frac{C_{p} - C_{v}}{C_{v}} = \frac{R}{C_{v}} \\
  \frac{C_{p}}{C_{v}} - 1 = \frac{R}{C_{v}} \\
  \frac{C_{p}}{C_{v}} = \frac{R}{C_{v}} + 1 \ .
\end{equation}
Now, you need to find $C_{v}$. Here, the degrees of freedom that you mentioned come into the game, since the equation that helps us out is the equipartition law, stating that each excited quadratic degree of freedom contributes $\frac{1}{2} R T$ to the molar internal energy $U$
\begin{equation}
  U = \frac{1}{2} R T (f_{\text{trans}} + f_{\text{rot}} + 2 f_{\text{vib}})
\end{equation}
with $f_{\text{trans}}$, $f_{\text{rot}}$ and $f_{\text{vib}}$ being the quadratic degrees of freedom for translation, rotation and vibration respectively. It is important to note that only excited degrees of freedom contribute to $U$. Which ones are active depends heavily on temperatur. Vibrations are usually not excited at room temperatur (there are exceptions, e.g. iodine) - they usually need temperatures in the order of $1000 \, \text{K}$ but that varies greatly. Rotations are much easier to excite - I know of no example where they aren't excited at room temperature. Translations take very little energy to excite and can always be considered active. If all degrees of freedom are active you get: For linear molecules $f_{\text{trans}} = 3$, $f_{\text{rot}} = 2$ and $f_{\text{vib}} = 3N - 5$, where $N$ is the number of atoms in the molecule. For non-linear molecules $f_{\text{trans}} = 3$, $f_{\text{rot}} = 3$ and $f_{\text{vib}} = 3N - 6$.
Since $C_{v} = \frac{dU}{dT}$ you get
\begin{equation}
  C_{v} = \frac{1}{2} R (f_{\text{trans}} + f_{\text{rot}} + 2 f_{\text{vib}}) \ .
\end{equation}
Substituting this into the equation for the ratio between $C_{p}$ and $C_{v}$ yields
\begin{equation}
  \frac{C_{p}}{C_{v}} = \frac{2}{(f_{\text{trans}} + f_{\text{rot}} + 2 f_{\text{vib}})} + 1 \ .
\end{equation}
Now, you only have to know how many degrees of freedom each of the molecules has and plug them into the equation.

Hey Philipp, I like your answer, but I think you were a bit to fast with what $n$ stands for (and whether $U$ is molar internal energy or internal energy)… in any case, your statement below the first equation (“$n$ is the number of atoms in the molecule”) is wrong.

@F'x You are right. It's been a while since I took the thermodynamics class and my memory seems to be worse than I hoped. Thanks for pointing that out. Usually I put a little more thought into my answers (as my longer ones hopefully show) but this time I thought I'd remember it correctly and thus didn't look it up again. I corrected my answer and hope everything is all right, now. If my answer still lacks something, please tell me.

Your answer only is correct for high enough temperatures. rotational and vibrational degrees are not active at low temperatures. So, the question does not make sense without giving the temperatures of the gases.

@Georg Of course, you are right that rotational and vibrational degrees of freedom are not always active. I will add that to my answer. But rotations are usually excited at quite low temperatures and since no temperature was given in the question I assumed room temperature. And for room temperature the choice is clear: $\ce{He}$ since it is the only gas that cannot possibly have rotational or vibrational degrees of freedom and thus $\frac{C_{p}}{C_{v}} = \frac{2}{3} + 1 \approx 1.67$.

@Philipp what I like to see is mentioning the problem at least. Whether rotation is active in Hydrogen at room temperature or not is not the important detail.

@Georg I updated my answer. Is this enough detail for you or would you add something.

@Philipp great answer now!

@F'x Thanks. One thing I like about this site is that one has some knowledgeable people around that point out problems so that answers can be improved on points that haven't been thought of initially.


text:
Is it dangerous to inhale the steam coming from electrolysis of salt-water solution
Is chlorine in there that accumulates (at the anode) and emerge out as chlorine gas? I was worried since I added a pinch of salt to the steamer in order to hasten the heating of tap water when I caught a cold (I should've just boiled a pot of water on the stove). I'm worried since chlorine gas is bad for the respiratory system. Also I'm curious to know what is this brown precipitate under the solution - hydroxides or cupric-oxides from the electrode?

Yes, it is possible to generate chlorine gas in an aqueous solution containing chloride ions.  Further, I wonder why adding a salt will "hasten" the heating of tap water, since we typically teach that addition of solutes results in the boiling point elevation of solvents.

Yes, but the conductivity increases and you didn't mention if it was dangerous.

Yes, inhaling chlorine gas is dangerous.  As for the conductivity of the solution, why does this matter if you are trying to boil it rather than electrolyze it?

So if I understand the question correctly, you heated up water with some salt by performing electrolysis.
If so, you would also generate chlorine gas. However, the heat capacity of water is really large, so I doubt that you would even be capable of bringing water to a boil by just jolting current through it.
To answer your question: Yes, when performing electrolyzation of any solution you should apply the usual precautionary measures, such as working in an open space or under a fume hood. Under no circumstances should it be possible for you to inhale the gases produced.

"However, the heat capacity of water is really large, so I doubt that you would even be capable of bringing water to a boil by just jolting current through it." Despite the high heat capacity of water,  generating steam results from  more energy added (over time) than is lost.

Some chlorine is is evolved  but much more of the chlorine will react with the sodium hydroxide forming sodium chlorate  (NaOCl), which is how a bleach is made, unless your using a membrane to keep the ions separate.

Your question is not very clear.
Did you just heat the salt water? Or you have put electrode to electrolysis water? What material is your electrode? Copper?
Let me assume you are using copper electrode to electrolyze brine. Then you will have the following reactions:
Cathode: $$\ce{Cu - 2e -> Cu^2+}$$
Anode: $$\ce{2H2O + 2e -> H2 + 2OH-}$$
Over all:
$$\ce{Cu + 2H2O -> Cu(OH)2 + H2}$$
So, when copper is your electrode, you will have no $\ce{Cl2}$ gas for sure. You will only breath in hydrogen gas.
If you use graphite as the electrode, you will get overall reaction:
$$\ce{NaCl + 2H2O -> NaOH + H2 + Cl2}$$
It is possible to have the $\ce{Cl2}$ gas. However, if the amount of $\ce{Cl2}$ is not really big, it will be absorbed by water especially when it is basic.
$$\ce{Cl2 + 2OH- -> ClO- + Cl- + H2O}$$
And over all:
$$\ce{NaCl + H2O -> NaClO + H2}$$
No gas emitted. Electrolysis of water is actually one way to make bleach.
The conclusion is: it is safe to inhale unless you are using extremely big battery.

Some Chlorine gas will likely evolve from the solution but your right in that your really making bleach.

Your also make a decent point that  it will require a lot of energy to make appreciable unsafe amounts, but i would recommend against inhaling bleach and chlorine gas if you can avoid it.  On a further note, the fact that steam is being created implies that a fair amount of energy is being added. Not knowing the applied voltage leaves us not knowing how efficent the electrolysis was versus resulting in waste heat.

Electrolysis:
Your question "Is it dangerous to inhale the steam coming from electrolysis of salt-water solution?"
The answer short answer is:  It is not safe.
The long answer involves relative safety, and depends on a few things.
First thing  to understand is that chlorine is produced at the anode and sodium hydroxide produced at the cathode. Most of the chlorine will combine with sodium hydroxide producing NaOCl, in other words bleach. But, some chlorine gas may evolve from the anode because sodium hydroxide has not yet migrated to the anode in sufficient quantity. Either way, it is not safe to breath bleach or chlorine.
The relative danger depends on:

How efficient is the electrolysis: Higher voltages though causes more current to flow, are less efficient at electrolysis because it produces more heat than actual electrolysis for each unit of energy applied. More heat, thus steam more steam per electrolysis product, is comparatively less harmful.
Quantity of salt: Adding only a pinch of salt means that either very little bleach or even less chlorine gas will be produced; In such low concentrations either product is likely to remain dissolved in the water rather than evolve.

Steamers:
On the other hand, you mentioned adding a pinch of salt to a steamer. Conventional steamers do not perform electrolysis.  Instead, electricity is passed through a restive load with the only product being heat. Adding a pinch of salt may raise the boiling point, and thus requires more heat to make steam, but it would be completely harmless, except for the dangers of heat from the steam.

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text:
Optical isomerism in Co-ordination Compounds
Optical isomerism in normal organic molecules can be found out by looking for chiral carbon, and the number of isomers can be found out by drawing different spacial orientations of the same.
But, I don't know how to check whether a given coordination compound is optically active or not, and how many optical isomers does it have. I don't know how to draw different spacial orientations with all those polydentate ligands.
I would like to learn the general guidelines for finding optical isomerism in co-ordination compounds. I would like to solve questions like how many no. of optical isomers are there for $\ce{[CoCl2(en)2]^+}$

Any 3rd-year inorganic chemistry textbook worth its weight in salt will provide some information about isomers in coordination compounds.  An optically active compound contains no rotation-reflection symmetry element.  Typically, this means that if a molecule has an inversion center or a mirror plane then it is not chiral.
It takes a systematic brute force and ignorance approach to determining the number of isomers a coordination compound has, and then a bit more visual thinking to determine if any of the isomers are enantiomers.  The following figure is taken from Miessler, Fischer, Tarr 5th edition and is useful for identifying isomers of 6-coordinate complexes.

Note that a 6-coordinate complex with 6 unique ligands in an octahedral field would be enantiomers. In fact, there are 15 different isomers, each of which is chiral resulting in a total of 30 isomers!
The method of determining isomers this way is to formulate a structure of the type M<ab><cd><ef> where <ab> indicates two ligands that are trans to each other.  Then, you start varying a, b, c, ... until you've come up with as many unique combinations as possible.  For your case, you have 2 instances of a bidentate ligand.  Bidentate ligands are typically designated with capital letters in this notation, so the formula for your molecule would be $\ce{M(AA)_2b_2}$.  Further, since bidentate ligands are too small to coordinate in a trans conformation, you will decrease the number of possible isomers.  One isomer, for example, would be M<Ab><A'b><A'A>, where I use A and A' to designate the different $\ce{(en)}$ ligands.
The answer to your question, then, is that there are two isomers, and they are enantiomers, since you can't rearrange the A, A' and b in the formula above to come up with another unique configuration.

There is an easy way of identifying optical isomers in coordination compounds.

Roughly draw the structures of all possible isomers of the compound by
interchanging the ligand positions.
Cross out those structures whose mirror image overlaps with itself.
The remaining are optical isomers of the given compound.

Once you do this a few times, you can draw this in your head using your imagination and figure it out faster.
To check if a coordination compound is optically active, just see if if the mirror image of the compound overlaps with itself (for the given isomer), if it doesn't, the compound is optically active.

This is correct, but if someone actually has significant problems with optical isomerism, then will stumble somewhere.

If you know the theory well , and have a basic understanding of chirality , enantiomers and optical activity in organic compounds , this should make perfect sense. :p

Of course, but guys who know it all don't need help. BTW there shouldn't be any space before your commas.

@Mithoron my answer is way less complicated than the first ! , I used this in my 12th grade exams a month ago...!

It's OK, maybe a bit like TL:DR, just saying that if you answer in the future, OP's may be nagging you about details etc., good luck though.

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text:
What is the structure of N₅P₃?
What is the structure of the molecule $\ce{N5P3}$? $\ce{N5P3}$ is not documented online, so could anyone please comment on the structure of the molecule:

How many ligands does each phosphorous atom have?
How many ligands does each nitrogen atom have?

I'm not sure why previous answer cited only the first paper on phosphorus(V) nitride $\ce{P3N5}$. A couple of years later the same team reported three polymorphs $\ce{α,β,γ-P3N5}$ with three-dimensional network. For $\ce{α-P3N5}$ and $\ce{γ-P3N5}$ crystal structures have been refined. Both modifications demonstrate vastly different structural types and coordination of pnictogens.
Low-pressure phase $\ce{α-P3N5}$
Data obtained via powder neutron diffraction with higher resolution by using synchrotron radiation. The crystal structure was determined by direct methods and refined by the Rietveld method. [1, 2]
Structure description from [1]:

In $\ce{α-P3N5}$ zweier single chains occur parallel to $[110]$ and
  $[1\bar{1}0]$, in which the $\ce{PN4}$ tetrahedra are connected alternately through edge- and corner-sharing. These zweier single chains are linked together through additional vertex-sharing $\ce{PN4}$ tetrahedra. [...]  The network structure of connected $\ce{PN4}$ tetrahedra in $\ce{α-P3N5}$ could be characterized by analogy to condensed silicates by the specific distribution of the occurring $\ce{P_nN_n}$ ring sizes.

Liebau formula: $_\infty^3\ce{[P_3^{[4]} N_3^{[2]} N_2^{[3]}]}$ *

Figure 1. Crystal structure of $\ce{α-P3N5}$, view along $[010]$. Purple coordination tetrahedra correspond to $\ce{[PN4]}$.
Low-pressure phase $\ce{β-P3N5}$
$\ce{β-P3N5}$ was reported in a polymorphic mixture with $\ce{α-P3N5}$ and is it's stacking modification (no further crystallographic investigation has been published) [1]:

On the basis of the solved crystal structure of $\ce{α-P3N5}$, we are now working on the determination of structural models for $\ce{β-P3N5}$, and disordered $\ce{P3N5}$. The HRTEM results show that similar to the situation in the polytypes of the different
  stacking variants are based on the same structural modules, which are linked in different ways.

High-pressure phase $\ce{γ-P3N5}$
Data obtained via powder X-ray diffraction. The crystal structure of $\ce{γ-P3N5}$ was determined by direct methods and refined using the Rietveld method. [3]
Structure description from [3]:

The novel high-pressure modification consists of a polymeric three-
  dimensional network structure of linked $\ce{PN4}$ tetrahedra and
  tetragonal-pyramidal $\ce{PN5}$ units. [...] Rods from trans-edge-sharing $\ce{PN5}$ units running along $[010]$ are linked by vertices, leading to the formation of layers perpendicular to $[100]$. [...]  In accordance with the pressure–coordination rule a partial increase of the coordination number of the phosphorus atoms from four in $\ce{α-P3N5}$ to five in $\ce{γ-P3N5}$ is observed. [...] The density of $\ce{γ-P3N5}$ is $32\%$ higher than the density of $\ce{α-P3N5}$. [...] Therefore the high-pressure phase could show interesting materials properties (e.g. great hardness or low compressibility).

Liebau formula: $\ce{P_5^{[5]}[ P^{[4]} N_3]N_2}$ *

Figure 2. Crystal structure of $\ce{γ-P3N5}$, view along $[010]$. Purple coordination tetrahedra correspond to $\ce{[PN4]}$; yellow coordination pyramids — to $\ce{[PN5]}$.
* using Liebau's nomenclature for silicates, where the superscripted number in square brackets after the element symbol denotes the coordination number.
References

Horstmann, S.; Irran, E.; Schnick, W. Synthesis and Crystal Structure of Phosphorus(V) Nitride α-P3N5. Angewandte Chemie International Edition in English 1997, 36 (17), 1873–1875. https://doi.org/10.1002/anie.199718731.
Horstmann, S.; Irran, E.; Schnick, W. Phosphor(V)-nitrid $\ce{α-P3N5}$: Synthese ausgehend von Tetraaminophosphoniumiodid und Kristallstrukturaufklärung mittels Synchrotron-Pulver-Röntgenbeugung. Zeitschrift für anorganische und allgemeine Chemie 1998, 624 (4), 620–628. https://doi.org/10.1002/(SICI)1521-3749(199804)624:4<620::AID-ZAAC620>3.0.CO;2-K.
Landskron, K.; Huppertz, H.; Senker, J.; Schnick, W. High-Pressure Synthesis of $\ce{γ-P3N5}$ at 11 GPa and 1500 °C in a Multianvil Assembly: A Binary Phosphorus(V) Nitride with a Three-Dimensional Network Structure from $\ce{PN4}$ Tetrahedra and Tetragonal $\ce{PN5}$ Pyramids. Angewandte Chemie International Edition 2001, 40 (14), 2643–2645. https://doi.org/10.1002/1521-3773(20010716)40:14<2643::AID-ANIE2643>3.0.CO;2-T.

$\ce{N5P3}$ is more commonly written as $\ce{P3N5}$, and known as triphosphorus pentanitride. It's a crystalline solid at ambient conditions and not a molecular compound. From the first publication that reported the pure compound and its structure [1]:

In the solid a three-dimensional cross-linked network structure of corner sharing $\ce{PN4}$ tetrahedra has been identified with 2/5 of the nitrogen atoms bonded to three P atoms and 3/5 of the nitrogen atoms bonded to two P atoms.

Only the combination of spectroscopical and diffraction methods, especially electron microscopy for structural analysis, enables a detailed structural and crystallographic characterization of $\ce{P3N5}$ as no conventional single-crystal data for structure determination have been available for this compound.
A representative for an unique binary structure type was found for the first time which combines a tetrahedral structure with the very rare molar ratio of 3:5 for the constituent elements.

References

Schnick, W.; Lücke, J.; Krumeich, F. Chem. Mater. 1996, 8 (1), 281–286 DOI: 10.1021/cm950385y.


text:
Cannizzaro or aldol condensation?

The reaction between p-methylbenzaldehyde and $\ce{NaOH}$ is an example of:
A) Cannizzaro reaction
B) Aldol condensation
C) Hydrolysis
D) Haloform reaction

I selected B, aldol condensation, as the given compound is an aldehyde and it reacts with $\ce{NaOH}$. But the given answer is A. Why is it so? The $\ce{NaOH}$ is not mentioned to be concentrated.

The aldol condensation needs $\alpha$-hydrogens next to the carbonyl group in order to proceed. Check out this reaction equation from the Wikipedia article:

One of these hydrogen atoms is removed to form the enolate. See this mechanism (also from the Wikipedia article).

Now, look at the structure of p-methylbenzaldehyde. This molecule does not have any $\alpha$-enolizable hydrogen atoms. The $\alpha$-position is an $sp^2$-hybridized carbon atom in the aromatic ring. The aldol condensation is thus not possible.

We can also rule out the haloform reaction because there are no halogen compounds given. There are also no hydrolyzable functional groups (halides, ethers, esters, anhydrides, amides, nitriles, etc.).
The only remaining possibility is the Cannizzarro reaction, which requires a benzaldehyde derivative or other aldehyde lacking $\alpha$-enolizable hydrogen atoms.

Is it possible to provide Pka of hydrogen of methyl group in paramethylbenzaldehyde?


text:
Can carboxylic acids be oxidised further?
I know that compounds like alcohols can be further oxidized to aldehydes (or ketones) and to carboxylic acids. But, is carboxylic acid the best oxidized form of a given compound?
What happens when carboxylic acids further reacts with oxidizing agents like $\ce{KMnO4}$?

The maximum oxidation product would be $\ce{CO2}$.

In carboxylic acid, the carbon atom of the COOH group is already in high oxidation state, namely +3. In order to reach a higher oxidation state (+4), it requires breaking the C—C bond, typically to form molecular CO2. Thus, oxidation of carboxylic acid in strongly oxidizing conditions is known as a decarboxylation reaction.

I guess theoretically you could oxidize a carboxylic acid ester to a carbonate ester.  Not sure if there is any practical way to do that of course.  But you could imagine some crazy probably impossible oxygen insertion reaction that turns methyl acetate to dimethyl carbonate.


text:
Entropy and enthalpy change on mixing liquids
Question:

A concentrated solution of copper sulphate, which is dark blue in colour, is mixed at room temperature with a dilute solution of copper sulphate, which is light blue. For this process:

Entropy change is positive, but enthalpy change is negative.
Entropy change and enthalpy changes are both positive.
Entropy change is positive and enthalpy does not change.
Entropy change is negative and enthalpy change is positive.

The final conc. of the mixture would be less than the initial concentration of the concentration $\ce{CuSO4}$ given.
Therefore  I think the entropy change is positive. But, what about enthalpy?  What factors should I look into to find answer to these types of questions?

Dilution in general is a process involving negative enthalpy change. Therefore, since the final concentration is lower than the initial, therefore the enthalpy change is negative, making the mixing favourable under any temperature.

The enthalpy change could be positive or negative, just like the enthalpy change when dissolving a solid in a liquid. Of course, for the specific case here, you could look it up.

This is difficult, because we do not know what is your current understanding and at what “level” of approximation you expect the question to be answered.
At a undergraduate level, or in the absence of any details (or thermochemical data or access to thermochemical tables), I would go with option (3).

Enthalpy: we are not given any indication at what is the enthalpy of mixing. In the simplest approximation, the enthalpy of mixing is zero, as in an ideal solution. It would correspond to a regime where the enthalpy of dissolution ($\Delta H_{\text{sol}}$) does not depend on the concentration of the solution. It's probably a good approximation, until you reach high concentrations.
Entropy: the entropy of mixture is positive, i.e. the process of mixing the two solutions is spontaneous. It can be explained by this reasoning.

You could argue that the copper and sulphate ions are (on average) better solvated in a more dilute solution. Furthermore, the ions with the same charge can "go out of each other's way" more easily in a dilute solution, so that the correlations between the ions are weaker. Both of these effects would lead to an energetically more favourable state, so that the enthalpy change should be negative (exothermic reaction).

To the extent that there is an effect, I don't think the sign of the enthalpy change can be argued so easily. It depends on the nature of the intermolecular interactions, and there may be other important effects than those you describe (pairing, for example).

@F'x Admittedly, my answer might be based too much on intuition. I certainly haven't considered all possible effects. Would you really say, that the enthalpy of dissolution doesn't depend on the concentration for the case at hand? Maybe, I should be more careful with my intuition then.

On first approximation, it's constant. Then comes a first-order term in $c$, but its sign is not necessarily clear-cut. Just like enthalpy of mixing in real life can be either positive or negative, thermochemistry is full of weird corner cases :)


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Bromination of cyclohexene

Cyclohexene is reacted with bromine in carbon tetrachloride in the dark. The product of the reaction is:

I know that reaction with bromine in $\ce{CCl4}$ will result in vicinal dibromocyclohexane. So the answer is either A or B. But how to decide between A and B? Since its an SN2 reaction it would lead to inversion of the reacting compound. Does this have to do anything with the answer?

Philipp's answer is good, but structures A and B are both drawn in misleading ways. Both appear trans (one bromine is up and one is down), but neither correctly shows the conformation (should be either diequatorial or diaxial).

I updated my answer.

I just feel compelled to add: the drawings of the chair conformers in that question are frankly appalling.

To answer this question you have to look at the reaction mechanism. Since the reaction is conducted in the dark, you don't have any radicals to worry about. Under these condition bromine reacts with an electron-rich alkene in a electrophilic addition reaction in the course of which a cyclic bromonium ion is formed. This bromonium ion is highly reactive and in the absence of other nucleophiles reacts with the leftover $\ce{Br-}$ ion in an $\text{S}_{\text{N}}2$ reaction which leads to the opening of the 3-ring and formation of the trans product.

Update:
As @BenNorris pointed out, the ring structure in the question are drawn wrongly. So, I will add the correct version: The dibrominated product will initially be formed in a diaxial conformation (this is always the case due to the ring geometry and the mechanism of the $\text{S}_{\text{N}}2$ reaction). This is usually a very unfavourable comformation so the ring will subsequently flip (if possible) to the diequatorial conformation.


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Thermodynamics behind Deliquescence?
The phenomenon of deliquescence wherein a water soluble substance absorbs water from atmosphere and dissolve in it forming a saturated solution, has thermodynamic roots. I have read the fact that the vapour pressure of the saturated solution should be lower than the the vapour pressure of water at that temperature. Why is this so?
Moreover, since deliquescence is a spontaneous process and therefore thermodynamically favourable, there must be a negative $\Delta G$ for the absorption of water vapour and conversion of solid substance into its saturated solution? How can we derive, understand or quantify various thermodynamic parameters ($\Delta G, \Delta S, \Delta H$ etc) for the process of deliquescence?
I am comfortable with basic thermodynamic manipulations, binary phase diagrams, multiphase systems and equilibria.

The vapour pressure lowering is a consequence of Raoult's law with a non-volatile solute, i.e. a solvent that has zero vapour pressure (or at least, lower vapour pressure than the solvent).
Regarding the second part (“How can we derive, understand or quantify various thermodynamic parameters (ΔG,ΔS,ΔH etc) for the process of deliquescence?”), I am not sure exactly what you are looking for. This is a complex phenomenon, and there is a lot of research literature of the topic. To give only one example, see

“Water-Solids Interactions: Deliquescence”, Lisa J. Mauer and Lynne S. Taylor, Ann. Rev. of Food Sci. Technol., 2010, 1, 41-63.

As for the first part of the question, how is the lowering of vapour pressure essential for deliquescence?                   And the paper you cited is behind a paywall. Can you provide a link to some other literary source?

@SatwikPasani I'm sorry, but I didn't find a decent open access source on that (although I may not have done an exhaustive search, of course)… your university library probably has access

“how is the lowering of vapour pressure essential for deliquescence?” — I am not sure, you quoted that as an established fact… do you have a source on this?

I did read it in a physical chemistry book long ago but can't seem to remember the name of the book. It was also asked in a chemistry olympiad mock paper.


text:
Halogenation of alcohols in the presence of ZnCl2
I know that $\ce{OH-}$ is usually a bad leaving group, but why does the following reaction occur via SN1 in presence of anhydrous $\ce{ZnCl2}$?

Is $\ce{HCl}$ really part of your reaction equation? Because if it is you have no problem with $\ce{HO-}$ as leaving group as $\ce{HCl}$ would protonate the hydroxyl group converting it into a good leaving group.

I guess I'll make an answer out of my comment then :)

You are right, $\ce{OH-}$ is a bad leaving group. But the $\ce{HCl}$ in the reaction mixture protonates it. Thus, the bad leaving group $\ce{OH-}$ is converted to the good leaving group $\ce{H2O}$. Anhydrous $\ce{ZnCl2}$ might additionally facilitate the reaction by acting as a Lewis acid and it binds the expelled water thus preventing the back reaction.

+1, good answer. I just want to make one minor point: since the mechanism is SN1 and the carbocation intermediate is planar, one should get a racemic mixture of products. The representation of stereochemistry is therefore somewhat inaccurate and misleading, I think.

@GregE. Oops, sorry. You are completely right. I answered a question about $\text{S}_{\text{N}}2$ four hours ago and seem to have mixed that in. I've corrected my answer.

@Philipp How does ZnCl2 acts as lewis acid ? Does it uses it's empty 4s orbital?

@Lalit  https://chemistry.stackexchange.com/questions/14028/why-is-aluminium-chloride-a-lewis-acid-if-its-an-ionic-compound


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Is it possible to react two distinct carboxylic acids to form a dicarbonyl ester?
By a dicarbonyl ester I mean something like this:

where R1 and R2 are alkyl substituents and the reacting carboxylic acids are:

since the byproduct is chemically stable (water), there should be a way even if by nanomachines (read: enzymes)

The standard term for such molecules is acid anhydride, as they can be viewed as the product of a condensation reaction between two carboxylic acids, with concomitant loss of $\ce{H2O}$. Asymmetric anhydrides (i.e., ones that decompose into two distinct carboxylic acids if hydrolyzed) can certainly be prepared. One possible route is to synthesize an acid chloride from carboxylic acid #1 (thionyl chloride, $\ce{SOCl2}$, being one common reagent used to make the acid chloride) and then react it with the carboxylate salt of carboxylic acid #2. In that reaction, the carboxylate functions as a nucleophile and attacks the carbonyl carbon of the acid chloride, with chloride anion expelled as the leaving group.
I don't think it's an efficient laboratory method to prepare asymmetric anhydrides by simple dehydration of stoichiometric mixtures of distinct carboxylic acids. I suspect that would probably yield a (roughly) statistical distribution of products, at least in the ideal case. In reality, various factors might skew the distribution one way or another.


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Is there a difference between imidazolidinone and imidazolidone?
I've had my nose in the early literature on GFP (Green Fluorescent Protein), which contains a 4-(p-hydroxybenzylidene)imidazolidin-5-one fluorophore, and in one paper the author refers to the "imidazolidone ring".
I assumed that this change of spelling was equivalent, but I can't find anything that looks reliable on Google (there are also pages referring to imidazolinone, which looks like it refers to an aromatic ring with Nitrogens at positions 2 and 4...)
I don't think I'd know quite where to look on the IUPAC website for the info so trying here...
Can anyone help me understand the variations?

According to the NIST ChemBook, 2-imidazolidinone and 2-imidazolidone are synonymous. The official IUPAC name is imidazolidin-2-one. 1,3-ethyleneurea is also another name for this same molecule:

But, as you note, tiny changes in spelling can sometimes be much more tricky than this. To take an example from this family: imidazolidinone and imidazolinone (note that the later ends in –inone, not –idone) are ketones formed from imidazolidine and imidazoline, respectively. Imidazolidine and imidazoline are partly saturated and fully saturated imidazole derivatives:

Imidazolidine is a proper IUPAC name, but the official name of imidazoline is 4,5-dihydro-1H-imidazole.

Ah that site's brilliant, I'm just used to trawling NCBI resources.
So the -idine refers to saturation, and the -one obviously refers to the ketone, I think I've got it -  thanks!

After a little more reading it looks like "oline" is a suffix for 4 & 5-membered nitrogen heterocycles, azole is for similar sized rings with 2 non-C atoms (but I kind of already knew that) and -idine is just a general "related to term" according to Wiktionary... I really need to discover some better info sources though :/


text:
Why do carbonates, oxides, and pure metals precipitate before metal sulfides?
Water tends to leach metals near the surface and deposit them deeper through precipitation in two layers. The upper layer is usually made of metal carbonates, oxides, and occasionally pure or native metals, while the second and deeper layer is metal sulfides. The following excerpt is summarized above.
Source: "Secondary Enrichment of Ore Deposits" by Samuel Franklin Emmons, which was published in the American Institue of Mining Engineers, Transaction 30, (1901), p. 177-201.
The paper does not go into detailed chemistry for the most part but mentions that the chemistry varies from one morphology of soil/rock to another. The author postulates a thermodynamic reason along the lines of reduction potential that facilitates this stratification:

The actual changes observed by me in a great body of pyrite ($FeS_2$) carrying galena (PbS or Lead(II)Sulfide) in a limestone ($CaCO_3$)country-rock, which had undergone partial decomposition from the periphery inwards, are as follows: the original fresh pyrite or marcasite crystals are first disintegrated and slightly pitted on the surface, then changed to  melanterite or hydrated ferrous sulfate and the galena becomes anglesite. In the outer or more fully oxidized zone, the iron-vitriol has changed in part to yellow, basic sulfate; in part to limonite with a separation of native sulfur.
The theoretical changes that are assumed to take place by the action of waters carrying oxygen or oxidizing agents are: first, an alteration of the iron sulfide ($FeS$) to ferrous sulfate (Iron(II)Sulfate or $FeSO_4$) with the formation of sulfuretted hydrogen (Hydrogen Sulfide? $H_2S$) and sulfur which may have oxidized to sulfuric ($H_2SO_4$)or sulfurous acid ($H_2SO_3$). By further oxidation the ferrous sulfate will become, in part at least, ferric sulfate (Iron(III)Sulfate or $Fe_2(SO_4)_3$), and this in its turn will react upon the remaining ferrous sulfate, or upon the sulfides, and form more ferrous sulfate or sulfates of the other metals which are present. By this cycle of reactions, a supply of both ferric and ferrous sulfates would seem to be provided in the oxidized zone, but the extending downwards of the ferric salts would decrease as the supply of oxygen in the waters became less abundant.
It may be assumed that the sulfates of the metals thus formed would be transported for greater or lower distances, generally in proportion to their solubility, the iron sulfates being the most soluble; next, those of copper and zinc; silver sulfate is less soluble and also more readily decomposed, while lead sulfate is extremely insoluble.

""The paper does not go into the chemistry of this and only mentions that the chemistry varies from one morphology of soil/rock to another."" This was a rather wise statement by the author, who obviously did not know a lot about chemistry.

Asked this question on a Geology forum.

Why Do Carbonates, Oxides, and Pure Metals Precipitate Before Metal Sulfides?

They don't. I cannot get to the particular paper you read, but I can provide both a logical and a quantitative argument that the metal sulfides precipitate first. The thermodynamics of this situation do not involve reduction potentials so much as simple solubilities. Metal sulfides are much less soluble in water than metal carbonates.
From your description, I imagine a system like the following image. There are two strata of sedimentation: an upper layer of metal oxides and a lower layer of metal sulfides.

There are three reasonable explanations why the sulfides are on the bottom.

The sulfides did not precipitate from the water; they were always there. The carbonates and oxides deposited on top of them.
The carbonates and oxides were deposited first, and the sulfides were deposited second, settling through the carbonates and oxides to form a deeper layer.
The sulfides were deposited first, and the carbonates and oxides were deposited second.

Explanation 1 is very reasonable, but counterproductive. We will ignore it.
Explanation 2 is the one supposed by the OP in the question, and perhaps it was also provided in the paper. Let us examine this idea a little. In order for this to happen, the metal sulfides would have to be denser than the metal carbonates and oxides to settle through. All it takes the counterexample of iron to see that this is not always the case. Iron (II) sulfide is denser than iron (II) carbonate but less dense than iron (II) oxide.
 - Iron (II) oxide, $\ce{FeO}$, 5.745 g/mL
 - Iron (II) carbonate, $\ce{FeCO3}$, 3.961 g/mL
 - Iron (II) sulfide, $\ce{FeS}$, 4.84 g/mL
Additionally, if the sulfides were settling through, the layers would not be so well stratified - the boundary between them would be indistinct.

Metal sulfides are much less soluble in water than metal carbonates.

That leaves with Explanation 3, which is the correct one. Metal sulfides are almost universally less soluble that metal carbonates. We'll get to where the oxides come from later. Both metal carbonates and metal sulfides are nearly insoluble in water, so the minute amounts present are hard to quantify in the "grams of solute per liter of solution" kind of way. The saturated solution may only have few parts per billion of the compound. Instead, we talk in terms of the solubility equilibrium, in particular, the solubility product (constant), $k_{sp}$, the equilibrium constant associated with the combination dissolution and dissociation.
We can determine the concentration from the $k_{sp}$. Consider barium carbonate $k_{sp}=5.1\times10^{-9} \text{ M}^3$. The solubility equilibrium is:
$$\ce{BaCO3(s) <=>Ba^{2+}(aq) + CO3^{2-}(aq)}$$
And the equilibrium constant is (the concentration of solid barium hydroxide is constant and is thus part of $k_{sp}=k_{obs}$.
$$k_{sp} = [\ce{Ba^{2+}}][\ce{CO3^{2-}}]=5.1 \times 10^{-9} \text{ M}^3$$
Since $[\ce{CO3^{2-}}]=[\ce{Ba^{2+}}]$ (mass balance):
$$[\ce{Ba^{2+}}][\ce{CO3^{2-}}]=[\ce{Ba^{2+}}]^2$$
$$[\ce{Ba^{2+}}]^2=5.1\times 10^{-9} \text{ M}^3$$
$$[\ce{Ba^{2+}}]= 7.14\times 10^{-5} \text{ M}=>0.0141 \text{ grams per liter}$$
Now, let's look at carbonates and sulfides. All $k_{sp}$ came from here.
Cadmium:

Carbonate $\ce{CdCO3}$, $k_{sp}=5.2\times 10^{-12} \text{ M}^2$, $[\ce{Cd^{2+}}]=2.28\times 10^{-6} \text{ M}$
Sulfide $\ce{CdS}$, $k_{sp}=8\times 10^{-27} \text{ M}^2$, $[\ce{Cd^{2+}}]=8.94\times 10^{-14} \text{ M}$

Cadmium sulfide is less soluble than cadmium carbonate by a factor $10^8$ (100 million). The same behavior exists for cobalt, copper, iron, lead, manganese, nickel, silver, and zinc (all metals for which data for both the carbonate and sulfide are listed).

Why might this paper suppose that the sulfides are deposited second when the sulfides are much less soluble than the carbonates?

The authors of the paper probably did not know that (and perhaps few others did either). That paper was published in 1901. Svante Arrhenius presented his thesis on the behavior of salts in water in 1884. It was not well thought of initially. He received a Nobel Prize for the work in 1903 (after the paper was published), but there is little reason to believe the authors (mining engineers) would have been taught controversial and cutting edge physical chemistry in whatever minimal introductory chemistry courses they had taken (if any). I may mention recent Nobel work in chemistry in my intro courses, but I certainly don't teach recent Nobel work in chemistry in into courses.

Where do the metal oxides come from?

Metal oxides form either by decomposition of the carbonates or by oxidation of pure metals (which probably occur from the decomposition of the sulfides):
Decomposition of sulfides:
$$\ce{FeS(s) <=> Fe(s) + S(S)}$$
Oxidation of metal:
$$\ce{6Fe(s) + 3O2(aq) -> 3Fe2O3(s)}$$
$$\ce{2Fe(s) + 3H2O(l) -> Fe2O3(s) + 3H2} \text{ (acid is helpful here)}$$
Decomposition of the carbonate. Since the carbon dioxide would be carried away by the water, this equilibrium is continually pushed toward the oxide.
$$\ce{FeCO3(s) <=> FeO(s) + CO2(aq)}$$

In marine environment, reducing conditions generally occur a short distance below the depositional interface and sulfur is available in solution and organic matter. The result is reduction of iron, first to mono-sulfides and then to pyrite (see Chap. 19). p.403.
Iron minerals are forming today in several environments:
(a) tidal flats and bog lakes
(b) oceanic shelves in tropical climates
(c) deep basins with restricted circulation, such as fjords and the Black Sea
The formation of pyrite in muds has been the subject of several theoretical, experimental, and field studies. Sulfur isotope analysis has helped to throw light on the origin of the sulfur. These studies have been summarized by Berner (1970)  and his conclusions are given below. He deduced that the fulfur in the pyrite of  modern muds comes from two sources (a) organic matter and (b) sulfate dissolved in seawater. Although organic matter forms up to 10% of some modern marine muds, this carbonaceous material contains only 1% sulfur. Yet the muds frequently contain more than 1% pyrite and so it is clear that an additional source of sulfur is required. The other source is the bacterial reduction of dissolved sulfate, which has been shown by sulfur isotope studies to be supplied by diffusion into the sediment from the overlying seawater. Where dissolved $H_2S$ exists, pyrite is generally the thermodynamically stable form of iron; pyrite is not formed directly however.

The first iron compounds to appear are black, finely disseminated
  iron mono-sulfides.

They can tribute much of the black color generally seen just beneath the surface of tidal flats and organic-rich lakes and stagnant basins. These compounds, which are thermodynamically meta-stable, are observed to change to pyrite a few inches below the sediment-water interface. Berner has presented evidence indicating that

the formation of pyrite takes place mainly by the reaction of
  mono-sulfides with elemental sulfur produced from the oxidation of
  $H_2S$ either inorganically or through the action  of
  sulfur-oxidizing bacteria.
If the bottom waters contain dissolved oxygen, elemental sulfur  is
  produced by the reaction of FeS and $H_2S$ with dissolved oxygen
  stirred into the sediment by storms, currents, and burrowing
  organisms.

The fact that free $H_2S$ is present in a sediment, or even in the deeper parts of basins with restricted circulation, does not necessarily mean that iron sulfides will be the only iron minerals present. In the Black Sea deeper waters have a  slightly higher salinity (22 ppt) than the surface waters (18 ppt). The result of the density stratification has been to prevent mixing between the surface, oxygenated waters, and the waters below a depth of 150m with the result that the waters from 150m to depths in excess of 200 m contains dissolved $H_2S$. Muds in the deeper parts of teh Black Sea contain 3 to 5% total iron, on a carbonate-free basis. Normally 30 to 50% of this iron is present as iron sulfide. Not all the iron is present as pyrite because over 95% of the iron supplied to the Black Sea is in the form of very fine clastic particles (magnetite, hematite, chlorite), which are not converted to sulfides as they  settle through the deeper waters despite the high content of $H_2S$ in these waters. Once buried below the sediment interface, the conversion of the iron compounds to sulfides is limited by the availability of organic matter, sulfur, and perhaps also by slow reaction rates. This situation occurs in spite of the high content of organic matter, which reaches values in excess of 5% organic carbon. (To me this is confusing because "organic matter" first causes pyrite and then doesn't... perhaps the word "normally" was meant to apply to pyrite formation and the latter to this special case of increasing salinity at depth.)
The "bog iron ores" of modern swamps and lakes represent one of the few examples of modern iron-rich sediments. They are found in pororly drained, recently glaciated regions or areas of older iron-rich rocks. Two general groups have been recognized: lake ores and bog ores. Lake ores consist of oolitic or pisolitic grains, cemented together into discs up to a foot or more across. They are commonly found in water only a few feet deep, at the border of the lake. Bog ores form thin layers of earthy-to-pisolitic iron oxides at the surface or below several meters of peat. Both types appear to be formed by the migration of acidic organic-rich groundwater. Precipitation of ferric iron takes place where the migrating soil or swamp-derived waters  enter a more oxidizing, less acidic environment.
p.603-605.

Blatt, Harvey, Gerard V. Middleton, and Raymond C. Murray. Origin of Sedimentary Rocks. Englewood Cliffs, N.J: Prentice-Hall, 1972. Print.


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Do you break the metallic bond when you break a metal into two pieces?
As we all know, metallic ions are surrounded by a sea of electrons. If we continually bend or stretch a metal, say, iron, it will break. Does that mean we break the metallic bond in subatomic level?

Yes, you are breaking metallic bonds. Similarly, when you crush a crystal of salt, you are breaking ionic bonds. There aren't that many good examples of breaking covalent bonds1, though breaking a polymer like nylon usually will do it.
The "physical" breaking of bonds is not a very exotic thing. Remember that while the energy required to break every bond in a gram of substance is a lot, the energy required to just break a slice of those bonds isn't much.
A very common question that arises is "what happens with the open valencies left behind when you physically break bonds?". Usually, gases from the atmosphere adsorb onto the metal surface, forming hydrogen bonds.
A note: When you do this, there is nothing going on at the subatmoic level, only at the atomic level.
1. This is not due to the strength of covalent bonds -- indeed, these are usually weaker than ionic bonds. This happens because most covalent substances are made up of smaller molecules (as opposed to macroscopic lattices) held together by hydrogen bonding/Van der Waals forces, and it is these bonds that break when you break a covalent substance.

What do you mean by "forming hydrogen bonds" ?

In addition to ManishEarth’s answer, I'll add that this is only strictly true if you take a pure single crystal of metal and break it. If you take a real-life piece of iron, it is not a pure crystal but contains a large number of crystalline defects. The most common (and easiest to think about) in a polycrystalline material is the presence of grain boundaries between the many crystallites that make up the metal particle:

Micrograph of a polycrystalline metal; grain boundaries evidenced by acid etching (from Wikipedia)

So, your metal piece is not a single crystal, and crystalline defects are key in the understanding of mechanical response to applied stress, including the case of fracture in which you are interested. So, by breaking a piece of metal in two, you may not actually break so many metallic bonds :)


text:
Does the temperature of water change during electrolysis?
Applying an electric potential to impure water splits water into its constituents H/O.
Does the temperature of the electrolyte change during this electrolysis?

""Applying an electric potential to impure water splits water into it's constituent H/O."" This is wrong, just "potential" does change nothing. One needs to have a current flowing to achieve electrolysis. And if you read a physics book first, you may learn about currents and heat.

Because the electrolyte has a finite resistance, heat will be generated based on the power, $I^2R$, the time of electrolysis (since $power=\frac{energy}{time}$) and the heat capacity $C=\frac{energy_{heat}}{\Delta T}$.  This would be a very rough estimate since it assumes the solution is the greatest source of resistance.

However, Wikipedia states that there are cases where the temperature drops because of the electrolysis (“in the electrolysis of steam into hydrogen and oxygen at high temperature, […] heat is absorbed from the surroundings”). So it may not be as easy as Joule effect…

@F'x Interesting point, although this is a case where a reference on the Wikipedia page would be beneficial since this is a fairly unexpected phenomenon.

@bobthechemist Why is this unexpected? Decreasing temperature by converting liquid to gas is how most air conditioners work. Or is this caused by a different mechanism?

@AlexanderGuyer The comments are mixing a variety of concepts.  A temperature drop during electrolysis of water is unexpected.  If we now complicate the problem to incorporate gas phase electrolysis with phase changes, then all bets are off.  My comment was in reference to the lack of decent citations in the Wikipedia page (9 years ago).

Since you have not given the voltage you are applying to the system I assume that there is excess voltage meaning to spilt water needs about 1.23 V (quick wiki)
Anode,
$$\ce{ 2H2O(l) → O2(g) + 4H+(aq) + 4e- -1.23 V}  $$
Cathode,
$$\ce{2H+(aq) + 2e- → H2(g) 0.00 V}$$
To get more current to flow more exactly to get more products you will need to supply more voltage thus the excess voltage and current for that voltage will convert into heat according to the fundamental law of nature the energy conservation law.
Total Electric Energy Input gets converted to Chemical Energy + Heat (if exceeds voltage is appied)
$$\ce{V_{input} * I = (1.23 V) * I + Heat}$$
If you take the excess heat energy out of the system (I.E heat sink or what ever means) keep the system temperature constant then $\pu{1.23 V}$ will remain constant otherwise the potential will change according to the Nernst Equation.
Please also note that $\pu{1.23 V}$ comes due to standard state $\pu{ T= 298.15 K }$

+1 Also, even if you do nail the reaction voltage, that voltage just refers to the potential difference across the interface between the electrode and the electrolyte. You'll necessarily have to apply a little extra voltage -- to compensate for resistance i.e. voltage drop in the rest of the circuit -- to reach the required interface potential. Some of that circuit will be outside the electrolyte but most will be submerged conductors and also the electrolyte itself. The heat dissipation is given by $P=I^2R$ where $I$ is the current and $R$ is the resistance of the rest of the circuit.

(Other forms of the dissipation equation can be readily derived from $V=IR$ and $P=IV$, too).


text:
Is real water made up of heavy water and light water?
I need to know, is real water made up of heavy water which is formed with deuterium, and light water which is water by protium?
We know that hydrogen has three major isotopes, including protium and deuterium.
We also know that water, $\ce{H2O}$, is made up by hydrogen and oxygen.
All this might mean that real water itself can naturally contain some heavy water - Deuterium oxide. Is this correct?

Water has formula H2O. Oxygen has 3 stable isotopes (99.76% 16O, 0.039% 17O, 0.201% 18O), and hydrogen has two (99.985% 1H, 0.015% 2H). Thus, there are 9 natural isotopic configurations for water: 3 possibilities for oxygen, multiplied by 3 possibilities for 2 hydrogens with 2 possible isotopes.
Out of those 9 possible configurations, only 4 have a natural abundance larger than 1 ppm:

So, pure water is a mixture of 9 different isotopic configurations. You will note that D2O, which is what is usually called heavy water, is not actually among the 4 more common isotopic configurations given by natural abundance.

Also of note: water that is not “heavy water” is sometimes called “light water”, but never “soft water”. “Soft water” is a term used for a chemical process of natural (non pure) water that has been depleted from certain metal cations. (It is used in contrast to “hard water”.)

I've submitted an edit to the question which, among other things, changes "soft water" to the more correct "light water".

Don't forget that water is highly dynnamic, an O-H bond exists for about 10exp-13 seconds only. This makes the question of molecules in water soemwhat philosophic.

@Georg yes, water is “highly dynamic”, but your order of magnitude for O–H bond lifetime (0.1 ps) is very wrong… even hydrogen bonds have lifetime of the order of 10 ps

Theoretically, you are correct; water, being chemically composed of hydrogen and oxygen, can contain any isotope of hydrogen (or oxygen) and so there are many isotopic combinations of water. However, if you look at hydrogen's molar mass (which is calculated based on a weighted average of the atomic masses of the two stable isotopes), you'll see that it's much closer to 1g than to 2. This is because Earth's hydrogen content is dominated by the protium isotope, with an atomic mass of 1 dalton. If there were a 50-50 mix between protium and deuterium we'd expect the molar mass to be 1.5g, but as it is, Earth's hydrogen content is only approximately 0.015% deuterium.
So, we could expect, on average, a slightly lower percentage of water to have at least one deuterium atom, because a very few have two. As shown by F'x's chart, the concentration of naturally-occurring deuterium oxide is measured in the parts per billion (22ppb). That's actually a lot of heavy water, if you were to get all of it in one place; there are about 31 trillion kilograms of naturally occurring deuterium oxide on the planet (31 km3). That's enough to fill Lake Mead, created by the Hoover Dam. However, to get one kilogram, or liter, of $\ce{D2O}$ by separating it from light water, you would need to process 45 million kilograms of water. Put another way, there is a little over one gallon of natural deuterium oxide (5L to be more precise) in a typical Olympic swimming pool.


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Why is AlMg3 called aluminum alloy?
In AlMg3, magnesium atoms outnumber aluminum atoms by a factor of three. In terms of mass, the ratio of aluminium to magnesium is a mere $27.0 / (3\times24.3) = 0.37$.
Shouldn't AlMg3 be better called magnesium alloy?

AlMg3 is not the chemical formula of the alloy, it's the product name for a wrought alloy. Its technical sheet specifies the following composition:

Product name
AlMg3
Class of product
Al-Mg alloy for MIG/TIG welding.
Corresponding standards
DIN 1732, SG-AlMg3, AWS A5.10, ER 5754
Nominal composition (weight %)
Al: Bal. Si: 0.4 Mg: 3 Mn: 0.5

Thus, the 3 in the product name refers to 3 % (in weight) of magnesium.


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Why is nicotinoyl chloride available as its hydrochloride salt? How can we separate them?
Nicotinoyl chloride is only available as its hydrochloride salt.
Why can't nicotinoyl chloride be synthesized in free form? How can we seperate nicotinoyl chloride from its salt form? Can we try any base for this purpose?

I suppose (but am far from sure) that it's because acyl chlorides are very reactive, and the hydrochloride is here for stabilization…

hydrochloride salts are typically made to address solubility issues, and that might be the case here as well.

Hydrochlorides are also often more crystalline and easier to purify: http://chemistry.stackexchange.com/questions/2627/why-are-solid-phase-organics-often-conjugated-to-hcl

Unfortunately, i don't have access to the original reference from 1928 Über die Einwirkung von Thionylchlorid auf die Pyridin-monocarbonsäuren by Meyer and Graf, but it's cited in The Preparation of Nicotinyl Chloride from 1953.
In this later note, where dried and grounded potassium nicotinate was reacted with oxalyl chloride in anhydrous benzene and the resulting nicotinyl chloride eventually destilled off under reduced pressure (75-90 °C, 10-12 mm), the authors state:

On exposure to the air, the oily acyl chloride soon solidified to nicotinic acid hydrochloride.

With other words, the combination of moisture and the basicity of the pyridine moiety results in the decomposition of the product. This is apparently prevented by deactivation of the pyridine upon conversion to its hydrochloride.


text:
Finding chiral carbons in aromatic or cyclic compounds
I know that for a carbon to be chiral four different groups should be attached to that particular carbon. But, what about in the case of cyclic compounds? What's the method to check chirality if the carbon is not $\mathrm{sp}^3$ hybridized and only three, or two groups are attached to it (through double or triple bonds)?

Please note that there are different types of chirality, not only center-based: http://en.wikipedia.org/wiki/Chirality_(chemistry)

If an atom is part of a double or a triple bond it is never a stereogenic center because then this atom and the atoms bound to it form something planar, so that you can always find a symmetry plane.
But an atom that is part of a double bond can be prochiral, if it can be converted from to a stereogenic center in a single step.

Allenes can be chiral

@K_P That's true, good point. But still their ($\ce{sp^{2}}$ carbon) atoms are no stereogenic centers.

But if the question is about chirality, a molecule having stereogenic centres is not a sufficient (eg. meso compounds) nor a necessary (as in the case of allenes or eg atropisomers) condition for it to be chiral.

@K_P He asked about chiral carbon atoms and since I had been quite busy at the time I only provided an answer within that scope. If you want to add an answer including axial chirality you are welcome. I'll happily upvote it.

I don't think a new answer is needed, you already provided a very good one and one that OP was happy with. I just added my comment as additional information.


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Potassium chlorate from potassium chloride and hydrogen peroxide
Why can't you make potassium chlorate using the chloride anion from the potassium chloride and having oxygen bond with it from hydrogen peroxide?

Do not attempt! Peroxide is not a strong enough oxidant to oxidize chloride ion to chlorate. Neither is oxygen. However, peroxide can oxidize chloride to poisonous chlorine gas.
Redox reactions are spontaneous if the two half reactions would produce a positive potential in a galvanic cell. Most redox half reactions have measured standard electrode potentials against a known standard so that they can be compared.
The reaction you want to do is:
$$ \ce{3H2O2 + Cl- -> 3H2O + ClO3-}$$
The two half-reactions are:

Oxidation:   $\ce{Cl- +3H2O -> ClO3- +6H+ +6e-}$
Reduction:   $\ce{H2O2 +2H+ +2e- -> 2H2O} \ \ \ \ E^ \circ =  +1.78 \text{ V}$

The reduction potential of hydrogen peroxide is on the table, but the reduction potential for the other reaction is not. However, the other reaction is the sum of two reactions on the table, so we can use Hess's Law.
$$\ce{ 2ClO3- + 12H+ + 10e- ->Cl2 + 6H2O} \ \ \ \ E^\circ = +1.49 \text{ V}$$
$$\ce{Cl2 +2e- -> 2Cl-} \ \ \ \ E^\circ = +1.36 \text{ V}$$
$$\ce{2ClO3- 12H+ + 12e- -> 2Cl- + 6H2O}  \ \ \ \ E^\circ = +2.85 \text{ V}$$
Since standard potentials do not scale by stoichiometric factor (it's in the Nernst equation instead), the value we just found also works for our oxidation reaction. However, since this reaction is reverse of the "reduction half-reaction", we need a negative sign:

Oxidation:   $\ce{Cl- +3H2O -> ClO3- +6H+ +6e-}\ \ \ \ E^ \circ =  -2.85 \text{ V}$
Reduction:   $\ce{H2O2 +2H+ +2e- -> 2H2O} \ \ \ \ E^ \circ =  +1.78 \text{ V}$
Overall: $\ce{3H2O2 + Cl- -> 3H2O + ClO3-}\ \ \ \ E^ \circ =  -1.07 \text{ V}$

This reaction is not spontaneous. Peroxide cannot oxidize chloride, at least not to the chlorate ion. Note that peroxide can and does oxidize chloride to poisonous chlorine gas in the presence of acid. This redox is spontaneous and will happen.  Do not attempt!

Reduction:   $\ce{H2O2 +2H+ +2e- -> 2H2O} \ \ \ \ E^ \circ =  +1.78
   \text{ V}$
Oxidation:   $\ce{2Cl- + 2e- -> Cl2} \ \ \ \ E^ \circ =
   -1.36 \text{ V}$
Overall: $\ce{ 2Cl- + H2O2 + 2H+ -> Cl2 + 2H2O} \ \ \ \ E^ \circ =  +0.42 \text{ V}$

Peroxide will oxidize chlorine gas to chlorate, but who wants to handle chlorine?

Potassium chlorate ($\ce{KClO3}$) has covalent bonds between the chloride and the oxygen, while potassium chloride ($\ce{KCl}$) lacks those covalent bonds to oxygen.
Hydrogen peroxide decomposes into water ($\ce{H2O}$) and oxygen gas ($\ce{O2}$) while oxidizing (LEO: Loss of Electrons is Oxidation) its environment. Peroxide is useful primarily for sucking electrons from a solution, which is very different from forming covalent bonds.
If you wished to dissolve solid potassium in water to make KOH peroxide would likely perform nicely as the peroxide would suck electrons from the solid K and thereby help it form ions ($\ce{K+}$) which would combine with $\ce{OH-}$ from the water to from KOH.

http://chemistry.stackexchange.com/questions/843/what-role-does-hooh-have-in-copper-ii-acetate-formation

http://chemistry.stackexchange.com/questions/87/how-may-copper-acetate-ligands-be-manipulated-to-change-colors

My mistake, I was unclear. Why doesn't the oxygen react with the chloride ion?


text:
Predicting bonds from properties of compounds

You are given compounds $x, y$ and $z$. $x$ conducts electricity in aqueous solution as well as in molten state. $y$ conducts electricity in aqueous form only. $z$ conducts electricity in neither the aqueous nor the molten state. Predict the intramolecular bonds in each compound.

Compound $x$ is clearly ionic. In water, the ions disassociate and form free charges for flow of electricity. In the molten state, the ionic bonds break, and again free charges can flow.
I'm not sure about the other two compounds.

""In the molten state, the ionic bonds break,"" This is wrong! What do You think holds the melt together?

As stated by the requirements, the words "molten" and "aqueous" refers to specific classes of compounds that are more likely to be solid near room temperature and have an acceptable solubility in water. If so I'd think $y$ can be an organic compound like maleic anhydride (covalent), but for $z$ the closest class is related to n-polyketones, because they are soluble in water but non-dissociated like the previous compounds.

A is an ionic salt like NaCl.  Z is covalent and does not solvate to ions when dissolved in water, e.g., sugar, poly(ethylene glycol).  Y is polar covalent and requires solvation to ionize, e.g., HCl.

Note that substances with mobile pi-electron systems can also conduct electricity. For example, graphite is a good conductor. Diamond, although made up of the exact same element (carbon), is not a conductor.


text:
Why aren't there any triangular molecules?
Look at the molecular structure of benzene:

It's a perfect hexagon. Why aren't there any molecules arranged in a triangular fashion with bonds forming the edges and the molecules the vertices? Or, for that matter, square molecules? And what about other n-gonal molecules? octagons?

There are triangular molecules: http://en.wikipedia.org/wiki/Cyclopropane

Molecules whose shapes are connected to squares or higher polygons than a hexagon are usually subject to deformations due to ring strain etc., so that they deviate from the perfect polygon geometry.

There exists triangular molecule, eg. cyclopropane, but they are highly reactive due to ring strain.

Ring strain results from a combination of angle strain, conformational strain or Pitzer strain, and transannular strain or van der Waals strain.

C-C-C bond angle in rings should be approximately 109.5° degrees which in case of small rings like cyclopropane, bond angles are 60° whereas tetrahedral 109.5° bond angles are expected. The intense angle strain leads to nonlinear orbital overlap of its sp3 orbitals. Because of the bond's instability, cyclopropane is more reactive than other alkanes. Since any three points make a plane and cyclopropane has only three carbons, cyclopropane is planar.
Due to this strain they exists in higher energy state and therefore Because of their high strain, the heat of combustion for these small rings is elevated.
Kindly see Ring Strain, it will help you to understand about these kinds of molecules.

You might want to comment on the actual underlying phenomenon behind ring strain, i.e. electron-electron repulsion.

Ring strain is an awful concept. It is purely built on phenomenology as in "stuff that one could be expecting from hybrid orbital angles." It has only little more scientific justification than the VSEPR model.

@Martin : But Ring Strain is the only reason that was explained to me in 12th grade.

Unfortunately topics like this will always be taught as something that appears to be a law, without questioning the reasons, that underlie this concept. Most of steric repulsion, attraction and other effects can be described and explained by more advanced theories, i.e. electronic structure theory. For someone in school it might be appropriate to conclude with models like these, but if the underlying theory is not too far off it should be included as well.

While ring strain may be phenomenological, I think it should be taught. The interactions of atoms and orbitals in tight rings does lead to a strained high energy structure, and opening the ring releases this energy, relieving the strain. It would be complicated to approach all those interactions from the ground up, the concept of ring strain wraps that up nicely. And with a good molecular model kit, you can "experience" ring strain by jamming a cyclopropane together.

white phosphorous allotrope (P_4) is a deltahedron (tetrahedron) consisting of four triangles fused by mean of each side. It doesn't bear that much angular strain, cause the measurements done on 2nd-row elements-based compounds (i.g. hydrocarbons and related) does not necessarily include 3rd-row and heavier elements. This also holds in non cyclic compounds. While the H-O-H in water is close to theoretically predictable 109,5°, the H-S-H angle in hydrogen sulfide is very close to a right angle, showing little or no hybridization to have occurred.
Moreover the rules with closed shell molecules and non-electron deficient ones, appear to be inadequate to descrive electron deficient compounds, spanning from closo-boranes (very often deltahedranes) to metal clusters
P.S. sorry for my english :)

The best example for relatively stable 3 membered rings are the epoxides. They are very reactive, yes, but they can be isolated and stored, and then used in synthetic routes, like those sold by sigma aldrich:
http://www.sigmaaldrich.com/materials-science/material-science-products.html?TablePage=20203736
You probably wouldn't have guessed that all of those guys above did exist, would you? :-D
You can also understand its stability trough the molecular orbital theory. It turns out that molecular orbitals of these 3 membered rings, like cyclopropanes or epoxides has quite similar molecular orbitals compared to those of alkenes, so their reactivity is somehow similar.
Also, square, penta-, hepta-, octa-....gonal molecules do exist!
You would be surprised about some macrocyclic compounds being stable like those over here:
http://www.google.com/patents/US6200254


text:
Is there any relationship between ultrasonication and temperature
In chemistry we use ultrasonication for cleaning vessels ,synthesis of chemical compounds (sonochemistry) etc.I know that during ultrasonication high temperature and pressure is formed "locally".But I want to know whether there is any equations which relates between ultrasonication and temperature .

I'm not sure whether I oversimplify things here but in the process of ultrasonification the atoms and molecules are highly accelerated leading to high velocities. And since the velocity is linked to the kinetic energy via $E_{\text{kin}} = \frac{1}{2} m v^{2}$ which is in turn linked to the temperatur, you have a direct connection between $v$ and $T$. Example: for a free particle inner energy = kinetic energy, so via equipartition theorem: $U = E_{\text{kin}} = \frac{3}{2} k_{\mathrm{B}} T$ and thus the average quadratic velocity is $\sqrt{<v^{2}>} = \sqrt{\frac{3 k_{\mathrm{B}} T}{m}}$.

@Eka, as you mentioned, pressure is also important. I can't remember the numbers now, but if you do some research there are some examples of average temperatures caused by cavitation bubbles bursting at specific frequencies / power. Not sure about a formula, though.

http://www.scs.illinois.edu/suslick/
Publications list, re "The Sonochemistry of Materials and Biomaterials"

The temperature is highly nonlinear with sound pressure and dependent on external pressure, tensile strength of liquid etc. Cavitation, http://en.wikipedia.org/wiki/Cavitation, can produce local shock waves and temperatures of thousands of K, causing sonoluminescence. See also http://www.scs.illinois.edu/suslick/sonochembrittanica.html.
Because of that nonlinearity and all the dependencies, I do not know if any equation for calculating the peak temperatures exists, and there is still research going on to see how much of sonoluminescence is due to black-body radiation, the recombination of radicals etc.
BTW, you can view a mantis shrimp with a sonoluminescent punch at https://www.youtube.com/watch?v=sNzI8K-lCqo and https://www.youtube.com/watch?v=ti2Uoc1RXuQ, and another species at https://www.youtube.com/watch?v=QXK2G2AzMTU.


text:
What's the technical term for supercharging an acid with an oxidizer? And does it always work?
In a recent Mythbusters episode, where they tested some Breaking Bad chemistry assertions, they debunked the hydrofluoric acid body-dissolving myth referenced elsewhere in Chemistry.SE in the question "Is it actually possible to dispose of a body with hydrofluoric acid?".
The answer is no, not really, and for the same reasons as my answer to the linked question; fluorine's just too electronegative to let the proton go do its acid thing.
To attempt to replicate the results (in this case dissolving out the bottom of the bathtub they were using), they moved to a much stronger acid, sulfuric, and then "supercharged" it with what they simply called their "special sauce" (as they said, they're not in the business of showing people how to dispose of bodies). They gave it away in the end, however, by saying the special sauce was "basically a lot of hydrogen and oxygen", and since it wouldn't be water the obvious candidate is hydrogen peroxide, a strong oxidizer (in high concentrations at least).
This seems to be a well-known trick of household chemistry; add household peroxide to cream of tartar or lemon juice, for instance, and watch those rust stains come right out. It also seems to be the bracingly direct way to form perchlorates in the lab; add hydrogen peroxide to hydrochloric acid, and behold perchloric acid (or a fuming crater, if you weren't managing the exothermic nature of this reaction properly).
So, first question, what's the technical term for this trick? It can't just be "supercharging".
Second and related question, does it work for all acids and oxidizers, or do only certain combinations of acids and oxidizers have this effect?
Lastly, for purely academic purposes, what would the basic reaction have been between sulfuric acid, hydrogen peroxide and organic tissue ($\ce{HxCyOz}$)? The sulfate anion is pretty highly oxygenated as it is, so I'd imagine the combination, unlike with a hydrohalic acid forming an oxoacid, would cause the reaction to be less acid-base, and more rapid oxidation and "combustion" of the organic tissues, leaving behind mainly carbon in water and off-gassing a large amount of steam and sulfur dioxide.

Apparently the combination of $\ce {H2SO4}$ and $\ce {H2O2}$ is known as "Piranha Solution" https://en.wikipedia.org/wiki/Piranha_solution

Thanks @Snipergirl, that article has a lot of good info about the reactions involved. It looks like the primary mechanism of action is that the sulfuric acid strips an oxygen from the peroxide, becoming Caro's acid (peroxymonosulfuric acid), which has similar acidic, oxidizing and explosive properties as perchloric acid.

Sounds like pretty scary stuff I must say! I guess the technical term in that case for the trick is "oxidising" or "super oxidising"?

Another term I found was "forced induction", but it seems to have been used in context of air-breathing combustion engines.

Peroxysulphuric acid (which is what you get with sulphuric acida and hydrogen peroxide) is very good at destroying carbon-containing things (and many others). Chemists sometimes use it in labs to clean things like sintered-glass filters that have picked up too much carbon (e.g. from activated carbon based catalysts). It is one of the the few things that works well for certain insoluble contaminants. But it is really dangerous.

I don't know if it's called anything, but when you use an oxidizer in solution, you usually need a source of hydrogen ions for the oxidized oxygen to be reduced.  In the case of hydrogen peroxide, $\ce{H2O2}$, the oxygen is in the $-1$ state rather than the normal $-2$.  The usual way $\ce{H2O2}$ acts as an oxidizer is to be reduced to water in the presence of 2 hydrogen ions, usually from a strong acid:
$\ce{2H+ + 2e- + H2O2 -> 2H2O}$
Notice that the reaction consumes 2 electrons on the left, which would come from other reactants that have been oxidized (i.e. lost electrons).
The point is that a strong acid by itself has only a corrosive effect, and a highly dehydrating effect in the case of sulfuric acid.  But if you add an oxidizer, then in the presence of a fuel (like an organic compound) plus sufficient energy (say from the heat of reaction), then you are effectively getting a combustion reaction as well.  This is the sort of "combustion" that goes on inside your body to use food as fuel, for example.  I supposed this could be called "supercharged".
EDIT: So to answer your specific questions:
1) No technical term that I know of;
2) Yes, assuming the acid is strong enough to donate protons;
3) In most organic molecules, carbon is more electronegative than hydrogen, so it can be oxidized by oxygen.  And so you get a "combustion" reaction where molecular carbon is oxidized to elemental carbon (i.e. burnt).


text:
Significant Figures in almost-scientific notation
For a homework assignment on significant digits, I have to find the number of significant digits in the number $0.40 \times 10^3$. I know that in scientific notation, the number of significant digits is equal to the number of digits in the stem. However, I've also learned that for a number to be in scientific notation, it must have a stem between $1$ and $10$. What is the rule for this situation? Is the number of Significant digits $2$ or $3$?

Most of the answers to your questions can be found in this wikipedia article. You are right if you say:

I've also learned that for a number to be in scientific notation, it
  must have a stem between 1 and 10.

This is the convention in scientific notation. The number format of $0.40 \times 10^3$ is called exponential notation (at least it was called that way in a book about the programming language Fortran I'm reading at the moment): Here the number has a stem between 0 and 1 followed by a power of 10. This is quite common in computer science.
As to the significant digits: This wikipedia article (especially the paragraph "Scientific notation") contains the information you need. According to its information I would say that the number $0.40 \times 10^3$ has $2$ significant digits.

"I've also learned that for a number to be in scientific notation, it must have a stem between $1$ and $10$".
We have: $0.40\times10^{3}$
$\hspace{13 mm}=4.0\times10^{2}$
In this form, the number fits the requirement of the stem being between $1$ and $10$.
So, we can see that the number has two significant figures.


text:
Notation "5% solution"
What does 5% aqueous solution of glucose mean? Note that it's just 5% and not 5%w/w. Do we assume it to be w/w?
The question is

If the osmotic pressure of 5% aqueous soultion of glucose is $\pi_1$ and that of 5% aqueous solution of urea is $\pi_2$ then:
A) $\pi_1=\pi_2$
B) $\pi_1<\pi_2$
C) $\pi_1>\pi_2$

Does 5% mean that the solutions are equi-molar/molal?

Generally speaking, percentage aqueous solutions like that are either weight for weight or weight for volume. Either way, in this example, you are comparing how many osmoles there for the same weight of urea ($CO(NH_2)_2$) and glucose ($C_6H_{12}O_6$).
Mole fraction as expressed in a percentage does exist though I have not seen that notation much myself and is usually designated n/n%
From the wording your question I'm guessing that it's a physiology or biochemistry question, in which case they're almost certainly referring to the 5% aqueous solution of glucose as the common IV hydration fluid 5% Dextrose Solution, which is w/w.
Otherwise, it's hard to guess as it's ambiguous!

It might be a biochemistry question(the lesson focuses only on solutions and colligative properties and not on their applications) but how would glucose be related to urea in intravenous solutions?

While urea is not found as commercially available IV solutions, one could see it being used as a hypothetical one in a physiology question. The reason being that urea is a constituent of both intra- and extra-cellular fluid. It also happens to be an ineffective osmole because due to its fat solubility it diffuses freely across the phospholipid bilayer of cell membranes, but obviously aside from that it it can exert an osmotic pressure, just one that cancels out.

However, if it's not a biochem question feel free to ignore the second paragraph!

Actually the second paragraph was more useful as the given answer was that the osmotic pressure would  be equal as 5% would refer to equal concentration. I have never heard of 5% moles(mole fraction as a percentage maybe?)

Yeah, I don't think that mole percentage is a very commonly used measure even though obviously it exists and can be used. Maybe the other guys might have some idea? I should probably amend my answer though!


text:
Is it possible to create an acid anhydride that's stable in water and if so what conditions must its structure satisfy?
I ask this question because if acid anhydrides are always unstable in water then my interest, as a student of pharmacology, would be greatly diminished. Acid anhydrides are, in my books, at least, the result of a dehydratation reaction between two carboxylic acids and hence their structures look like this:

Where R1 and R2 are organic substituents.

Is silica (SiO2) an acid anhydride?

Nope, acid anhydrides have a minimum of 3 oxygens due to the fact they are composed of two carboxylic acids joined at the OH group with H2O as a by-product of said joining (hence the term anhydride).

Would you consider ATP?

@F'x not in this life time since I see no carboxylic acids in its chemical structure

ATP is an anhydride from the phosphoric acid, and sometimes called a “acid anhydride” for this reason… but I left it as a comment, not proposed it as an answer :)

What you’re interested in is a carboxylic anhydride. Acid anhydrides in general are a greater class of compounds that include silica, ATP/ADP and others because they all give acidic hydroxy groups after the addition of water (if slowly).

I don't know of any acid anhydride that has completely stability in water, but I think the best way to stabilize them would be to choose a cyclic one, which would therefore favor the anhydride form because it makes a closed ring structure. Looking in that direction, it seems that phthalic anhydride in particular has a  rather slow reaction with water (to the point that solubility values are available!).

Or, if you want a larger ring, homophtalic anhydride (also here):

but I don't find any data on its stability/solubility in water.

Hmm, good idea. Any ideas about an acid anhydride that has a 6-member ring with the carbonyl groups hanging off it? Like sort of like phthalic anhydride but with an extra carbon in the 6-membered ring? I ask because I would imagine, based on the higher stability of 6-membered rings that it might just be even more stable in water.

This answer is years later than the original question, but I add it for anyone who come across it now or in the future...
All acid anhydride and acid chlorides are thermodynamically unstable in water.  They will eventually hydrolyze, but that time could be seconds, minutes, hours, days, or years. Moreover, in the absence of a catalyst, the rate of hydrolysis (kinetics) of any acid anhydride or any acid chloride (carboxylic, phosphoric, sulfonic, etc.) is strongly temperature dependent (hotter = faster, colder = slower, as you would expect).
For example, the classic hydrolysis of acetic anhydride is quite slow until temperatures exceed 55—60 °C.  Similarly, sulfonyl chloride hydrolysis is very slow until temperatures exceed about 70 °C.  This is in part driven by solubility - the substrate needs to be in the same phase as the water for fast rates.  Phosphoric anhydrides such as ATP or ADP can be extremely stable in water at physiological temperatures (37 °C).  The presence of an protic acid, or particularly  a nucleophilic catalyst (hydroxide, amines, thiols), can change the kinetics dramatically.  Thus, you have to pay attention to the temperature, the phasing/solubility, and the identities/properties of the other components in your reaction mixture.

In the inorganic sense of anhydride, carbon dioxide will do.  It forms carbonic acid through the reaction $\ce{CO2 +H2O <=> H2CO3}$, but the equilibrium is strongly on the side of the reactants.  Thereby carbon dioxide is an acid anhydride because it can react with water, but remains predominantly in the anhydride form because that reaction is limited in extent.