text:
High quality phase diagram for ethanol
I know the more general question has been asked here:
Open Database of Phase Diagrams?
And this has a link to slightly more specific question with more answers:
Phase diagram for first 10 elements
And this in turn has led me to Wolfram Alpha which has the best answer I can find so far:
http://www.wolframalpha.com/input/?i=ethanol+phase+diagram
But it's still pretty poor compared to water:
https://commons.wikimedia.org/wiki/File:Phase_diagram_of_water.svg
Are good phase diagrams really such a rare occurrence? What reference standard exists for industry?  Is there a famous book people buy that has all this data in it?

@Mithoron, I'm happy ethanol is treated as a common rather than proper noun but sometimes it's useful to have the CH3CH2OH to search by.

That makes some sense, add it back if you wish.

Actually part of the motivation for my comment was just to add CH3CH2OH to the page for search engine indexing :)  I know it probably won't be weighted as highly as if it was in the title but I think it'll be fine now.  Thanks for your input!

Computerizing certainly makes sense... The problem is that there isn't a "phase diagram" markup language. So all phase diagrams would have to be reduced to some common format that a particular program could handle. The "problem" is that such work is "grunt" work. Not much glory in following the parade of elephants with a wheelbarrow and shovel.


text:
proving an unknown organic compound using reactions
I have been given three spectras (MS, IR, H NMR) to analyse, from them I have found that the compound is chlorobenzene (and I've confirmed online). The task I have gotten asks me to further justify my findings by using reactions and predicting and analysing the spectra of the products. However, I am quite unsure how I'm going to predict and analyse the spectra of the products from the reaction to prove my compound.

You could think of some characteristic reactions that chlorbenzene would undergo and then evaluate your hypothesis based on product spectra. For example there's one reaction through aryne intermediate. I wonder then if you could also test for chlorides in the reaction mixture.

As comment on language: The (single) "spectrum", and in plural several "spectra" (well, Webster lists "spectrums" as an alternative of modern times, too).  For future reference, please refrain from thinking /saying / writing  three "spectras" which were two plurals in one word.

The task you had being assigned is called identification by derivatization. It's one of the strongest proof for the confirmation of the structure of a molecule.
If you have an hard-to-identify compound X, you can make it react with Y and then analyze the product of the reaction, X-Y, instead.
I once synthesized a polysubstituted aromatic compound and wasn't sure whether or not a iodine atom was present on the ring. So I performed a Heck reaction on the molecule obtaining the corresponding substituted olefin, which was a known compound, present in the literature and could be identified by its NMR spectra.
Chlorobenzene is quite an anonymous compound and its derivatization does not lead to significative compounds. Just an example to answer your question, you can perform a nitration to get o-chloronitrobenzene which is a solid and can be identified by its melting point.

+1 For the example in the end, nitration will give a mixture of o- and p- nitrochlorobenzenes. You might further nitrate this mixture or use stronger conditions in the initial reaction to get 2,4-dinitrochlorobenzene which is also a known compound.

The key is to unequivocally form something (I'm presuming these are thought experiments)  so there can be no ambiguity.  The aryne example to get only one product is a good one.  Could you also use deuterium compounds e.g. form an organometallic and quench in one case with water (to give benzene)  and in the other with D2O as a comparison? Gentle reduction with H2 or D2 would have the same effect. Forcing conditions would give cyclohexane.

Not bad in theory. In practice Cl-metal exchange doesn't generally work with chlorobenzene. The reduction would work but then you create benzene or d-benzene that would not be easy to isolate and characterise.

Mass spec, boiling point and NMR, proton and carbon

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text:
Is it possible that atoms with 120 protons are possible, but that atoms with 119 protons aren't possible?
We currently know that there are atoms with atomic number up to 118 are possible.
Is it possible that atoms with 120 protons are possible, but that atoms with 119 protons aren't possible? Or are there theoretical arguments or maybe heuristic arguments why this can't happen?
I suspect the answer to be yes, because technetium and promethium have no stable forms while the elements surrounding it in the periodic table do have stable forms. A similiar thing might happen here. The more general question: Does there exist $k > l > 118$ such that atoms with $k$ protons are possible, but that atoms with $l$ protons aren't possible?
It is not a duplicate of The last element's atomic number. I wonder whether there could be 'gaps' in the periodic table, not what the last element's atomic number is.

Please define 'possible' - do you mean a short lifetime, or completely unbound?

@JonCuster What I meant: An element is possible if there is at least one isotope with a half live greater than the Planck time. I'm not sure if it really works, and whether there are isotopes with half live smaller than the Planck time, but I think it will do for this question.

Your 'definition' doesn't make much sense, becuse for sth to decay in time shorter than Planck time, it would need to be about Planck radius small, and even single proton in comparison huge.

This is of course more of a nuclear physics question rather than a chemistry question. I'd expect just about any atomic number is possible, but that all elements above lead z=82 are radioactive. So for z>82 then it is a question of how long is the half-life.

Look at "Island of stability" that might help too

According to the Wikipedia page element 119 synthesis has been attempted indicating physicists think it is possible.  There is some speculation as to whether our current technology could even detect such short lived elements such as elements 119 or 120.  There is also speculation that the technology today may not be able to produce element 120.  It may be possible for such gaps (who knows maybe element 562 is stable and element 561 cannot form?) but at the current time no gaps are known with certainty.
One of the bigger problems for higher atomic number atoms is that as the nucleus becomes more charged the electrons orbiting the nucleus must go faster and faster to stay in orbit.  However nothing can exceed the speed of light and thus if an atom were to have electrons near the speed of light (say 99.99999% of c) at rest. The mere revolving of the earth around the sun would require the electron to move faster than light which presents and interesting case/thought experiment. Because of this it is reasonable to deduce that there is a penultimate element atomic number where anything larger would required electrons to exceed the speed of light though I cannot say what the magic number would be.
I will say though your question has a quite lot of merit as the Oddo-Harkins rule holds that the abundance of elements with odd atomic numbers in the universe is far less than that of elements with even atomic numbers except for hydrogen of course.  Further radioactive nuclei with even atomic numbers tend to be more stable and have a longer half life than those with odd atomic numbers.

There is no magic limit related to speed of light. (If it would be, it would occur around the number 137). Relativity does not work that way. As electrons move faster, they grow heavier, so it's still totally possible to have a stable orbit for any atomic number. Also, in fact they don't literally "move" at all, and there are no orbits - just the clouds of probability density.

Hey Ivan, I don't follow this train of thought: "As electrons move faster, they grow heavier, so it's totally possible to have a stable orbit for any atomic number." Could you elaborate more?

Imagine that electrons follow Bohr's orbits (which they do not, but let's forget about that for a moment). Then we have to balance the centripetal force $mv^2\over r$ against the electrostatic force $kZe^2\over r$. As Z exceeds ~137, the necessary $v$ exceeds the speed of light, so we might be tempted to think this is the end. But no, this is derived in the assumption that $m$ does not depend on $v$, which is not quite so (as per special relativity).

@A.K.  what about the Viola-Seaborg relationship, that odd numbers of protons and odd number of neutrons provide longer half lives for alpha decay. If alpha decay is the limiting factor, 119 should be favored over 120 as far as stability.  http://arxiv.org/ftp/arxiv/papers/1504/1504.00872.pdf

Thank you for your nice answer. There is however one thing about relativity, which is that you may not add speeds. So if something moves with the speed 0.9999999c, and then the Earth rotating with 0.00001c, the total will be something like 0.999999901c.

"the abundance of elements with odd atomic numbers in the universe is far less than that of elements with even atomic numbers except for hydrogen of course." I would make anexception for nitrogen, which is formed in the CNO cycle and is the only odd element other than hydrogen in the top ten.

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text:
Fluid Separation
I've been trying to find a way to prevent cornstarch and water from separating. If I'm right, emulsifiers are used to prevent fluid seperation with oils, but will an emulsifier fix my problem too? If so, will it break the non-Newtonian properties of the solution?


text:
In the Wittig reaction, why do stabilised ylides produce E-alkenes and unstabilised ylides produce Z-alkenes?

Wikipedia says that with unstabilised ylides they react very quickly in step 1-2 and step 4-5 is the rate limiting step. But with stabilised ylides step 1-2 is the rate limiting step and this is why stabilised ylides give E-alkenes. But there is no explanation as to why this makes E-alkenes and where in the mechanism it occurs.
I would have thought stabilised and unstabilised ylides would give E-alkenes to reduce steric hinderance but this is not the case. Any help explaining this would be very appreciated. :)

The Wittig reaction is generally considered to proceed via a concerted [2+2] cycloaddition, to form the oxaphosphetane 5 directly, without the intermediacy of the betaine 3. Philipp's answer (based on thermodynamic oxaphosphetane formation) is a mechanism that was accepted and taught many years ago (Clayden 1st ed was published in 2002). Nowadays, the generally accepted explanation is different; please see the duplicate questions.

I have given some justification on why unstabilized ylids give Z-alkenes in this answer of mine.
A possible explanation why stabilized ylids show E-selectivity is given in the organic chemistry textbook by Clayden, Warren, Wothers and Greeves on p. 817:

Just a disclaimer: this explanation, given in Clayden et al. Organic Chemistry 1ed, is generally no longer considered to be correct. I've written up a summary of the experimental evidence and the currently accepted model for (E)-selectivity: Why do stabilised ylids lead to trans alkenes in the Wittig reaction?


text:
Wood Hydrolysis
According to this article on Live Strong,

children have an average saliva pH of 7.5, while adults tend to be more acidic, with a saliva pH of 6.5 or lower.

Even though a pH of 6 is not very acidic, would someone be able to burn a hole in wood, for example, by applying enough saliva to it? Would the low acidity of saliva do nothing to the wood? I would assume, using common sense, that this would not be possible or else we would all have holes in our mouths (ew). However, would applying enough of a sufficiently concentrated solution for enough time to wood make it possible to burn a whole through it? If this is somehow possible, what would be the necessary conditions to make it happen in real life?
Thanks a lot in advance for any responses. All are greatly appreciated!

Given enough time (thousands of years), wood mostly would disintegrate by itself. Saliva will hardly make that happen any faster. 6.5 is not that acidic, after all. Water which we call "pure" is more acidic than that.

@IvanNeretin depends on where you are and what the water contains. Otherwise true, though.

The bacteria in your saliva and the general presence of moisture would do far more to degrade the wood than any acidity of the liquid.  It's still going to take years.  Mechanical damage from licking (ouch ouch ouch splinters) would likely be far faster still, and you can imagine what a slow process that would be.

Yes, wood is composed primarily of cellulose, which is composed of acetal linkages, and acetals are unstable in the presence of acid. Acetals are hydrolyzed in acidic conditions.
How long this would take is dependent on temperature and acid concentration, among other factors.
Also, cellulose in general is a very reduced carbon compound. Carbon prefers to be oxidized as carbon itself is not very electronegative. Carbon dioxide has a much more negative free energy of formation as opposed to cellulose. So given enough time cellulose will eventually decompose to carbon dioxide and water. This reaction must be spontaneous because the oxidation of cellulose to carbon dioxide and water is both enthalpically and entropically favorable. Breaking C-H bonds to make much stronger C=O bonds is enthalpically favorable, and the transformation of a solid to a gas and liquid are entropically favorable.

Cellulose has an average oxidation state of $\pm 0$. Do you honestly call that ‘very reduced’?

@Jan - acetic acid also has an average oxidation state of 0, yet its heat of combustion is -871 kJ/mol. Average oxidation state has no utility.

Ok, let me be more specific: the carbons in cellulose have $-\mathrm{I}, \pm 0$ and $+\mathrm{I}$. Acetic acid’s have $-\mathrm{III}$ and $+\mathrm{III}$. My standard deviation is smaller. But the point I was trying to make: I consider ‘very reduced’ to be $-\mathrm{II}$ or lower, i.e. saturated alkane.

@Jan what about carbon dioxide? It has an average oxidation state of 0 too. Would you argue that carbon dioxide is not highly oxidized? Average oxidation state is not helpful. An individual oxidation state analysis helps. And yes I agree that while the individual carbons in cellulose are not the most reduced, they are somewhat reduced and there are a lot of carbons per monomer of cellulose.

Also @Jan wouldn't the average oxidation state for any neutral compound be 0?

No, I was referring to the oxidation state of carbons averaged across all carbons. I mean, I generally agree with your answer, but the words very reduced don’t fit a sugar and put me off from upvoting. (Carbon dioxide’s carbon, of course has $+\mathrm{IV}$)

Absolutely not.  Wood is a plant and plants routinely have pH's between 5.5 and 6.1  So the wood would just see it as a protein rich relatively alkaline water solution.  That said, the water in the saliva would help the wood rot faster if it was otherwise dry.

Please elaborate as to why.

Your aliva is a higher pH than woods internal pH, therefor the wood is acidic relative to the saliva, so the saliva is alkaline relative to the wood.  Both are with in less than 1 pH of each other though.  So the pH difference would not cause the cellulose to denature in and of itself.
The OTHER component of spit though, the water, and possibly some of the enzymes, would damage the wood by allowing microorganisms that normally break down wood to do so more quickly.
So don't spit in the house.

In the question.

Elaborate on what you are asking and why you are downvoting.  My answer was clear and concise, i even went into more detail upon request, but your response "In the question".  Makes no sense.

I'm sorry I meant to add the information to the answer. That said concise answers are fine, but just giving the answer without the rationale is not what we strive for. Anyone is welcome to contribute answers but the aim of this site is quality and usefulness to future users (essentially we aren't Yahoo Answers). Please take a minute to look over the help center and tour page to better understand our guidelines and question policies.

I gave the answer "Absolutely not".  I then explained the answer is fairly precise detail.  unless you expect a full narrative on what pH means , what proteins are, what saliva is as well.

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text:
Determine the mole fraction
Under certain conditions, the well-known fertilizer $\ce{NH4NO3 (s)}$ explodes according to the following balanced reaction:
$\ce{2NH4NO3 (s) -> 2N2 (g) + 4H2O (g) + O2 (g)}$
Exactly 8 g of $\ce{NH4NO3}$ is placed in a 1.18 L constant-volume container containing dry air at 56.01 kPa and 15 ºC. The explosive reaction goes to completion and the temperature rises to 527 ºC. The container remains intact after the explosion. Determine the mole fraction of N2 in the gas phase after the explosion.
Data: Assume that the composition of dry air is 79% N2 and 21% O2 by volume.
So I know that $\ce{N2}$ takes 0.9322 L of volume. I calculated it by 0.79*1.18 = 0.09322
The only thing that I don't know is the number of moles, so do I just use the formula of ideal gas law which is $PV = nRT$,
 P = 0.5528 atm( 56.01kPA) V = 0.09322
However, which temperature should I use?
And am I using the right approach for this problem?

The temperature is provided because you have to make sure that the water becomes steam. Otherwise, the mole fraction is independent of temperature.

If it helps, you may assume that the remainder of dry air only consists of one inert gas. It will not affect your final answer.

@orthocresol I'm still confused about the question. So the volume of N2 (0.009322) which i got by multiplying 79%*8g/100% is the volume of N2 before reaction or after reaction? And if it is of  volume after reaction, can i just plug it in formula PV = nRT (T = 527 ºC) to search for number of moles of N2?

You have to separate the "before reaction" (8 g ammonium nitrate + 1.18 L dry air) and "after reaction" (decomposition products + 1.18 L dry air). Don't just do the calculation first and then later wonder about what you're doing. You should first think about what you want to find, then do the appropriate calculation to find whatever it is.

@orthocresol So i tried analyzing the problem i come up with this solution: First, i searched for the moles of NH4NO3 which is 0.1 moles which equals to the number of moles of N2 produced. So is 0.1 a solution for this problem?

Not really. You're asked to find the mole fraction of N2 after the reaction. That means the number of moles of N2, divided by the total number of moles of gas. Now bear in mind that some N2 existed before the explosion (it's 78% of the dry air = 0.093 L; how many moles is that?) and some of it was created by the explosion (0.1 mol, as you said). Also bear in mind that the total number of moles of gas = total number of moles PRIOR to the explosion, plus whatever gases were created during the explosion.

So you have four quantities to find, one of which you have already found. a) Number of moles of N2 before explosion. b) Number of moles of N2 created during explosion = 0.1 mol. c) Total number of moles before explosion. d) Total number of moles produced during explosion. (a+b) gives you the number of moles of N2 after the explosion, and (c+d) gives you the total number of moles after the explosion. So the mole fraction of N2 is (a+b)/(c+d).

You're asked to find the mole fraction of $\ce{N2}$ after the reaction. That is defined by:
$$x_{\ce{N_2},\text{f}} = \frac{n_{\ce{N_2},\text{f}}}{n_{\text{tot},\text{f}}}$$
where $x$ indicates a mole fraction, $n$ the number of moles and the subscript $\text{f}$ means "final". Furthermore, since the container is sealed, no gases can escape the container. This means that:
$$\begin{align}
n_{\ce{N_2},\text{f}} &= n_{\ce{N_2},\text{i}} + \Delta n_{\ce{N_2}} \\
n_{\text{tot},\text{f}} &= n_{\text{tot},\text{i}} + \Delta n_{\text{tot}}
\end{align}$$
where the symbol $\text{i}$ indicates "initial" and $\Delta n$ indicates the change in the number of moles caused by the decomposition of $\ce{NH4NO3}$. The four quantities on the right-hand side of these two equations are the quantities that you will have to find.

$n_{\text{tot},\text{i}}$ - This can be found by using the ideal gas law, $pV = nRT$ with the data given before the explosion. Substituting $p = 56.01\mathrm{~kPa}$, $V = 1.18\mathrm{~L}$, and $T = 286{~K}$, you will find that $n_{\text{tot},\text{i}} = 0.0276\mathrm{~mol}$.
$n_{\ce{N_2},\text{i}}$ - Since 79% of pre-explosion air is $\ce{N_2}$, $n_{\ce{N_2},\text{i}} = (0.79)(0.0276\mathrm{~mol}) = 0.0218\mathrm{~mol}$.
$\Delta n_{\ce{N_2}}$ - This is related stoichiometrically to the number of moles of $\ce{NH4NO3}$. In fact, it is numerically equal, as the coefficients of $\ce{NH4NO3}$ and $\ce{N2}$ are the same. The molar mass of $\ce{NH4NO3}$ is $80.04\mathrm{~g~mol^{-1}}$, so $\Delta n_{\ce{N_2}} = (8\mathrm{~g})/(80.04\mathrm{~g~mol^{-1}}) = 0.100\mathrm{~mol}$.
$\Delta n_{\text{tot}}$ - This is also related stoichiometrically to the number of moles of $\ce{NH4NO3}$. Each mole of $\ce{NH4NO3}$ leads to the formation of $(2+4+1)/2 = 3.5\mathrm{~mol}$ of gaseous products (observe that water will be gaseous after the explosion due to the rise in temperature). Therefore, $\Delta n_{\text{tot}} = (3.5)(0.100\mathrm{~mol}) = 0.35\mathrm{~mol}$.

Putting it all together, we have:
$$x_{\ce{N2},\text{f}} = \frac{0.0218 + 0.100}{0.0276 + 0.35} = 0.323$$
Note that the mole fraction is dimensionless. The exact final temperature is also of no use; its only purpose is to tell you that water will exist as a gas (and therefore contribute to the total number of moles). It could be any other number above $100^\circ\mathrm{C}$ and the final answer would be the same, as the mole fraction is independent of temperature.
I hope I didn't mess up the arithmetic.


text:
Hydrates In Aqueous Solutions
Say you had copper(II) sulfate pentahydrate as a solid.
In an aqueous solution, is it still referred to by its hydrated form? Is it still called copper(II) sulfate pentahydrate (aqueous) or simply copper (II) sulfate (aqueous)?
I guess what I'm asking here is does the "hydrate" label apply exclusively to solids?

Yes, the "hydrate" label applies exclusively to solids. Once you got your salt dissolved, there is absolutely nothing special about those water molecules that came from the solid.

Yes, the ‘hydrate’ label applies exclusively to solids. Consider the following experiment: Dissolve both anhydrous $\ce{CuSO4}$ and $\ce{CuSO4 . 5 H2O}$ in water. You will get the same, blue solution.
What happened on a microscopic level is that the crystal lattice which once included the water atoms is broken up and the ions dissolve separately. What used to be copper(II) sulphate pentahydrate is now turned into a hexaaquacopper(II) complex.
$$\ce{CuSO4.5 H2O + H2O -> [Cu(H2O)6]^2+ + SO4^{2-}_{~(aq)}}$$


text:
Effect of substituents on aniline C-N bond length
Which compound has larger $\ce{C-N}$ bond length, 4-aminobenzaldehyde (1) or aniline (2)?

I think it's the second compound. $\ce{-CHO}$ is an electron withdrawing group. This means the electron density of the ring decreases. So the carbon attached to nitrogen becomes more electronegative. Hence its p-character decreases and the $\ce{C-N}$ bond length decreases.

You think right, but for the wrong reason. The inductive effect is not that great and does not quite reach that far. It is the effect of conjugated $\pi$-system that really matters. See, we may draw an alternative resonance structure for I with minus on O, plus on N, and rearranged double bonds (quinone-like ring instead of aromatic). This means the C-N bond in I is kind of "partially double" (i.e., has a noticeable $\pi$ component), hence it is shorter.

Indeed, there is some conjugation between the aromatic ring and nitrogen's lone pair in unsubstituted aniline as well. That's what makes aniline a weaker base than ammonia, and its aromatic ring more reactive towards electrophiles than benzene. It is just that the $\ce{-CH=O}$ group in para position makes that conjugation stronger yet, as we may see qualitatively by drawing the resonance structures, or (supposedly) quantitatively by means of quantum chemistry.


text:
Axial vs equatorial attack of Grignard reagent on dioxan-5-one and dithian-5-one
How to account for the diastereoselectivety (axial vs equatorial attack) of these reactions?

Isn't that very advanced for 1st year? (from your profile). Think about the outcome when S or O are plain carbons and then try to find which one O or S doesn't "behave" as expected.  Clearly the 2 different results have to come from different transition states. How can $\ce {MeMgI}$ interact with the starting materials?

Oops, that's a mistake. The second year started this week. MeMgI can coordinate to the oxygen and then attack from above. On the other hand it doesn't coordinate so strongly to sulfur and the Me- adds to the sterically less demanding equatorial position?

Nevertheless, the addition to dioxan-5-one doesn't seem to be in agreement with Felkin-Ahn's model, and chelation is not an option in this rigid system.

https://en.wikipedia.org/wiki/Asymmetric_induction - so it's Anti-Felkin product?

Also, you should add info like this in your comments to your question

I don't see how it is Anti-Felkin.

I've found the solution in Fleming: Molecular Orbitals-Reference edition:

The dioxan-5-one is attacked by nucleophiles in the gas
  phase from the equatorial direction, because of electrostatic repulsion from the fraction of negative charge
  carried by the two oxygen atoms in the ring, as Houk had predicted. The selectivity is not particularly high
  (67:33), because the usual torsional strain is still present, enhanced in this case by the shorter C—O bond
  lengths making the ring less puckered, and increasing the dihedral angle by which the carbonyl oxygen atom
  is pushed below the neighbouring C—H bonds. In solution, however, the ketone is attacked
  from the axial direction, just like cyclohexanones. Evidently the solvent insulates the charge on theincoming nucleophile from the electrostatic repulsion, the torsional strain remains high and there are no axial
  substituents hindering axial attack.The sulfur atoms in the dithia analogues are less electronegative than the oxygen atoms in the acetals but the axial lone pairs bulge further up, and are not heavily solvated. However, the carbonyl group is
  displaced even further down by the changes in bond length and bond angles from having two C—S bonds in
  the ring. The one effect makes axial attack less favourable, and the other makes equatorial attack less
  favourable. In the event both solution and gas phase reactions take place with a strong preference for
  equatorial attack.

You couldn’t have presented that in any less readable way, could you? (Sorry if I sound aggressive here but remember that your question (and answer) will remain as references for future visitors, and this type is likely to put people off.)

Maybe it could be better presented but still deserves an up-vote


text:
What is the reaction that takes place in the cathode of a Mg-Fe electrochemical cell?

Setup
A magnesium electrode dipped in a magnesium sulfate solution and an iron electrode dipped into iron (III) sulfate solution

Am I right in thinking that you need to first identify the redox reactions to determine to determine overall voltage of the cell at standard conditions?
I have identified $\ce{Mg}$ to be oxidized to $\ce{Mg^2+}$ at the anode, but I am not sure which is reduced at the cathode. Would it be the sulfate ions?

@Mithoron Oxidize to sulfurous acid?

Sulfate can indeed be reduced by magnesium as magnesium is very strongly reducing, and you would get $E^\circ = +2.54\mathrm{~V}$ as you said, but there's probably a stronger oxidising agent than sulfate in that $\ce{Fe2(SO4)3}$ solution.

[...] I am not sure which is reduced at the cathode. Would it be the sulfate ions?

orthocresol has already pointed out in his comment that there's more in $\ce{Fe2(SO4)3}$ than just the sulfate.
Take a look at the cation:
$$\begin{align}
\ce{Fe^3+& + e- &<=>&Fe^2+\quad &+0.77V} \\
\ce{Fe^3+& + 3e- &<=>&Fe(s)\quad &-0.04V}
\end{align}$$
(Standard electrode potentials are cited from wikipedia.)

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text:
Do you know a simple fast front-end MD software for OS X
I need a GUI front-end simple MD software for OS X, something similar to ascalaph maybe.
What I need to do is a rigid optimization of waters around a protein. I don't have many water molecules around, so a few tips about how to not let them fly away would be nice too.

Could you put your protein an waters in "box" so they're forced to stay within some distance of your protein?

yes, I still would like to do an energy minimisation.

I'm biased, but I recommend Avogadro which can do simple force field optimization using MMFF94 or GAFF. It's not MD, but it would certainly work. It runs on Mac.
The water molecules shouldn't "fly away" during minimization, since there should be hydrogen-bond and Van der Waals terms attracting them to each other and to the protein.
I'll also point out that VMD runs on Mac and are quite easy to use. It can set up MD calculations with NAMD, GROMACS, etc.

I agree with VMD. That's a great choice!


text:
Finding partial pressure of a gas in a mixture

A mixture of argon and carbon dioxide was initially at a total pressure of $\pu{1.72 atm}$ in a volume of $\pu{1.72 L}$ at $\pu{17 ^\circ C}$. The carbon dioxide was then completely removed from the mixture by passing it through a solution of sodium hydroxide ($\ce{NaOH}$). The argon was collected in a $\pu{1.1 L}$ vessel and its pressure found to be $\pu{1.55 atm}$ at a temperature of $\pu{28 ^\circ C}$. What was the partial pressure of carbon dioxide in the original gas mixture?  Assume that argon and carbon dioxide behave as ideal gases.

So, initially I found the moles of argon in in a $\pu{1.1 L}$ vessel which is $\pu{0.06896 moles}$
Then, i subtracted $\pu{1.72 L}$ from $\pu{1.1 L}$ in order to get a volume of carbon dioxide and used it in $PV = nRT$ and i got $\pu{0.04474 moles}$ of carbon dioxide .
Finally, I used the formula for partial pressure
$\frac{\text{moles of carbon dioxide} \times\text{total pressure}}{\text{moles of argon} + \text{moles of carbon dioxide}}$ and got $\pu{0.6771 atm}$.
However, my answer is not correct according to the textbook. Where did i make a mistake?

Your intial step is correct, finding the number of moles of argon in the 1.1 vessel. The next step however is incorrect. Because the pressure and temperature are both in different conditions for argon in the end and argon and carbon dioxide together in the beginning, you cnnot simply subtract the volume 1.72 - 1.1 L and then find the number of moles of carbon dioxide there. Remember, subtracting the volume and finding the moles of co2 would only work if the conditions are the same, but in this case, the pressures and temepratures are different. Thus, you can just simply go about finding the total n moles of Argon and carbon dioxide through PV=nrt.
So, 1.72 (1.72L) / 0.08206 x (273 + 17) = 0.1243 total moles. Subtract 0.06869 moles from earlier from 0.1243. You will get 0.055356 moles of CO2. Now, you can use your partial pressure method.
Hope this helps.


text:
Why are glycosidic bonds always 1->4 in polysaccharides?
I know 1->6 bond exists too, in amylose for example, but that one seems logical, since the 6th carbon is easier to access.
I understand why the bonds originate from the hemi-acetal carbon, but I don't understand why they are 1->4.
Why not 1->3 or 1->2 ? Since there are hydroxyles on every carbons in the ring?
While I was writing this, I just had an illumination, maybe someone can confirm:
Could it be that, the formation of glucose chains in plants and animals always make use of enzymes and that it is these enzymes which, by the shape of their active site, dictates what the bonds will be?

Why are glycosidic bonds always 1->4 in polysaccharides?

The premise of the question is false.

Why not 1->3 or 1->2 ?

1,3-linked glycosidic bonds certainly exist.  In fact there's a whole class of molecules called beta glucans of which the 1,3-linked variety forms a major part.

Could it be that, the formation of glucose chains in plants and animals always make use of enzymes and that it is these enzymes which, by the shape of their active site, dictates what the bonds will be?

I think what you say here is true, but you should keep in mind that there are many different enzymes, and some will specifically form one linkage, and others will form alternate linkages.


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Finding the mass of NaNO3 in original mixture from preciptate
A 1.61-g sample of a mixture of $\ce{AgNO3}$ and $\ce{NaNO3}$ is treated with excess $\ce{Na2S}\:(\mathrm{aq})$. The precipitate is filtered off, dried and weighed. The dried precipitate weighs 0.44 g. What is the percentage by mass of $\ce{NaNO3}$ in the original mixture?
I attempted this problem by using two equations: one with $\ce{AgNO3}$ and $\ce{Na2S}$, and the other with the other reactant and $\ce{Na2S}$. Then I wrote an equation in moles, using m and 1.61-m as the mass of the reactants. My answer was 2.34 g, which is greater than the original so I know I did something wrong.

The dried precipitate is $\ce{Ag2S}$.
The number of moles of $\ce{Ag2S}$ is:
$$n=\frac{m_{\ce{Ag2S}}}{M_{\ce{Ag2S}}}=\frac{0.44}{32+108\times 2}=\frac{0.44}{248}=0.00177\,\mathrm{mol}$$
The number of moles of ion $\ce{Ag(I)}$ is$$n'=2n=0.00355\,\mathrm{mol}$$
On the other hand, the percentage by mass of $\ce{AgNO3}$ in the original mixture is: $$\frac{m_{\ce{AgNO3}}}{m_{\ce{(AgNO3 + NaNO3)}}}\times 100=\frac{n'\times M_{\ce{AgNO3}}}{m_{\ce{(AgNO3 +NaNO3)}}}\times 100=\frac{0.00355 \times (108+14+16 \times 3)}{1.61}\times 100=37.48\%$$
So, the percentage by mass of $\ce{NaNO3}$ is: $100-37.48=62.52\%$


text:
Finding moles of water evolved
How many moles of water are released on gentle heating of 1 mole of $\ce{Al2Cl6•12H2O}$?
Since the heating is gentle, the water molecules that are not present in inner orbital complex of compound and not directly attached to aluminium will be evolved, but the structure of $\ce{Al2Cl6}$ is a bridged tetrahedron. Then again, water is a better ligand than chlorine, so overall I cannot decide what the structure of the compound should be.

Anhydrous aluminium chloride ($\ce{AlCl3}$) exists; however, it cannot be obtained by heating of the hexahydrate ($\ce{AlCl3.6H2O}$).
$$\ce{AlCl3.6H2O \rlap{\ \ \not}->[\quad] AlCl3 + 6 H2O}$$
The crystalline hexahydrate $\ce{AlCl3.6H2O}$ actually is the complex salt $\ce{[Al(H2O)6]^3+ 3 Cl-}$.
The acidic hexaaquaaluminium(3+) ion is susceptible to hydrolysis.
$$\ce{[Al(H2O)6]^3+ <=> Al(OH)3 + 3 H3O+}$$
Therefore, the hexahydrate $\ce{AlCl3.6H2O}$ hydrolyses upon gentle heating and releases three equivalents of water and hydrogen chloride.
$$\ce{AlCl3.6H2O -> Al(OH)3 + 3 HCl + 3 H2O}$$

And vigorous heating of the $\ce{Al(OH)3}$ would yield $$\ce{2Al(OH)3 -> Al2O3 + 3H2O}$$


text:
Thermochemistry Problem - How did they get that volume?
The chemistry problem in my textbook is : Calculate the work one when 2.0 L of methane gas, $\ce{CH4}$ (g) undergoes combustion in excess oxygen at 0ºC and 1.00 bar. Assume the volume of water formed is negligible.
I know that we should use the work equation $\mathrm{Work=-Pressure \cdot \Delta Volume}$, and that the pressure is 1.00 bar and we must convert this into pascals. All i need is the volume and convert that to $\mathrm{m^3}$. I got the equation $\ce{CH4 + 2O2 -> CO2 + 2H2O}$, but what I don't get is that how do we find the change in volume?
The book says let $\ce{O2}$'s volume = $\mathrm{V_{initial}}$, and that the total volume of gases for initial is $\mathrm{V_{i}}$ + 2.0 L. It says the final volume is 2.0 L of $\ce{CO2}$ (which I get because it should be the same amount of liters for carbons), but then it says that the water's volume is $\mathrm{V_{i}}$ - 4.0 L of $\ce{O2}$.
Can someone please explain the water's volume? How did they get that? Why is then the total final volume $\mathrm{V_f}$= $\mathrm{V_i}$ - 2.0 L. I thought the number of moles for initial should be same as the number of moles according to Gay-Lussac's law, so then the number of volume total should be the same for both sides, but this is clearly not correct. Am I applying the Gay-Lussac's law wrong? Why is the final volume not the same as the initial volume even if the initial and final have the same amount of moles on each side?

I think most of your issues regarding this question will be solved if you actually put the states in your chemical equations, so it becomes: $$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}$$ It is important to note that all the substances are in the gaseous state except for water which is a liquid. Therefore your application of Gay-Laussac's law is incorrect here as the number of moles of gaseous reactants is not equal to number of moles of gaseous products. Thus the initial and final volume shouldn't be the same. This is also why in the question they have stated to assume that the volume of water is negligible as water is a liquid which has very little volume compared to a gas.
The change in volume can be calculated by noting that 3 moles of gas reacts to form only 1 mole of gas. Since there is 2L of methane, that means that the volume is going to decrease by 4L.
Therefore: $\mathrm{V_f = V_i - 2.0L}$


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How does a water molecule bond to an anhydrous salt to form a hydrate?
Some sites suggest that water molecules bong to anhydrous salt via a "loose bond". What does that even mean? $\ce{H2O}$ is a polar molecule and salts are made up of ionic bonds. Is it a hydrogen bond between the anions and hydrogen atoms?

"Loose bond" is really a loose term which engenders loose understanding. The bonds that hold water molecules in the structure are usually dative bonds, which is a variety of polar covalent bonds. They are formed between water's O and the metal ion, using spare electron pairs of the former. There is nothing unusual about having covalent bonds in an ionic crystal. Think, for example, of $\ce{S-O}$ bonds in the sulphate anions within $\ce{MgSO4}$ or any other ionic sulphate.
Alternatively, in some relatively rare cases water molecules may be held by hydrogen bonds to the anions, like you suggested. This is the case with that fifth water molecule in $\ce{CuSO4\cdot5H2O}$, as explained by @ringo.

Water is often necessary in crystal formation, and in some cases is incorporated into the structure of the crystal. Though these waters of hydration can be removed by heating the crystal, some of the properties and structure of the crystal are lost. For the most part, water acts as a ligand in hydrates, surrounding the metal cation. The ionic attractions occurs between the coordination complex and the anions, and hydrogen bonds do occur between $\ce{H2O}$ molecules and the anions, like you stated.
$\ce{CoCl2.H2O}$, for example, exists as $\ce{[Co(H2O)6]^2+}$ and $\ce{Cl-}$.
Water does not have to necessarily surround the metal, however. $\ce{CuSO4.5H2O}$, a very popular hydrate, exists as $\ce{[Cu(H2O)4]^2+}$ and $\ce{SO4^2-}$. Copper is surrounded by 6 oxygen atoms, 2 from the $\ce{SO4^2-}$, 4 from $\ce{H2O}$. The other $\ce{H2O}$ molecule is in the structure of the crystal, but does not bind directly to copper.
[Source]


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Zn-Cu galvanic cell
In a Zn-Cu galvanic cell, if both electrodes connected by a wire are placed in the same container with a mixture of both electrolytes, will electrons still flow through the wire?

Of course they will. But there also will be another process we'd rather avoid: metallic Zn will start reacting with Cu electrolyte directly. That's the reason why we usually want to separate cathode and anode spaces.


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On the notion of a component in thermodynamics
Wikipedia article on a component (in thermodynamics) first mentions that:

In thermodynamics, a component is a chemically-independent constituent
  of a system. The number of components represents the minimum number of
  independent species necessary to define the composition of all phases
  of the system.

A bit later it is said in the same article that:

The number of components is equal to the number of distinct chemical
  species (constituents), minus the number of chemical reactions between
  them, minus the number of any constraints (like charge neutrality or
  balance of molar quantities).

How does the second statement follows from the first one?

Well, a reaction comes with an equilibrium constant, i.e., mathematically speaking, some equation on constituents. Now we have N unknowns (constituents) and M equations (constraints and equilibrium constants). Supposedly N>M. This means we can't determine all unknowns. To do so, we must fix N-M free variables. These are our components.


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How to prove that the orientation of the atomic orbitals in the superposition $\psi= a\psi_{1} + b\psi_{2}$depends on the coefficients $a$ & $b$?
Hybridisation is a quantum superposition of orthogonal atomic orbitals; for instance
$$\newcommand{\k}[2]{\langle #1 \lvert #2 \rangle}
\psi_{sp^3}= c_1\psi_{2s}+ c_2\psi_{2p_{x}} + c_3\psi_{2p_y} + c_4\psi_{2p_{z}}.$$
It is a quantum superposition which implies the coefficients $c$ to be the probability amplitudes for finding the electron in respective atomic orbital provided that it is prepared in $\lvert\psi_{sp^3}\rangle$(provided $\sum_{i=1}^4 c_1^2 =1$) that is, $$c_1= \k{\psi_{2s}}{\psi_{sp^3}}, \\ c_2= \k{\psi_{2p_x}}{\psi_{sp^3}},\\ c_3= \k{\psi_{2p_y}}{\psi_{sp^3}},\\ c_4= \k{\psi_{2p_z}}{\psi_{sp^3}}.$$
Up to this, I could easily understand this as it was just like some other superposed state often encountered in QM like $\lvert\phi\rangle = \lvert i\rangle c_i + \lvert j\rangle c_j + \lvert k\rangle c_k$ where $c_i = \k{i}{\phi}.$
But, here in the case of hybridisation, they are much more than mere comples numbers that are the amplitudes of the electron to find it in one of the base states that is the associated atomic orbitals when being prepared at the hybridised state. It is evident from Peter Atkins' words:

The coefficients in the hybrid have been chosen to give the correct directional properties of the hybrid. The squares of the coefficients give the proportion of each atomic orbital in the hybrid.

Also, the snapshot$^1$ below shows how for different values of coefficients, the spatial orientation is changed.

That means the coefficients  direct the spatial orientation of the atomic orbitals that are the base states for the superposition or hybridisation.
Is there any mathematical deduction that shows the orientation of the atomic orbitals in the superposition depends on the coefficients $c_i$?
I'm basically not understanding how these coefficients, which are rather probability amplitudes, would involve in determining the spatial orientations of the atomic orbitals. Can anyone explain me why these probability amplitudes influence the orientation of the atomic orbitals? Also, what is the mathematical proof of this?

$^1$The snapshot, above, is taken from Principles of Physical Chemistry
 By Hans Kuhn, Horst-Dieter Försterling, David H. Waldeck.

Do not cross-post same question on different SE sites. You already asked the question on Physics.SE, so wait for an answer there (or ask the mods there to migrate it here).

I'm already deleting it. So, no offense @Wildcat .

In quantum mechanics states/orbitals are defined within the mathematical framework of the Hilbert space. A Hilbert space is a vector space that generalizes the notion of Euclidean space -  one example of the latter would be the normal two- or three-dimensional spaces you might be used to from basic geometry. So, orbitals are nothing more but abstract vectors in a Hilbert space and you can draw many analogies between them and the more familiar two- or three dimensional vectors you learn about in high school. So, let's think of quantum mechanical states as vectors for now. If you add two vectors, $\vec{a}$ and $\vec{b}$, you get another vector, $\vec{c} = \vec{a} + \vec{b}$, which if $\vec{a}$ and $\vec{b}$ are linearly independent, will have a direction different from either $\vec{a}$ or $\vec{b}$. You can also choose different coefficients for the superposition, e.g. $1$ for $\vec{a}$ and $-1$ for $\vec{b}$ yielding a vector $\vec{d} = \vec{a} - \vec{b}$ which now has a direction different from either $\vec{a}$, $\vec{b}$ or $\vec{c}$. So, the direction of a sum vector changes with the coefficients.

The same principle holds true for orbitals. If you for example think of the $\mathrm{p}_{x}$ orbital as a vector pointing in the $x$ direction, i.e. in two dimensions $\bigl(\begin{smallmatrix}1 \\ 0 \end{smallmatrix}\bigr)$, and of the $\mathrm{p}_{y}$ orbital as a vector pointing in the $y$ direction, i.e. in two dimensions $\bigl(\begin{smallmatrix}0 \\ 1 \end{smallmatrix}\bigr)$, then a superposition of both (with equal coefficients of $\frac{1}{\sqrt{2}}$ to ensure normalisation) will give you a $\mathrm{p}$ orbital pointing in the $\bigl(\begin{smallmatrix}1 \\ 1 \end{smallmatrix}\bigr)$ direction.

But as with vectors you are by no means confined to simply add them together with equal coefficients. For example, if you add $2$ parts of $\mathrm{p}_{x}$ and $1$ part of $\mathrm{p}_{y}$ you get a $\mathrm{p}$ orbital pointing in the $\bigl(\begin{smallmatrix}2 \\ 1 \end{smallmatrix}\bigr)$ direction, and that is clearly different from $\bigl(\begin{smallmatrix}1 \\ 1 \end{smallmatrix}\bigr)$.

I appreciate your effort, sir but Feynman always in his lecture cautioned that state-vectors are rather abstract stuffs & can't be comparable to three-dimensional vectors. Hmmm....Quantum superposition has no analogy with vector-addition:(

There certainly are differences but in your case you only need to use the basic properties that all vector spaces share and that thus allow you to make analogies from three-dimensional euclidean to hilbert spaces: for example every linear combination of members of a vector space must also be a member of the vector space. And those linear combinations can be done for euclidean and hilbert spaces alike. Also, if two members of a vector space are linearly independent then they have a different "direction". Those are general concepts that don't depend on the kind of vector space.

@user36790 Well, quantum mechanical states are complex functions rather than real vectors. That alone gives you some important differences. But I don't have any striking pitfalls in the back of my mind where it is actually dangerous to draw the analogy between three-dimensional vectors and quantum states.

@user36790 Btw, what you do when forming hybrid orbitals is not a quantum superposition but a change of basis or rotation of your coordinate system, in the same sense as that you can use the vectors $\bigl(\begin{smallmatrix}1 \ 0 \end{smallmatrix}\bigr)$ and $\bigl(\begin{smallmatrix}0 \ 1 \end{smallmatrix}\bigr)$ as a basis for the two-dimensional euclidean space but you can just as well use linear independent combinations of the two, e.g. ...

... $\frac{1}{\sqrt{2}}\bigl(\begin{smallmatrix}1 \ -1 \end{smallmatrix}\bigr)$ and $\frac{1}{\sqrt{2}} \bigl(\begin{smallmatrix}1 \ 1 \end{smallmatrix}\bigr)$.

Hybridisation is indded quantum superposition. Check the answers of  this post.

@user36790 Ok, I am not sure then if quantum superposition might not simply be the same as linear combination. I thought they might have subtly different meanings, but my QM course was quite some time ago. If it is nothing more than a linear combination then I don't understand Feynman's statement that Quantum superposition has no analogy with vector-addition because both are simply a manifestation of one of the basic axioms of vector spaces in general (see here).

Quantum superposition & linear combination are both same things, but in QM, they are bit-abstract like the possibility of electron being at two places at the same time but that is another topic.......

Feynman, I think, wanted to distinguish the subtle abstractness of quantum superposition from the mere linear-combination of vectors.

Ok, I am still convinced that my answer is correct and doesn't use any wrong analogies. If someone can show me where my analogies fail I would be interested to see that. It would actually be interesting to know in what sense Feynman sees a fundamental difference between vector additon and quantum superposition because the only difference I can think of is the possibility of complex coefficients.

You are overestimating his statement by saying  fundamental. Also, I've not in any way tried to offend you that your answer is incorrect; I was just ... just a bit confused, that's it. I am again saying thanks for your effort. Sorry for those arguments:(

@user36790 I am not offended. No worries. Nor is there anything to feel sorry about. Maybe you could ask the question what Feynman ment by "Quantum superposition has no analogy with vector-addition" on Physics.SE. Because I think I know some quantum mechanics and I don't really see his point. Maybe it is really just about the abstractness and the complex nature of quantum states but maybe there is actually a fundamental difference that I simply don't see. Would be an interesting question.

I am quite sure he is talking about the abstractness, though. He is an excellent explainer; I've learnt QM(introductory) from his lectures but really sometimes his English does annoy me a lot.

As a German I think I am in no position to question anyone's English (our dialect can be really terrible) :) If you want a more mathematical yet very well explained exposure to QM I can recommend this book by Shankar. The first two chaperts give you a really nice résumé of the linear algebra and classical mechanics basics you need for QM.

Yes, this book is quite famous in India after Griffiths but I preferred Feynman to them as it is more diverse & above all it was his valuable lectures at CALTECH; who would want to miss that?  His lucid, unorthodoxical way of  explanation made these lectures a classic. Though these lectures are over 45 yrs old, still they are not outdated. Every physicist has studied his lectures at least one time in their career. Thanks for the recommendation, nevertheless.

@Philipp - re: ""Quantum superposition has no analogy with vector-addition" // I would take that to be a comment on the multibody problem. So "real" atoms don't form bonds using two dimensional vectors. The "real" problem is a 3D 3-body problem.  So we use it because it provides useful insight, not because the method is "real."


text:
IUPAC naming of double salt
My book gives the IUPAC name of $\ce{KAl(SO4)2.12H2O}$ which is nothing but emperical formula of potash alum as
"Potassium aluminium (III) sulphate -12 water"
But Wikipedia gives it as "Aluminium potassium sulfate dodecahydrate" , though there could be change in name because of number of water molecules but there is significant difference between the two names
So, my question is how to give IUPAC name to a double salt (potash alum is a double salt), and why is there difference between two iupac names I.e the one which is given in my book and the one in wikipidea.

Aluminium potassium sulphate (or sulfate) dodecahydrate is correct. Cations in alphabetical order. However, the reversed form is tolerable.

According to the current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book), alphabetical order is used.

Alphabetical order is used in formulae as follows.
(a) Within the group of cations and within the group of anions, respectively, in formulae of salts and double salts. Deviations from this rule may be acceptable if it is desired to convey specific structural information (…)

and

Alphabetical order is used in names as follows.
(d) In compositional names, the names of the formally electropositive components and the names of the formally electronegative components are each arranged alphabetically with the former group of components preceding the latter. (…)

Note that the order of components in the name may deviate from the order of the corresponding components in the formula. For example:
$\ce{KMgF3}$ (since K < M)
magnesium potassium fluoride (since m < p)
$\ce{ZnI(OH)}$ (since I < O)
zinc hydroxide iodide (since h < i)


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Mass number, (relative) atomic mass, average mass
What is the difference between mass number, atomic mass and average atomic mass?
I know the mass number is the amount of protons + the amount of neutrons. The average mass is the weighed average of the isotopes that occur in nature. But then what is the (relative) atomic mass?

Related: Quick and simple explanation of molar mass, molecular mass and atomic mass and Units of mass on the atomic scale

It is relative to the mass of another element.  So, you may set the mass of carbon to be 12.000 and get all the masses of the elements relative to that; or, you may set the mass of oxygen to be 16.000 and get the masses of the elements relative to that.  These two mass systems will be slightly different but internally consistent.  Hope that helps.

It’s pretty simple:

The mass number is the integer you get if you count (and add up) the neutrons and the protons of a given element. Thus, a hydrogen atom of the $\ce{^1_1H}$ isotope has a mass number of $1$ (only proton), $\ce{^12_6C}$ has $12$ ($6$ protons, $6$ neutrons) and $\ce{^81_35Br}$ has $81$ (of which $35$ are protons, the remaining $46$ neutrons).
The atomic mass is what these atoms actually weigh in atomic mass units. For reasons that boil down to $E = mc^2$ (or so I believe) and the nonzero mass of an electron, this is not an integer except for one exception:

$\ce{^1H}$’s atomic mass is $1.007825032 1(4)$
$\ce{^12C}$’s atomic mass is exactly $12$. This is because $1~\mathrm{u}$ was defined as exactly $1/12$th of the mass of a carbon-12 atom.
$\ce{^81Br}$’s atomic mass is $80.916289(6)$

The average mass takes into account an elements different isotopes and their natural abundance and calculates an overall average. Thus, this is no longer defined on an isotopal basis but on an elemental one. The average masses of the elements discussed above are:

$\ce{H}$ has $1.00794 (7)$ (this is larger than the atomic mass of $\ce{^1H}$ due to the heavier isotopes.)
$\ce{C}$ has $12.0107 (8)$
$\ce{Br}$ has $79.904 (1)$ (this is lower than the atomic mass of $\ce{^81Br}$, because about half of the naturally occuring bromine atoms are $\ce{^79Br}$.)

well lets start it off with the example of a pizza, first imagine a pizza that is divided into 12 slices where 6 slices are meat free and the other 6 with meat and you were told that different pizzas have different amount of slices and different proportions of meat:non-meat ones the total number of pizza slices here is what we could refer to as "pizza number" now the seller tells you that this pizza(12 slices) weighs exactly 1200 grams and the other one (mini-pizza)weighs 600 grams now there weighs here is what we will refer to as pizza mass, now someone asks how much does this pizza weighs relative to the other we do bunch of calculations and we find out that the relative pizza mass of the mini should be on the scale of 1,and so the other pizza should be two{its double the mass} now what we have is relative pizza mass
now to sum it up lets consider the atom the atom nucleus does have proton and neutron and different atoms have different number of protons and neutrons in different proportions the number of both proton and neutron number is mass number it is practically helpful if we have different isotopes because each isotope have different mass number and so could be identified by their numbers,the atomic mass which is measured by mass spectrometer is the real mass of the atom which also varies between atoms if they were to sum up all those isotopic masses and take the average the result will be average atomic mass considering how small an atom is it will be awkward to use units like kg because the total mass will be small that you will need to write the number in standard form of such 1*10^-26 so the use the relative atomic mass to make it easier by using the scale of 12 amu(atomic mass unit) for a carbon 12 atom, and by that they can make other relative mass such as hydrogen(1.008),oxygen(15.999) and so on
 summary using an example

nitrogen atom with 7 neutron and 7 protons have mass number of 14 and thus isotope name is nitrogen-14
the atomic mass of nitrogen-14 is 2.32525265e-26
average atomic mass of all isotopes is  2.325x10^-26
relative atomic mass of all isotopes of nitrogen is 14.007 amu

*note:this is based on my personal experience if someone with any suggestions and arguments please consider a comment
*that is all what I have, thanks


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Pressure in Ideal Gas Law
In Boyle's law ($p$ inversely proportional to $V$ for constant $n$ and $T$), $p$ refers to the applied (external) pressure on the gas.
However, in the ideal gas law, my textbook says that $p$ refers to the pressure of the gas, which I think refers to the pressure applied by the gas. How is this possible, since the ideal gas law is derived from Boyle's law (and Charles' and Avogadro's laws)?
Shouldn't the $p$ in the ideal gas law then refer to the applied (external) pressure like in Boyle's law, not the pressure of (i.e. applied by) the gas?

Everything should refer to the pressure exerted by the gas. The line on Boyle's law that you read was probably written with the assumption in mind that the internal and external pressures are equal.

But doesn't Boyle's law make more sense when P is the applied pressure? I thought the P in the ideal gas law needed to be changed, not the one in Boyle's law.

"Make more sense" is very vague... It makes more sense from an experimental point of view because the only way you can vary the volume is by varying the external pressure. However in that case the internal pressure will increase to match the external pressure, and Boyle's law will hold equally well regardless of which pressure you use. Which is why I said, it was probably written with the implicit assumption that the internal and external pressures were equal.

That is in the case of a container with movable walls, which will allow for equilibration of the internal and external pressure. In the case of fixed walls, you cannot experimentally verify Boyle's law anymore; however, the ideal gas law still holds true as long as you use the internal pressure. The ideal gas law relates properties of a system to each other, and in this case where the system cannot "see" anything outside since the walls are fixed, the external pressure cannot have any influence on the volume or temperature of the gas inside.

At equilibrium (that is, if volume/temperature/mass is stable and not changing over time), the external pressure applied to the gas will equal the internal pressure of the gas.
This is (more or less) a consequence of Newton's third law. The external environment pushes on the gas with a certain force, and the gas pushes back with an equal and opposite force. At equilibrium, no work is being done on the gas or on the environment, so the forces have to balance. (This is very crude and hand-wavy - a more thorough description would require much more math.)
So (at equilibrium), there really isn't any difference between the two. In the Boyle's law case, the value is described as the external pressure, because the typical experimental setup in that case involves using something like a piston to control and vary the external pressure applied to the system. In the ideal gas case, you may not explicitly be controlling the external pressure, instead you're controlling temperature, volume, or particle numbers. In those case, you're normally measuring the internal pressure of the system, rather than (directly) controlling it or the external pressure. But (at equilibrium) the pressure of the gas pressing out is the same as the pressure of the container pushing in.
It's all just a matter of perspective, and how the system is set up.


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Formation of polymers - addition polymerisation
In addition polymerisation, a double bond in the monomer is broken and a long chain of polymers is formed. How is this bond broken (where is the energy input) and how is the long chain formed?
(something to do with free bonds?)

During radical polymerization, we use molecules called initiators $I$ for forming free radicals. The molecules of initiators contain  at least one fragile bond that can easily dissociate upon heating, and yield free radicals.
$$I\rightarrow2A^.$$
The as-formed radicals  have enough energy to attack the monomer molecule $M$ and transfer the radical to the monomer molecule:$$A^.+ M\rightarrow  AM^.$$
More precisely, one electron of the double bond in the monomer forms a covalent bond with the free electron of the radical $A^.$. The other electron of the double bond is now free. We have  a radical  on the monomer  $AM^.$.
The radical monomer  attacks another monomer molecule, and the polymerization reaction propagates:
$$  AM^. + M \rightarrow AMM^.  $$
In the above reaction, one electron of the double bond in the monomer molecule forms a covalent bond with the free electron of the radical $AM^.$.The other electron of the double bond is now free $AMM^.  $.
and so on..

Addition polymerisation is usually classed as chain growth. Whilst free radicals are usual and the mechanism is by homolytic fission, cationic and anionic initiation are both valid where propagation is by heterolytic fission. For example, the polymerisation of cyanoacrylate to give superglue is usually initiated by trace water and would generate an anionic chain in the process. The chain would only grow by encountering and reacting with monomer.
Step growth polymerisation e.g. condensation type, have multiple reactive oligomers or chains which can combine without the reaction terminating. This cannot happen in chain growth radical polymerisation as termination would occur and simply wouldn't happen for ionic polymerisation mechanisms.


text:
Why is phenyl weakly activating?
The susceptibility of a benzene ring to electrophilic attack depends on the type and number of groups bonded to the ring.
Activating groups donate/release electrons and increase the electron density in the benzene ring and so render it more vulnerable to electrophilic attack.
Deactivating groups do the opposite, withdrawing electrons and reducing electron density in the ring.
The phenyl group is said to be weakly activating, suggesting that it donates electrons to a benzene ring bonded to it.

One possible explanation I can think of is that phenyl group donates electrons through resonance with the other ring and it therefore increases electron density to a greater extent relative to hydrogen, which is the reference point for activating ability.

"...the phenyl group (having the same structure) should have the exact same electronegativity as the benzene ring itself and hence have no activating or deactivating effect whatsoever." The activating/deactivating effects of substituents are not measured relative to the benzene ring, they are measured relative to hydrogen.

It donates electron density to the ring via the resonance (or mesomeric) effect, which negates the electron withdrawal (through the inductive effect) to some extent.

In the following diagram, which carbocation is more stable, A or B?

The answer is A because the positive charge on the positively charged carbon can be delocalized into the attached benzene ring (I've drawn one of the possible resonance structures to illustrate this).
The same thinking can be applied to your question. Look at resonance structures C and D (we can draw many more) for the attack of an electrophile on a benzene ring.  If there is a second benzene attached and it is attached ortho or para to where the electrophile attacks, then the positive charge can be delocalized into the second benzene ring.
This additional delocalization will stabilize the system. Hence, the second benzene is an ortho-para activating substituent.  However, since this stabilization involves electrons involved in pi bonds, the stabilization (activating effect) will be weaker than when a substituent with a "more readily available" lone pair of electrons (amino, hydroxyl, etc.) is involved.

By your last sentence, do you mean an activating substituent on, say, the para position, would donate electrons into the ring and towards the cation/electrophile? Wouldn't that increase the strength of the stabilising effect?

@HarryHolmes In my last sentence I'm just saying that a lone pair on, say, a methoxy group is more available then the pi electrons in an aromatic ring.  Hence the methoxy group would more strongly activate (and direct) than a phenyl ring.  If the biphenyl has a p-methoxy substituent, and since the phenyl ring and methoxy don't activate the same available positions, the methoxy group will "win out" and control the location of the attacking electrophile.


text:
What role does sulfuric acid play in the titration of oxalic acid and potassium permanganate?
The titration is between oxalic acid and potassium permanganate, with $\ce{H2SO4}$ added to oxalic acid. My question is what role other than a dehydrating agent and maybe catalysing does the $\ce{H2SO4}$ play in this titration.
The question came to me while reading a book (Advanced Chemistry by Philip Matthews - Cambridge university press), I was reading about titrations and this question came to me, but it isn't mentioned in the book.

Write down the balanced equation of the reaction and you'll see right away that you need some extra acid. That's what it is for.

@IvanNeretin Why not HNO3 or HCl?

@ReeshabhRanjan HNO3 itself may oxidize something. HCl itself may get oxidized by permanganate.

how does it act as a dehyradting agent in the titratuion?

The used analytical reaction of purple permanganate to colourless $\ce{Mn^2+}$ occurs under acidic conditions:
$$\ce{MnO4- + 8H+ + 5e- <=> Mn^2+ + 4H2O}$$
Under neutral conditions, permanganate would be reduced to dark brown manganese(IV) oxide:
$$\ce{MnO4- + 4H+ + 3e- <=> MnO2 + 2H2O}$$
Therefore, sulfuric acid is added to make the solution acidic. Actually, excess acid is required since $\ce{H+}$ is consumed during the reaction (8 mol $\ce{H+}$ per 1 mol $\ce{MnO4-}$).
Sulfuric acid is used because it is stable towards oxidation; whereas, for example, hydrochloric acid would be oxidized to chlorine by permanganate.

Thanks to the comment by Ivan Neretin, I figured out that H2SO4 is actually to provide surplus H+ ions, to keep the reaction going
And also it acts as a medium which doesn't interfere with the oxidizing agent to keep the quantitative purpose of the reaction

Sulfuric acid acts as a catalyst in the reaction
$\ce{MnO4- + 8H+ + 5Fe2+ -> 5Fe3+ +  Mn2+ + 4H2O}$
It initiates a 5e transfer in the ionic equation
It used rather than $\ce{HCl}$ because $\ce{HCl}$ is volatile acid which will oxidize to $\ce{Cl}$ in presence of $\ce{KMnO4}$ (strong oxidizing agent). $\ce{H2SO4}$ is stable to oxidization. Also excess $\ce{H+}$ is used because for every 1mole of $\ce{MnO4-}$ , 8 moles of $\ce{H+}$ is consumed.

There are three reasons:

to provide acidic medium
to oxidize $\ce{Mn^{+2}}$ to $\ce{Mn^{+3}}$
to speed up the reaction

H2SO4 increase the acidic content of the solution so as to prevent MnO4 (purple) to reduced to MnO2(dark brow). Hence sulfuric acid is stable in the present of strong oxidising agent. If HCl acid is use it will be oxidised to Cl and make the end point much higher. Furtherore sulfuric acid increase the H+ as it consume.


text:
How to prepare iron(II) carbonate from iron(III) chloride?
Now, this is an awkward reaction first of all. I know how to prepare $\ce{Fe2(CO3)3}$ from $\ce{FeCl3}$:
$$\ce{2FeCl3 + 3Na2CO3 -> 6NaCl + Fe2(CO3)3}$$
But the question is asking for $\ce{FeCO3}$. It seems to be an impossible task but since this was asked in a public examination, does anybody know how to prepare $\ce{FeCO3}$ from $\ce{FeCl3}$ (in one, two or three reactions)?

I can't see how reducing it would be a problem

Also, $\ce{Fe2(CO3)3}$ does not really exist.

@Mithoron could you please explain how I could reduce it?

@IvanNeretin It doesn't really exist? How? Why?

Being a salt of a weak acid and a weak base, it hydrolizes instantly and completely upon mixing. But that shouldn't bother us, for we don't need it anyway. We need $\ce{FeCO3}$; it is more or less OK.

$$\ce{2FeCl3 + 2H2O + SO2 -> 2FeCl2 + H2SO4 + 2HCl}$$
$$\ce{FeCl2 + Na2CO3 -> FeCO3 + 2NaCl}$$

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help]. At the moment this reads more like a comment than an actual answer - could you elaborate a little more. With a bit more rep, you will be able to post comments on any question/answer.

This is the simplest method I can think of:
$\mathrm{2FeCl_3+Fe\rightarrow3FeCl_2}$
$\mathrm{FeCl_2+Na_2CO_3\rightarrow FeCO_3+2NaCl}$
It only uses iron metal and sodium carbonate, which are readily available. Sodium bicarbonate (baking soda) can be used instead of sodium carbonate, but a greater mass would be required. The iron metal should have a high surface area, perhaps in the form of iron powder or steel wool.


text:
Why are the d-orbitals so named?
Why are the $d$-orbitals so named?
The naming of $p$-orbitals is fine.
I can easily name any p orbital according to the axis along which it is oriented.
But what about $d$- orbitals?
Why are there subscripts such as $x^2-y^2$ and $z^2$ ?
Are we supposed to read it like "x squared- y squared"?
Also why does the $d_{z^2}$ orbital has a ring like structure around it?
Is there any order in why the degenerate d orbitals are filled? Like for $p$-orbitals we have an order of filling: $p_x$ then  $p_y$ then  $p_z$.(Not too sure of this.My friend told me about this.)
My friend also told me that for d-orbitals, $d_z$ is conventionally taken as the principal orbital.(Again not too sure what he meant by that).
Although I wouldn't want to get into knowing about f orbitals,but information on that might help someone looking for answers on that.
I wonder why this hasn't been asked on CSE before.
NB: I have visited this link , but couldn't understand much.
Thank You.

That answer should have told you everything you needed to know already, so strictly speaking this is a duplicate. The whole point is, the $p_x$, $p_y$ and $p_z$ orbitals are so named because the corresponding wavefunctions are of a form $xf$, $yf$ and $zf$ respectively where $f$ is a function of the the coordinates $r, \theta, \phi$. The same thing can be said of the $d$ orbitals.

Well there is much more to the question than that.And also, I didn't understand much from that as stated in the question.I really don't have much knowledge on wavefunctions as well. :(

Exactly the same thing can be said of the $d$ orbitals. They are of the form $xyg$, $yzg$, ... where $g$ is also a function of $r, \theta, \phi$. Same goes for the $f$ orbitals. There is no correct order for filling. I could call the x-axis the y-axis, and I could call the y-axis the x-axis. That would make the $p_y$ orbital the $p_x$ orbital and vice versa. But obviously the molecule does not know which axis I have called which. Lastly, I have absolutely no idea what a principal orbital is, and I don't think that such a thing even exists.

The idea about wavefunctions is that each orbital is not a real object, it is a mathematical function obtained by solving the Schrodinger equation. The diagrams that you see are just graphs of these functions, just like how you can draw a graph for $f(x) = x^2$. It just so happens that the graph of $d_{z^2}$ has a ring. Of course before you are introduced to quantum mechanics, the way you view an orbital is like a container, but that is not really the case. Ultimately, there is no good way to explain the reason behind the naming of the orbitals without quantum mechanics.

Well that definitely helped.Thank you.You should go ahead and post this as an answer.

Technically, not even $\ce{p}x$, $\ce{p}y$ and $\ce{p}z$ are accurate names, they should be called $\ce{p}{-1}$, $\ce{p}{+1}$ and $\ce{p}{0}$. The first two are, in fact, complex orbitals (i.e. have a contribution with $i$ in it) and only their linear combinations gives us $\ce{p}_x$ and $\ce{p}_y$.

Of course, and so are the d orbitals, they are $\mathrm{d}{-2}$, $\mathrm{d}{-1}$, $\mathrm{d}{0} (= \mathrm{d}{z^2})$, $\mathrm{d}{+1}$ and $\mathrm{d}{+2}$. On top of that, I went back to check and it turns out that $f = f(r)$ and $g = g(r)$, i.e. they are independent of $\theta$ and $\phi$. And the $\mathrm{d}_{z^2}$ orbital is of the form $(3z^2-r^2)g(r)$, but apparently we just say $z^2$ because it is easier to say.


text:
Inductive effects of halogens
The Wikipedia page on inductive effects states that the order of inductive effects of halogens is as follows:
$$\ce{-F} > \ce{-Cl} > \ce{-Br} > \ce{-I}$$
From this, and from the fact that electronegativity of fluorine is higher than that of chlorine, I guessed that the acidity of p-fluorophenol is more than p-chlorophenol.
But the $\mathrm{p}K_\mathrm{a}$ values are:

p-fluorophenol: 9.8
p-chlorophenol: 9.4

Where did I go wrong? Why does this discrepancy occur? Detailed explanations are very much welcome.

The inductive effect is not very strong when the halide is so far away from the phenol group, so you have to look at the resonance effect. The halides all donate electron density. Which do you think is the best at donating electron density?

@orthocresol So does the +R effect for all halides at the para position become more significant than the -I effect?

Well, it depends on the context. The fact that the halides are deactivating overall should tell you that the -I outweighs the +R in general. My point is that, the decreasing trend in -I is not important in this question because 1) it's too many carbons away, so the trend is very weak; 2) as you said, it obviously leads to the wrong prediction regarding the $\text{p}K_\text{a}$'s. And you're not answering my question.

The less electronegative halides would more easily donate their lone pairs.

Not just that, but also the overlap with the benzene ring. Think about the Lewis acidity of the boron trihalides. If you're still confused, here's the answer: http://chemistry.stackexchange.com/questions/10288/ Replace "the central boron atom" with "the benzene ring" and there's the answer to your question.

This is due to the fact that we also consider +M effect . In fluorine it’s 2p orbital that participated in overlapping but in chlorine it’s 3 p orbital. 3p of chlorine does not overlap efficiently with 2p of benzene carbon so +m decreases in chlorine thus acidic nature increases

All your observations are correct.But one thing you seem to be missing is the presence of vacant d orbitals in Chlorine.I am sorry but I couldn't find pictures as such but I am sure you will understand after drawing out the resonance structures.
So when the lone pair on oxygen is delocalized and is at para position,then the d orbital of chlorine can "take up" and stabilize the compund.Flourine can't do the same.

This is the picture you want. http://i.imgur.com/p3CWF47.jpg But d orbitals aren't a good explanation.

@Karan Singh If it was at the ortho position then I would have agreed but at such a long distance it doesn't seem very accurate.

@orthocresol I am a student myself and I recently learned this.But can you tell why d orbitals aren't a good explanation?

http://chemistry.stackexchange.com/a/5242/16683

Order of -I in Halogens is −F > −Cl > −Br > −I
Order of +M in Halogens is −F > −Cl > −Br > −I
So whenever halogens are in resonance (as in p-fluorophenol and p-chlorophenol), we take combined effect as:
−F < −Cl < −Br < −I
and consider halogens as an overall EWG.
Hence, p-chlorophenol is more acidic than p-fluorophenol.

This is a wrong answer, your second line contradicts itself. Halogens are considered overall as electron-withdrawing groups. Right. Then the withdrawing effect is inductive effect, so shouldn't the order of -I in halogens be considered, giving F > Cl? Think it over. Also, when you answer, including a source, from a good textbook or online resource, is a great idea.


text:
Why laboratory glassware shatter in big pieces?
What are laboratory glassware made up of? This question came up in mind because once I accidentally dropped a test tube and it shattered into big pieces. I observed that it didn't broke into pieces on first impact to the floor rather it bounced and broke into pieces on second impact. Any ordinary glass would shatter into small pieces On the very first impact to the floor. Does this type of glass possess elasticity?

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

Gee, a Google search doesn't hurt.

Trust me that you can shatter lab glassware into very small pieces …

This question would be more suited for physics.stackexchange.com

Kitchen glassware is made to contain large internal stresses, so it shatters into small, non-dangerous pieces. That is hard to achieve with Duran etc., because that stuff has a small thermal expansion coefficient, just so that it does not break under large temperature changes.
I expect that the kind of kitchen glassware that is supposed to withstand large temperature changes also breaks into rather large pieces.
And yes, glass is highly elastic. Elasticity is effectively the only mechanical property it does possess, before breaking. ;-)


text:
What is the correct way to draw the end groups of polylactic acid?
I came across two different ways to draw the end groups of polylactic acid:
 (source)
AND:
 (source)
Which one is the correct way?

Why do you think it would make a real difference in polylactic acid?

@Mithoron I guess it wouldn't; but then why do people bother drawing redundant information?

Well, maybe simpler form would be confusing for newbies...

As @Mithoron said, both are correct but the first is more compact.

The graphical representation standards given in the IUPAC Compendium of Polymer Terminology and Nomenclature (Purple Book) are not really clear about end-groups. However, note that the subscript n should be printed in italic type.

@bon this reduces the degree of polymerization by 2...


text:
Suitable solvents for column chromatography
Why are 100% hexane or 100% ethyl acetate never used as eluents in column chromatography?

You can. Why do you assume that you can't?

I’ve done it. Many times.

Why, of course you can, but most of the time one of them would be "too strong" and the other "too weak", so neither would work well for you. That's why you have to mix them to the right proportion (which depends on your system, and is found by trial and error).

Not "trial and error", please. Organic chemistry is still science!

@Karl Trial and error is a scientific method. It’s how we arrived at today’s chemistry. There would be no Wittig reaction (just to name one) without trial and error.

But you can do a lot better that trial and error for a column chromatography. Do TLC, check polarity of expected products.

@Karl That is exactly what I understood by trial and error, and precisely what I assume Ivan meant by trial and error.

@Jan Well, I think you're misusing the expression. You do one educated guess for a TLC solvent mixture, measure the Rf value, and adjust the mixture accordingly. Probably repeat the last two steps, then run your column. I see no possible "error" here.

@Karl Of course, that depends on your definition of error. I consider an incorrect solvent mixture (i.e. one that does not give an $R_\mathrm{f}$ of $0.3$) to be an error and the next step to be a readjusted trial. So repeat trialling solvent mixtures until no error.

"Error" means you're doing something stupid. "Stupid" is an experiment that gives you one answer, "negative", but no clue as how to  proceed further. ymmv

Your comment is not entirely true. The solvent system used depends upon the behaviour of your (crude) product on silica gel.
Neat hexane (or a substitute such as petroleum ether or cyclohexane) is often used to wash 'grease' (non polar compounds) off the column, whilst neat ethyl acetate (or ether) is often used to elute highly polar compounds.
Ideally, you choose a solvent system in which the compound you want to isolate has an Rf of around 0.3 (please read the original flash column chromatography paper for more details), however, in practice, many chemists run 'gradient' columns.  This entails starting with neat hexanes, and then slowly increasing the polarity of the solvent until the compound of interest elutes. This is also the way HPLC and automated flash (eg. CombiFlash/Flashmaster) works, using gradient elution.

‘many chemists run gradient columns’? I haven’t seen too many ‘many chemists’ then …

Anyone involved in serious organic synthesis intuitively runs some form of gradient column if they're doing anything more complicated than a simple reduction or protecting group manipulation. Starting with a low polarity solvent to flush off any grease then slowly working up to something strong enough to elute the product. In total synthesis, this is very commonplace, though potentially not so much if you were doing methodology.

I can’t agree. I go looking for the solvent mixture that gives my product an $R_\mathrm{f}$ value of $0.3$ (approx.) and just go with that. Then I find pure product at around the seventh-ish fraction. And yes, I do natural products chemistry and yes, most of my colleagues here do the same thing. The only exception is if I really don’t know what happened and I get both unpolar and polar components.

Running a gradient column provides additional understanding.  To only focus on the compound of interest is to miss a trick. This would not be normal practice in an industrial setting whether the chromatography is normal phase, as in flash or biotage , or reverse phase preparative or analytical. Collecting and identifying the other components helps build up a mass balance or mechanistic understanding as required.

Simply speaking you would often only use pure hexane (or better cyclohexane, as the former is somewhat toxic, or (cheaper) isohexane) to wash something completely unpolar off the column before further elution steps.
Pure ee you would use to just wash out everything that is still on the column after your chromatography.
Of course there can be the (i think rather rare) case where the polarity of either one is just what you want.


text:
Shock induced illuminated material
Any suggestions for a shock induced material that would illuminate and be capable of being visible for a minimum of three seconds? Non-toxic and minimal heat production would be optimal.

You're asking about mechanoluminescent material?

And what kind of money would you expect to earn from this stuff? ;-)


text:
Trigonal bipyramidal arrangement with 2 lone pairs
So basically according to university textbooks, it is said that the second lone pair in a trigonal bypyramidal arrangement would also reside on the trigonal plane, causing the two lone pairs to be beside each other and compressing the other 3 bonded pairs (resulting in T-shape).
However, my question is: Why doesn't the two lone pairs move to the two axial groups so that the the three bonded pairs reside on the trigonal plane? Surely this would result in the less repulsion within the arrangement? Wouldn't the repulsion between a lone pair and 3 bonded pairs at 90 degrees be less than the repulsion between a lone pair and another lone pair & 3 bonded pairs?

http://chemistry.stackexchange.com/questions/9623/lone-pair-more-repulsion-vspr-theory

The 90° interactions are the most significant (highest energy).  How many lone pair - lone pair, lone pair - bonding pair and bonding pair - bonding pair 90° interactions does each of the two possible structures have?

For the one accepted in the text book there is one lone pair - lone pair and 3 lone pair - bonded pair. And for the I propose there are just 3 lone pair - bonded pair. However, the difference is that mine is three 90 degree interactions whereas the ones in textbook are unstated

If both lone pairs are in the equatorial plane, then there are no 90° lone pair - lone pair 90° interactions; your count of the 90° lone pair - bonded pair interactions for this structure is also incorrect (should be 4).

Yes, the 90° interactions are the most important. It's also worth pointing out that the differences in energy are small, and a Berry pseudorotation interconverts axial and equatorial in these species.

Which species are you talking about? I know of no main-group compound with an electron dectet which you seem to assume (5 electron pairs). And transition metals usually go for 18, not 10 either.


text:
Can you extract specific compounds from infused oils?
I heard a discussion when sitting on the train this morning by a couple of cannabis folk on the way into Denver. This is in regards to coconut oil (for baking).
For example: if someone extracted THC into coconut oil, is there anyway to then separate the THC back out of the coconut oil so you end up having coconut oil in one cup, and THC in the other without affecting degradation? I'm not a smoker by any means, but I thought it was a rather interesting conversation. Feel free to be as technical as possible, as I would like to test this theory among many chemical compounds known in homeopathic medicine.

Can you elaborate on what you mean by "chemical compounds known in homeopathic medicine"?

THC is often extracted from cannabis by mashing up the plant with a solvent such as hexane. The slurry obtained may then be filtered off, and the solvent evaporate to give a crude extract from the cannabis which will contain the THC amongst other organic molecules.  Depending on the extraction solvent, the components of the cannabis plant are extracted in different concentrations, depending upon their relevant affinity for dissolving in the solvent.
Coconut and olive oil have both been used in extractions, as they are safe for human consumption and so can be used directly once isolate without the need to remove potentially toxic (hexane) or flammable (hexane) solvent.
In general, the solvents used are inert, that is, they do not react to degrade the extraction products as you mentioned.
With regard to homeopathy, this too involves the extraction of active components from plant/biological matter. However, in homeopathy, it is generally less of an exacting science (for instance specifically getting THC out of cannabis) and more a case of extracting ALL of the components of a plant, for instance by making a tea. If you are interested in this, you might like to look at the recent Nobel prize winner Youyou Tu (2015, Physiology or Medicine). She realised that boiling the plant in water to create a tea was destroying the functionality of the molecule that chinese medicine was trying to extract, limiting its potency. By exploring alternative extraction methods she was able to get the biologically active compound without degradation leading to a useful drug.

This is not what I meant. I meant to ask if it is possible to extract the THC OUT of the coconut oil. For instance, could you procure someones infused coconut oil and extract the THC back out of it?

If you had THC in coconut oil, and then you added a solvent which was immiscible with coconut oil, and in which the THC had a higher affinity than for coconut oil, then yes. This is similar to common aqueous-organic workups after chemical reactions.

What solvent would you might recommend then? It must leave nothing in the end product, thus that it is safe for human consumption.


text:
Using Cv For Non Ideal Gasses
So we are doing calculations with the first law of thermodynamics:
$$U = q + w$$
And we have been studying $\mathrm dU(T,V)$.
So it is true that for some equations of state other than the ideal gas equation, $\mathrm dU(T)$ only, so therefore it follows that $U= C_V(T_2-T_1)$. So I am wondering is it even possible that $C_V$ is used for non-ideal gasses? And would it also be equal to $\frac 32R$?
And if $C_V$ can be used, why?
(Also is this true for $C_p$?)

You can still use the heat capacities as they are defined by the equations: $C_V = (\partial U / \partial T)_V$ and $C_p = (\partial H / \partial T)_p$. The fact that $U = U(T)$ for an ideal gas is just a special case, so in general, the heat capacities of a real gas will not be the same as for the ideal gas.

By the way: it's $\Delta U = q + w$ with the triangle.

We usually don't use the real gas $C_V(T,p)$ directly in calculating the change in U in going from thermodynamics equilibrium state A to thermodynamic equilibrium state B, if both these states are beyond the ideal gas region.  Starting from state A, what we do first is use the real-gas p-v-T equation to determine the change in U in going from the initial pressure $p_A$ to a low pressure (or large specific volume) in the ideal gas region $p^*$ at constant temperature $T_A$.  Then, our second step is to hold the pressure constant at pressure $p^*$ and raise the temperature from $T_A$ to $T_B$, using the ideal gas heat capacity to determine the change in internal energy for this step (since we are now in the ideal gas region).  Finally, we hold the temperature constant at $T_B$ and raise the pressure from $p^*$ to $p_B$ using the real-gas p-v-T equation again to determine the change in U.  The sum of the changes in U for these three steps adds up to the change in U in going from $(T_A,p_A)$ to $(T_B,p_B)$.


text:
How to determine and derive the protonation state of a functional group based on its pKa?
In our biochemistry class we were introduced to a rule of thumb, which goes something like this:
"If the pH of a solution is one or more units below the $\mathrm{p}K_\mathrm{a}$ of a group, then said group is protonated and if it is one or more unit above, then the group is de-protonated"
However, we weren't provided with a mathematical derivation of this fact.
I attempted one by myself using:
$$\ce{HA + H2O <=> A- + H3O+}$$
thus, $$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{\ce{A-}}{\ce{HA}}$$
or, $$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{1-\alpha}{\alpha}$$
and then define $\mathrm{p}K_\mathrm{a}$ as the point where ${\alpha=0.5}$ i.e proton occupancy is 50%
$$\alpha(\ce{pH}) = \frac{1}{1+\exp\left\{-\ln10(\mathrm{p}K_\mathrm{a}-\ce{pH})\right\}}$$
where $\alpha$:= degree of proton occupancy.
By looking at this function, I can sort of get that if pH goes up, $\alpha$ goes down i.e case of protonation, and if pH goes down  $\alpha$ goes up i.e case of deprotonation.
This seems a little bit "back of the envelope" to me. Is this sufficient, or am I missing something.

You're on the right track.  The Henderson-Hasselbalch equation is where this rule of thumb comes from; the key thing to remember is that, in biochemistry, your $\mathrm{pH}$ is fixed by the environment (there's always something acting as a buffer).  This is in contrast with typical H.H. problems in general chemistry where the amount of molecule and its $K_\mathrm{a}$ is changing the $\mathrm{pH}$ of the solution.
Anyhow, given that $\mathrm{pH}$ and $\mathrm{p}K_\mathrm{a}$ are fixed values, we're free to simply compare the ratio of $[\ce{A-}]$ and $[\ce{HA}]$:
\begin{align}
\mathrm{pH} &= \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{A-}]}{[\ce{HA}]}\\
\log\frac{[\ce{A-}]}{[\ce{HA}]} &= \mathrm{pH} - \mathrm{p}K_\mathrm{a}\\
\frac{[\ce{A-}]}{[\ce{HA}]} &= 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}}\\
\end{align}
If $\mathrm{p}K_\mathrm{a} = \mathrm{pH} + 1$ then we have $\frac{[\ce{A-}]}{[\ce{HA}]} = 10^{-1}$.  In other words, we have 10 times as much $\ce{HA}$ as we have $\ce{A-}$ (i.e., we have $\approx91\%\ \ce{HA}$) as we should expect for something that is mildly basic.
If $\mathrm{p}K_\mathrm{a} = \mathrm{pH} - 1$ then we have $\frac{[\ce{A-}]}{[\ce{HA}]} = 10^{1}$.  In other words, we have 10 times as much $\ce{A-}$ as we have $\ce{HA}$ (i.e., we have $\approx91\%\ \ce{A-}$) as we should expect for something that is mildly acidic.
Clearly, if the $\mathrm{pH}$ differs from the $\mathrm{p}K_\mathrm{a}$ by only 1 $\mathrm{pH}$ unit, we don't really have a completely protonated or deprotonated molecules, but it's certainly mostly protonated or deprotonated.  As the $\mathrm{pH}$ and $\mathrm{p}K_\mathrm{a}$ get further apart, then we continue to change this ratio by orders of magnitude.
Let's look at a concrete example.  Physiological $\mathrm{pH}$ is 7.4 and the side chain of aspartic acid has a carboxylic acid group with a $\mathrm{p}K_\mathrm{a} = 3.9$, so we predict
$$\frac{[\ce{RCOO^-}]}{[\ce{RCOOH}]} = 10^{7.4 - 3.9} = 10^{3.5} = 3200$$
In other words, we still have some protonated sidechain, but we have >3000 times more deprotonated sidechain than protonated sidechain (did you notice that? the difference was >3 so we have >1000 difference in concentration ratio).  For all practical purposes, the aspartic acid side chain is "completely deprotonated" at physiological $\mathrm{pH}$.

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Is there an agent that changes colour when exposed to air?
And if so, does it come in a liquid form?
Thank you for your time

Need it be reversible?

It's too broad, there are lots of things that react with oxygen, changing color.

Many compounds change colour when exposed to 'air', however, this is usually due to the presence of either water or oxygen, rather than 'air'.
To give a few examples:
*when distilling a solvent such as THF over sodium/potassium, benzophenone is used as an indicator. In the absence of air, the indicator leaves the solution a deep purple colour, however if an excess of air is present, the indicator turns a clear/brown colour.
*drierite is an indicator often used in inert gas drying tubes. In its dry form, it is either deep blue or orange, however when wet it turns white/pale orange.
Several other indicators exist, however its worth noting that air is merely a mixture of multiple things, and as such any indicator will indicate for a specific component of air rather than air as a whole- with water and oxygen being the most common things to indicate for.


text:
Does the electrolysis of water produce chlorine gas?
http://orbitingfrog.com/2014/11/02/electrolysis-of-water-with-pencils-and-a-9v-battery/
This article seems to indicate that "a small amount of Chlorine may be produced as you [perform electrolysis on water]". Is this true? I thought it was just H2 and O2 produced.

Water chloration is used to kill some bacteria and microbes in tap water. More info https://en.wikipedia.org/wiki/Water_chlorination
that menas, that chlorine can be produced.

If you use tap or lake water, it will contain a noteable amout of $\ce{NaCl}$ whose chloride ion will be electrolysed to $\ce{Cl2}$. This has nothing to do with water chlorination. (Actually the article says to ‘use a spoonful of salt’, so you’re actively adding more chloride.)

If you use salt for the electrolyte, you won't get much oxygen... mostly chlorine (and hydrogen at the negative electrode, of course).

It's the tablespoon of salt he added, giving the chlorine when electrolysis is done. I have done electrolysis with salt-free water (though a bit acidic to get the water to electrolyze faster) and chlorine was nowhere to be found.
Electrolysis of saltwater produces hydrogen and chlorine; and leaves sodium hydroxide behind.

You get chlorine gas as a product only when NaCl is added, as mentioned by @Nihilist_Frost. Even on addition of NaCl, oxygen gas must be the product(expected) because it has slightly lesser reduction potential than chlorine. But this is not the case because of over-potential of oxygen, which makes this half-reaction unfavorable. This is specifically termed as Bubble Overpotential.
P.S.: I tried to find a link for bubble over-potential, but I am afraid it is not available on the internet.
P.S.S.: Electrolysis of pure water produces only H2 and O2, and that too of very high purity (in the range of 99.5%)

If you don't want to have chlorine gas as byproduct in your electrolysis add sodium or potassium hydroxide - you would get O2 and H2.
Water electrolysis is even possible with distilled or deionized water, because due to the CO2 in the air, there are always some HCO3 ions in solution that will aid electrolysis (given the voltage is high enough).

If you pass current through salt water (sodium chloride solution), you will produce chlorine preferentially to oxygen.
In practice, this chlorine will attack your anode if it's made of copper, or steel, or most any other metal. Since the linked demonstration uses pencils (graphite), that doesn't happen. Instead, the anode produces free chlorine, which dissolves in the water. The cathode produces hydrogen, leaving hydroxide ions.
So, now you have an aqueous solution with sodium ions, chloride ions, hydroxide ions, and free chlorine. The dissolved chlorine reacts with the hydroxide ions, disproportionating into chloride and hypochlorite. (If the solution is hot enough, it can go all the way to chlorate.) This is how bleach (sodium hypochlorite) is manufactured industrially.
In other words, this process won't produce bubbles of oxygen. It will produce chlorine, but probably not as bubbles of gas. If you want to produce oxygen, use sodium or potassium hydroxide solution, not salt water.

I've read the question and read the replies. Firstly the question is misleading as water does not conduct electricity, it is only the impurities in the water that do which is why an electrolyte is always added to water during the process of electrolysis. Secondly when an electrolyte is added to water the water becomes a solution of the electrolyte, for example, add sodium chloride to water it will become a sodium chloride (salt) solution. So when we look at the products of the electrolysis of a salt solution we can see that we are splitting the sodium from the chloride producing chlorine at the anode and a sodium based gas at the cathode (which is palmed off as hydrogen). This is characteristic of any electrolysis process involving a solution. So to answer your question, one should always produce chlorine by electrolysis when there is a chloride present in the solution.