text:
Chromic acid oxidation of aldehydes and primary alcohols

The answer has to be both A and D. Chromic acid is a strong oxidizer and will take both aldehydes and primary alcohols to carboxylic acids.
However the computer thinks otherwise. Am I just missing something really simple here?

There are many $\ce{CrO3}$ oxidation reagents (Jones, Sarett, Collins, PCC, etc.), however, most are selective and stop the oxidation of a primary alcohol at the aldehyde stage.  On the other hand, the Jones reagent is relatively unselective and will oxidize a primary alcohol to the corresponding carboxylic acid - if enough reagent is present. It takes two-thirds of an equivalent of Jones reagent to oxidize one equivalent of the alcohol to the aldehyde and an additional two-thirds equivalent to oxidize the aldehyde to the acid.  Also the Jones oxidation is run under acidic conditions (typically dilute sulfuric acid).
I would have though that A, B (the gem-diol will equilibrate with the aldehyde under acidic conditions) and D were correct answers.  It seems like the computer answer is either in error or it is a "trick" question (seems unlikely for computer-based learning).  The "trick" would be that since an excess of reagent was not specified only 1 equivalent is assumed.  Therefore there is not enough reagent to carry out two sequential oxidations; as a consequence only the aldehyde or gem-diol could be completely oxidized with one equivalent of reagent. But then A or B is not an acceptable answer either.


text:
How is tea affecting water's color?
Black tea is an amazing product of the green tea leaves. A recent study has proven that;

The anion profile for the tea solution indicated that oxalate and citrate were the major anions detected.

This resulted in the fact that a $\pu{1\!\% w\!/\!v}$ solution of normal (World Blend – Tetley USA Inc; you decide it for yourself!) tea has a pH of $\rm 4.9$. Non of these pieces of detailed info were an answer to my question.
My question is, definitely somehow oxalate and citrate managed to dissolve into water as we put tea leaves in it;

Where are they stored? (How does tea plant make them? I'm looking for the reactions that occur in the leaf)
How and why do they get released from tea leaf? Since you don't usually see any "big" differences between tea leaf and a soaked tea leaf? (I'm not looking for the how of solution)
What makes the color of water change? Why does the color vary with pH alteration? (There must be a reaction, what is it?)
Pt: These questions are additional questions to the title.

The colour varies with pH because a major contribution to the colour profile are polyphenols known as tannins, which will have -OH groups which are not that hard to protonate/deprotonate. As the phenolic rings are the main chromophores the resulting changes in the electronic levels will slightly change the wavelength at which light is absorbed.

Where are they stored? (How does tea plant make them? I'm looking for the reactions that occur in the leaf)

Both citrate and oxalate are associated with ATP production via the citric acid cycle. As is obvious from the name, citrate is a critical component of the process. Oxalate is a hydrolysis product of oxaloacetic acid, which is also part of the cycle. Both of these ions are present within the plant cells.

How and why do they get released from tea leaf? Since you don't usually see any "big" differences between tea leaf and a soaked tea leaf? (I'm not looking for the how of solution)

If your tea leaves are crushed or broken in any way, this really is just simple dissolution. If you have whole tea leaves, the oxalate and citrate are migrating into and out of the plant cells. The cell wall is permeable, so the ions (or corresponding acids) can simply flow through it. The cell membrane is semipermeable, and the diffusion of the ions across it is a biologically controlled process that I honestly don't have the background to explain.

What makes the color of water change? Why does the color vary with pH alteration?

The pH and color are not related properties. The pH is affected by the citric acid and oxalic acid associated with the anions previously discussed, as well as other minor components that are likely present. The color comes from some of the many other compounds that are extracted into the water, which number in the hundreds, if not thousands. Some of these compounds are inherently colored - no reaction is required to produce color in the solution.

To expand: the colored compounds in tea and many other plants often change color depending on pH. This is more or less typical example of pH indicator. http://en.wikipedia.org/wiki/PH_indicator.

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text:
Draw a simplified MO diagram for the pi system of Methyl vinyl ether

My attempt at the answers:
i) 2 bond pairs and 2 lone pairs on oxygen atom therefore bond angle of slightly less than 109.5 degrees, around 104 degrees.
ii) 120 degrees bond angle therefore $\mathrm{sp^2}$ hybridization.
iii) From what I understand, the sigma-pi separation means that you can neglect interactions between the $\mathrm{p_z}$ and $\mathrm{p_x,p_y}$ orbitals?
The $\ce{CH2}$ carbon would form $\mathrm{sp^2}$ hybrid orbitals.

The $\ce{CH3}$ carbon would form $\mathrm{sp^3}$ hybrid orbitals and would then form 3 sigma bonds with the hydrogen leaving only one orbital to interact with the oxygen. I assume that the oxygen $\mathrm{s}$ orbitals do not match the energy level of the carbon and will form non-bonding orbitals. Therefore the $\ce{C-O}$ sigma bond would be $\mathrm{sp^3-p}$.

I am unsure how to use this knowledge to actually draw the MO diagram? I have tried an attempt below. I can draw MO diagrams for simple diatomic molecules. And how would the MO diagram explain the observed 120 degree bond angle?

iv) I haven't used the MO diagram to draw these resonance structures but I have come up with two based on the octet rule:

Drawing MO diagrams is strictly only possible for two entities, hence the term is rather unfortunate in the question. I guess what was really meant is, that you should draw the MOs and order them according to their energy.

I am unsure how to use this knowledge to actually draw the MO diagram?

First, chemists used Lewis structures to explain molecular properties, but eventually this method was replaced with resonance structures (like you've drawn in part iv above).  Resonance structures focus primarily upon p orbitals in a molecule and usually do not consider the sigma system.  Molecular Orbital (MO) theory was a further improvement on resonance theory.  MO theory still focuses on p orbital connectivity and typically disregards the sigma system, but it lets you view the actual patterns that the molecular orbitals generate.  It is very similar to the orbital diagram you included above, but we leave out the sigma system and focus on the p orbitals and the electrons they contain.
In part ii) above you conclude that the ether oxygen is $\ce{sp2}$ hybridized.  This means that there is one p orbital on oxygen that contains a lone pair of electrons (there are also 3 $\ce{sp2}$ orbitals on the oxygen, 2 are used to form sigma bonds to carbon and the third holds the second lone pair of electrons on the oxygen).  This p orbital can interact with the 2 p orbitals involved in the carbon-carbon double bond.  These 3 p orbitals can interact to produce 3 molecular orbitals (see pictures below).  These 3 molecular orbitals are produced by combining (adding and subtracting) a p orbital and the 2 ethylenic molecular orbitals (the bonding and antibonding pi orbitals for the double bond).  All of the orbitals shown in the MO diagram below are p orbitals.  We will fill the MOs with 4 electrons, 2 from the pi double bond and 2 from the oxygen p orbital that contains one of oxygen's lone pair of electrons.
So, we have determined that we have 3 p orbitals (one on oxygen, 2 on the olefinic carbons) that can overlap and form 3 molecular orbitals.  The system we have created with 3 p orbitals from methyl vinyl ether is similar to the all carbon allyl analogue with 3 p orbitals, one p orbital on each of the 3 carbons.  Specifically your system is analogous to the ally anion with 4 pi electrons (2 from your double bond and 2 from the  oxygen lone pair).

Image Source
For more information on how to construct molecular orbitals in organic molecules see here and here.

Why is it 3 p-orbitals that overlap @ron, I thought oxygen has 2 p-orbitals as it have 4 p-electrons? Oxygen has two lone pairs and the C=C carbons have one p-orbital each that overlap to form the pi bond so why does the system have 4 pi electrons? I don't understand the diagram, are they all pi-orbitals? If so then how did you derive them?

My attempt of a MO diagram of methyl vinyl ether looks nothing like yours @ron http://i.imgur.com/Lo045gY.png?1

@StackExchangeLover  I've revised my answer, let me know if you have other questions.

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text:
Using MO theory, give an explanation for the C-C bond length in cyanogen
1. The problem statement, all variables and given/known data

2. The attempt at a solution
I know that the MO diagram for CN- is this:

I am unsure how to draw the MO diagram for cyanogen. I know that the higher the bond order, the shorter the bond length. I have tried to draw a MO diagram for cyanogen by combining two of the MO diagram for CN:

Not sure if this is necessary but I have drawn the following resonance structures:

I think that structure (I) contributes the most as structures (II) and (III) have a positive charge on the nitrogen.
I think the C-C bond in cyanogen is shorter than the "usual" C-C bond lengths due to delocalisation across it but I don't know how to explain this using Molecular Orbital theory.

is the C-C bond length in cyanogen longer or shorter than "usual" C-C bond lengths? 2) Start by listing all the possible molecular orbitals and populate them from the lowest energy up. I.e. you know there will be two C-N sigma orbitals and two C-N sigma stars; a C-C sigma and a C-C sigma star, etc.

@Dissenter The C-C bond length in cyanogen is shorter than usual C-C lengths
6 electrons are in sigma bonds, the 2 C-N sigma bonds (so two C-N sigma stars) and the one C-C sigma bond (so one C-C sigma star). Each nitrogen has a lone pair of electrons, this accounts for (2x2) 4 more electrons, so these will be non-bonding orbitals in the MO diagram? That leaves 8 electrons to fill the 4 C-N pi bonds (so 4 C-N pi stars?).
How do I know the order of the molecular orbitals and which orbitals form these molecular orbitals e.g. do two sigmas bond together from CN to form a new sigma in NCCN?

I think the answer is hidden in this question: http://chemistry.stackexchange.com/q/21973/4945 At least an overview of all valence orbitals is given there.

See my answer to your question on methyl vinyl ether for some background on molecular orbital (MO) theory.
In this case, your molecule is analogous to the all-carbon system, butadiene.  In cyanogen you have 4 p orbitals that can align, overlap and mix together to generate 4 molecular orbitals and you have 4 electrons (1 from each p orbital involved in a pi bond) to place in your molecular orbitals.  Since the carbons in cyanogen are $\ce{sp}$ hybridized you actually have a second set of molecular orbitals that are identical to the first in all respects except that the 4 p orbitals in the second set are rotated 90 degrees with respect to the other set of p orbitals.
Let's count electrons. We have 4 electrons in each of these orthogonal molecular orbitals that each contains 4 p orbitals, so we have 8 electrons in our two orthogonal sets of molecular orbitals.  We also have 2 lone pairs remaining on each nitrogen for a total of 4 electrons.  Finally we have the sigma system; we have 2 C-N sigma bonds and 1 C-C sigma bond each containing 2 electrons for a total of 6 sigma electrons.  For the whole molecule we have 8+4+6=18 electrons as expected.
As I mentioned in my other answer, we typically focus on the delocalized molecular orbitals formed by combining the p orbitals and disregard the sigma system (usually) when we create our MOs. Disregarding the sigma system usually simplifies matters.
There are a number of ways to create the MO diagram for butadiene or cyanogen, but all of them will have the common theme of mixing 4 p atomic orbitals and generating 4 molecular orbitals.  Here is one way to do it, let's start by mixing 2 p orbital to create the MO's for ethylene (or nitrile).

image source
Now we can mix 2 ethylenes (or nitriles) to generate butadiene (or cyanogen)

image source
We label molecular orbitals $\ce{\Psi_1 ~- ~\Psi_4}$ in order of increasing energy (so $\ce{\Psi_1}$ is the molecular orbital at the bottom of the figure and $\ce{\Psi_4}$ is at the top).  Don't worry about the "S"'s and "A"'s in the diagram, they are just telling us about the symmetry properties of the molecular orbitals which we don't need to focus on right now.
Notice how $\ce{\Psi_1}$ has overlap between the orbitals on the two middle carbons.  This indicates overlap and a higher bond order between these 2 carbons, just as your resonance structures II and II indicate.  Just like in butadiene, this carbon-carbon bond has some double bond character and, therefore there is a higher barrier to rotation about it then you might otherwise expect.

Sorry @ron, I don't understand your diagram for cyanogen. What does the S and A stand for? Cyanogen has 18 electrons but your MO diagram only has 4 electrons? Where is \ce{\Psi_1} in your diagram? How did you go from the MO diagram for CN to this?×Comments may only be edited for 5 minutes×Comments may only be edited for 5 minutes×Comments may only be edited for 5 minutes

@StackExchangeLover  I've revised my answer to address the questions in your comment.

This feature of cyanogen was already mentioned in the comment by Geoff to the Question "How many electrons are there in the pi-system of cyanogen?". (And I was very tempted to vote to close as a duplicate.)
Your resonance structures are correct and it is noteworthy, while I might be the predominant one, the other two are still considerably influencing the electronic structure. The problem with this is, that you cannot sufficiently express delocalisation in Lewis terms. A molecular orbitals approach is necessary to understand the bonding situations. Below you can find the complete valence MO scheme of cyanogen, which I repost updated from the linked question.

You can clearly see, that the eight $\pi$ orbitals are delocalised over the complete molecule. Half of them include the bond between the central carbon atoms, i.e. the degenerate $\pi_\mathrm{u}$ orbitals.
For a carbon carbon single bond you would expect, a bond length of $\mathbf{d}(\ce{C-C})=150~\mathrm{pm}$ and for a double bond $\mathbf{d}(\ce{C=C})=134~\mathrm{pm}$. The calculated bond length is about $\mathbf{d}(\ce{C\bond{~-}C})=138~\mathrm{pm}$ (DF-BP86/def2-SVP) and clearly between a single and a double bond. This has of course an effect on the neighbouring carbon nitrogen bonds. You would expect the triple bond to be $\mathbf{d}(\ce{C#N})=108~\mathrm{pm}$ and the double bond $\mathbf{d}(\ce{C=N})=120~\mathrm{pm}$. The calculated bond length is $\mathbf{d}(\ce{C\bond{~=}N})=118~\mathrm{pm}$ and only little shorter than a double bond.
Covalent radii are taken from Pekka Pyykkö and Michiko Atsumi, Chem. Eur. J., 2009, 15 (46), 12770-1779.

Thank you for not closing my question @Martin. Sorry I don't understand your diagram. How does the diagram tell me that the pi orbitals are delocalised over the complete molecule? You need 4 atomic orbitals to form one molecular orbital? What is the significance of the black and non-black parts? What do the subscript u and g mean?

@StackExchangeLover I am currently on my tablet and cannot answer properly. What do you know about symmetry and point groups? The circles on the bottom represent s orbitals. Black and white represent the different phases of the orbitals. U is fot ungerade and g is for gerade. Best you read a little more about how orbitals combine. Since i am out of the office, i will not be able to update this post for at least a week, sorry.

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text:
How to find the volume of oxygen release by the decomposition if potassium chlorate?
In the reaction:
     $$\ce {2KClO3 -> 2KCl + 3O2}$$
What is the volume of oxygen released under NTP conditions when $36.75~\mathrm{g}$ of $\ce {KClO3}$ is heated?
I tried to find the answer this way:

$2~\mathrm{mol}$ $\ce{KClO3}$ form $3~\mathrm{mol}$ $\ce{O2}$
$0.3~\mathrm{mol}$ $\ce{KClO3}$ form $0.45~\mathrm{mol}$ $\ce {O2}$, i.e. $36.75~\mathrm{g}~\ce{KClO3} = 0.3~\mathrm{mol}~\ce{KClO3}$
I don't know how to convert moles into litre. Can we convert mass directly to volume in $\mathrm{L}$. I read somewhere that $1~\mathrm{L} = 1~\mathrm{kg}$.

In fact, under NTP conditions, each mole of an ideal gas occupies the volume $22.4\ce{L }$.
In your case, the volume of  $0.45\ce{mol }$ of  $\ce{O_2 }$ is  $22.4\times 0.45=10.08\ce{L }$.

You should guide the student to the answer, not provide it on a silver platter. Otherwise, the student doesn't learn.

I totally agree with you. But here the student must know this fundamental property of an ideal gas under NTP conditions. Apparently, he doesn't know it and it is frequently used in solving chemistry problems. So, I allowed for myself to solve the problem this time, as a role model problem.

1 kg for 1 liter is, in fact, liquid water at room temperature.
You need to use the ideal gas law $\ce {PV=nRT}$.

@Martin You know, you could have added that yourself.

That's right, thought a comment would be appropriate to do so.

Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules", ie at constant temperature and pressure, 1 mole of any gas will occupy the same volume. At NTP, this volume is ~$24dm^3$. Therefore the volume of $0.45mol$ of $O_2$ will be $0.45 × 24 = 10.8dm^3$.  $1dm^3 = 1L$ so your answer would be $10.8L$


text:
Dimers and polymers of electron deficient compounds
Why does $\ce{BeH2}$ polymerize into $\ce{(BeH2)_{n}}$ in solid state whereas $\ce{AlCl3}$ dimerises into $\ce{Al2Cl6}$ in the vapour state ?

First of all, you have it all backwards.
You sound like the $\ce{AlCl3}$ says, "Oh, I am in solid state now! Let me turn to crystal (polymer) structure! [a while later] Oh, I am in vapour state now! Let me turn to dimer structure!"
Where in reality it says, "I don't have enough kinetic energy to break the bonds faster than they form, so I can only be in the crystal structure, in which I cannot move around, so I am in solid state now. [a while later] I have enough kinetic energy now! I can now break bonds faster than they form, but still not enough to break the bonds between the dimer, so let me exist as dimer. Also, I can move around now, since the intermolecular forces are not that strong. I am now a gas."

Almost everything polymerizes when the kinetic energy is low, forming a solid. The difference is only whether the bond responsible for the polymerization is van der Waals' forces or covalent bond or ionic bond.
For $\ce{BeH2}$, the bond is dative covalent as that makes the most sense.
For $\ce{AlCl3}$, the bond is ionic as that makes the most sense.

When the energy is high enough, $\ce{BeH2}$ decomposes, so all discussions henceforth are meaningless.
For $\ce{AlCl3}$, dimer has less energy than monomer, so dimer is formed. Remember that things like to be in states of low energy. Dimer enables it to move around, hence it is a gas.

When the energy is even higher, even the dimer bonds break and you have $\ce{AlCl3}$ monomers.

Should I elaborate on "makes the most sense"?

I would say the bonds in $\ce{(AlCl3)2}$ are all dominantly covalent with some ionic character. Also there will be $\ce{Al2Cl6 <--> AlCl2+ + AlCl4-}$

@Martin-マーチン Did I say they are not covalent?

I think so: "For AlCl3, the bond is ionic as that makes the most sense."

@Martin-マーチン I was talking about the crystal...


text:
Does the protective layer of Aluminum oxide form on the surface of aluminium alloys (such as Al-Zr) as well?
Aluminum is an ideal element for usage in airports. Unlike Iron, when oxygenating, there is a protective layer formed that does not allow the rest of the Al metal to oxygenate.Does the protective layer $\ce{Al2O3}$ form on the surface of aluminum alloys (such as Al-Zr) as well as when it happens to Aluminum in its pure state?

Welcome to chemistry.SE! Good Questions always give a background of what is being asked so that the answerer is fully determined on what to tell. The info. you provide will be helpful for other viewers of this post, even if it's something very rudimentary. Anyway, you might want to take a look at help center for usual questions about this site or this link to see how to beautify your posts.

Short answer: Yes.
Long answer: Any alloy containing 'passivating' elements such as aluminum, which have the tendency to form a protective oxide as you describe, has the potential to also form the same sort of passivating oxide layer as the pure material.  It's not guaranteed, though.
Pure zirconium also forms a passivating oxide $(\ce{ZrO2})$$^{(1)}$, so the AlZr alloy you mention is almost certain to form a similar protective oxide.
More broadly, a major area of research at present is in the 'alumina-forming' alloys ("alumina" is another word for aluminum oxide, $\ce{Al2O3}$), where aluminum is included in, e.g., steels or nickel superalloys, to try to achieve this passive aluminum oxide layer on the surface in high-temperature, corrosive environments$^{(2-5)}$.  With materials like a stainless steel-plus-aluminum, the passive oxide scale only forms if the alloy has the right percent range of aluminum -- otherwise, the aluminum oxide ends up staying spread throughout the metal and not concentrating at the surface.  If the alumina doesn't end up spread evenly over the surface, you don't get nearly as good of a protective effect.
(1) http://www.mtialbany.com/metals/zirconium/
  (2) http://topaz.ethz.ch/function/web-het-secured/pdfs/Prescott-92.pdf
  (3) http://www.metalle.uni-bayreuth.de/de/download/publications_downloads/Bensch__Sato__Warnken__Affeldt_______Reed__Glatzel__Acta_Mat_.pdf
  (4) http://www.netl.doe.gov/File%20Library/Events/2010/ods/Gordon_Tatlock_Corrosion.pdf
  (5) http://www.ornl.gov/science-discovery/advanced-materials/science-to-energy-solutions/alumina-forming-austenitic-alloys-licensed


text:
How is N to C ratio measured in forming austenitic stainless steel?
Nitrogen is the future hope of stainless steel industry; since in the production of austenitic steel it can even completely replace nickel (which is relatively very expensive) in some cases and for another reason: it decreases the temperature at which carbides should form.
As a result, there is always an optimum carbon to nitrogen ratio. My problems are:

How is N decreasing the temperature of the formation of carbides? Is it a catalyst or something?
Resources, however, didn't mention how the N to C ratio is measured? How? Am I missing something very easy in here?

Update: I just found this paper in which almost all of the info above can be found; but it's not my main reference.


text:
What is the difference between "bromine" and "bromine water" in their reaction with an alkene?
In question 16 of the A-Level Chemistry 2013 exam, there is this reaction scheme:

$$\ce{CH3CH=CH2 ->[?] CH3-CHBr-CH2Br}$$

The question asks for both the name of the reagent and the product. I know that the product is 1,2-dibromopropane, but am confused about the reagent. I was told that the correct answer was "bromine", whereas "bromine water" was a wrong answer. Why is this so?

From ChemGuide:

Using bromine water as a test for alkenes
If you shake an alkene with bromine water (or bubble a gaseous alkene
  through bromine water), the solution becomes colourless. Alkenes
  decolourise bromine water.
The chemistry of the test
This is complicated by the fact that the major product isn't
  1,2-dibromoethane. The water also gets involved in the reaction, and
  most of the product is 2-bromoethanol.

In essence, the electrophilic pi bond in the presence of bromine and water attacks diatomic bromine as usual, forming the usual bromonium ion intermediate.
After this step, however, we have two competing nucleophiles which can open the ring: bromide ion and water. The water is likely present in great excess, making it the more likely attacker (statistically).

Of course, the desired reaction below to form a dibromide also occurs. But it is in competition with other reactions such as the one listed above.

So usually electrophilic halogenation of alkenes takes place in an inert medium such as methylene chloride or carbon tetrachloride. These aren't inclined to act as nucleophiles, so they don't participate in ring opening, leaving the $\ce{Br-}$ ion to do its thing unhindered.


text:
Is the aromaticity broken in some resonance structures of para-nitro-aminobenzene?
According to my course of organic chemistry, para-nitro-aminobenzene has to break his aromaticity to delocalize the electrons of the amino-group in the nitro-group.
I don't really see this, when I draw the resonance structures.

@Simon Ravelingien - why does your diagram use equilibrium arrows to show resonance contributors?

As far as I am aware the aromaticity is not broken

You need three conjugated pi bonds in the ring to conserve aromaticity in this system. The molecule's "true" structure is a weighted sum of the resonance structures; you therefore would expect a partial loss of aromaticity compared to benzene.

What do you mean by loss of aromaticity?

A broken conjugated system or 4 instead of 6 electrons in the conjugated pi bonds

I agree with Dissenter, the convention is to use a double headed arrow for resonance structures. It helps people be aware of the difference between resonance structures and different species in equilibrium.

@Bon Aside equilibrium arrows,the intermediate 5 structures do not seem aromatic.

*diImine quinone would undergo a hydride-shift, not simply "quinone would undergo a hydride shift."

One could argue  the delocalization and resonance stabilization of the pi electrons is extended beyond just the ring.  Aromaticity is not really broken, but superseded.

Your teacher is correct, some of the resonance structures in your figure are not aromatic.  Still though, as resonance structures they contribute to the overall description of the molecule, just less than the structures that are aromatic.  The resonance structures at the right and left end of your figure are aromatic. The fourth resonance structure (the one in the middle) is referred to as being a "quinone" rather than an aromatic.
o- and p-benzoquinone are examples of quinones; they are not aromatic compounds (ref, note the sentence "Quinones are conjugated but not aromatic").

Likewise, their carbon analogues, o- and p-xylylene are also non-aromatic quinoid-like compounds.
For a compound to be aromatic it must be

cyclic
contain 4n+2 pi electrons
have a continuous loop of p-orbitals
and you must be able to draw double bonds - internal to the ring - from each ring carbon atom.

It is this last point that serves to differentiate quinones and quinone-like molecules from aromatic compounds.
Finally, aromatic compounds need not be planar, deviations in excess of 15° from planarity can be tolerated and the molecule can remain aromatic (see, for example, [6]-paracyclophane)

Can benzoquinone be called anti-aromatic since it has 4n pi electrons?

@ChemExchange  No, benzoquinone is not anti-aromatic, it is not a simple cyclic array of 4 pi electrons.  In fact, anti-aromatic compounds distort to some other geometry so they are no longer anti-aromatic and hence, more stable.

@ChemExchange ron is right, but your point is valid - you can indeed write some anti-aromatic mesomeric structures for quinones, but as their are destabilizing, they aren't major contributors - in tropone it's the other way round.

@Mithoron But all resonance structures have 4 pi electrons so all have destabilizing effect.

@ron So Huckel's rule fails here?

@ChemExchange What do you mean by "here"?

@ron Here in the case of benzoquinone. It has 4 pi lectrons inside the ring.

@ChemExchange  Huckel's rule applies to cyclic systems where all of the p-orbitals and electrons are contained within the ring.  With quinones we have some orbitals and electrons external to the ring, so Huckel's rule doesn't really apply in such cases..

@ron So the 4 bulleted points in your answer are not Huckel's rule but a modification of it?

@ChemExchange  I never mentioned Huckel's rule in my answer.  The 4 bullets represent a few points, likely incomplete, on what I think is needed for aromaticity.

@ron Yes, you never mentioned it. I meant whether they were a modification or not. Anyway, I got the answer to it. Thank you.

Let us continue this discussion in chat.

@ChemExchange  I need to go somewhere now, I'll see if you're around later.

Alright. I will leave my message in that chat.

I believe Quinones are aromatic according to the goldbook definition of aromaticity. I cannot find evidence for the statement that they would not be other than the wikipedia page, and it has been disputed there unfortunately without citation.

@ChemExchange related http://chemistry.stackexchange.com/questions/15272/aromaticity-of-cyclic-ketone, http://chemistry.stackexchange.com/questions/27360/resonance-of-aromatic-compounds

@Martin Why? Can you contribute your reasoning here?

@ChemExchange I might post an answer in a while, but I am not going to participate in an extensive chat about it. Sorry I am too busy for that. In the meantime you can have a look at this answer of mine. Aromaticity is not a very well defined concept and I also believe that it is hard to tell apart from overall resonance stabilisation. Hückel's rules are nice guidelines, but they are not the definition of aromaticity. This is always taught wrong. They also do not explain Y-aromaticity and sigma aromaticity...

Aromaticity is not a very well defined concept (there are no strict rules) and I also believe that it is hard to tell apart from overall resonance stabilisation. The IUPAC goldbook states for aromatic:

aromatic

In the traditional sense, 'having a chemistry typified by benzene'.
A cyclically conjugated molecular entity with a stability (due to delocalization ) significantly greater than that of a hypothetical localized structure (e.g. Kekulé structure ) is said to possess aromatic character. If the structure is of higher energy (less stable) than such a hypothetical classical structure, the molecular entity is 'antiaromatic'. The most widely used method for determining aromaticity is the observation of diatropicity in the 1H NMR spectrum.
  See also: Hückel (4n + 2) rule, Möbius aromaticity
The terms aromatic and antiaromatic have been extended to describe the stabilization or destabilization of transition states of pericyclic reactions The hypothetical reference structure is here less clearly defined, and use of the term is based on application of the Hückel (4n + 2) rule and on consideration of the topology of orbital overlap in the transition state. Reactions of molecules in the ground state involving antiaromatic transition states proceed, if at all, much less easily than those involving aromatic transition states.

Hückel's set of rules are often used as nice guidelines, but they are not the definition of aromaticity. In fact they are very, very rigid and strict and most of the compounds you'd usually consider being aromatic in this context, do not comply with these rules. The IUPAC goldbook defines:

Hückel (4n + 2) rule
  Monocyclic planar (or almost planar) systems of trigonally (or sometimes digonally) hybridized atoms that contain (4n + 2) π-electrons (where n is a non-negative integer) will exhibit aromatic character. The rule is generally limited to n = 0–5. This rule is derived from the Hückel MO calculation on planar monocyclic conjugated hydrocarbons (CH)m where m is an integer equal to or greater than 3 according to which (4n + 2) π-electrons are contained in a closed-shell system.

In this framework only the first and the last structure obey Hückel's rule. The proton and carbon NMR clearly proof that p-nitroanniline is an aromatic compound, the spectrum is very close to the data of benzene.
Resonance structures are just a crutch to explain a very complicated and complex electronic structure in simple and understandable terms, i.e. Lewis structures. In highly delocalised systems a superposition of all resonance structures has to be considered to paint a somewhat conclusive picture of the bonding situation. There are contributions of different structures and not all of them may be equal in the overall description, however, they do not change the fact, that the experimental property is observed. This is again a case, where you can see the limitations of the resonance concept.

All the resonating structures follow Huckel's Rule and so they are not losing their aromaticity.

Planar Cyclic Ring
4n+2 pi-electrons (the exocyclic pi electrons are not considered)1,2.
All atoms of the ring in conjugation

A classic example of an aromatic cabanion is cyclopentadiene ion.
The negative charge or the lone pair is in resonance and is in an unhybridised orbital. Hence it gives $sp^2$ and not $sp^3$. So its planar.

Are the structures with a negative charge at a carbon atom really planar?  I would expect tetrahedral bonding angles for those.

@CurtF. the phenyl ring is a hexagon (or near enough) so it cant have tetrahedral angles

Can you really say that any of the resonance structures in the middle have 6 pi electrons in the ring? Each (ring) double bond counts for two electrons, the lone pair counts for two electrons so we're up to 6 electrons right there. The exocyclic pi bond counts for zero? One? Two?

@CurtF.  asked, "Are the structures with a negative charge at a carbon atom really planar?"  Yes.  The cardinal rule for resonance structures is that nuclear positions cannot change.  So since the aromatic ring is planar, so must it be in the resonance structures.  You're right that carbon with a negative charge would prefer to be non-planar.  Since it can't deform in the resonance structures, that tells us that those resonance structures contribute less to the "real" structure.

@CurtF. A classic example of an aromatic cabanion is cyclopentadiene

The negative charge or the lone pair is in resonance and is in an unhybridised orbital. Hence it gives sp2 and not sp3. @ron Am I correct in saying this?

@ChemExchange  Yes, the carbon is sp2 hybridized.

@ChemExchange - thanks for the extra info.  I was wrong to downvote you, but I can't reverse it unless you make some kind of edit to your initial answer.  It would probably be great to add in the extra info about cyclopentadiene etc into your answer.  But if you don't have time, just add in a whitespace or something and I'll reverse my vote.

@jerepierre I don't understand why the exocyclic bonds have to be included in the 6 pi electrons.

@ChemExchange My point is just that depending on your decision there, the aromaticity changes: zero=aromatic, one=non-aromatic, two=anti-aromatic. Without addressing the electron count, you can't say it follows 4n+2.

@jerepierre True. I need to mention that in my answer too.


text:
Can fully-hydrogenated oils be mixed with unsaturated oils for use in food (margarines, etc)?
Would it be feasible to mix fully hydrogenated oils with unsaturated oils (eg. soybean oil) to create an edible fat with the consistency of margarine?
If so, what is the motivation for interesterified oils? If not, then what is the constraint that prevents this? Do the fats not form a solution?
I'm trying to understand how palm oil came to replace partially hydrogenated oils in margarines in response to trans-fat bans, and what possible alternatives might be.

The purpose of hydrogenation is to convert liquid oils into semi-solids at room temperature.  Palm "oil" is unusual among natural vegetable oils/fats in that it is naturally fairly unsaturated and is thus a semi-solid at room temperature.
Economics is probably the driving factor preventing significant research into the solution (pun intended) you propose with mixing saturated and unsaturated fats to get the desired consistency.  Hydrogenation techniques are cheap, well established, agency approved, etc.  Same idea with palm oil.
I hope I hit the points of your question, comment if I missed anything.


text:
Clean silver with aluminum foil and *
I already have two questions touching on this subject and there is a third question as well.
The thing is that all the other question talk about cleaning silver with aluminum foil and carbonated water. The recipe I use though is to soak the silverware and the aluminium foil in salt water (simply mix table salt and cold water). I got the idea many years ago watching a commercial for a product called "SilverStar" or something like that on TV-Shop which was essentially a 0.5 cm thick aluminium plats which you were supposed to but in a bowl with saltwater and all your jewellery and it would magically become polished.
The explanations for the "recipes" using carbonated water are all along the lines that the acid dissolves part of the aluminum and then the aluminum ions in solution somehow react with the silveroxide.
The question then is if I only use table salt and (still) water will the chemistry be more or less the same but only much slower (relying on CO2 dissolving in the water from the air) or is there something else going on?
My understanding of what was happening with the salt water was that the salt water acted as a "wire" and the aluminium acted as the "sacrificial metal" in a Cathodic Protection. I always thought that you could essentially put the aluminium in one bowl and the silver in another and connect them with a copper wire.
So the question is twofold

What is the purpose of the table salt in my recipe?
Is there any (chemical) difference between "my" recipe with table salt and the more mainstream one with different flavours of carbonated water?

If you look at Ben's answer on the third question, he explains the role of the acid in removing the passivation layer on the aluminum foil. Plain table salt doesn't do this, so the reaction may be slower in a neutral solution.

Since you want to remove the tarnished silver ($\ce {Ag2O}$), you would convert it into silver metal (or reduce it). Therefore, you would oxidize the aluminum ($\ce {Al -> Al^3+}$). The silver and aluminum need to be in physical contact and the sodium chloride or the carbonated water is just providing the medium to transfer the ions; the aluminum does not dissolve first.


text:
Is 1000ml of 'hot-ice' solution dangerous?
So I was trying out the sodium acetate trihydrate 'hot ice' experiment, and my 300ml test worked great. So, naturally, more is better right? (cringe)
Now I have a 1000ml in a pyrex beaker cooling down, and I am belatedly wondering if this is going to generate too much heat when I set it off. I'd imagine the beaker itself is safe, I don't imagine it would do anything to the granite countertop, but it is resting on a plastic cutting board, which could conceivably melt.
How much of this can I make before I'm doing something risky?

I bet it was fun to watch - to much fun to think in measuring the temperature change ;) Small volumes (<< 300 mL) are actually used as pocket warmers and usually considered to be safe. I'd say the 1000 mL experiment is safe too, but you might want to skip the cutting board and place the beaker in a larger glass bowl ("Kristallisierschale"). In the end, it's your granite table, not mine.

According to About Chemistry: it states The heat generated by the crystallization of a refrigerated sodium acetate solution should not present a burn hazard. I hope nobody would pour it on their skin to test the validity of the statement. http://chemistry.about.com/od/homeexperiments/a/make-hot-ice-sodium-acetate.htm

Ahh. I remember the "more is better" idea we had at uni when we were told to destroy some leftover sodium from an overfilled and un-reglementary storage that had been "discovered". It turns out there is a limit, and that limit is way before tossing 3 kgs of sodium into a lake from a dingy. It is not fun trying to row a dinghy that is on fire, with oars that are burning, on a lake that is locally very much like concentrated lye..

The heat generated increases directly with the amount of sodium acetate trihydrate used, but the maximum temperature, ignoring surface effects, should not change at all. After all, if the heat in 300 ml is used to raise the temperature of 300 ml of substance, or if 300 kg is used to raise the temperature of 300 kg, it's being spread out proportionately. No matter how much you use, it won't exceed ~58 C
However, do not try this in a container that would not allow room for expansion, e.g. a glass bottle, as in crystallizing, sodium acetate trihydrate might take up more room. See warning at http://chemmovies.unl.edu/chemistry/dochem/DoChem058.html. You could safely do this on a polyethylene cutting board, though, which melts at ~100 C

I would like to point out that water at 55°C can cause burns in a few seconds: http://www.hotwaterburns.org/#!temperature-vs-time/c1f8e


text:
How to know whether s-p mixing will happen in heteronuclear molecules?
How do I know whether s-p mixing would occur in some diatomic heteronuclear molecules like CN, CO, NO etc?

I don't think you can know, a priori, whether or not s-p mixing occurs without first having some experimental data.  I'll use water, a triatomic molecule, as my case in point.
For over 50 years students have been told that water is roughly $\ce{sp^3}$ hybridized.  The general description is that there are two equivalent O-H $\ce{sp^3}$ sigma bonds and two equivalent lone pairs, also in $\ce{sp^3}$ orbitals.  The lone pair - lone pair repulsion is greater than the sigma bond - sigma bond repulsion, so the lone pair-O-lone pair angle opens up slightly and the $\ce{H-O-H}$ angle closes down to the observed 104.5 degrees.
With the advent of photoelectron spectroscopy it was found that the two lone pairs in water were not equivalent (2 signals were observed for the lone pairs).  Now, the hybridization of water is described as follows:

2 $\ce{sp^3}$ O-H sigma bonds
one lone pair in a p orbital
and the second lone pair in an $\ce{sp}$ orbital

If chemists knew how to predict s-p mixing, it wouldn't have taken them over 50 years to understand how they mix in water.

what are the implications of water's hybridization? Is one lone pair more reactive than the other?

Is there any handy rule for some simpler cases, like the molecules I mentioned?

@Marko  Not that I'm aware of.

@Dissenter  The two lone pairs are not interconvertible by a symmetry operation.  Therefore they are different, this includes reactivity.

Muliken's 1935 "Electronic Structures of Polyatomic Molecules. VII. Ammonia and Water Type Molecules and Their Derivatives" givens energy values for water MOs, with the two non-bonding (lone pair) orbitals being 32 eV and 12.7 eV.  UV and "electron impact" data are cited.  http://scitation.aip.org/content/aip/journal/jcp/3/8/10.1063/1.1749715   The degree of s-p mixing for each orbital is calculated in his 1955 Electronic Population Analysis on LCAO–MO Molecular Wave Functions. I http://scitation.aip.org/content/aip/journal/jcp/23/10/10.1063/1.1740588 Non-equivalent lone pairs are long known.

@DavePhD  I guess it depends on what you mean by "known".  Saying that the two lone pairs on water are separated by 19.3 eV (1,862 kJ/mol !!) doesn't convince me that Mulliken's analysis came anywhere close to describing reality.  In fact, Mulliken was so far off the mark that when Nyholm and Gillespie came out with their take on VSEPR some 25 years after Mulliken's 1935 paper, they chose a completely different hybridization model for water.

@ron Muliken 1935 has the 4 molecular orbital energies as 32, 18, 17 and 12.7 eV (with the 17 and 18 eV values being two separate features in a 1933 spectrum), compare to the modern values of 32.2, 17.0, 13.8 and 12.6, but he incorrectly assigned the experimental data thinking the highest and lowest energy orbitals were non-bonding.  By 1955, it was realized that the 32eV orbital was bonding and experimental values were 16.2, 14.5, 12.6eV http://scitation.aip.org/content/aip/journal/jcp/23/12/10.1063/1.1741879 So by 1935 4 non-equivalent orbitals were known, but not bonding vs non-bonding.

Preliminary.
Hybridization is a questionable model. It was originally introduced to consider organic molecules, and there it works brilliantly. However, it is hard to use outside organic chemistry with molecules with simple bounding. For example, adequate description of cyclopropane requires fractional indices in orbitals.
Mathematically, hybridization process is transformation of several orbitals using linear transform. Since molecular orbitals are described as linear combinations of atomic orbitals, there is no problem to use any imaginable set of atomic orbitals as long as their linear combinations can produce all 'original' atomic orbitals. Basically, hybridization is similar to change of basis vectors in geometry: it is OK as long as the transformation follows simple rules, and it may be built to preserve values of some scalar vector expressions.
Since from mathematical viewpoint hybridization is nothing more than a special transformation of basis set, it does no change 'actual' molecular orbitals, meaning that calculations using hybridized and unhybridized orbitals must provide same result (actually, the very transformation in question is purposefully built this way). So, your question does not have a proper answer as the question itself makes wrong assumption. Hybridization does not occur, it is a model people use, it does not correspond to any real-world phenomenon
Now, the proper question would be: when hybridization model is applicable and provides valuable insight into electronic structure of considered molecule?
When you are considering electronic structure of isolated 2-atom molecules it is usually better to not use hybridization. It does not provide any simple conclusions for inferring preferred geometry, but leads away from more 'accurate' MO LKAO model. Considering interaction of such molecule as a ligand with metal center, however, returns some use to hybridization for simple molecules, because it simplifies geometry consideration. For example, using electron count rules it is possible to some extent to predict geometry of $\ce{NO}$ fragment, inferring preferable hybridization state of the nitrogen atom. It is also much easier to consider interactions with central atom for $\sigma sp$ lone pair  than several dinuclear molecular orbitals.
When considering geometry of organic and some inorganic molecules ($\ce{BCHNOF}$) with little angular stress and no hypercoordinated atoms, hybridization works well enough as long as you do not try to use it to predict one-electron properties, like ionisation potentials. However, stressed molecules like [1,1,1]-propellane or dehydroadamantane and hypercoordinated atoms, like methanium cation, require separate, case-by-case analysis to be integrated into framework of hybridization model. For most other atoms hybridization model cannot provide valuable insights.
Again, hybridization does not work for 1-electron properties. For example, methane, a perfect example of ideal $sp^3$ hybridization has TWO peaks in photoionization spectrum, implying it has 2 unequal sets of valence orbitals (roughly -2.25 and -1.5 eV energy levels) despite hybridization model predicting it to have 4 perfectly equal bonds. MO LKAO model, on the other hand, predicts several bonding orbitals, one significantly lower in energy than three others.

I didn't think about hybridisation, but s-p mixing that causes the pi MO to be above sigma MO in some molecules (like nitrogen) when I asked the question.

@Marko It occurs in ALL molecules, the question is to what degree. Unfortunately, there is no way to predict if it will change relative position of orbitals in any particular case without some complex calculation. If relative position of orbitals is important, use some free package (say, us-gamess, free for personal use) to calculate orbitals and view them in some free viewer, say, avogadro. Fortunately, it is quite easy for 2-atom molecules these days.


text:
Alternative PCB etching solutions - Is KMnO4 or NaClO suitable oxidizer?
I need to etch small PCB for my project, since I do this once in a blue moon I would like to try some easy approach, preferably with household items. I'm aiming at the use of acetic acid with hydrogen peroxide to aid the reaction. But this got me wondering what are some other household items usable in the process?
I was able to find potassium permanganate ($\ce{KMnO4}$) which is a strong oxidizer, but I'm not sure if this still holds for water soltuion mixed with acetic acid.
My other idea for strong oxidizer was sodium hypochlorite ($\ce{NaClO}$) found in bleach, but I'm not really sure how safe this is. Even though I will etch only small PCB, 1 by 2 inches, I would rather not endanger myself or anyone else.
Does any of the proposed ideas make sense? Is $\ce{KMnO4}$ or $\ce{NaClO}$ suitable oxidizer for etching PCB using acid solution?

I only added the [tag:everyday-chemistry] tag. Otherwise, I think that asking for cheap and safe ways to etch PCBs without killing oneself is a legitimate question.

Yes, it is. But I thought, maybe the answerer has no idea what range of chemicals @jnovacho has available. You had any ideas? Wow! :D

@MARamezani I've edited the question to narrow it down. I'm interested in use of the two chemicals I've mentioned.

While being extremely strong oxidizer, potassium permanganate tends to produce solids as byproducts. Also, no matter what etching agent you plan to use, you will need acidic solution to avoid producing solids. I highly recommend to search for hydrochloric acid (likely to be available via DIA and pharmaceutic store) or sulfuric acid (available via car service stores, used in lead accumulators). Than add a proper oxidizer, like hydrogen peroxide.

If you don't want to use the classical solid etchants, such as

Iron(III)chloride ($\ce{FeCl3}$) or
Ammonium persulfate ($\ce{(NH4)2S2O8}$) or
Sodium persulfate ($\ce{Na2S2O8}$)

there's the mixture of hydrogen peroxide ($\ce{H2O2}$) and hydrochloric acid ($\ce{HCl}$) in various ratios.

Typical recipes use

water
30% $\ce{H2O2}$
37% $\ce{HCl}$

in a volume ratio of 1:1:1.

Hydrochloric acid can be considered a "household chemical", it was/is available at DIY stores as muriatic acid and typically used to clean brickwork.
I would advise against mixing concentrated (glacial) acetic acid and hydrogen peroxide solutions of higher concentrations. You might end up forming peracetic acid, which is unhealthy and tends to decompose spontaneously (explode) at higher concentrations.
There's another method worth to be mentioned, although the etchant is usually not a household item.
Cupric chloride $\ce{CuCl2}$ in aqueous solution can be used at room temperature to remove metallic copper from the PCB too:
$$\ce{CuCl2 + Cu -> CuCl}$$
Thinking in terms of transition metal waste management, this seems to be pretty nice, since the resulting copper(I) solution can be regenerated (oxidized) to copper(II) (= the etchant) using hydrochlorid acid and hydrogen peroxide. This conference proceeding provides further information.
As far as the permanganate is concerned, it seems that it has been used to desmear PCBs in industry, but it has been superseeded by plasma treatment to avoid damage to the boards.

In my case, the acetic acid would be vinegar, the one I have at home is 8% water solution. Also the hydrogen peroxide is of medical kind 3% water solution. Thanks for the tip.

@jnovacho I think that you're safe with these concentrations. On the other hand, I can't tell how long the etching will take.

@jnovacho I highly doubt that 3% acetic acid will do much good.

@permeakra You're right! I updated my answer and added a more realistic recipe.

I’m using $\ce{CuCl2}$ for three years now. I like it a whole lot better than $\ce{FeCl2}$ – it’s cleaner, uses domestically available chemicals, and has less of a disposal problem.
It’s related to “mixture of hydrogen peroxide ($\ce{H2O2}$) and hydrochloric acid ($\ce{HCl}$)” because that produces the same result, ultimately, though there is more than one way to get there. I have in the past tried $\ce{HCl/H2O2}$ and ended up stripping the entire board in a few seconds: but those same chemicals can be used to make $\ce{CuCl2}$ which is far more controlled. I’m not much of a chemist, but my understanding is that $\ce{Cu + H2O2 + HCl -> CuCl2 + H2O}$: the initial reaction eats copper fast but the result is $\ce{CuCl2}$ and the reaction with that is slower.
Here is a practical guide to making and using $\ce{CuCl2}$:
http://techref.massmind.org/techref/pcb/etch/CuCl2.htm


text:
Where does the energy of formation of covalent bonds come from?
Consider two $\ce{H}$ atoms. Since the proton in one attracts the electron in another, they attract each other, and form a covalent bond ($\ce{H-H}$). Bond forming requires energy (436 kJ/mol for an $\ce{H-H}$ bond) – where does that energy come from?
If I put two $\ce{H}$ atoms in a container, would they form an $\ce{H-H}$ bond by themselves, or would I need to add energy to the system in order for them to bond?

What do you mean where does the energy come from?

Energy is conserved, so something in there has got to lose joules to form the bond. Who loses it? (Or correct me if my physics is wrong)

Nah, merry christmas. :) It doesn't really have to be something; it's just the thermodynamic system that will gain energy. Don't get me wrong, energy is released when new bonds form. You're only talking about something like H to H bonds, right?

Merry Christmas. I'm talking about forming bonds in general. What I imagine is that the proton and electron are moving through the air, each one having kinetic energy. However, an electron near them (that belongs to the second H atom) draws the proton, and thus the proton loses some kinetic energy. Another thing i'm confused - if bond forming is exothermic - would I sense heat (even just the tiniest bit) when bonds form near me?

You might find Terry Bollinger's answer to a similar question helpful.

Two hydrogen atoms forming a bond doesn't require an input of energy, it releases energy which is why most actual elemental hydrogen under earthly conditions is found as H2 hydrogen molecules.

"If I put two H atoms in a container, would they form an H−H bond by themselves, or would I need to add energy to the system in order for them to bond?" - If the two hydrogen atoms had NO kinetic energy (temperature is 0 K) and sat in place far apart from each other then no bond will be formed. If you raised the temperature then you increase the likelihood of the two Hs finding each other.

Your view of bonding is a little simplified and confused. For the details it is worth reading the great answer referred to in some comments already: Fundamental forces behind covalent bonding .
I'll stick with a simpler view that focusses just on the energy involved and not the mechanism of chemical bonding. Firstly, you are wrong that boding two hydrogen atoms needs an input of energy. It actually releases energy. This is a thermodynamic view of the reaction. This tells us whether there is more or less energy in the product of the reaction being described than in the chemicals being combined to create the product.
But a lot of chemical reactions are thermodynamically favourable (that is the reaction releases energy) don't happen easily (this is probably not true for atomic hydrogen which should react very easily to form H2). But there are plenty of other reactions that release energy according to thermodynamics but that don't happen easily. For example, elemental carbon plus oxygen will release energy if converted to CO2. But neither the diamond in jewellery or the coal in your fireplace (which is basically impure carbon) reacts easily to give carbon dioxide despite being in an environment with a lot of Oxygen gas in it. Here the problem is that room temperature oxygen molecules and carbon just don't have enough energy to kick off the process that creates the product. This is where we have to look at the kinetic view of the reaction rather than just the thermodynamics. Although the overall reaction is favourable according to thermodynamics, we need to add energy (firelighters for coal fires, a blowtorch or diamond rings) to the mixture to give the atoms in the reactants enough energy to overcome the kinetic barriers to the reaction. Think of having to climb a tall mountain to get to a lower valley on the other side. Or think of giving oxygen molecules enough energy to break carbon atoms apart from their normal solid structure.
Once coal or diamond are warm enough, the reaction will happen spontaneously and you have either destroyed your engagement ring or you have a nice coal fire. In this case you can feel the energy released from the reaction as a pleasant warming of your room.
Back to the question. The energy involved in a reaction is an accounting of the amount of energy used in the bonding of the products versus the reactants. The energy released in reactions such as burning carbon goes into the environment as heat. Other reactions happen spontaneously (try dropping a small lump of sodium into water). Some reactions require a tiny amount of encouragement (oxygen and hydrogen in certain concentrations will react explosively to form water when encouraged with the tiniest spark).
We can drive some reactions the other way using inputs of energy. For example we can electrolyse water into a mixture of hydrogen and oxygen. This requires us to add energy to the system (about 286kJ/mol) because the products are in a higher energy state then the reactants.
If you just focus on the thermodynamics of a reaction the numbers (kcal or kJ/mol) tell you how much the energy difference between the products and reactants is. But many reactions that should release energy require some input to kick off the reaction. This is confusing, but chemistry wouldn't be interesting and we wouldn't exist if thermodynamics were the only thing driving reactions.
Also energy is not created or destroyed here, just moved around. You can put energy in to a system to drive a reaction or the environment gets energy back from reactions that release it.
So, some reactions require energy to drive them because the products have higher energy, some release energy because the products have lower energy but many release energy but, confusingly,  require some energy input to kick off the actual reaction process.

As others have mentioned energy is released on bonding, which is why $H_2$ is more stable that two H atoms. To answer your last question first, H atoms in a box would eventually form $H_2$ but only if some third body (another H atom or wall of vessel) can take away some of the energy released before the H atoms fly apart again. This leaves $H_2$ in a highly vibrational state, further collisions will remove the remaining energy until it is at equilibrium with its surroundings.
Now as to bond formation, its easier to examine $H_2^+$, this has two protons and one electron; the simplest possible bond. The calculation involves evaluating the interaction between the two protons and either proton and the electron. These are of the coulomb type, that is product of charges divided by separation. (No complications from Pauli principle here as there is only one electron.)
Solving Schroedinger's equation involves terms with kinetic and potential energy (this is the Hamiltonian $H_0$). The terms for electron proton interaction are negative ($-e^2/r_1 - e^2/r_2$, where e is the electronic charge and $r_1$ the distance of proton 1 to the electron and similarly for $r_2$)) and the potential energy between protons is $+e^2/R$ where R is the internuclear separation).
The calculation proceeds by technical steps, say by using the variational theorem, and integrating over all distances and angles and distances using a basis set of atomic wavefunctions, call them $\phi_1$ and $\phi_2$.(for details see Cohen-Tannoudji, Diu & Laloe, Quantum Mechanics vol II for a full calculation).
We find that there are Coulomb integrals of the form $\int \phi_1\frac{e^2}{r_2} \phi_1 d\tau$, and $\int \phi_2\frac{e^2}{r_1} \phi_2 d\tau$. ($d \tau$ simply means integrate over all coordinates). These give the energy of the electrostatic interaction between proton 2 and charge distribution of the electron were it in the 1s orbital of proton 1 and vice versa.
In addition there are integrals of the form $A=\int \phi_1\frac{e^2}{r_1} \phi_2 d\tau$ which have no classical equivalent, i.e. they would not arise in a classical calculation. These are sometimes called 'resonance integrals' or 'exchange integrals' although this latter term can mean something else to physicists. The integral A is difficult to explain, other than to say that it arises out of the mathematics used when making a linear combination of atomic orbitals. The fact that integral A is not zero expresses the possibility of the electron ' jumping' as it were from the neighbourhood of one proton to that of the other and hence the possibility of 'quantum resonance' (The term 'resonance' arises by analogy with oscillating pendulums, see Coulson, Valence chapter 4 but does not imply actual motion).
The combination of these integrals (and the overlap integral $S=\int \phi_1\phi_2 d\tau$) produces a minimum in the potential energy profile as the calculation is repeated at different internuclear separations R and hence describes the energy of the $H_2^+$ molecule vs. separation of the nuclei.
The total energy comprises kinetic and potential energy terms, so is it the decrease in kinetic energy or potential energy that is responsible for the chemical bond? The overwhelming effect is due to the decrease in potential energy. This is also confirmed by Virial Theorem calculations.
In the $H_2$ molecule we can carry over what has been done for $H_2^+$. Now we have two valence electrons instead of one, they will both occupy an orbital similar to that in $H_2^+$ and the electrons will have opposite spins by Pauli Principle. The main difference is that there are now two electron contributing to binding and the bond if stronger. The essential principles outlined here can be carried through to forming bonds between other atoms although the calculations become tremendously more involved as there are many more electrons.


text:
Elimination of bromide and acetate under reducing conditions
Here is the reaction that I didn't find the mechanism of:

The conditions are zinc dust, acetic acid (solvent) at $\pu{11 °C}$
for 1 h.
The reagents are sodium acetate; 2,3,4,5-tetraacetoxy-1,6-bromohexane.
The product is 3-4-diacetoxy-1,5-hexadiene.
There is also some evolution of gas (I don't know what it is,
$\ce{HBr}$ or $\ce{CO2}$).

What reaction can lead to this elimination when it seems that no proton is leaving the molecule (instead of bromide) and acetate is leaving?

Welcome to chemistry.SE! You might not see it a bad idea if you take some of your time to take a tour or visit the help center for common questions about our community. I suggest you see this link and this link in order to learn how to make your posts better.

Pardon me? I edited your Q but it seems that there are little typos you got there. Did you really use Actinium Hydroxide as your solvent?! Even if you did so, formula shouldn't be AcOH; am I wrong?

i meant an acetic acid sorry

What reaction can lead to this elimination when it seems that no
  proton is leaving the molecule (instead of bromide) and acetate is
  leaving?

This is a reductive elimination reaction.  The classic textbook example is the reductive elimination of trans-1,2-dibromocyclohexane to cyclohexene in the presence of zinc dust and acetic acid.

It is basically a redox reaction where the zinc metal is oxidized and the dibromocyclohexane is reduced.  As with many reactions involving electron transfer from metal surfaces, the mechanism of this reaction is not fully understood.  Perhaps the zinc inserts itself into the carbon-bromine bond as a first step, but in any case, electron flow proceeds in the direction shown above.
The reaction has found many uses and has been widely modified to include cases such as yours where a bromine and another group are eliminated to form a double bond.  Here is a link to a paper where one of the bromines is replaced by an alkoxy group and the same zinc-mediated reductive elimination occurs.  Here is a link to another paper where zinc is used to perform a reductive elimination in a series of compounds where both bromines have been replaced, one by acetate and the other by hydroxyl.


text:
Why should the product of the normality of the acid or base and its volume be constant?
$$\ce{N1V1 = N2V2}$$
 where, N is the Normality of the Acid / Base &
 V is the volume in litres.
We ask you to give a scientific justification or derivation to the above equation
i.e. Why must the product of Normality & Volume be equal for Acid & Base ?
This is a question I found online and would greatly appreciate an answer.

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$$\ce{N1V1 = N2V2}$$ The equation is used basically for dilution of solution and determining one of the parameters of acid or base.
In the case of dilution of solution, the equation holds true because the no of gm equivalents of the solute remains constant in the solution even the solution is diluted.
Initial no of gram equivalents of solute = Final no of gram equivalents of solute ....................eq (1)
And we know, $$\text{Normality} = \frac{\text{No of gram equivalents}}{\text{Volume in Litre}}$$
$$\text{No of gram equivalents}=\text{Normality}\times\text{Volume in Litre}$$
Now lets consider,
Initial Normality =$\ce{N1}$
Initial Volume=$\ce{V1}$
Final Normality=$\ce{N2}$
Final Volume = $\ce{V2}$
Then , eq (1) becomes
$$\text{Initial Normality}\times\text{Initial Volume}=\text{Final Normality}\times\text{Final Volume}$$
$$\ce{N1V1 = N2V2}$$
In case of acid and base reaction,
No of gram equivalents of acid will always react or consume equal no of gram equivalents of base and vice versa
No of gram equivalents of acid = No of gram equivalent of base .......eq (2)
Normality of acid = $\ce{N1}$
Volume of acid =$\ce{V1}$
Normality of base =$\ce{N2}$
Volume of base =$\ce{V2}$
Then eq (2) becomes,
$$\text{Normality of acid}\times\text{Volume of acid}=\text{Normality of base}\times\text{Volume of base}$$
$$\ce{N1V1 = N2V2}$$
The number of grams of a chemical element or compound equal to the weight of its chemical equivalent. The gram equivalent of a chemical element is numerically equal to the quotient resulting from dividing the element’s atomic weight by its valence. For example, the gram equivalent of iron (atomic weight 55.847) with a valence of 2 is 27.92 g and with a valence of 3, 18.62 g. The gram equivalent of acid is that amount of the acid in grams containing one gram equivalent of hydrogen capable of being displaced by metal with the formation of salt. The gram equivalent of a base is that amount of the base in grams that is necessary for its complete reaction with one gram equivalent of acid. The gram equivalent of salt is that amount of the salt in grams containing one gram equivalent of metal. The concept of gram equivalent is widely used in volumetric analysis.
http://encyclopedia2.thefreedictionary.com/Gram+Equivalent

First, you need to know when you can use this equation. It is used to find the concentration of the solution when the volume of solvent is changed (when diluting the solute). Since the only thing that changed is the volume of the solvent (the amount of the solute is the same before and after), you use the equation to find the new concentration.
E.g. 100 mL of a 0.2 N solution of sodium chloride is diluted to 0.1 N. The new volume is:
$$\ce {N_{f} \times V_{f} = N_{i} \times V_{i}}$$
$$0.1\ce {N\times V_{f} = } 0.2\ce {N \times 100mL}$$
$$\ce {V_{f} = 200mL}$$


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Unitary Transformation of the HF equations
I am working on a code to preform a HF procedure and I am a bit confused about the unitary transformation of the basis. My instructions are as follows.

specify the basis and geometry (done)
Evaluate and store overlap, Kinetic energy, nuclear attraction, and electron repulsion integrals (done)
diagonalize $ \mathbf{S}$ and find $\mathbf{S}^{-1/2}$ which is the unitary transformation matrix.
Use the core hamiltonian to find the initial Fock matrix:
$$ \mathbf{S^{-1/2}}^{T} \mathbf{H}~\mathbf{S^{-1/2}}\equiv\mathbf{F}_0$$
Diagonalize the Fock matrix to find the mo coefficients:
$$\mathbf{F}_0 \mathbf{C}^{'}_0=\mathbf{C}^{'}_0\epsilon_0$$
Back transform the coefficients.
$$\mathbf{C^{'}_0}~\mathbf{S^{-1/2}}=\mathbf{C}_0$$
compute the density, use the density to evaluate new fock matrix, find new coefficients repeat until convergeance.

Now step 7 kinds glazes over a lot of the work but I understand that part, so It isn't important to the question.
I am confused by the $\mathbf{S}^{-1/2}$ matrix.  Is this literally one over the square root of each element of $\mathbf{S}$? Or is there some decomposition I should preform that I just do not understand.  I can carry out the whole procedure but without orthonormalization my results are nonsense.

You are going to want to check out Szabo and Ostlund page 143, they explain all the nitty-gritty for this matrix. The math itself is on page 22 and 23. But if you don't have a copy...
The $\mathbf{S}^{-1/2}$ comes from the need to orthogonalize your basis so that you get your equation (5) instead of the $\mathbf{FC}=\epsilon\mathbf{SC}$. It's one method of several to achieve this orthogonalization, and it's called symmetric orthogonalization.
So how do you compute this function of a Hermitian matrix? Well, functions on diagonal matrices are super easy; you just apply the function to each diagonal element. For non-diagonal matrices, this is not true. More to your question, you can't just take one over the square root of each element of $\mathbf{S}$. But you can if it's diagonal. And you can make any Hermitian matrix diagonal via a unitary transformation.
The trick, then, is to diagonalize $\mathbf{S}$ first, apply the inverse square root to each eigenvalue, then transform back using the eigenvectors. So you're going to want to

Solve $\mathbf{SU} = \mathbf{sU}$. Where $\mathbf{s}$ are the eigenvalues of $\mathbf{S}$, and $\mathbf{U}$ are the eigenvectors.
For each $s$ in $\mathbf{s}$ do $s^{-1/2}$.
Transform back to original basis by $\mathbf{S}^{-1/2} = \mathbf{Us}^{-1/2}\mathbf{U}^{\dagger}$

I read this and said to myself I really need a copy of that book (Szabo and Ostlund). I just looked it up on Amazon and I had no idea it was so cheap. Thanks for your help

S&O is really tremendous for understanding the mechanics of the matrix math behind SCF computations.


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What happens when an explosion can't lead to volume increase of the thermodynamic system?
What if an explosion is not able to exit the container?
Take a pipe bomb for example; assuming that the pipe bomb was hypothetically made out of such a strong material that the explosion is not able to rip the material apart, what would happen?
Asking this because I was told: NO MATTER WHAT the explosion has to and WILL "exit", but what if it just can't?

The explosive force is caused by the production of large quantities of hot gas (often $\ce{N2}$) which expands rapidly and cools. If there is no room for expansion then the gas will remain very hot and the pressure will increase. The effects of this will depend on what is inside the box

If the reactor withstands the pressure, you may collect your diamonds - nanodiamonds.

Well, take a look at what edits I made to your posts. I assumed that was what you meant. You will be able to reach results now. The "exit", just as @bon mentioned, is the release of some kind of gas (depends on what's exploding); if it doesn't result in the increase in mass it must result in the increase in pressure. I leave it to you to find the related formulas.

I'm not an expert in thermodynamics and gas physics so there might be some errors in this calculation but here goes:
According to several sources (wikipedia included) TNT decomposes according to this equation:
$$\ce{2C7H5N3O6 -> 3N2 + 5H2O + 7CO + 7C}$$
If 100kg of TNT was placed in an indestructible $1m^3$ container containing an inert gas (we will say helium) at standard temperature and pressure here is a rough estimate of what might happen if it was detonated.
Using a Hess's Law cycle we can calculate the enthalpy of decomposition of TNT from formation enthalpy data: $$\Delta_{decomp}H = 63.2*2 - 285.8*5 - 110.5*7 = -2076.1~ kJmol^{-1}$$
Then we can calculate the energy released by the decomposition: $$M_r~of~TNT = 227.13~gmol^{-1}$$ $$n = \frac{100}{0.22713} = 440.3~mol$$ $$Energy~released = 440.3 * 2076.1 *1000 = 914.1~MJ$$
I'm not sure how to go about specific heat capacities for gas mixtures so I won't put a figure on the temperature rise that this will cause but I suspect it will be in the thousands of degrees due to the amount of energy and so the resultant pressure rise will also be very significant, probably enough to liquefy the gases. Also as @Klaus Warzecha hinted at in his comment these kind of temperatures and pressures will probably cause some interesting chemical reactions to take place.
http://courses.chem.indiana.edu/c360/documents/thermodynamicdata.pdf

You divide by 1000 (not multiply) in the last equation.

@LDC3 2076.1 is in $kJmol^{-1}$ so to convert to $Jmol^{-1}$ you multiply by 1000. I then prefixed the answer with mega to make it a nicer number

${440.3 mol \times 2076.1 kJ/mol = 914100 kJ}$ or $914.1 MJ$

correct which is what I wrote

and I multiplied by 1 MJ/1000 kJ.

I did a calculation and came up with an increase in pressure of about 280 atm. You can see my calculations here. Most gas cylinders can handle 100 to 150 atm (one is as high as 400 atm), so getting a container to hold 300 atm should not be a problem.
Added:
On this page is list of cylinders with their working pressure (which is at least 1/2 the maximum pressure it can handle). 2000 psig is about 130 atm and 6000 psig is 400 atm.

interesting. i suppose 100kg of TNT in a $1m^3$ container isn't actually that much. it has a density of $1654~kgm^{-3}$

@bon Water has a density of $1000kg/m^3$.

yeah but it doesn't fill much of the container is what I meant

@bon Yeah, I noticed that too when I was doing my calculations, but I omitted the volume of TNT to simplify the calculation.

I would say you know better, but its hard to imagine for any Gas cylinder to not just be ripped into shreds.


text:
Why the second ionization energy of helium is greater than the first?
maybe because the first electron has a lower energy. Why does this happen?

Do you have any ideas about electron energy states?

Opposite charges attract. The greater the difference in charge, the more the attraction.

The second ionization energy of any element will be greater than its first ionization energy. One way to rationalize this phenomenon is by considering that the first electron is being removed from a neutral atom and the 2nd electron is being removed from a positively charged ion that is already electron deficient.  Additionally, Slater's Rules for calculating effective nuclear charge can be used to provide a semiquantitative reasoning for this observation.


text:
What is the proper name for the but-1-ene's isomer methylpropene?

Give the name and draw the graphical formula of an alkene that is an isomer of but-1-ene and that has a different carbon skeleton.

I think it's 2-methylpropene, but I'm not sure whether the correct name is 2-methylprop-1-ene or 2-methylpropene.

The terms "correct"/"proper" are ambiguous. The preferred IUPAC name (PIN) of the same is: 2-Methylprop-1-ene, although the names Isobutylene, 2-methylpropene are often used alternatively.

When considering whether numbering is necessary you must see whether there are alternative structures which have the same basic name but different numbering.
In this example putting the methyl group on the 1 or the 3 position of the longest carbon chain changes its name from prop- to but-. Since the only other position that the methyl group can go is the 2 position then both numbers are unnecessary and so the shortest unambiguous name is methylpropene.

Generally, the complete information about a structure is explicitly given by the name and does not rely on any implied information. Nevertheless, the practice of omitting locants when there is no ambiguity is widespread in general nomenclature. For preferred IUPAC names, however, locants are omitted only in a few exceptional cases described in Subsection P-14.3.4 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book).
Ignoring the rules for omission of locants, the complete systematic name of the parent compound would be ‘prop-1-ene’. According to Rule P-14.3.4.2, however, the locant ‘1’ is omitted. (Note that no other isomer can be generated by moving the double bond from its position to another.)

P-14.3.4.2 The locant ‘1’ is omitted:
(…)
(d) in unsubstituted dinuclear and trinuclear alkenes and alkynes and monounsaturated cycloalkenes and cycloalkynes; (…)

Thus, the preferred IUPAC name for the unsubstituted parent structure is propene.
Ignoring cis/trans isomerism, the substitutable hydrogen atoms of propene have three different locants (‘1’, ‘2’, and ‘3’). Therefore, the locants for substituent prefixes (e.g. ‘1-chloro’, ‘2-chloro’, or ‘3-chloro’) cannot be omitted in substituted propene.
Since, the locants for the substituent prefixes cannot be omitted, also the locant ‘1’ of ‘prop-1-ene’ can no longer be omitted for preferred IUPAC names.

P-14.3.3 Citation of locants
In preferred IUPAC names, if any locants are essential for defining the structure of the
parent structure or of a unit of structure as defined by its appropriate enclosing marks, then all locants must be cited for the parent structure or that structural unit. (…)

For example, (1E)-1-chloroprop-1-ene, (1Z)-1-chloroprop-1-ene, 2-chloroprop-1-ene, and 3-chloroprop-1-ene are the preferred IUPAC names for some substituteded propenes.
This principle applies to any substituent. The special situation that replacing the chloro substituent in (1E)-1-chloroprop-1-ene, (1Z)-1-chloroprop-1-ene, and 3-chloroprop-1-ene with a methyl substituent would actually lead to the names (2E)-but-2-ene, (2Z)-but-2-ene, and but-1-ene instead of ‘(1E)-1-methylprop-1-ene’, ‘(1Z)-1-methylprop-1-ene’, and ‘3-methylprop-1-ene’, respectively, does not change the preferred name for 2-methylprop-1-ene. (Note that the substitutable hydrogen atoms of propene have three different locants irrespective of the kind of the substituent.)
Therefore, the preferred IUPAC name is 2-methylprop-1-ene.


text:
Is it likely that increased understanding of quantum physics will change our understanding of chemistry?
Reading that the large hadron collider will be up and running with twice as much energy in March 2015, I was curious whether our understanding of subatomic particles has changed our understanding of chemistry.
Edit: To attempt to narrow down the question to something with a definitive answer: has the discovery of subatomic particles changed experimental chemistry?  If so, what is an example?

Welcome to chemistry.SE! This question of yours is mostly based on hypothetical predictions of the future; thus, in spite of the fact that these predictions are somehow "science-based", good answers could be too many. Chemistry.SE is based on conceptual chemistry, rather than chatty predictions. You might need to take a look at this page to find out which questions are not appreciable in our community, or be kind enough to take a tour of the website.

In my opinion, nowadays' science, specially chemistry, is not something that can be blown away with a single wind, if that's what you meant. There couldn't be principal errors today since the science now is mostly experimental, not theoretical.

Well, you asked for an example, I'll give you one.
"Quantum" is undeniably the most effective keyword our science has seen, it's changed our 'understanding of chemistry and physics' drastically over the past century.
For example, between 1900 to 1925 (years vary from source to source, but this era exists, that's what that is important), in which Bohr's model was the main focus of study (as wikipedia puts it), is known to be the era of "old quantum theory". There were series of theories, which started by Max Plank, and partially came to results with the final modifications of Sommerfeld; and ended in Erwin Schrodinger's famous model. New efforts led to new results and therefore the "old quantum theory" was never consistent.
What was all of this about? Scientists are always eager to find out the differential behavior of matter and energy. Sub-atomic particles (in first level, n, p & e) were discovered in order to explain the phenomenons that Roentgen saw (radioactivity of Uranium and other heavy elements in 'pitchblende'); Faraday experimented (the $\ce{SnCl2}$ solution experiment); Greek Leucippus and Democritos tried to explain; and so on. After every discovery, since the 'rules' that were discovered usually were up to a point in explaining what's around us, we change them and make new discoveries. And that was the reason of what Einstein, Bell and people like those did. They corrected the quantum for it to be what we know today.
So you see, the science was always about understanding; so why wouldn't a big discovery like quantization that led to quantum mechanics be able to change our direction of thought, either we are chemists, physicists or simply humans?
Links you might find helpful: About the old quantum theory in wikipedia; About the old quantum theory 2; Bell's theorem(which linked classical quantum to quantum mechanics) and finally quantum mechanics according to wikipedia.

I find this a strange question, because all contemporary chemical bond theory is quantum mechanical. So quantum mechanics already influences chemistry a great deal (look here)! I suppose you could ask whether deeper quantum theory gives us more knowledge, and I'd guess yes! For example, look at this Periodic Videos analysis of the chemistry of seaborgium (link). We need more experimental data to test all theories, including the quantum mechanical ones.


text:
How does UV affect skin colors in dark-skinned people?
Skin color is one of the things one would rather not ask anything about! Only in humans, it can vary from very dark brown to pale pink. In darker-skinned people, the color is mainly due to melanin, which is produced by melanocytes.
There are three types of melanin: eumelanin, pheomelanin, and neuromelanin. Eumelanin is the most common; so it must be the reason of the dark skin of the dark-skinned. And that's true, since the two major groups of eumelanin are black and brown species.
The first ray produced by sun that comes to mind when you think about affecting the attributes of skin is UVA, since it penetrates most into the skin (Even though its energy is lesser than UVB). So, when I attach these two together I reach the fact that if UV rays do not cause free radicals they will eventually change the skin's color. (This is what is mostly believed and its effect on natural selection is impossible to hide)
Now, I give up. How does UV do to the skin that makes it darker? This must be very easy, huh? What am I missing?

when you are exposed much to UV, the body creates pigment melanin(dark coloured) that darken the skin as a side effect of protecting our precious body from the harmful UV

Nature published a related article recently: Skin colour: No hiding in the dark

There is a journal article almost exactly about the title question, Mechanisms of Skin Tanning in Different Racial/Ethnic Groups in Response to Ultraviolet Radiation Journal of Investigative Dermatology (2005) 124, 1326–1332.
However, the body of the question assumes that eumelanin changes color.  Instead, the article finds that melanin gets closer to the surface of the skin, and the degree to which this effect occurs varies with race/ethnicity.


text:
Why did the salt on the skin of my turkey react with the aluminum foil it was touching?
I sprinkled salt on my turkey skin and left it to dry out over night. It was covered with aluminum foil. The next morning the aluminum foil had holes ‘burnt’ in it and the turkey skin turned silver at the spots where the salt crystals had touched the aluminum foil. Was my turkey poisoned with aluminum at this point? I did not eat it to find out.

If you don't mind, I'll start with a more colloquial and hopfully not to lengthy explanation:
Precious metals, such gold, do not corrode under "household conditions", others, such as iron or aluminium do. Rusting of iron is a typical example. Here, the corroded surface often is very porous and the corrosion carries on until the whole piece of metal is turned to rust (oxidized).
In the case of aluminium, the initial corrosive layer is very tight and serves as a protection for the rest of the aluminium foil. Such a very thin and transparent protection layer often is artificially applied. The item are sold as Eloxal (= electrolytically oxidized aluminium).
Now comes the table salt, which is sodium chloride ($\ce{NaCl}$). Sodium chloride is hygroscopic, it "attracts" the water and that's why you sprinkled it on the turkey. It is also very soluble in water. When it dissolves, sodium $\ce{Na+}$ and chloride $\ce{Cl-}$ ions are formed.
The chloride ions attach and move into the protective oxide layer of the aluminium foil and finally induce the corrosion of the aluminium metal. Aluminium ions are released to the water (= the wet surface of the turkey).
For a more scientific, in-depth explanation of this corrosive processes known as pitting, you might want to have a look at this article or this review.

Was my turkey poisoned with aluminum at this point?

There seem to be some crazy myths concerning the "toxicity" of aluminium (ions), often spread by opponents of immunization and other quacks.
Fact is that

a healthy human body is capable to excrete aluminium ions though the kidneys
the British Alzheimer Society does not see any convincing relation between the uptake of aluminium and Alzheimer's Disease
the cancer information service of the dkfz (Deutsches Krebsforschungszentrum, German Cancer Research Centre) refutes the aluminium induced breast cancer myth
many antacids do contain aluminium hydroxide as one ingredient. The effect of "poisoning" is not known.

In summary:

Yes, aluminium ions were released to the wet surface of the turkey.
There's no reason to assume that the turkey is "poisoned".

You asked:

Was my turkey poisoned with aluminum at this point?

I don't know. But I do know that it's not recommended for you to combine salty foods and aluminum foil in the future.
In Germany, the Federal Institute for Risk Assessment (BfR) wrote an article named "FAQs about aluminium in food and products intended for consumers". I shall quote a bunch from that article; the discussion of aluminum and salt is at the very end of my post.

What health risks does aluminium absorption pose?
Any assessment of the hazardous potential of aluminium focuses is on its effects on the nervous system and the fact that it is toxic to reproduction (effects on fertility and unborn life) as well as the effects of aluminium on bone development.
When aluminium is ingested with food, its acute toxicity is low. ... But even in healthy individuals, the light metal accumulates in the body in the course of a lifetime, especially in the lungs and the skeletal system.
...
What quantities of aluminium can be absorbed ... without any health risks?
For oral intake from food, the European Food Safety Authority (EFSA) has derived a tolerable weekly intake (TWI) of 1 milligramme (mg) of aluminium per kilogramme of bodyweight.

Finally, later in their article, they touch upon mixing aluminum with salty foods:

Why can aluminium be transferred to food from packaging or tableware?
Aluminium is soluble under the influence of acids or salt. For this reason, packaging and containers used for food such as beverage cans, yoghurt cup lids and aluminium containers for fruit juice are coated on the inside to prevent transfer of aluminium ions to the food or drink.
Aluminium from tin foil can be transferred to foods containing acid and salt. For this reason, the BfR recommends that acidic and salty foods are not wrapped in aluminium foil.

You might want to look at what happened to people involved in the mass Aluminium Sulphate poisoning event in Camelford, UK.
http://www.bbc.co.uk/news/uk-england-cornwall-17367243
Alzheimer's-like death and illness.


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How does one export a PDB structure to SVG?
This question is one that I'm wondering whether it belongs here or on Graphic Design SE. All I want is some software that can export a PDB structure of a protein (displayed in ribbons style) or DNA double helix, or a combination of the two, to a high-quality SVG file. I know that several pieces of software for 3D molecular modelling can create raster images of PDB files quite easily (e.g., Accelrys DS Visualizer, Jmol, PyMOL, etc.). But SVG files have several advantages over PNGs, now I don't mind if the colours are all bland (i.e., each shape having one colour that is not subject to any shading or other gradients; I am saying this is OK as I realize that vector images often have problems with intricate colour gradients) just so long as the end result is a decent-looking SVG file. I have tried using Avogadro for this purpose but the resulting SVG was 264 MB in size -- a little too big for my taste. If you'd like an acceptable size range it'd be <5 MB.

You could try some cartoon settings in Pymol that replace the gradients with solid colors (I don't remember how they are called exactly), and trace the image with Inkscape or similar programs to get an SVG.

The Avogadro form attempts to generate an SVG from the spherical mesh - not necessarily what you want. I feel like I saw something using JavaScript, so I'll flag this and come back later.

Another way is to save VMD files as postscript (.ps), also a vector format, and then convert to SVG. PyMol does not have the ability to save as vector graphic yet.

@BH2017 Not about exporting to yield a .svg, however taking into account your other profiles close to computing: there are bridges to enter .pdb into blender, e.g. bioblender (http://www.bioblender.org/), BlendMol (https://durrantlab.pitt.edu/blendmol/, https://doi.org/10.1093/bioinformatics/bty968), proteins-blender (https://dmnfarrell.github.io/bioinformatics/proteins-blender) of varying comfort to interact with, too.

If you haven't found answer till now, you can use VESTA (Visualization for Electronic and STructural Analysis) package. It is a freeware and can be downloaded from
http://jp-minerals.org/vesta/en/
It is available for both Windows and Linux. It can be used to import any many structural files (pdb included) and get an output in any supported format(which includes SVG). From their website:

VESTA is a 3D visualization program for structural models, volumetric data such as electron/nuclear densities, and crystal morphologies.

However, if you do not strictly require SVG format them my personal favorite is "QuteMol". It gives way too many options to adjust visual representation
It is an open source (citeware, i.e. you have to cite their paper for any academic publishing) molecular rendering software available from:
http://qutemol.sourceforge.net/
It can generate PNG, JPEG anf GIF(animated) images of maximum dimensions up to 8192x8192 pixels so you can use them for high resolution printing/posters etc. too. One small flaw is that it is rather ill maintained and considerable amount of bugs are there.For e.g. using default options is fine but in case you want to use customized colour scheme then their example ART(atomic rendering table, just a table of atomic symbols, their radii, and colour in RGB codes) file gives following example:

X    0       0.853  204    0    0
H    1    0.930  199  199  199
He    2  1.085  107  115  140
Li    3  3.100    0  128  128

However in reality you will require 2 colums of radii for programme to recognize it (core and vander waal radii). Similarly atomic symbols with single alphabet are not rendered properly!!! But all these problems have work around and results are worth it (specially for large protein molecules).
sample this:

(source: sourceforge.net)

Updated as per suggested. I have tried to restrict answer to free/ open source software only as i believe thats what the user wanted to ask. As in case of paid softwares lot many options creep up

Also available for Mac now.


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Negative ions and health: pseudo science or something more?
Over the last year or so I've heard from friends, who subscribe to holistic medicine, talk about negative ions, saying that they are good for your mental/physical health. Being the only one among them with a chemistry background, I was suspicious when i heard this. Below are examples I've seen that use negative ions in a medicinal context:
Negative Ions Purify Air

Virtually all particles in the air have a positive charge, while negative ions have a negative charge. In which case, negative ions and particles magnetically attract to one another. When there is a high enough concentration of negative ions in the air, they will attract to floating particles in large numbers. This causes the particle to become too heavy to remain airborne. As a result, the particle will fall out of the air, and will then be collected by normal cleaning activities, such as vacuuming or dusting. - Ion-Imbalance

Negative Ions from the Beach

You may have experienced the power of negative ions when you last set foot on the beach or walked beneath a waterfall. While part of the euphoria is simply being around these wondrous settings and away from the normal pressures of home and work, the air circulating in the mountains and the beach is said to contain tens of thousands of negative ions -- Much more than the average home or office building, which contain dozens or hundreds, and many register a flat zero. -WebMD

Apparently they Help You Breath

People travel from all over the world to visit salt mines like the ones found in the Himalayas, Austria, and Germany. In Germany, the hot salt springs are so revered, companies often send overworked employees there to rejuvenate.
  Why are the salt mines so beneficial?
  Quite simply... the very dry, negative ion environment of a salt mine is key.
  Well... even if you're not able to travel to Asia or Europe to take advantage of salt mines, you can still take advantage of these negative ions to help improve the air you breathe. The solution involves pure Himalayan crystal salt...- Mercola

Well it's not surprising that two thirds of the above sources are trying to sell products (salt lamps, air purifiers).
Since we're dealing with ions here, I figure the chemistry world must have some sort of input. What's the chemical or biochemical support, or the evidence against, for the above claims? If there are negative ions present from waterfalls/oceans/salt rocks, what ions are we talking about?

To address @dissenter, one of these sites claims tourmaline is a source of negative  ions...

I think some relevant questions are: 1) are gas-phase ions easily formed at STP and 2) what are these gas-phase ions that your sources speak specifically of.

Tourmaline is definitely not a source of ions, either positive or negative.

Charge separation at waterfalls with airborne ions, resulting in a potential between the base of the waterfall and the surrounding air, is a phenomenon examined by Philipp Lenard (Über die Elektrizität der Wasserfälle, Ann Phys (Leipz), 1892, 46, 584–636).
The effect is real and more recent field studies, such as Characterization of ions at Alpine waterfalls, as well as studies on the formation of charged droplets, such as Charge Separation in the Aerodynamic Breakup of Micrometer-Sized Water Droplets should be taken seriously.
However, the sources that you cited are pseudoscientific marketing bullshit.
The first quote links to a web shop that offers marvels like magnetic bands and ion-balance bands. While the silicone band probably isn't hamful to the skin, it will not emit anything. Nada, nothing, nix!
The webshop behind the third link sells Himalayan salt lamps. It's a matter of taste to have a lamp housing made from unrefined sodium chloride with traces of iron in it (hence the colour), but I wouldn't expect any health effects through the emission of whatever from these lifestyle items. Himalayan salt is being sold for quite a while, even for nutrition but if at all, the stuff has seen the Himalaya mountains only from a distance.
There are no halite mines in the Himalaya mountains.
Instead, the material is likely to come from the Salt Range in the Punjab province in Pakistan, or from Poland.
In summary, it's business as usual: Scientific facts are mashed up with esoteric mumbo jumbo to sell overpriced and useless stuff to well-meaning but naive clients.
Oh what a tangled web we weave,
When first we practise to deceive!
-- Sir Walter Scott