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What are physical properties and chemical properties?
As the question title says, I would like to know what the physical and chemical properties are. If possible, please include examples. Since it seems to be related, I would also like to know what physical and chemical changes are. I believe that things such as crushing a can is a physical change, but I am not sure what would constitute a chemical change (perhaps doing something that rearranges the molecules?)

Edit: I originally wanted to ask if cooking an egg constitutes as a chemical or physical change, but I thought that that would be closed as too vague. If possible, please explain which one it is too. Thanks!

This is a tricky question.
I have to concur with Paul - it's difficult to give a truly hard and fast definition of the distinction between a chemical change and a physical change as there is a fairly smooth continuum between 'physical' and 'chemical' interactions. However, a good rule of thumb is that chemical changes involve changes in electronic structure, usually in the form of chemical bonding or redox. The problem is that traditionally 'physical' properties may have big implications for chemistry - I'll give some examples.
The Van der Waals and dipole-dipole interactions are mediated respectively by 'instantaneous' and permanent polarisations of electron density with respect to positive charges. Dipole-dipole interactions such as those found in water are quite significant and give rise, for instance, to water's frankly freaky liquid range as well as water's ability to mediate a wide range of chemical processes. These interactions don't involve significant changes in electronic structure however they have chemical implications in, for example, making a choice between liquid water or liquid nitrogen as a solvent.
The atomic mass (and hence the isotope) of an element is typically considered a physical property, however it has implications for reaction kinetics. This can be illustrated (expensively) by giving living organisms nothing but heavy water to drink. They will eventually experience toxicity due to the slowing of key biochemical reactions.
The magnetic properties of an atomic nucleus are typically considered to be physical, however they can be exploited in the observation that e.g. different isotopes of uranium have slightly differing hyperfine splittings and hence different ionisation energies. This allows isotopes to be enriched by selective ionisation.
Regarding the egg:
The dramatic changes that occur when you cook an egg are the result of denaturation of the proteins in the egg. Proteins are long chains of amino acids that take on their characteristic active conformations through a process called 'folding', so called because that's what it looks like (if you've heard of Folding@Home, that simulates the folding of proteins). If you raise the temperature much higher than physiological temperature, the protein degrades through unfolding into a different conformation. This involves the Van der Waals and dipole-dipole interactions holding together the secondary/tertiary/quaternary structure of the protein being broken en mass as well as the formation and/or scission of covalent bonds.

Paul's comment and Richard's answer are correct that that chemical and physical change span a continuum. However, at the introductory level, there is a difference between physical and chemical properties.
Physical changes do not involved the conversion of one substance into another. They can be reversible (ice melting and freezing at 0 °C) or essentially irreversible (free expansion of a gas).
Chemical changes are those that involve the conversion of one substance into another. They can be reversible (acid-base equilibria) or essentially irreversible under normal conditions (combustion).
Physical Properties are those properties of a substance or mixture than seem to match with physical descriptions of objects and with the physical changes, especially changes of state.
Intensive physical properties include such characteristic data as densities, specific heat capacities, boiling points, freezing points, Henry's Law constants, etc.
Extensive physical properties vary with the sample (are not characteristic to the substance): temperature, pressure, volume, mass, size, shape, etc.
Chemical Properties are those properties which, as Richard mentions, seem to have to do with electronic transitions (aka chemical changes). These include all of the known reaction, reactivity, thermochemistry, kinetics, and equillibrium data.
In addition, (and non-intuitively) most of the properties of substances and mixture that interact with our five senses are also chemical properties: hardness, color, taste, smell, etc.
Finally, the following properties (and others like them) are also chemical properties: toxicity, flammability, conductivity, nutritional value, etc.

The problem with your division is that it takes phenomenon governed by identical laws and separates them into two categories, physical and chemical.  The ice-water equilibrium is governed by the same laws that govern the acetic acid ionization equilibrium.

@PaulJ.Gans - True, but the distinction has historically been made that way, and it is in most General Chemistry texts that way.

Physical changes are reversible, whereas chemical changes are irreversible. For example, boiling water is a physical change because you can reverse it by cooling it down, but setting fire to wood or boiling an egg cannot be undone.

I'm not sure that this is right.  Many chemical changes can be undone and a number of physical ones can not be.  In my view there is no distinction between chemical and physical changes, they both obey the same laws.  They clearly differ at the extremes, but for example, the conversion of ice to water can be looked at as a chemical change just as well as a physical one.  So can the solution of salt in water.


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What is a "lagged bath"?
First question, and this is one I've searched for in vain.
In the context of setting up cooling baths, I've come across the term "lagged" bath.  I cannot extract what this means from any available context.  I assume it has something to do with time passing (i.e. a lag), but I'm not certain.  If I had to guess, I'd assume it means allowing the cooling agent to come to equilibrium temperature, but again, not certain.
Any direct technical experience that can explain this term would be greatly appreciated.

I'm not 100% certain on this, but my best guess on this would be referring to 'lagged' as in 'lagging', as in insulation material. So, a lagged bath is an insulated bath.

Duh! I'll bet that's right.  I'll wait a bit for potential additional commentary, but that jives with the context I read.  So I'll likely accept this--thanks!

Yes, this term is also used to describe insulating material for pipes:  https://www.environmentcentre.com/wp-content/uploads/2016/08/5_Pipe_lagging_new_branding.pdf

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Which materials/compounds expand in response to cold temperatures?
Which materials or compounds expand in response to cold temperatures?
This material or compound should expand at a slow (but constant rate), from about 0–20 months when in the following temperature range −25 °C to −10 °C.
Also, the compound should not react with plastic or be toxic to humans or the environment.
Please explain why the compound does this as well.

This seems to me to be a chemistry question. I found one compound http://pubs.acs.org/doi/abs/10.1021/ja106711v ,  Cubic ScF3 from the references in the wiki article http://en.wikipedia.org/wiki/Negative_thermal_expansion#cite_note-greve-7

Isn't the typical example is water and ice?

In old time printing using type, the alloy used for the type was designed to have a zero coefficient of thermal expansion over a wide range so that they type did not change shape when cast at a high temperature and then cooled off.
It is very likely that slight modifications of that formula would yield an alloy with either a positive or negative thermal coefficient.

Perhaps one of the most studied materials to exhibit negative thermal expansion is cubic zirconium tungstate $(\ce{ZrW2O8}).$

It's already been mentioned that certain liquid-solid phase transitions are accompanied by negative thermal expansion (NTE). This also includes elements ($\ce{Si, Ga, Ge, Sb, Bi, Pu}$), alloys (Wood's metal) and numerous materials (zeolites, quartz, tugstates etc.) that obey virial expansion due to the way atoms are packed in a crystal lattice:
$$\frac{p}{RT} = \frac{1}{V_m} + \frac{B_{2V}(T)}{V_m^2} + \frac{B_{3V}(T)}{V_m^3} + \ldots$$
where $p$ - pressure, $R$ - gas constant, $T$ - temperature, $B_{iV}(T)$ - $i$-virial coefficient, $V_m$ - molar volume.

There is also another physicochemical phenomenon called Gough–Joule effect. If mechanical stress is applied to the elastomer or rubber string which is subsequently heated, the polymer contracts as a result of the intensive growth of the role of the entropy factor (the maximum entropy in polymers occurs in the coil-like ordering of the molecules). This effect is also utilized in the rubber-band heat engine (Feynman wheel).

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Chemical software for solving reactions
I'm looking for a software which can solve chemical equations. For example something like this (given with result):
\begin{align}
    \ce{CH4 + 2O2 & → CO2 + 2H2O} \\
    \ce{C3H6 + HCl & → C3H7Cl}
\end{align}
Is there are such software? I'm studying chemistry as a hobby and want to check if I'm doing the reactions correct.

I'm sorry, but this question is of the "make-a-list" type and thus needs to be closed. Please read the [faq] before posting :)

Other questions like this one were not closed and seem to be helpful, see: http://chemistry.stackexchange.com/q/24/1208 Why not re-open this question?

No, there is no such thing as chemical reaction are not that easily predictable. You can get result right in some cases, but it is impossible to count all factors.


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Are all the phenyl protons in benzyl alcohol equivalent in the ¹H-NMR spectrum?
Below is a $\ce{^1H}$ Nuclear Magnetic Resonance (NMR) spectrum for benzyl alcohol. The hydrogen in the alcohol group gives rise to a unique signal. The two hydrogens in $\ce{CH2}$ give rise to a unique signal with around twice the signal area. What I don't understand is why all hydrogens in the phenyl group are equivalent. My guess would have been there would be 3 signals from the phenyl ring. One pair of $\ce{H}$ that are equally distant from the methanol, another pair that is also equally distant, and last, a single hydrogen that is the furthest away from the methanol. Clearly this is not the case, and the 5 hydrogen give rise to a single signal only.
Why? What is a good way to think about equivalent and non-equivalent atoms in NMR spectrometry? I would rather avoid rules like the one that says "if you can substitute the hydrogen with another atom and it gives the same compound, the hydrogens are equivalent". I would rather like an intuitive way to think about it, if it's possible!

You're right, those five hydrogens are not magnetically equivalent. You would expect to see three signals for the aromatic protons.
If you look closer you'll see that the aromatic signal is not a perfect singlet, but is rather more complicated with an additional small peak and a broader base, it is a multiplet. You have three distinct signals, but they are so close in chemical shift that they overlap and result in this multiplet.

The three signals are very close in chemical shift, and have a strong coupling. If you have heard about $AB$ spin systems, where the CS difference is similar to the coupling constant, which make the outer two lines of the two doublets become weaker ("roof effect"), and the inner two stronger and move closer to one another: This is an $AB_2C_2$ system. The inner lines coalesce completely, and the outer lines become an unshapely foot around the central line.
If you had a stronger magnet, where the CS difference (in Hz) is proportionally larger, you would get multiple peaks and might even be able to integrate and identify them.

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Why does attractive forces of nucleus increase more than shielding across groups?
When you study the electronegativity of the elements, the general trend is that it rises with increasing group number, and decreasing period.
Supposedly this is because the attractive forces of the nucleus increase quicker than the shielding effect of the electrons.

Is this shielding effect simply a repellant force, or is there another phenomenon at play?
Why does the shielding effect not increase as rapidly?
Why does the difference between the attractive forces and the shielding effect become less pronounced with increasing period?

(source: wikispaces.com)

Question (1)
To at least a very good approximation, the shielding effect is merely due to a repulsive force.  Like much of chemistry at the atomic level, electrons experience a balance of counteracting forces due to the nucleus and other electrons.
Question (2)
This is usually explained in the following way.  As one goes across a period from left to right, the charge on the nucleus increases at each step, increasing the attractive potential for electrons (which you already noted).  In addition, an electron is added to the valence shell for each step.  Since the added electrons are in the same shell as the preexisting valence electrons, they do not contribute as strongly to shielding as one might initially expect.  Electrons that are in lower shells contribute to shielding more effectively since they are "more likely" to be between a valence shell electron and the nucleus.
Question (3)
One way to justify the increasing effectiveness of shielding in later periods is an extension of the answer above to question (2).  There it was stated that valence shell electrons do not effectively shield other valence shell electrons since they are at the same level.  An natural extension of this would be that the further apart two electronic shells are, the more the lower shell screens the higher shell.  Compared to the valence electrons in Pb, the 1s electrons are essentially right up against the nucleus and, from the view of Pb's 6p electrons, should almost completely neutralize the charge of two protons in the nucleus (i.e. very effective shielding).
Another way of looking at this is that as the principle quantum number (n) increases, the orbitals become less penetrating.  This means they spend less time near the nucleus (where the shielding is weak) and more time further from the nucleus (where shielding is strong).


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Reaction mechanism for oxidation of primary alcohol to carboxylic acid
Can someone help me identify the steps for the following reaction? I'd like to keep track of everything that happens. I can't figure out what happens with oxidation in this context, and thus can't get started.
It's 2-methylpropan-1-ol oxidized with potassium dichromate(VI) in aqueous sulfuric acid to yield 2-methylpropanoic acid (also known as isobutyric acid).

A dichromate salt and aqueous sulfuric acid are used to prepare chromic acid in situ, which is among the most powerful oxidizing agents commonly available. A primary alcohol is first oxidized to an aldehyde (via a chromate ester intermediate). Under acidic conditions, the aldehyde's oxygen acts as a base and abstracts a proton, the carbonyl carbon is attacked by water (whose oxygen acts as a nucleophile) and simultaneously the carbonyl π-bond breaks, and then finally the resulting molecule is deprotonated by another water molecule, yielding a diol. That diol is then subject to further oxidation by the remaining chromic acid, which occurs by the same mechanism as the initial oxidation to an aldehyde. The ultimate result is a carboxylic acid.


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Why does carbocation stability increase in the order 1°, 2°, 3°?
I'm reading about $\:\mathrm{S_N1}$ and $\:\mathrm{S_N2}$ reaction mechanisms. 1° carbocations are unstable to the point of not having been observed in solution, ever. 2° are more stable, and 3° carbocations are the most stable.
I know that higher stability carbocations do also require less activation energy in their formation. Why is that? Why exactly are they more stable? And is this another way of saying they are also less reactive? Does this in turn mean they require a better nucleophile compared to less stable carbocations?
A small side question. How should I think about resonance structures? You don't view it as changing between the forms, spending more time being the more stable form, but rather as a seamless blend of the forms, with more characteristics of the more stable form, right?

It may be better if you post your question about resonance in a separate post. Short answer: yep, it's a blend. More specifically a quantum superposition, but no need to get into that. Basically, the electrons of all the resonating bonds are "shared" over the given area. The probability of finding an electron is higher in the places where stability is more, but that doesn't mean that the electron is zipping back and forth in the area. You need to understand a bit of quantum mechanics to understand this.

One word: hyperconjugation! The more carbons bound to the carbocation, the more bonds can take part in hyperconjugation. This effect happens when a bond between two atoms donates electron density to the electron-deficient carbocation, making it more stable.

Why doesn't this work with hydrogen, considering hydrogen is less electronegative already to begin with?

@Brian: It is not the carbons attached to the carbocation center that donates electron density via hyperconjugation but the bonds attached to it (free electron pairs would also do). If you consider the $\cẹ{(CH_3)_3C^{+}}$ cation the carbocation center contains an empty p orbital. This can interact with vicinal $\ce{C-H}$ $\sigma$ bonds, that happen to be parallel to it, thus lowering its energy.

From what I have read, the increasing stability of carbocations is attributed to the combined inductive and hyperconjugative effects, at least in case of alkanes. In other organic molecules, the mesomeric effect can also come into play. These effects are easy to understand, so i suggest you read up on them.
Subsequently, your next question deals with why more stable carbocations have a lower activation energy in their formation. In my opinion, you're getting the cause-effect mixed up. The carbocation is stable BECAUSE the energy required to form it is lower, not the other way around. And why is the energy lower? Because of the aforementioned stabilization effects!
This is from the point of view of a 17 year old kid who's been studying organic chemistry for about 3 years, and faced the same problems as you're facing. I won't be able to explain it as ManishEarth did though, using quantum mechanics.


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How can a group be both a good nucleophile and a good leaving group?
I'm reading about $\:\mathrm{S_N1}$ and $\:\mathrm{S_N2}$ reaction mechanisms, and I have a few questions.
My book has a couple of tables, one lists a bunch of substances grouped as good, moderate and poor nucleophiles. Among the good are $\ce{Br-}$ and $\ce{I-}$. So a couple of halogens. I'm figuring chlorine is in that group too.
Then it has a table showing ability to function as leaving group. Iodine, bromine and chlorine anions are at the top of that list too.
My question is: How can these be quite attracted to the positive charge AND quite easy to remove at the same time?

 I'm not entirely sure of this, comments appreciated
Leaving group tendency is a thermodynamic variable. So, it depends more on the equilibrium and energies of the reaction. We calculate it from the basicity of the group (weak base -> more stable while solvated -> better leaving group).
Now, iodide is a weaker base than $\ce{OH-}$, since its conjugate acid $\ce{HI}$ is a stronger acid than $\ce{H2O}$. So, iodine is a better leaving group.
On the other hand, nucleophilic tendency is basically "how willing to sacrifice a lone pair is the group?".  It's more dependent on forces, and less on energy, as it is a kinetic phenomenon. The outer shell electrons in $\ce{I-}$ are more loosely bonded to it than those in oxygen in $\ce{OH-}$. So, it is a better nucleophile.
Another way to look at it would be via the "accessibility" of the group in solution. This basically attempts to explain the difference between a nucleophile and a leaving group. In a leaving group, you have to "give up" the ion to the solution. This means that solvation doesn't really matter here -- once you've given it up, it's gone. The process will have a similar rate for different ions. However, when looking at the reverse process, you need to check how solvated the ion is. If the ion is surrounded by water molecules (like $\ce{F-}$, it will be much harder for the electrophile to "extract" it. On the other hand, an ion like $\ce{I-}$ is less solvated, and thus more "accessible".

To further contrast with the "leaving group tendency", nucleophilicity is a kinetic phenomenon.

@BenNorris: ? That's already there, see the third paragraph :)

Doesnt the size of the nucleophile play a vital role?  Because iodide is huge wouldnt it have more of a basic nature than nucleophlic nature.

Why is a weak base more stable when solvated? Shouldn't it be the other way round?


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What chemicals are going to be used to make the smoke black or white?
For the papal election, white and black smoke will be used to signal success or failure in the voting process.
From what I understand, the ballots are burnt and chemicals are added to create the color.
But what chemicals are used to get black or white smoke?

I think Titanium Tetrachloride http://en.wikipedia.org/wiki/Titanium_tetrachloride would be a good option for white smoke (I've heard it's used quite often in smoke bombs).

This link could also be useful: http://thehappyscientist.com/science-experiment/black-and-white-smoke

@kaliaden, I read that link before posting, and it was an interesting read. But wouldn't Titanium tetrachloride be ruled out as poisonous?

It is quite dangerous because of the corrosive fumes released. But the link mentions that they have experimented with lots of (not especially safe) stuff before, so I wouldn't rule it out completely :)

I think Titanium Tetrachloride would be a good option for white smoke (I've heard it's used quite often in smoke bombs). But it releases quite a lot of $HCl$ gas and so it might not be very safe.
A quick Google search comes up with a lot of answers to your question though. This site even answers your question in the same context (The elections going on at the Vatican):
http://thehappyscientist.com/science-experiment/black-and-white-smoke
Apparently you can make white smoke using Magnesium powder as well. (The white colour is due to fine particles of Magnesium Oxide $MgO$ )
It seems the Vatican has experimented using many sources before (Military flares, Smoke bombs and so on) though apparently they didn't turn out too well...
EDIT
The Wikipedia article on Coloured smoke gives another possibility. You can add Titanium Dioxide ($TiO_2$) to a mixture of Potassium Chlorate ($KClO_3$) and Lactose for white smoke. $KClO_3$ is a powerful oxidizing agent which reacts with Lactose (which is a reducing sugar) to create an explosion. $TiO_2$ is safe and is used as a pigment in most white paints and as a food colouring as well.
EDIT2
The official recipes have been revealed by the Vatican press office. See this story on the New York Times website.
The previous edit was partially correct. They use a mixture of $KClO_3$, lactose and pine rosin for white smoke. For black smoke, they use Potassium perchlorate ($KClO_4$), anthracene (an aromatic compound made of three fused benzene rings) and sulfur.
The process (quoted from the site):
"The chemicals are electrically ignited in a special stove first used for the conclave of 2005, the statement said. The stove sits in the Sistine Chapel next to an older stove in which the ballots are burned; the colored smoke and the smoke from the ballots mix and travel up a long copper flue to the chapel roof, where the smoke is visible from St. Peter’s Square. A resistance wire is used to preheat the flue so it draws properly, and the flue has a fan as a backup to ensure that no smoke enters the chapel."

I just saw that NYT article, and was about to post that information as an answer. I'm glad to see someone beat me to it.

I suppose I was just in time.. I never thought we'd see the white smoke recipe in action so soon. :) We now have a new Pope (who already has a Wikipedia page within half an hour after being announced!) http://en.wikipedia.org/wiki/Pope_Francis


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How to rationalise the resonance structures and hybridisation of the nitrogen in a conjugated amine?
I was given the first structure, and then drew the other 5 resonance structures:

First of all, are they correct? ChemBioDraw had some complaints, but as far as I can see there's the same number of electrons, and no valence orbitals exceeding capacity.
My reasoning is that the first structure contributes more. There's no electron deficiency on any of them, so we can disregard that. There is charge separation on the last 5 though, and in combination with this, the negative charge is on the least electronegative atoms, thus making the last 5 structures noticably higher in energy, and thus they contribute less to the true electronic image of the molecule.
Hope I'm correct so far; here's my question:
In the first structure, nitrogen is sp³ hybridised, but on all others it's clearly sp² hybridised. So what does that mean? It's somewhere in between, but closer to sp³-hybridised?

First of all, are they correct? ChemBioDraw had some complaints, but as far as I can see there's the same amount of electrons, and no valence orbitals exceeding capacity.

Yes, these are the six most important resonance structures for this compound. The reason ChemDraw complains is that it is trying to act smarter than you, and it most certainly is not. It interprets that negative formal charge on the carbon atom as implying a lone pair, since carbon can only have the negative charge if it also has a lone pair. When you add the lone pair and the charge, ChemDraw is suddenly stupid and thinks you have exceeded the octet on carbon (3 bonds + 1 explicit LP + 1 implied LP from the charge).

In the first structure, nitrogen is $\mathrm{sp^3}$ hybridised, but on all others it's clearly $\mathrm{sp^2}$ hybridised. So what does that mean? It's somewhere in between, but closer to $\mathrm{sp^3}$-hybridised?

If the lone pair on a nitrogen (or on any atom)) participates in resonance, then that nitrogen (or whatever) atom must be $\mathrm{sp^2}$-hybridized so that the the lone pair is in a $\mathrm{p}$-like orbital to ensure appropriate symmetry for $\pi$-overlap. Put another way, if an atom is $\mathrm{sp^2}$-hybridized in one resonance structure, then it is $\mathrm{sp^2}$-hybridized in all of them. Atoms that are $\mathrm{sp^2}$-hybridized and $\mathrm{sp^3}$-hybridized have differing geometries, which is not permitted in the resonance phenomenon.
In truth, hybridization is an approximation we make to make quantum mechanics jive with molecular geometry. Geometry is real (empirically determinable) and hybridization is fictitious (not empirically determinable), but hybridization makes QM behave better conceptually and mathematically. Hybridization is also a useful predictor of chemical reactivity of various bonds and functional groups in organic chemistry. Experimentally, I would guess that the nitrogen atom is trigonal planar (or very close to it). Trigonal planar implies $\mathrm{sp^2}$-hybridized (not the other way around).
In molecular orbital theory, we do away with the need for both resonance and hybridization. This molecule would have 7 $\pi$ orbitals formed from linear combinations of the p orbitals on the 6 participating carbon atoms and the nitrogen atom. The probability density function plots of these 7 orbitals will be more complex than you might be used to, but they will suggest the same electron density and charge density as a resonance hybrid assembled from your six resonance structures. See the Wikipedia article on conjugated systems, which is not very good but will give you the idea.

My reasoning is that the first structure contributes more.

The five charge-separated resonance structures are more important than you think, since they imply that the five-membered ring is aromatic like the surprisingly stable cyclopentadienyl carbanion. However, you might not yet be this far along in your studies of organic chemistry. At the introductory level of understanding of resonance in organic molecules, the first structure is most important for the reasons you list.

Thanks for the comprehensive and lucid answer. Is M.O. more descriptive? Why use resonance and hybridisation at all?

Hybridization and resonance represent a "quick and dirty" or "back of the envelope" method that you could do anywhere. Generating good MO diagrams and representations for molecules even as complex as yours requires calculus, linear algebra, and a computer with appropriate software. The answer generated generated by hybridization/resonance is almost always the same as the MO theory answer, so choose the quick and easy option first. Use the more difficult theory only if the first approximation is not good enough.

@BenNorris If an atom participates in resonance why its hybridization must be $\mathrm{sp^2}$? I mean shouldn't be the weighted sum of the hybridizations in each resonance structure?

"Trigonal planar implies sp2-hybridized (not the other way around)"- did you mean that we see this trigonal planar atom, and we rationalize it by calling it sp2? And not that "sp2" (or any hybridisation for that matter) is an intrinsic property of atoms?

@HarryHolmes - Orbitals are mathematics constructs defining regions of space with high electron probability density. Geometry is an empirically measured property.

So ... That's a yes, right?

@HarryHolmes - That's a "yes."

I wanted to add a perspective to this since this can be a contentious issue, and I would still say knowing whether a nitrogen is $sp^2$ or $sp^3$ hybridized is not straightforward. I was originally hoping to add this under this question specifically about aniline, but that question is now closed and marked as duplicate, with a link to this question.
In a simple molecule like aniline, one might think that the hydrogens of the amine would be planar with the ring. However, we know from crystal structures of aniline that it is indeed pyramidal. If you have access to the Cambridge Structural Database (CSD), you can see this in structure BAZGOY. The improper dihedral angle of the hydrogens is ~26 degrees in that structure.
I was also curious to see what QM had to say about this. I performed some optimizations of analine with the improper dihedral angle of the hydrogens constrained at various angles. Here is a drawing of the improper dihedral I constrained.

I used the MP2 method and cc-pVTZ(-F) basis set, in implicit solvent (PBF) using the Jaguar software from Schrodinger. You can see in this plot and gif that the lowest energy angle was at ~35 degrees. That's pretty much completely pyramidal (in ethane, an equivalent angle is 33.4).

However, that being said, I also wouldn't say this is dogmatic. This is an implicit solvent calculation, and in real water with real hydrogens and oxygens to hydrogen bond to, things may be still different. NMR might be able to help figure this out, or it could further confuse if the timescale of flipping smears the signal to give a flat angle when it is not.
I should also point out that there are other crystal structures that contradict this. If you look at the crystal structure VOMFOS, which is of an adenosine analog, the amine hydrogens are planar with the ring. Is that just an artifact of the crystallization packing network, or a true representation of the hybridization of that nitrogen? Hard to say.
I think my main point here is to be careful what you assume. If you asked me on a test in undergrad whether the amine on aniline is $sp^2$ or $sp^3$, I would have said $sp^2$ with confidence. Experimental and computational evidence suggests that might not be the case.


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Why is this atom sp² hybridised?
I'm doing exercises on hybridisation, and I was given this molecule:

I'm wondering about this (electron deficient) oxygen atom. My intuition says it should be ${sp^2}$ like the answer says, but honestly I only see that it's $sp$ hybridised. Is it because you ask what it would look like if it was not electron deficient?

I might add, that since oxygen has a lone-pair I somehow missed, that means it is of course not electron deficient afterall.

This would become much clearer if you show the lone pair on the oxygen atom.
One $p$ orbital of the Oxygen atom forms the $\pi$ bond with Carbon, while the other three (remaining one $s$ and two $p$) orbitals are $sp^2$ hybridized.
Two $sp^2$ orbitals are used to form the $\sigma$ bonds while the last $sp^2$ orbital is occupied by the lone pair.

Note: I always like to stress that the s and p orbitals don't actually combine in any way to form the $n$ $sp^n$ hybrid orbitals (This is a pretty common mistake). Hybridization is just an approximation used in quantum mechanics to explain the geometry of the compound better.

Ah of course. Somehow missed the lone pair.

You can find the hybridization of the atom by finding its steric number:
steric number = no of atoms bonded (to the atom you are finding the hyb. of) + lone pairs with that atom.
if Steric no comes to be 4, there is $\ce{sp^3}$ hybridization in the atom.
if steric no comes to be 3, there is $\ce{sp^2}$ hybridization in the atom.
if steric no comes to be 2, there is $\ce{sp}$ hybridization in the atom.

and this certainly is not true for any period different than 2nd. Try to use it on PH3 and you will end with sp3, however the free electron pair is s-type orbital.

Well, it helps in basic studies.

+10 to kaliaden's comment.  The "reason" is because linear molecules are $\ce{sp}$, trigonal-like molecules are $\ce{sp^2}$, and tetrahedral are $\ce{sp^3}$.  Hybridization rules are completely bogus, and basically you know the answer by first observing the molecule experimentally, and post-hoc hammering it into these resonance structures.
This is nothing but a bunch of memorization. It is interesting to note that if you made $\ce{O}$ into $\ce{S}$, then it _may_not_ have the same shape.  (Actually, it probably does, but where it is different, we sound off the d-orbital participation mumbo-jumbo)


text:
Will borosilicate glass absorb 808nm laser light?
Will borosilicate glass absorb 808nm laser light? If it instead refracts, then what is the refractive index? Also, will borosilicate reflect lower frequencies of light like those used in non contact temp. measurements?

The amount of reflection depends on the polarisation and angle of incidence, look up Brewster angles and Fresnel equations. Refractive index depends on wavelength, usually we use Sellmeier equations to preduc=ict this

Borosilicate glasses are a family that has many members with differing properties. You can look up the provider for your glass, and check whether it is transparent in the near infrared. I think it would be in most cases (see here for one example).

SO you have to get that data on a per manufacturer basis.

@kurtnelle yes!

Unless doped, glasses are effectively 100% transparent in visible and near ir but not usually below approx 300nm in uv.  The amount of reflection depends on the polarisation and angle of incidence, look up Brewster angles and Fresnel equations. Refractive index depends on wavelength, (greater at shorter wavelengths in visible region) usually we use the Sellmeier equations to predict this if a value at exactly the wavelength you want is not available.


text:
Mol file of Morphine
I am using the Morphine mol file and I'm attempting to create a matrix which contains a table of the atoms and how they bond similar to a ball and stick format.
In the mol file bond block listed bellow what do the type place holders mean?
  7  1  1  0  0  0  0
  1  2  1  0  0  0  0
  2 12  1  0  0  0  0
  1  3  2  0  0  0  0
  3 14  1  0  0  0  0
  2  4  2  0  0  0  0
  4 18  1  0  0  0  0
  3  5  1  0  0  0  0
  4  6  1  0  0  0  0
  5  6  2  0  0  0  0
  7  8  1  0  0  0  0
  7  9  1  0  0  0  0
  7 22  1  1  0  0  0
  8 10  1  0  0  0  0
  8 11  1  0  0  0  0
  8 20  1  1  0  0  0
  9 12  1  6  0  0  0
  9 13  1  0  0  0  0
 10 15  1  1  0  0  0
 10 14  1  0  0  0  0
 11 16  2  0  0  0  0
 13 17  1  6  0  0  0
 13 16  1  0  0  0  0
 15 19  1  0  0  0  0
 21 15  1  0  0  0  0
 22 21  1  0  0  0  0

This is the format I'm trying to create but not all the atoms have the full amount of bonds. What is wrong?
C   7   2   3
C   1   12  4   12
C   1   14  5   5
C   2   18  6   18
C   3   6   3   6
C   4   5   4   5
C   1   8   9   22
C   7   10  11  20
C   7   12  13  12
C   8   15  14  15
C   8   16      16
O   2   9   2   9
C   9   17  16  16
C   3   10      10
N   10  19  21
C   11  13  11  13
O   13
O   4   4
C   15
H   8
C   15  22  22  22
C   7   21  21  21

The bond block has the number of the atoms followed by the number of bonds mol files support up to three bonds. Not all the atoms are supposed to have all their bonds filled.

Ah, so you had a problem and solved it yourself? That's great! (also, welcome to Chemistry.SE). It would be nice if you elaborated your answer though.. :)

The self-accepted answer given here is wrong.
Each line in a bond block of a MOL file represents one bond between two atoms.
A line in a bond block has the following compact format:
111222tttsssxxxrrrccc
With other words, each entry (=different number or letter) in such a line can take up to three spaces. If it needs less. e.g. when the atom number is smaller than 100, the data is right-aligned.
111 is the number of the first atom
222 is the number of the second atom
These atom numbers refer to the position of the atom in the atom block in the same file, where the cartesian coordinates and the atom type (element) are given.
ttt is the number decoding the bond type: 1=single, 2=double, 3=triple, etc.
sss describes the stereochemistry, 1 is a single bond pointing up, 6 is a single bond pointing down, etc.
The latter are used in the flat 2D MOL file of morphine cited in the question, e.g. for the only hydrogen atom in the file: a bridgehead H-atom.


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Products for the hydrogenation of the cycloalkene in (3R,4R,5S)-4,5-diamino-3-(pentan-3-yloxy)cyclohex-1-ene-1-carboxylic acid

Write the products (more than one) for the hydrogenation of the cycloalkene ring in (3R,4R,5S)-4,5-diamino-3-(pentan-3-yloxy)cyclohex-1-ene-1-carboxylic acid. Reagents are molecular hydrogen with Pt catalyst.

My problem here is that the question implies there are several products for hydrogenation of the cycloalkene. I only see one product, where each end of the alkene obtains a hydrogen, and the pi bond is gone. How can a hydrogenation of this alkene result in more than one product?

Hydrogenation of the alkene introduces a stereo centre, thus the products are:


text:
synthesizable functional groups
I was told in this question: https://chemistry.stackexchange.com/questions/4789/synthesizability-and-energy-minimization-of-a-structure/4791#4791 to
[Count the] "number of accessible functional groups (alkenes, amines, alcohols, esters, etc) which make it easier to prepare, 5) number of inaccessible functional groups (aromatic rings, polycyclic systems, quaternary carbons, complex or long alkyl groups),"
But when I googled accessible functional groups and inaccessible functional groups no pages had a longer list. I found a SMARTS file which lists some common functional groups. Could someone help me determine which are readily synthesizable?
Thank you

What you are asking amounts to the collected knowledge, wisdom, and intuition of the organic chemistry community. Your best bet will be to find an organic chemist, preferably one with a graduate degree, to work with you on this project. Said organic chemist can apply his/her knowledge/wisdom/intuition to each functional group in less time than it would take you to look up the answer. Doing so frees you to work on the algorithm you describe in your other question.

The beginning of the answer to this question takes me two semesters to get through with my introductory organic chemistry students, and that is just for the common functional groups. After covering the basics of organic structure and reactivity, most texts have 15-20 chapters organized by functional group containing the reactions of and syntheses of said functional groups.
The common functional groups covered in most introductory sequences:

alkanes
cycloalkanes
alkyl halides
alkenes
conjugated dienes and other unsaturated systems
alkynes
aromatic ring systems
alcohols
glycols and other diols
phenols
ethers
epoxides
aldehydes
ketones
carboxylic acids
esters
anhydrides
acyl halides
amides
amines
nitriles
thiols
sulfides
acetals
imines
enamines
heterocycles
nitro compounds
diazo compounds
carbohydrates
amino acids
nucleic acids

If you want to know about every functional group, you will need an advanced, probably graduate level, synthesis course. However, even I have not heard of some of the designations on your list, like semithiocarbazide.

Some good ideas on where to start, if you choose not to consult with an organic chemist.

Check an undergraduate level organic chemistry book out of your local library and get some sense of the subject. In your case, it does not look like you need to know how the reactions work, just that they exist, so you can use the text as a reference source.
Sign up for a free MOOC in organic chemistry. Again, you just want to know that the reactions exist, so you can pick and choose the content from the course.
Acquire a copy of one of the seminal books in organic chemistry, March's Advanced Organic Chemistry. This book is not for the fainthearted but it gives good treatment to the functional groups, including the less common ones.
Find (do not buy - these books are not worth the price tag) a copy of either Comprehensive Organic Transformations: A Guide to Functional Group Preparation by Richard Larock (which is now 14 years old) or the set of Comprehensive Organic Functional Group Transformations, which appears to be many volumes (now 9-10 years old).


text:
Why does molten plastic (PLA) stick to metals (AL)?
I know this is broad so I will narrow:
Why does molten PLA stick to Aluminum?
I ask because I'm looking for a way to prevent/reduce the friction of the plastic inside the aluminum barrel of a 3D Printer. I believe that I can treat the surface of the aluminum with a High Temp enamel however I need to be sure of the physics of why the plastic sticks in the first place. Is it the surface finish, or is the molecular attractiveness? If it is the finish, then I can make it smooth. If it is static attraction then I can charge the barrel etc.

Aluminum reacts with oxygen in the air to produce aluminum oxide, which forms a fine coating on any exposed aluminum surface protecting against further oxidation.
$$\ce{ 4Al + 3O2 -> 2Al2O3}$$
Aluminum oxide is polar, amphoteric, and hygroscopic: it absorbs moisture from the atmosphere. The surface of aluminum oxide is likely coated with $\ce{Al-OH}$ groups. The polar aluminum oxide and aluminum hydroxide surface is attracted to the polar ester functional groups in PLA.
Aluminum surfaces can be rendered nonpolar by silanization. This article may be behind a paywall for you. I apologize. I could not find a good open access reference. The Wikipedia article on silanization is pretty terrible. The surface is rendered nonpolar, which may decrease the interaction with PLA.

So its the oxide layer that is attracting the plastic. Therefore if the oxide layer is removed with vinegar and a polyurethane layer is added immediately afterward, the plastic will not stick as well as it did before?

I don't know if polyurethane sticks to bare aluminum. It might stick to aluminum oxide better. Most polyurethanes are also polar, which means PLA might stick to polyurethane as well.

I did a material test, so far the PLA didn't stick to the polyurethane coated aluminum.

Another common cause why your plastic gets stuck in your hot-end is simple thermal expansion. Try to up the cooling on your hot end to reduce the length of the temp gradient in your metal-barrel hot-end.


text:
How hot can Thermal Grease get?
Thermal Grease is used to remove the air gaps between CPU and heatsink. However, computing aside, what would determine the max temp that the grease can handle before it burns or breaks down?

There are different kinds of thermal grease, and the temparature they can handle will vary.

Your best bet is to contact the manufacturer or consult the labeling information for your brand.

Since most thermal greases are ceramic and/or metal suspended in a silicone or organic oil, the temperature the grease remains stable at will likely primarily be determined by the properties of that oil -- either its ease of evaporation, or decomposition temperature (or, I suppose, its flashpoint, but that seems unlikely).
As Ben Norris says in the comment above, your best bet is to contact the manufacturer or check their data sheets: for common CPU uses, most manufacturers will give you stable and extreme temperature ranges which are likely to be fairly conservative (for example, here), and they may be able to help with more esoteric data if you're using it for other purposes.

I tried burning thermal grease; it catches fire pretty easily.

@kurtnelle - That's to be expected; both silicone and other organic oils are organic compounds; get them hot enough, like exposing them to a naked flame, and they'll combust just like any hydrocarbon. Most fuel gas flames burn at between 1000-1500 degrees, with some of the crazier ones like acetylene going up to 2500. That's well above the flash point of just about any organic substance, as well as a few metals.

However, without true combustion as an ignition source, very few things on the surface of the Earth can sustain those kinds of temperatures through other means. You could likely put a dollop of silicone thermal grease on a cookie sheet and stick it in the oven at 450*F for an hour and it wouldn't burn (it would likely melt). The circuitry that grease protects by conducting heat away from it would melt down at a third that temperature.


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

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

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

I guess he used fluoroantimonic acid. :)

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

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

Related: article from chemistryworld.com.

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

"asking for a friend"

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


text:
Butane Burning Color
Why does Butane burn blue? From my (very) basic understanding of chemistry, carbon and hydrogen do not burn blue, their radiation spectra are different-so why should the combination burn blue?
Please elaborate as much as possible, about bonds, orbitals, electrons, and the like!

The blue flame from butane (or many other hydrocarbons) is due to the complete combustion of the gas.  From the Wikipedia article Flame color it mentions,

In the most common type of flame, hydrocarbon flames, the most important factor determining color is oxygen supply and the extent of fuel-oxygen pre-mixing, which determines the rate of combustion and thus the temperature and reaction paths, thereby producing different color hues.

Specifically, making the link to the common tool in the laboratory, a Bunsen Burner, which typically has two 'types of flame (as seen below):

Image Source - which states that:

The Bunsen Burners we use at school use a mixture of alkane gases like propane and butane

The yellow flame is from an incomplete combustion, producing $\ce{CO}$ and the colour is due to (according to the Wikipedia article):

incandescence of very fine soot particles that are produced in the flame.

With more oxygen, a complete combustion of butane is possible (without the soot):
$$\ce{2C4H10 +13O2 -> 8CO2 + 10H2O}$$
Which provides

enough energy to excite and ionize gas molecules in the flame.

and the blue flame colour is due to the

emission of excited molecular radicals in the flame, which emit most of their light well below ~565 nanometers in the blue and green regions of the visible spectrum.

The blue color is the Swan lines' emission of $\ce{C2}$.  Rather than acetylene diradical it is more likely to be the olefin singlet dicarbene,


text:
Why won't water freeze if you put ice in it, despite ice being frozen?
Just now I got some room temperature water and added some ice to cool it, and it just came to me, like the scientist I wish I was ... why doesn't frozen (freezing point) ice added to higher temperature liquid compound force the water molecules to solidify?

Why won't water freeze if you put ice in it?
It will, even at room temperature. You just need a big enough, cold enough ice cube.
Don't believe it? Add a few drops of water to an ice cube in an ice cube tray (which is the same as adding an ice cube to a few drop of water). Wait a few seconds, turn the tray upside down. No water will fall, presumably because it has frozen. Repeat, but immediately turn the tray upside down. You'll get a wet floor. You can conclude that, given enough time, the water will be frozen by the ice. To the whys and hows:
From the second law of thermodynamics we know that:

An isolated system, if not already in its state of thermodynamic equilibrium, spontaneously evolves towards it.

And

Heat cannot spontaneously flow from a colder location to a hotter location.

So if you put ice and water together in a cup, heat will flow from the water to the ice until they reach thermal equilibrium, which will be a temperature between the water's initial temperature and the ice's initial temperature. If the temperature of the ice was low enough or if there was enough ice for the water to keep giving away heat, then the water will freeze. That's considering our cup is an isolated system.
Why don't you usually see that happening? Well, regular ice cubes are probably around −5 °C, and there usually is more water than ice in a cup, so before the water has a chance to reach 0 °C all the ice has already melted. Besides that, the ice is receiving heat from the room, unless you're in a place where's below 0 °C. In that case, the water WILL eventually freeze, but mostly because of the environment.

Water, like most substances, requires more energy gain or loss to undergo a state change (from solid to liquid, liquid to gas) than to be warmed or cooled by a degree of temperature without changing state. Liquid water requires about 4.2 Joules per milliliter per degree Celsius to raise or lower its temperature. However, the Latent Heat of Fusion (Lf) of water (the energy that must be added to ice to melt it or removed from water to solidify it at the freezing point) is 334 Joules per milliliter.
Therefore, for ice to freeze the surrounding water, the ice must be cold enough that it can afford to gain enough energy to invoke this state change, without melting itself. Now, solid ice has a lower volumetric heat capacity; it only takes about 1.9 J/(mL * K) to change its temperature. So, for an ice cube, let's say 10mL in volume, to freeze 1mL of the surrounding water (at 0°C) without melting itself, the ice must be able to absorb 334J of energy without melting. That means the ice must be -16.7°C, or about 2°F.
That's well within the range of your average freezer (which in the US is normally kept around 0°F to ensure everything stays good and frozen), but that back-of-the-envelope calculation assumes the energy is gained evenly throughout the entire ice cube. This doesn't happen; it's common knowledge that ice melts from the outside in, and that's because it's a relatively poor conductor of heat compared to substances like metals, which sometimes seem to instantly liquefy when they reach their melting point. For the ice to be able to cool that kind of water volume without melting, therefore, it must be much colder; cold enough that the ice can afford to gain enough heat just in a few millimeters of its outer surface layer to freeze the water.
Regardless of this, if you have a fridge with in-door ice and water, and you fill a cup with the ice and then pour the water over it, the ice will clump; the combination of two cubes in close proximity, a small volume of already-cold liquid water in a thin layer between them, and the ice being colder than freezing, results in the ice freezing the water, joining the two cubes. This happens throughout the ice. If you go with crushed ice, which has smaller pieces but more surface area and a larger volume of frozen water to total water, the effect is even more pronounced. But, eventually, the overall thermal energy of the room adds more energy into the glass of water than the ice can absorb without melting. So, it melts.

The latent heat of vaporization (now normally referred to as the enthalpy of vaporization) and the latent heat of fusion (enthalpy of fusion) are different. The former is boiling, the latter is melting. For water, the enthalpy of vaporization (boiling) is 40.7 kJ/mol (2300 J/mL) whereas the enthalpy of fusion (melting) is 6.01 kJ/mol (334 J/mL) - but aside from the "typo", good points.

@R.M. Thanks for the comment. The edited answer fixes the "typo".

At room temperature, water is a liquid. So the ice will gradually turn into liquid.
You may ask your question in a different way:  if you add water to ice, why doesn't the ice turn into liquid?  It does, at room temperature.
If you add water to ice, at freezing temperature (Antarctica), the water turns to ice.
I think it is mostly due to the surrounding air temperature.

Well, yes and no. Ice will freeze cold water in a glass sitting on a table, as the colder ice and warmer water exchange energy to reach a rough equilibrium. But, the overall equilibrium of the larger system, at room temperature, eventually adds more energy than the ice can absorb without melting, so it does.

The question did not specify what the ambient temperature is (it said room temperature water, not system at room temperature). It might be best to first look at the question for an insulated system (no heat transfer), and then perhaps say what happens if you allow heat transfer.

Now if we have a look on values of sp heats of ice and water.
L(ice)=334 kJoule per kg
C(water)=4200 Joule per kg per degree Celsius
C(ice)=2100 Joule per kg per degree celsius
If we want 1 kg of water at 1 degree celsius to first reach zero degree Celsius and then be converted to ice the amount of heat released will be greater than 4200 joules
On the other hand,  If we want 1 kg of ice at minus one degree celsius to first reach zero degree Celsius and then be converted to water the amount of heat required will be around 2100 joules
Therefore in general if we want water to be converted to ice the ratio of ice to water should be more than 2 isto 1.
(Assuming temp of both are at same distance from 0degree celsius)


text:
Why can't you see air but can feel it
Is there a device that can measure the air in the room in lbs?
I would appreciate the anseer

A scale.  Consider an experiment where a container of "air" is submerged into liquid helium (just about the coldest stuff known.)  This will turn the air into solid $\ce{N2}$ and solid $\ce{O2}$ which can then be weighed.
If you knew the area of the room, then you would know the "pounds per square inch" or "p.s.i."
But the above answer, I think, is a little cheeky.  I think it is usually done via the ideal gas law:  $ P V = n R T$, and solving for $n$, the number of moles of gas. Then through Avogadro's number, one can deduce the number of molecules.  The molecules (e.g. $\ce{N2}$) have a known molecular weight, which can be expressed in pounds.
A barometer, a thermometer, and a ruler are the only three devices one would need to get the number of pounds of air.

Thanks for taking the time to answer.  Please, though, consider flagging such "nonsense" (or nearly so) questions for closure.


text:
Chemical methods to extract Oxygen from Rust
Let's say I have a large quantity of rust and want to extract the oxygen from it in a way that allows me to collect the oxygen in a container.
What chemical process can I use to do this and what equipment does it require?
It's preferable if the process is doable at home so I can demonstrate the process but it's not mandatory.

Good question; unfortunately, no good answers. The problem is that oxygen, pure oxygen, doesn't like to exist as such; it will form a diatom like many gases, but it's very electronegative, and so it will prefer to get the extra electrons from something more willing to donate them, like the iron. This is why "oxidation" has that name; until fluorine was isolated in the mid-1800s, oxygen was the strongest known electron acceptor ("oxidizer"). It's easier to run a reaction that gives the oxygen something else to oxidize, producing the raw metal, than to get the iron to complex with something else and release the oxygen. In at least one circumstance, this is very good for humans and other aerobic organisms; oxygen complexes with the iron in your hemoglobin for transport to your cells. In most other circumstances, we could do without this behavior.
Rust is insoluble in water but will dissolve in strong acids. It will react with hydrochloric acid to form iron chloride and water (sometimes producing hydrogen gas if there isn't enough oxygen to go around). The easiest "household" method to obtain oxygen from iron oxide, therefore, is to dissolve the rust in muriatic acid forming iron chloride and water, neutralize any remaining HCl by adding lye to form sodium (and/or potassium) chloride and water, distill the water from the chloride salts by boiling and then cooling the vapor, then electrolyze the water with a battery to produce oxygen and hydrogen. That's cheating, though. First, water's water, and you're adding far more to the party with the acid and base solutions than you would ever get from the rust. Second, muriatic acid and lye, while available in relatively pure form for pool maintenance and for soapmaking respectively, are not exactly everyday household chemicals in those forms (both are components of other household chemicals, such as drain cleaner and lime scale remover, but are somewhat difficult to isolate from them).
Theoretically, if you heat the rust enough, the components will dissociate and you'll be left with pure iron and oxygen. Unfortunately, even with carbon in the mix (producing elemental iron metal and carbon dioxide, which is the reaction behind iron ore refinement and recycling), the minimum temperature required for this reaction to become favorable is 842K = 569*C ~= 1056.2*F. I can't find data on iron-oxygen bond enthalpy or entropy, which is needed to calculate the spontaneous decomposition temperature of the compound, but logic states it's much higher than with added carbon, because you're not getting any energy release by forming the carbon-oxygen bond, so every Joule of energy needed to break the iron-oxygen bonds has to come from outside the material. Theoretically it would cost you about 2.5kWh/kg of heat added to the iron; in practice the amount of heat you'd have to generate (and lose to various inefficiencies) to get the metal that hot is beyond any household device's capabilities, unless your hobby happens to be glassblowing.
The "thermite reaction" between iron oxides and aluminum under heat is another well-known method of separating the iron and oxygen. The rust and aluminum powder, given some initiating heat, will exchange the oxygen to form alumina (Al2O3) and pure iron, and enough heat to keep the reaction going (with plenty to spare; a thermite reaction is too bright to watch without a welder's mask and in fact is used in some situations to weld iron). Two problems with this; first, it's illegal in the US to make thermite without BATFE approval, and second, the reason it works at all is because aluminum oxide is ridiculously stable, more so than iron oxide, and so it takes even more energy to reduce the oxide back down to the raw metal (about 8.2 kWh/kg for aluminum, theoretical, versus 2.25kWh/kg for iron). This is evidenced by the highly exothermic nature of the thermite reaction itself; if creating aluminum oxide produces far more than enough heat to smelt the iron to get the oxygen, think of how much would be needed to go the other way.
Pretty much the only direct route to releasing oxygen gas from rust without a blast furnace is to introduce an even more powerful oxidizer. There are very few known to modern science, and none of them are "household chemicals" to say the least. The easiest to work with would be elemental fluorine gas, and the experienced chemists in the crowd will know that for the extremely relative statement that it is. Fluorine is the most electronegative element in the known universe, and so pretty much any substance that isn't already highly fluorinated (like some polymers, such as Teflon) will react pretty vigorously with it. That includes the water, proteins, enzymes and other compounds in your own body; a fluorine gas leak is a "drop everything and run" type of emergency. As used on rust, the fluorine will simply barge in and displace the oxygen from the rust to form a white to pale green iron fluoride, liberating the oxygen as a gas. You then have to separate the oxygen from the excess fluorine, and one of the standard ways in industry to catch excess fluorine is to pass the exhaust gas over or through a powdered sodium oxide filter, which will form sodium fluoride and liberate more oxygen in a very similar reaction as you got with the iron. Sodium oxide isn't a pleasant chemical to work with either; add water (such as by breathing in any loose powder) and it forms the strong base sodium hydroxide, which eats flesh about as efficiently as anything else you could think of.

You can also go via the same process used for aluminium: electrolytic smelting, either directly (probably), or using a solvent. Not exactly doable at home, though!

A long read but a fantastic answer, seems it wouldn't be too easy to do even given the right industrial equipment!

@Aesin - Electrolytic smelting is pretty much the same basic  iron-for-carbon redox as in a blast furnace, but using electrical arc plasma to generate the necessary heat in a more localized fashion. And no, not exactly the type of thing you could do at home unless you leveraged the main 200A trunk line into a monster DC converter.

Two variations on the acid dissolution:
a. Dissolve the rust ($Fe_2O_3$) in 100% sulfuric acid, producing ferric sulfate with some enneahydrate $Fe_2(SO_4)_3.9H_2O$. When the reaction is done, evaporate the $H_2O$ and electrolyze it.
b. Use 100% phosphoric acid, get $FePO_4.2H_2O$. When the reaction is done, evaporate the water and electrolyze it.
And a third (pure speculation):
c. This way probably won't work, but  organometallics do all sorts of crazy stuff. Ferrocene can be made by heating iron with cyclopentadiene at 300 C (https://en.wikipedia.org/wiki/Ferrocene). I think it might also work with iron oxide, perhaps even better (or not).
$Fe_2O_3$ + 6 $C_5H_6$ --> 2 $FeCp_2$ + 3 $H_2O$ + $(C_5H_5)_2$
or something like that. Collect the water and electrolyze it to get the oxygen. Quite doable in a lab - if it works.


text:
Calculating bond disassociation energy and bond energy for ATP
I'd like to calculate Bond Dissociation Energy and Bond Energy for ATP to a similar molecule. What free (at least for academic use) computer programs which natively support and document BDE?
If you don't mind, could you provide a tutorial for using those programs to perform the calculations?

I think your other question goes into this and more, so I'm closing this one for the time being.

An example program is GAMESS.  It is free. But be warned: the calculation of BDEs is an extremely difficult thing to do.  I am presuming that you are a beginner, so you have to be careful for issues surrounding which BDE:
\begin{align}
\ce{A-B &-> A+ + B-}   \tag{ionic}\\
\ce{A-B &-> A. + .B}  \tag{homolytic}
\end{align}
Most electronic structure theory programs assume gas-phase, absence of solvent.  We know that homolytic dissociation is favored in the gas phase, as charge separation into ions is unfavorable in a Coulombic energy.
That being said, the run-of-the-mill electronic structure theory calculations (e.g., HF, DFT, MP2, CCSD, etc. etc.) will not break the atoms apart in a homolytic manner.  They tend to break apart atoms in the ionic fashion.  But this is a problem, as ions are abhorrent in the gas phase!  Basically, simple electronic structure theory methods over-estimate the bond strength, as the product of the dissociation (ionic) is erroneous.
You must use multiconfigurational methods to get it right (e.g. CASSCF, MCSCF). But these are very, very difficult unless you have had a lot of training.  I did them for years, and they were still hard for me.  But they will give you $\ce{A. + .B}$ as a product!  That is the gas-phase BDE.
You may be able to study closed-shell, closed-shell dissociations without much problem, using the HF and DFT methods.  Examples include dimers, host-guest interactions where the products look like good-old fashioned molecules that can stand by themselves. (Think water dimers, etc.)
Also, beware of Solvent Effects (favoring ionic dissociation $\ce{A-B -> A+ + B-}$) and solvation energy.  In fact, the solvation energy is why $\ce{Na-Cl}$ breaks apart into $\ce{Na+}$ and $\ce{Cl-}$.  In the gas phase, it would be $\ce{Na.}$  and $\ce{Cl.}$ radicals.
In summary, BDEs are very difficult things to compute to qualitative accuracy. Quantitative accuracy is even harder.  You may want to have an expert look over your procedure and results before you trust them completely.

Ah yes, bond breaking with electronic structure theory... what a massive headache.


text:
Regioselectivity in Kharasch addition of bromotrichloromethane
When hex-1-ene is treated with bromotrichloromethane in the presence of a peroxide initiator (e.g. dibenzoyl peroxide), what is the regioselectivity of the addition?
I know that the radical addition of HBr to an alkene is considered anti-Markovnikov, but is this reaction considered a Markovnikov or anti-Markovnikov addition?

This looks like a typical Kharasch (chain) reaction. The reaction is initiated by the formation of peroxide radicals (by homolytic cleavage of the weak O–O bond). These peroxide radicals then go on to abstract Br• from the starting material:

Note that the alternative generation of Br• is unfeasible, as radicals tend to attack at less hindered sites: hence the phenyl radical attacks the bromine which only has one attached atom, and not the carbon which has four flanking halogens. The weaker C–Br bond is also preferentially broken over the other three C–Cl bonds, so there is no chance of generating a CCl2Br• radical.
The trichloromethyl radical then adds to C-1 of the alkene, resulting in the more stable secondary radical. This can now abstract a bromine atom from unreacted bromotrichloromethane to give the product and a new Cl3C• radical.

Unfortunately, with additions of X–Y across a C=C double bond (as opposed to additions of H–X), the designations Markovnikov and anti-Markovnikov lose their meaning. A Markovnikov addition of H–X would be one where the hydrogen adds to the less substituted carbon. However, here it is not clear whether to define "Markovnikov" as the regioselectivity where X adds to the less substituted carbon or where Y adds to the less substituted carbon.


text:
Why does heat solidify cake batter?
Increased vibration causes the atoms in an object to spread out farther away. Thus, with added heat, and object goes form a solid to a liquid to a gas (and then to plasma, with enough heat). Why is it then, that when a heat something like cake batter, it solidifies?

It is caused by physical and chemical reactions in the cake batter of two types:
1) When heated and mixed with water, starches present in the flour will undergo a process called gelatinization in which water is absorbed in the structure. This will cause the batter to 'set' from a liquid into a solid.
2) Additionally, reactions of the proteins from the egg that you added can also work to 'set' the batter. At first the proteins will unfold (denature) after which they start to form bigger proteins through cross-linking reactions that turn the liquid into a solid.
The baking soda in the batter turns into sodium carbonate, water and carbon dioxide. The latter will make your cake nice and fluffy.

Mostly wrong! 1st there are numerous doughs which do not contain eggs, nevertheless they set on heating. For that the swelling of starch with water and heat is resposible. But both processes are physics, not chemistry!

@Georg Thanks for the remark. I was indeed still wondering how it was possible for cake batter without eggs to work out and starch gelatinization is indeed the answer. I'll edit it in

@Georg wrong, the swelling of starch doesn't set the dough. It thickens it, but it doesn't make it "solid" - cook a pudding to see what happens. The eggless doughs set by denaturing of the gluten network, which arises when two wheat proteins bind in the present of water. As for cake, both the egg and the gluten are at work, and depending on the recipe, the egg may prevail.


text:
Why can't rust form without water?
Shouldn't iron oxide be able to form without water? It is just iron and oxygen. I don't really understand what the dot followed by the $\ce{H2O}$ means either. I was reading on wikipedia, but I have a rather terrible understanding of chemistry.
What purpose does water serve in rust formation, and what does it mean for a molecule to be hydrated-how is that different then just being surrounded by water molecules?

We could explain it from an electrolytic point of view.
Iron has a certain tendency to dissolve in water according to the following equation:
$\ce{Fe (s) -> Fe^2+ (aq) + 2e^-}$
The release of electrons causes a small current to flow in the metal (Remember that metals conduct electricity). This turns the point where iron dissolves into an anode, and the region around this area becomes rich in electrons, a cathode. Now, we have to remember that water dissociates to small extents according to:
$\ce{H2O (l) <=> H^+ (aq) + OH^- (aq)}$
If it was an acid, the $\ce{H^+}$ would have quickly taken all the electrons liberated from the dissolution of iron earlier, but the concentration of hydrogen ions is not large enough in water, so that we get another reaction taking place at the cathode [The hydrogen ions actually 'gather up' near to the surface but cannot do anything, creating a thin protective layer of hydrogen ions around the cathode, but not strong enough to prevent further reaction]. This reaction involves oxygen and other water molecules:
$\ce{2H2O (l) + O2 (g) + 4e^- -> 4OH^- (aq)}$
Now this is the $\ce{OH^-}$ that reacts with the earlier liberated iron (or ferrous) ions:
$\ce{Fe^2+ (aq) + 2OH^- (aq)->Fe(OH)2 (aq)}$
Now more oxygen will react to oxidise the iron (II) hydroxide...
$\ce{Fe(OH)2 (aq) \xrightarrow[O] Fe(OH)3 (aq) \xrightarrow[O] Fe2O3 (s)}$
You probably can recognise the last product.
As the $\ce{Fe^2+}$ are consumed, more iron will get dissolved (Le Chatelier's Principle) and keep the whole process going.
Maybe a little picture to go along with this :)

Note that water is not required if the temperature of the iron and oxygen are sufficiently high. I do not know the precise temperature, but from experience with welding I can say that noticeable amounts of orange-colored iron oxide begin to form as steel approaches its melting point, which is around 1370 degrees C (2500°F). This high-temperature oxidation in air, without water, is similar to what we usually call charring or burning.

@RalphDratman That sounds accurate. Heat basically prompts iron to react with oxygen, first giving FeO, and with more oxygen in the fray, Fe2O3 (rust). We could say that's the actual way that rust is formed, and water acts as catalyst (makes the same reaction possible with lower energy (heat) requirement.

The dot means that the compound is a hydrate of the parent chemical; the parent compound forms a weakly-bonded structure with one or more water molecules per parent molecule, because either the hydrogen atoms of the water, or the partial positively-charged region of the water molecule caused by its "bent" molecular shape, is attracted to partially negative regions of the parent molecule. It's not a "full" traditional ionic bond but it works in somewhat of a similar way. In solid form, this mixture of the parent compound and water molecules forms a crystal structure like ice (it doesn't have to be a well-ordered crystal like you typically get with salts), with properties that differ from a form of the substance that has more, less or no water in its structure.
An everyday example of the difference between a dehydrated and hydrated compound is Portland cement, used in concrete. In its raw form, it's "dehydrated" calcium carbonate along with a few other components, made by heating limestone until the water is released from the rock's structure. The remaining solid forms a fine powder that can't hold much of a shape. Add the water back in while mixing concrete, and the water molecules are re-incorporated into a solid structure with the calcium carbonate to essentially reform a rock in the desired shape.
As far as why rust only forms in the presence of water, water's "polar" nature is at work here as well. Being a polar solvent, it has some affinity for electrons, which are easily given up by the iron as a transition metal, and are attracted to the hydrogen atoms in the water. If a hydrogen atom successfully "captures" an electron, it balances its own electric charge, and is "liberated" from the bond it has with the oxygen in the water molecule, instead pairing with another liberated hydrogen to form a diatomic gas. Now, this hydrogen gas molecule won't get far, because water usually has some "dissolved" oxygen gas in it, and there's more oxygen just beyond the water, waiting to re-oxidize this hydrogen gas back into water. The net result is a positively-charged iron atom (typically in the +2 or +3 oxidation states, having given up 2 or 3 electrons to the water, respectively), negatively-charged hydroxide ions ($\ce{OH-}$), and water. These form a mixture of iron hydroxides:
$$\ce{2(Fe - 2e^{-}) + 4(H_2O + e^+) + O_2 \\ \to 2Fe^{2+} + 4OH^- + 2H_2 + O_2 \\ \to 2Fe(OH)_2 + 2H_2O}\\$$
$$\ce{4(Fe - 3e^{-}) + 12(H_2O + e^+) + 3O_2 \\ \to 4Fe^{3+} + 12OH^- + 6H_2 + 3O_2 \\ \to 4Fe(OH)_3 + 6H_2O}$$
That second equation typically occurs "stepwise", as you'll notice that's quite a lot of electrons and ions floating around at once. Typically, iron(II) hydroxide is formed first, and the iron will then readily oxidize further to its +3 state, either taking on an additional hydroxide to form iron(III) hydroxide, or liberating a hydrogen that will oxidize to form water, resulting in the monohydrate of iron(III) oxide-hydroxide:
$$\ce{4(Fe(OH)2 - e^{-}) + 2H_2O + O_2 \to 4FeO(OH)*H_2O}$$
Iron(III) hydroxide will readily rearrange itself into much the same state (it's not so much decomposition, as the water stays weakly bonded to the oxide-hydroxide to form the monohydrate):
$$\ce{Fe(OH)3 \to FeO(OH)*H_2O}$$
Finally, two of these iron oxide-hydroxide molecules are generally considered to "share" an oxygen from one of the hydroxide ions between them, with the hydrogen instead bonding to the oxygen of the other hydroxide, reforming water as a hydrate (possibly with some of it being released):
$$\ce{2FeO(OH)*H_2O \to Fe_2O_3*xH_2O + yH_2O}$$
Various hydrate structures of this iron oxide exist, which result in different colors of the compound from dark brown to deep red to red-orange. Other oxides are also possible by more direct action of the oxygen on the iron itself, such as iron(II) oxide ($\ce{FeO}$). Heating rust releases the water and forces the iron oxides to rearrange, which can also produce the third oxide of iron, iron (II,III) oxide (known in mineral form as magnetite).
You'll notice that the water participates in the overall reaction without really being consumed by it to form the product; it's destroyed and then re-formed in equal amounts, creating intermediates in the process. As such, water is a catalyst for these reactions. You'll also hopefully notice that this "ideal" reaction chain is just one of the possibilities; "rust" is an ill-defined term in chemistry, with the everyday compound consisting of a mixture of various hydrates of the three iron oxides, all of which are produced depending on subtle localized differences in the temperature and ratios of water, oxygen and iron.

the initial reaction in water liberates H2 gas. Ferrous ion, hydroxide, oxide is further oxidized by dissolved or interface elemental oxygen to give mixed iron oxides and eventually hydrated ferric oxide. Water prevents the formation of an adherent protective oxide layer.

By definition, rust includes water in its structure. Its not rust if it doesn't contain water.
Rust is porous- it has holes so corrosion of iron keeps going a lot longer if it rusts rather it forms a straight oxide.
Iron forms oxides instantly which are more protective than rust- not a lot more. Hot iron in oxygen environments corrodes by forming iron oxide.

Water is the catalyst in the process of rusting, therefore you cannot get rust without water.
Taken from http://www.chemicalformula.org/rust:

The overall chemical equation for the formation of rust is
Iron + water + oxygen --> rust
$$\ce{4 Fe(s) + 6 H2O(l) + 3 O2(g) -> 4 Fe(OH)3(s)}$$
Iron(III) hydroxide, $\ce{Fe(OH)3}$ then dehydrates to produce $\ce{Fe2O3.nH2O(s)}$
  or rust

That doesn't help my understanding much-All you did was introduce a new word.

Let me put it this way - the reaction that causes rust requires iron, oxygen, and water.

The OP already knew that; the question was why.

What purpose does water serve in rust formation?"
First, one should ask what is the temperature of the water and the source of the water? Cold from a faucet implying possible oxygen content along with say free $\ce{Cl2}$ or even $\ce{NH2Cl}$, or natural aerated waters containing transition metal salts (including those of $\ce{Fe, Mn}$, some $\ce{Cu}$, …), along with possibly gases other than $\ce{O2}$ like $\ce{N2O}$ from nitrate (present in well water) decomposition, or perhaps boiled distilled water without any dissolved gases or minerals. The composition of the iron alloy, I will ignore and assume pure $\ce{Fe}$, along with a pH range for the $\ce{H2O}$ of between 6 to 8.
I will attempt a brief depiction of the chemistry which could actually occur in a few scenarios to give a closer perspective of real world chemistry. For natural waters, as a starter, the chemistry described below is certainly relevant. However, with the distilled water in open air contact with submerged powdered iron, I will expect a comparatively much longer induction period till any rust is observed per the reactions outlined above. This is due to the underlying electrochemical nature of the corrosion process and no electrolyte (from dissolved salts, but instead awaiting dust particles) and no dissolved $\ce{O2}$ or an acid (source of $\ce{H+}$ could arrive via air containing $\ce{CO2}$). So iron rusting proceeds much faster in those cases containing dissolved oxygen, an acid source (free chlorine), and high mineral content (electrolyte).
There is also a possible direct action of water on iron via the following path without any oxygen presence:
$$\begin{align}\ce{2 [ H2O &<=>  H+ + OH- ]}\\[1.3em]
\ce{Fe + 2 OH- &-> Fe(OH)2 + 2 e-}\\[1.3em]
\ce{2 [ H+ + e- &<=> H^. ]}\\[1.3em]
\ce{H^. + H^. &-> H2}\end{align}$$
$$\begin{align}\text{Net:} &&\ce{Fe + 2 H2O  -> Fe(OH)2 + H2}&&\end{align}$$
This could then be followed by the so called Schikorr reaction, which details the conversion of iron(II) hydroxide into the mixed oxide iron(II,III), also written as Fe3O4 (see Wikipedia https://en.wikipedia.org/wiki/Schikorr_reaction ).
Interestingly, there is also a possible action on any formed ferrous in water containing $\ce{N2O}$ in the further presence of cupric (see Chemical Reduction of Nitrite and Nitrous Oxide by Ferrous Iron, by J. T. Moraghan and R. J. Buresh,in SSSAJ, Vol. 41 No. 1, p. 47–50, 1976.).
The reason for mentioning an acid source is due to the following reaction converting ferrous to ferric in the presence of oxygen:
$$\ce{4 Fe^{II} (aq) + 4 H+ + O2 -> 4 Fe^{III} + 2 H2O}$$
(see http://corrosion-doctors.org/Experiments/rust-chemistry.htm)
So, depending on various sources of the water, in addition to the mechanism with water and oxygen described above, the chemistry could be a bit more complex.

"Rust" can form without water, if by rust you mean an iron oxide. Fe2 O3 ( the red oxide) can form in air at ambient temperatures. The only example I know is fretting corrosion where the characteristic red powder forms around two rubbing steel/iron surfaces. Of course you can make all sorts of iron oxides at elevated temperatures.