text:
Why is hot concentrated sulfuric acid preferable to cold?
Why is hot concentrated sulfuric acid sometimes used to react with metals instead of concentrated sulfuric acid which is at room temperature?
I have ascertained that concentrated sulfuric acid has quite strong oxidizing properties.  Still, some metals which are below the hydrogen in electrochemical series do not react with concentrated sulfuric acid which is kept at room temperature.  When the acid is warm/hot, the reaction takes place.
All I want to know is how does temperature affects its oxidizing properties?

A really short answer is kinetics. Heating up chemicals adds more energy to the system, allowing molecules to colide with more force increasing the probably of a reaction taking place.

Firstly these metal are not non-reactive with concentrated sulfuric acid- rather they form protective layers whether sulfate or oxide which reduce the speed of the reaction (kinetics).
The effect of heating the sulfuric acid depends on the metal (e.g. steel is good for up to 25C 60-95% sulfuric acid) but there the trans-passive region where metals such as iron no longer have a protective layer. This is due to instability of the oxide/sulfate layer so it dissolves or falls off.

When we say that a metal falls below hydrogen in the electrochemical series, we typically refer to the oxidizing power of hydrogen ion at 1 molar concentration in water solution. The oxidizing power of concentrated sulfuric acid, forming sulfur dioxide and water when it reacts, could be greater than that of the aqueous hydrogen ions; if so, then thermodynamically some noble metals could react with concentrated sulfuric acid. Then it's a matter of  kinetics: getting the redox reaction to go reasonably fast and getting enough solubility of any passivating films that might form, which is where heating the acid comes in.


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Why are there free electrons in a metal?
Recently, we covered metallic bonding in chemistry, and frankly, I understood little.
I understand that:

Metals bond to each other via metallic bonding
Electricity can flow via free or delocalized electrons

But, I do not understand why the metal atoms turn into ions and delocalize the electrons, why don't the metal atoms stay as atoms?

When electricity flows, the electrons are considered "free" only because there are more electrons than there should be, and because the transition metals, such as iron, copper, lead, zinc, aluminum, gold etc. are willing to transiently accept and give up electrons from the d-orbitals of their valence shell.
Transition metals are defined in part by their stability in a wide range of "oxidation states"; that is, in several combinations of having too many or too few electrons compared to protons. This is thought to be because of the d orbital in their valence shells. Compared to the s and p orbitals at a particular energy level, electrons in the d shell are in a relatively high energy state, and by that token they have a relatively "loose" connection with their parent atom; it doesn't take much additional energy for these electrons to be ejected from one atom and go zooming through the material, usually to be captured by another atom in the material (though it is possible for the electron to leave the wire entirely). This impetus can be caused by many things, from mechanical impact to chemical reactions to electromagnetic radiation (aka light, though not all of it visible); antennas work to capture radio frequencies, because the light at those frequencies induces an electric current in the wire of the antenna. Now, in the absence of a continuous force keeping the electron in this higher energy state, the electron (and the metal atoms) will naturally settle into a state of equilibrium. Electricity is generated when just such a force is acting on the metal, giving energy to the electrons in the d orbital and forcing them to move in a certain direction.
This impetus can come from many sources, as discussed, be it the movement of a magnet within a coil of wire, or a chemical redox reaction in a battery creating a relative imbalance of electrons at each of two electrodes. The end result is that the electrons, given additional energy from this voltage source, are ejected from their "parent" atom and are captured by another. The "holes" left behind by these electrons are filled by other electrons coming in behind them from further back in the circuit. Thus, the energy provided by the voltage source is carried along the wire by the transfer of electrons.
The analogy typically made is to the flow of water, and it generally holds in many circumstances; the "voltage source" can be thought of as being like a pump or a reservoir, from which water flows through pipes, and the amount of water and the pressure it's placed under (by the pump or by gravity) can be harnessed to do work, before draining back to a lower reservoir. The pipes are similar to wires in many ways; the larger the diameter, and the smoother the inside of the pipe, the more and the faster water can flow through it (equivalent in many ways to the thickness and conductivity of the metal wire), and when under enough pressure (high enough voltage), the pipes will actually expand slightly and hold more water than they would at low pressure (this is a property of wires and other electrical conductors called "capacitance"; the ability to store a charge while under voltage and to discharge it after the voltage is released).

If it loses an electron, "usually to be captured by another atom in the material (though it is possible for the electron to leave the wire entirely)," where does it go?  Has it been "captured" by some other element we just don't know which one at that time?

Hard to say; it's difficult but not impossible for the electron to leave the Earth entirely and go zooming out into space. That would be just fine; the Sun bathes the Earth in bajillions of charged particles every second. Much more likely, our ejected electron will be captured by other materials within a rough line of sight of the atom from which it was ejected. That will affect the relative electron balance of that material alongside everything else, creating a static charge, but sooner or later the charges will equalize and the excess energy is released as a photon, likely heat.

KeithS's explanation works well with transition elements. But it does not explain why non-transition metals like aluminum or magnesium are good conductors.

This doesn't answer the question. It explains why electrons might flow but not why why metals contain "free" electrons which was the question.

Molecular orbital theory gives a good explanation of why metals have free electrons
The best way to explain why metals have "free" electrons requires a trek into the theory of how chemical bonds form. Molecular orbital theory, or, at least, a simple view of it (a full explanation requires some fairly heavy quantum stuff that won't add much to the basic picture) can explain the basic picture and also provide insight into why semiconductors behave the way they do and why insulators, well, insulate.
The first step in getting to a useful intuition involves picturing how small molecules form and how their bonds work. If you start from isolated atoms, the electrons form 'orbitals' of different shapes (this is basic quantum mechanics of electrons). There are plenty of pictures available describing what these look like. But, when atoms come together to form molecules, the simple view of what the clouds of electrons look like gets a lot more complex. If the two atoms form a molecule, they do so because the energy levels of the orbitals in the molecule are lower than those in the isolated atoms for some of the electrons. This is what causes chemical bonding.
The important insight from this picture of bonding is that molecular orbitals don't look like atomic orbitals. When a bond forms, some of the orbitals will fill up with electrons from the isolated atoms depending on the relative energy levels. There may also be other orbitals (some might, were there enough electrons to fill them, form anti-bonding orbitals, weakening the strength of the bond). In some molecules those orbitals might cover a number of atoms (archetypally, in benzene there is a bonding orbital that is shared by all the atoms in the six-membered ring occupied by two electrons and making benzene more stable than the hypothetical hexatriene with three isolated double bonds).
In some solids the picture gets a lot more complicated. In graphite, for example, the bonding orbitals are like benzene but might cover trillions of fused hexagons. What makes the solid hold together is those bonding orbitals but they may cover a very large number of atoms. In metals it is similar.
It is also worth noting that in small molecules you can often get a good idea of the shape of the discrete molecular orbitals, each containing two electrons, when you start dealing with large networks of atoms joined together, the simple, discrete, picture of individual two-electron orbitals becomes pretty useless as there are too many similar ones to make reasonable distinctions.
So solid state chemists and physicists start thinking of the picture as consisting of "bands" of orbitals (or of the energy levels of the orbitals). And this is where we can understand the reason why metals have "free" electrons. When metal atoms come together in a solid, the bonds between the atoms form lower energy orbitals than the isolated atoms. But the orbitals corresponding to the bonds merge into a band of close energies. And those orbitals might not be full of electrons. In metals these orbitals, in effect, form a bond that encompasses the whole crystal of the metal and the electrons can move around with very low barriers to movement because there is plenty of free space in the band.
In insulators, the orbitals bands making up the bonds are completely full and the next set of fillable orbitals are sufficiently higher in energy that electrons are not easily excited into them, so they can't flow around. In semiconductors the same happens, but the next set of orbital bands is close enough to the bands filled with electrons that thermal energy is enough to excite some of them into a fairly empty orbital where they can move around.
Wikipedia give a good picture of the energy levels in different types of solid: . The picture shows both the spread of energy levels in the orbital bands and how many electrons there are versus the available levels. In reality there is a continuum of band widths and gaps between insulators and metals depending on how the energy levels of all the bonding orbitals work out in a particular solid and how many electrons there are to fill them up.
In short, metals appear to have free electrons because the band of bonding orbitals formed when metals atoms come together is wide in energy and not full, making it easy for electrons to move around (in contrast to the band in insulators which is full and far away in energy to other orbitals where the electrons would be free to move).
This is, obviously, a very simple version of reality. But it links the easier theory or chemical bonding and molecular orbitals to the situation in network solids from insulators to metals.

Re: Why the metal atoms turn into ions and delocalize the electrons, why don't the metal atoms stay as atoms?
This is sort of asking why is water wet?
By definition if the atoms in an elemental sample have delocalized electrons (so that the sample will conduct electricity) then the element is a metal. If there are no delocalized electrons, then the sample won't conduct electricity  and the element is a nonmetal.

I'm more asking why Salt doesn't give up its electrons but steel does.
(I know Salt is an Ionic compound and behaves differently to a metal, it was just an example, but the point still stands)

The valence electrons in the outermost orbit of an atom, get excited on availability of energy. They overcome the binding force to become free and move anywhere within the boundaries of the solid. Thus they contribute to conduction

Though a bit different from what is asked, few things are worth noting:

Electrons barely move in metal wires carrying electricity.

Electrons do not carry energy, the electric and magnetic fields
around it (outside the wire) carry and transfers energy.

A great video to explain it:
https://www.youtube.com/watch?v=bHIhgxav9LY

Rather, the electron net velocity during flowing electrical current is very slow. Their random momentary thermal velocity, causing resistor thermal noise, is not so small.

If you want to comment rather than answering, I recommend you use a comment. I agree that the video is great.

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Tools to draw molecular structure (wireframe) along with atom numbering
I am looking for open-source tools which can generate 2d molecular structure from the MOL or SMILES file. Further, it SHOULD number the atoms in the order of occurrence of atoms in the connection table of MOL file or SMILE string. I used MarvinSketch 6.1.2 but the position of numbering is not very good, as it get cluttered with the bonds.
Please refer some tools with which I can draw molecule similar to the one shown in figure but numbering is also clear. In the figure, all numbering except 1-7 are ok.

To answer the question, I assumed you would like to stick with Marvin if possible. Likely, the options described and found functional in version 5.2.4 met at this site are functional at later versions, too.
So I drew from scratch a similar representation of your molecule.  And indeed, there is overlap by labels and the drawing.

Yet the software allows you to print the drawings.  So, thanks to the installation of HP-1102P printer and printing into a file I got a postscript file.  This vector format is of course beneficial, you may import it in inkscape, for example.  Note two potential pitfalls: select "import" (instead of "open") and "import text as text".  Twice you have to ungroup the drawing before being able to move the labels around as you prefer.  Or to change their colour (vide infra), or substitute them.  Eventually, you may either save it in numerous file formats or like in this case, export the resulting bitmap.

This seems to be one good work-around. However, I am not aware of image manipulation schemes. If you could provide some pointer to some tutorials on how to do this, it would be helpful. Thank you.

@DurgaDatta Just as an example and sticking to the same tools already mentioned, a short silent video was placed here.

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How can aluminium oxide be called an acid?
Aluminium oxide is amphoteric. It is easy to see that it is a Brønsted-Lowry base through the following reaction:
$$\ce{Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O}$$
The aluminum oxide splits and the oxygen accepts a proton, forming water.
But what about the reaction with a base? In my textbook, they say:
$$\ce{Al_2O_3 + NaOH \rightarrow 2NaAlO_2 + H_2O}$$
Now, the textbook claims that Aluminium oxide is an acid merely because it reacts with a base to form a salt and water, as is characteristic of a neutralization reaction.
But I'm not satisfied with this definition. I mean, acids aren't defined as 'things that neutralize bases', we have well-established definitions for them.
I tried to figure out for myself how this could be. Clearly, the Brønsted-Lowry theory cannot be applied here since the compound in question has no protons to donate. Therefore, the only alternative is the Lewis concept. I cannot see how that is applicable in this case.

The most basic definition of "acid" is that it is a proton donor (or one which accepts a lone pair)
All of this stuff is done in an aqueous medium, so we can assume that all aqueous ions and molecules are present. With this assumption (in this case, we are assuming that $\ce{OH-}$ is available to react), we get the following equation:
$$\ce{Al2O3 + OH- -> 2AlO2- + H+}$$
Similarly, we get:
$$\ce{Al2O3 + 6H+ -> 2Al^3+ + H2O}$$
where it is acting like a proton acceptor (base).

We could extend the acidic/basic nature of $\ce{Al2O3}$  to the Lewis acid/base theory. For more about this see this question which has been asked before: Is there such a thing as an acid without a hydrogen?
Recall the Lewis Theory of acids and bases which states that an acid is an electron pair acceptor and a base is an electron pair donor.
Applying the theory to  $\ce{Al2O3}$:
$$\ce{Al3+(aq) + 6 H2O(l) <=> Al(H2O)6^3+(aq)\tag{1}\label{a}}$$
Thus, in $(\ref{a})$ the  $\ce{Al(H2O)6^3+}$  ion is formed when an  $\ce{Al^3+}$ ion acting as a Lewis acid picks up six pairs of electrons from neighbouring water molecules acting as Lewis bases to give an acid-base complex, or complex ion
Similarly:
$$\ce{2NaOH(aq) + Al2O3 (s) + 3H2O (l) ->  2Na[Al(OH)4](aq)\tag{2}\label{b}}$$
In this case it  has donated its lone electron pairs to form sodium tetrahydroxoaluminate.
The Brønsted-Lowry theory is also observed in:
$$\ce{Al2O3 + 6H+ -> 2Al^3+ + H2O} \qquad{(3) - \text{a typical Brønsted-Lowry base}}$$
This compound somewhat behaves as both a Lewis acid/base and Brønsted-Lowry base (as noted in the question) as well.

This should be an accepted answer as it answer the question. In the reaction with NaOH, Al2O3 is an acid because it accepts electron pair, thus lewis acid.


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Boiling points of v. similar organic molecules
Given the similarity in formula between pentane ($\small\ce{CH3(CH2)3CH3}$, b.p. 36.2 °C) and neopentane ($\small\ce{C(CH3)4}$, b.p. 9.5 °C), why are their boiling points so different?

Welcome to Chemistry.SE. Would you be willing to share what level of explanation you are looking for (i.e. what level of chemistry you understand)? Do you want a High School level answer, a General Chemistry level answer, an answer using quantum mechanics, etc.? The more you put into a question like this the better answers you will get.

Solids and liquids are hold together by cohesion. Intermolecular interactions contributing to cohesion depend on the geometry of the molecules in interaction,too.  Here, n-pentane with an elongated molecular structure exhibits a much larger molecular surface as compared to neo-pentane of a more sphere-like structure*).  As a result, other criteria kept constant (like the molecular mass), the lesser the molecular surface area of molecules of a substance, the lower the boiling temperature of this substance.
*)True, neo-pentane's methyl groups point towards the vertices of a tetrahedron. Actually I recommend to build a molecular model of both compounds and place them side-by-side to see it.


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Why do people still use the mole (unit) in chemistry?
I know that the mole is widely used in chemistry instead of units of mass or volume as a convenient way to express amounts of reactants or of products of chemical reactions.
I'm wondering why people in chemistry still excessively use it for their measurement? To be backward-compatible and consistent with traditional textbooks? Why they don't simply express their quantity measurement per atom, per unit volume, per molecule or etc.?

It's similar to why people use words like "dozen", "score", "gross", "ream", etc.

Simply speaking, because it's an appropriate unit to use.
Let's imagine I wanted to measure the length of a rope. What would be an appropriate length to use? Inches? Centimeters? Feet, maybe? It would really be awkward to express it as 0.000189393 miles, or as 304,800,000 nanometers.
(Note: if you can't see why these units are awkward, take any page discussing things like this (e.g. biomass of certain species) and change all the units so that they're nonsense like this. Then put it away for a week and try to read it later.)
Now let's say I've changed my mind and I'd like to measure all the rope created in the world in a year. Now an appropriate unit is almost certainly miles or kilometers, and not inches or centimeters.
Let's consider something else: so far, I've been using length to measure ropes. Would it make sense to measure their combined mass instead? Maybe not for small amounts, since I think I would throttle you if you told me to cut off half a pound of rope, but for global-scale things, tons or metric tons may make sense.
On the other hand, using measurements like the average density of the rope or the combined diameters of all the ropes really wouldn't be much use at all.

What we've seen here is that when we're measuring things, there are measures that make sense (for rope, length, maybe weight) and some measures that really don't (color, average diameter, etc.). In those measures, there are units that are convenient (inches, feet), and some that aren't (nanometers).
This is exactly the problem with chemical units, but much more magnified. When I measure the energy released on hydrolysis of a sample, I'm not measuring the energy of one bond, or two bonds, or a thousand bonds, or even a million. I'm measuring a collection of so insanely many molecules that there's really no number in common language for it.
What is an appropriate way to measure things in these molecules? If I'm only interested in how much I have and not what's in it, per gram or per volume is often a good way to do it. Since when I'm measuring the temperature change of something, I don't particularly care what it's made of, measures of this (specific heat, for instance) are done in units of something per gram.
On the other hand, if I do care what my sample is made of, then I need a better way to measure it. For example, sodium chloride and calcium chloride look similar, but will have very different energies since there's three ions in a $\ce{CaCl2}$ unit and only two in an $\ce{NaCl}$ unit.
Since you can't have half a molecule, the simplest way to do it is to count the darn things. Unfortunately, measuring things per atom is a really awkward way to do things because of the problems I described above. The energy it takes to melt ice is 0.000000000000000000001029 J/molecule.
What we need then, is some count of molecules that's convenient. We could go by 1000 molecules, or a million molecules, but it's a pain in the butt to convert between that and macroscopic units (how much does 1 million calcium atoms weigh?) Now it just so happens that $6.022 \times 10^{23}$ atoms of carbon-12 weigh 12 grams, and a single carbon-12 weighs 12 amu. A single molecule of water weighs roughly 18 amu, $6.022 \times 10^{23}$ molecules of water weigh 18 grams.
Let's take a look at what using $6.022 \times 10^{23}$ molecules offers us over other units:

It is appropriate. $6.022 \times 10^{23}$ molecules will typically
fall in the gram to kilogram range of substance, which can easily be
measured out on a balance. It's also within the range of what you'd
usually use in a lab.
It allows for easy conversion. You can easily tell when you have
$6.022 \times 10^{23}$ molecules because your atomic mass is equal to
your macroscopic mass.
It makes sense. Intuitively, we'd like to measure things per atom or
per molecule, but doing so leads to some ugly units. Instead, if we
measure per $6.022 \times 10^{23}$ molecules, we still are doing a
counting-based measurement, but the numbers themselves are a lot
nicer.

You probably know this already, but we call $6.022 \times 10^{23}$ molecules a mole. These are the advantages of using the mole as a unit, and not for backwards-compatibility. I am of the opinion that if we shed all our units today and started from scratch, we would start off with grams and liters, but we would probably start using the measure of a mole again within a year. It's just a very powerful, useful way to measure things.
P.S. This is much longer than I was initially intending and was not written while I was in the most awake state of mind. Please let me know in the comments if you think this answer is unclear, or stupid, or just plain wrong.

I would summarize the above by saying that, when choosing a unit, we want one that is convienient, which typically means that you want its associated numerical values to be relatively close to 1 (ideally between 1 and 10, but if that's not possible then between .1 and 1000).  So 235 grams is fine, but at 2350 grams you might want to switch to 2.35 kg.  And .235 grams is fine, but at .0235 grams you might want to switch to 23.5 mg.  And obviously, as chipbuster explained, it's more unweildy to talk about  9.033 x 10^23 molecules than 1.5 moles.

Your statement that the "mole is widely used in chemistry instead of units of mass or volume.." is misleading.  There are seven base units of measurement in the SI system and your question alludes to two, mass and amount.  The other 5 are length, time, temperature, electrical current, and luminosity.
Consider going out for a typical (American) breakfast of bacon and eggs.  There are two diners that you like to visit; one serves small-sized eggs and bacon strips while the other uses jumbo eggs and offers canadian bacon.  At each place, you order "Two scrambled eggs with three slices of bacon."  But do you get the same amount of food at each diner?
A rash of bacon weights about 22 grams and a slice of canadian bacon is a bit closer to 50 g, so three slices of each weigh 66 and 150 grams, respectively.
A small chicken egg weighs about 43 grams while a jumbo egg weighs 71 g.  These weights are with the shell, and you visit good diners that give you scrambled eggs without the shell, but the point should be clear: while the amount of items ordered is the same, the weight of items is different.
Colloquially, we use the term amount when we are interested in answering a question beginning with "How much...".  The units of our response oftentimes depends on practical considerations about how we measure the substance.  From the deli, I'd order a pound of ham and two balls of mozzarella.  In this case using units of weights or amount could be valid (I'd like 12 slices of ham and 100 grams of mozzarella.)  If I'm waiting in line, however, I would say "there are 3 patrons before me" not "I have to wait for 425 pounds of people before I can be served."
Last a comment about the mole.  The International Bureau of Weights and Measures has a good definition of the mole that doesn't need to be repeated here.  To carry out the food analogy just one more step, the mole is the same type of word as a dozen.  When you go to the store to prepare for a large breakfast, do you think to yourself, "I'm going to buy twenty four eggs" or "I'm going to by 2 dozen eggs".  The word dozen, which stands for 12, is a way to simplify or group our numbers.  Along the same lines, it's much easier to say 12 grams of carbon contains 1 mol of carbon atoms instead of $6.022 \times 10^{23}$ atoms.
In summary The mole is used extensively in the sciences because we need a unit that describes an amount of substance, which is different from the mass of a substance or how much space a given amount of substance occupies.

Haha, one (1) is a good enough unit of amount I guess, but I'm no expert.

The mole is a really good bridge, in a manner of speaking, between MASS  vs.  NUMBER, when talking about things that are really really small !
Once you balance several dozen equations, especially for a practical lab use----like, to make something----you readily see that , in order to determine the proper masses for a reaction to go forward,  you need some kind of 'bridge' to convert a huge sum of microscopic energies [ molecule to molecule ] into a quick, readable format.
The mole does just that :  It gives you a way to quickly read the "energy ledger book" for reactions.  If  $\ce{C2H6  +  O2  ->  CO2  +  H2O}$,  you know that it takes  1 mole of C2H6  to do that reaction ,  and that your product will be 1 mole of $\ce{CO2}$  and 1 mole of $\ce{H2O}$.  IT ALSO MEANS  you can weigh out the molar mass of 1 mole of $\ce{C2H6}$,  which is ______  (you do the math, it's on the periodic table for you for each element---how convenient is that !),  and you will have assurance that you are using the right amt of reactant !
From this standpoint it does not take long to see that the mole is a pretty indispensable tool.

Consider this: you are comparing the effectiveness of acetic acid vs stearic acid (solid) as a drug. The former is a liquid and per molecule weighs about 5x less.
If you used 1 g of each, you would be adding different numbers of molecules. If you added 1 mL of each, same problem and compounded by differences in density. Since mole is a per molecule measurement, it allows you to make real comparisons between compounds. For example 1 mole of acetic acid will have the same number of molecules as 1 mole of stearic acid even though the weights, volumes, and phases might be different.
Why biology journals still report concentration as ppb or ppm is beyond me, because if there is a large difference in MW weight between compounds, you cant make comparisons.

It actually makes perfect sense to report some concentrations as ppm or ppb. Technically, it is only a very small molar fraction — but nonetheless measureable. And not only biologists do it, chemistry journals do so to. (Remind me to link a JACS paper using it, as soon as I run across one again.) Of course, one should never confuse ppm for a weight per volume measurement; it is always molar fraction. Other than that, $+1$.


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Heat transfer dependence on temperature gradient?
Is the rate of heat transfer between two bodies brought into thermal contact dependent upon the different in temperature between them, assuming they each have the same mass?
Imagine two bodies of water connected to each other by an impermeable glass barrier. If the difference between the temperatures is 10 degrees, will the system reach its equilibrium state faster per degree than if the difference is one degree?

Yes, it is very much dependent on temperature gradient. The stefan-boltzman law is only for radiative dissipation of heat and not for general heat transfer by conduction(contact).
The equation for steady state heat flow is
$$\vec h=-k \vec{\nabla T}$$ where $\vec \nabla$ is the gradient operator which is based on the temperature difference between the bodies and $k$ is the thermal conductivity. $\vec h$ is the heat per unit area.
If the system is not in steady state, then the heat diffusion equation becomes relavent:-
$$\frac{dT}{dt}=\frac{k}{c}\nabla^2 T$$ where $c$ is the specific heat of the body.
Lastly, a to a crude approximation, the rate of cooling in atmosphere at low difference of temperature of atmosphere and the body is given by Newton's law of cooling (partly by radiation):-
$$\frac{dT}{dt}=-K(T_{body}-T_{surrounding})$$ where $K$ is a constant.
In all these equation, it is the temperature gradient between the bodies in contact, which drives the rate of heat transfer (or temperature change).

Yes,
I believe according to Stefan-Boltzmann's law the hotter something is, then the more energy it will radiate per surface area, causing more energy to be lost faster at higher temperatures.
http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law


text:
Is there a element that shows this pattern of boiling point?
Is there a compound that turns to gas at a certain temperature but if heated more turns back to liquid and if heated more than turn to gas?
Does it exist naturally? Has a such a substance be found? Can such a substance exist? if yes, then why?
Why couldn't at some temperature molecules interact in such a harmony that they actually cling together?

No. If you take a molecular system in the gas state, and you heat it up (at constant pressure), there is no reason for it to go into a denser, more organized state, such as liquid or solid. As you increase temperature, the interaction energy between molecules becomes smaller compared to thermal energy.

More a condensed matter physics problem. I could imagine some very strange conditions, maybe involving superfluid helium, adsorbed on something, with magnetic fields...
Definitely not something you can make at home.
So for ~100% of cases, I agree.

@ssavec the question asks for an element, hence the simple answer… for other cases, yeah, you could imagine a system frozen where temperature increase would “unlock” some vibration modes/conformations which would in turn alter the molecular interactions… but I think this would have to be extraordinary (i.e. I cannot think of any real example)

no i meant anything..


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What happens to the structure of tetracyclines when they expire?
I need to explain (chemically) why expired tetracyclines should not be used, relating to the changes in tetracycline structure that occur.
All I can find is that they cause nephrosis, but not why.

What is an “expired” molecule?!

Expired teracycline, if taken can cause hepatotoxicity or Fanconi syndrome. I am not sure of the molecular basis of this effects. see http://en.wikipedia.org/wiki/Fanconi_syndrome and http://en.wikipedia.org/wiki/Hepatotoxicity

All drugs have an expiry/beyond use date.  Apparently, there is some sort of structural change in the molecule at this point, as our prof has asked us to find out what that is.

Apparently the culprits are anhydro-tetracycline (ATC) and 4-epianhydrotetracycline (EATC) - see Determination of tetracycline and its major degradation products by liquid chromatography with fluorescence detection. J. Pharm. and Biomed. Anal. 18 (1998), 839-845 for structural information and references to toxicity of those byproducts as known at the time of publication.
It is worth noting that renal tubular acidosis (RTA) - the condition that was reported in 1963 to have been precipitated by ingestion of expired tetracycline - is not presently considered as much of an issue as it was at the time (see this review). Finally, this piece in the New York Times from 2014 (entitled Do some drugs become dangerous after expiration?) has more information and references concerning tetracycline and RTA.


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How do 1s and 2p orbitals overlap?
In the following figure
we can see that the p-orbitals overlap 1s orbital (though relatively very little). How can an electron in p-orbital, be simultaneously in the 1s orbital at any given point in space and time?

The question is interesting, but slightly ill-defined.
If you take s and p orbitals as a solution of Schrodinger equation of hydrogen-like atom, then they are eigenfunctions of Hamiltonian, therefore are orthogonal. That means - no overlap. Do not be misled by the figure, if you do the math properly, $\langle1s|2p\rangle = 0$.
If you speak of just one electron, the solution is reasonably simple as stated in the article and the electron does not simultaneously reside in both orbitals. Even if you would think of interaction with excited states, it would not help, as single excitations do not interact with ground state. You can of course excite it, but than it is not "simultaneous".
In many-electron system, the picture gets more complicated, but this was not a question.

It is important to know that the overlap must be performed with the wave functions $\psi_j(\vec{r})$, not the density, $\rho_j(\vec{r}) = \psi_j(\vec{r})^* \psi_j(\vec{r})$. Wave functions have signs; they oscillate, as these oscillations encode the momentum distribution $\rho_j(\vec{p})$.
When you calculate the overlap between $\psi_s(\vec{r})$ and $\psi_p(\vec{r})$, $\int_V \psi_s(\vec{r}) \psi_p(\vec{r}) dV$ ($V$ is the volume) you should remember that the lobes of the "p" wave function have different signs, while the spherical $s$ wave function keeps the same sign over the surface of the sphere. Therefore, the integral has two identical but opposite (different sign) contributions that cancel the overlap.

Remember that the orbitals are really just approximations showing the most likely places for an electron with a given set of quantum numbers to be. If you don't apply a cutoff (like less than a 1/1000000 chance), all orbitals would encompass the entire universe. We just make pictures like this by identifying the places that an electron could be within some margin of probability.


text:
why there is no simple rule to predict the possible ionic charges of element in middle of periodic table
I know element in left two columns have 1+,2+... and right ones 1-,2-.
But in middle is all over the place. And with multiple charges.
What defines that if this element is gonna have this charge. All i have been told is that elements want to be close to as much possible to noble gasses because they are stable. Which doesn't explain much as to why such behavior is present. What does it mean to be stable. And why elements in middle find it stable just after 3 +/- charge even though they are far from noble gases.
Thanks, there are few questions in here but all very related i think.

TL;DR - The upshot is that there just happens to be a simple approximation governing small atoms but as the atoms get bigger and more complex, those approximations become less good and there is not another simple rule that works.
In more depth, the issue is a combination of two factors: distance from the nucleus and full/half-full valence orbitals.
In the first place, valence (outermost) electrons that are further away from the nucleus are more weakly bound than valence electrons that are closer. This means that the energy required to add or remove an electron is less, and so the charge given by a given electron matters less to the stability of the atom. Because of this, the further down the periodic table you go, the less the "normal" charges matter. As an example, Germanium is two rows below Carbon and is naturally found to have charges of +4 and +2 (as well as sometimes having +1, +3 and -4!).
The second factor is that elements find a special stability when their valence orbitals are either full (preferable) or half-full (less preferable, but still better than neither). So an element like Cr, which has almost no hope of gaining or losing enough electrons to look like a noble gas, "wants" to gain lose the 4 3d electrons, getting it down to a full 4s orbital. The tricky bit is that unlike the p orbitals, not all of the d orbitals have the same energy, so there are some that are more likely to be removed than others. There just isn't a convenient rule of thumb about which is which (nature is governed by physics, not approximations).


text:
Difference between conformational, constitutional, and structural isomers and the same and different molecules
Consider the following sets of problems:

I was hoping someone could clarify what each checkbox means exactly. All of these answers are incorrect (except the first one with the check marks signifies the correct answers).
For example, why are the first pair different molecules? What makes something a conformation/constitution/structural isomer?
What does it mean to say something is an "identical conformation of the same molecule" vs "different conformation of the same molecule?"
When can I definitiviely tell if something is an enantiomer or just a diastereomer? It seems like (for an enantiomer) you can't just do the check of "switch all dashes to wedges and all wedges to dashes" because, as in the third pair of newman projections, they are enantiomers of each other but I don't understand why that is.

This isn't an answer as I can't summarize it, but this site is really useful for that: Isomers explained via metaphor with cats. http://www.masterorganicchemistry.com/2010/11/11/on-cats-part-1-conformations-and-configurations/

Which book is this ??

Can you mention the right answers to the questions in the picture?

I'm going to assume that you don't need an explanation for the checkboxes the same molecule and difference molecules, so that leaves the following:
The terms constitutional isomers and stereoisomers make up two broad categories of isomers (molecules with the same chemical formula).  Constitutional isomers typically have different connectivities and stereoisomers have the same connectivities but differ in spatial arrangements.
enantiomers are stereoisomers that are non-superimposeable mirror images of one another.  Your hands are (roughly) enantiomers.  both are optically active is true of enantiomers.
diastereomers also known as geometric isomers are also stereoisomers that are not mirror images of one another. cis- and trans-dichloroethene are diastereomers.
If you do a web image search of isomer flowchart you'll see a lot of informative information on isomers.  Linked below is the public domain version from Wikipedia which is an example (although not the best in my opinion.)


text:
How do I find the thermal effect of this reaction?
I have graphite turning into diamond.I need to find the thermal effect of this reaction.How would one write that?I'm not sure how to proceed here.The teacher told us to use oxygen?

Can you please elaborate on what do you mean by "thermal effect"?

Look at the difference between the enthalpies of formation. Not sure why you'd need oxygen, unless there's something I'm missing.


text:
Why is the nomenclature glucose 6-phosphate, not 6-phosphateglucose?
What exact nomenclature rule define the naming of glucose 6-phosphate?

Reading this closely, technically, the compound should be 6-phosphonoglucose but glucose 6- phosphate is acceptable usage, probably derived from considering it as a salt analogue rather the ester that it is.

Usually, many esters are named by functional class nomenclature. They are systematically named by placing the name of the hydroxylic component denoted by an organyl group (alkyl, aryl, etc.) as separate word in front of the name of the acid component denoted by the anion name derived from the appropriate acid, for example ‘ethyl acetate’. However, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) contains various detailed nomenclature rules for special cases.
In this case, the relevant rules are described in Subsection P-102.5.6.1.2 as follows:

Esters of sugars with phosphoric acid are generally termed ‘phosphates’. In biochemical usage, the term ‘phosphate’ indicates the phosphate residue regardless of the state of ionization or the counter ions present. However, systematically the names must differentiate between a true phosphate, $\ce{-O-PO(O- )2}$, and an acid phosphate, i.e., $\ce{-O-PO(OH)2}$, called a (dihydrogen phosphate). The prefixes ‘phosphono’, for $\ce{-PO(OH)2}$, and ‘phosphonato’, for $\ce{-PO(O- )2}$, are also used, to denote O-phosphonic acid derivatives.
(…)

Therefore, correct systematic names for the compound that contains the group $\ce{-O-PO(OH)2}$ are ‘ᴅ-glucopyranose 6-(dihydrogen phosphate)’ and ‘6-O-phosphono-ᴅ-glucopyranose’.

Correct systematic names for the corresponding compound that contains the group $\ce{-O-PO(O- )2}$ are ‘ᴅ-glucopyranose 6-phosphate’ and ‘6-O-phosphonato-ᴅ-glucopyranose’.


text:
Epimerization in peptide synthesis
How can epimerization during peptide synthesis be detected? And how can we change or modify the coupling conditions to stop epimerization occurring?

The technique of Circular dichroism (CD) is used to detect changes and the presence of chirality in peptides and can detect both helical chirality in polypeptide chains that form helices and in individual chiral centers.
Polarimetry. The optical activity of a polypeptide (oligopeptide) has a specific rotation. A change in the optical rotation of such a peptide. Chirality is a property of a molecule as a whole. Such a change will alter the optical rotation of the peptide.


text:
What is Polymer Environmental Resin (PER)?
I'm no chemist and I probably wouldn't understand any technical answer but...
I'm investigating the environmental impacts of various materials as compared to PVC and this is one that crops up regularly as a more environmental alternative.
Unfortunately there does not seem to be any authoritative information on the web concerning this material.
I'm beginning to think that this is a green-washing propaganda move by the phthalate/heavy metal free PVC industry as a move avoid mentioning PVC in product advertising.
Anyway I would welcome any information on what exactly PER is and what it is made from.
Thank you.

PER is essentially PVC which has been plasticized and stabilized with acetyl tributyl citrate, instead of the host of phthalate-based plasticizers such as bis-diethylhexyl phthalate (DEHP) whose endocrine-disrupting properties give PVC its well-deserved bad reputation.
Citrate salts tend to be readily water soluble, and therefore this material is touted to be more biodegradeable than phthalate-plasticized PVC. I certainly cannot find any information to refute that claim. ATBC will further degrade into other acetates and butyrates, most of which are relatively common and harmless in the environment (though butyl compounds are infamously stinky; butyric acid is a favorite less-lethal offensive-odor weapon of law enforcement and groups like the Sea Shepherds). All PVC will eventually degrade as the plasticizer leaches out over time, unlike other types of plastics that are so ridiculously inert they'll likely be the fossils of our modern era, but again, the phthalates just aren't a great idea to have in the environment.
Of course, neither is the vinyl chloride monomer. What many proponents of PER fail to mention is that the monomer is worse for your health than the plasticizers. It's carcinogenic in relatively small doses, for one thing. It's also toxic to the liver in even smaller doses, measured in the low parts per million of body mass for an acute exposure. Lastly, the monomer is extremely flammable, though this is mainly an industrial hazard; the polymer is actually among the more flame-retardant commercial plastics, due to its chlorine content (in other words it's already highly oxidized, as opposed to something like polyethylene). None of this changes when the plasticizer does.
In short, while PER is a step forward, there are still plenty of inherent hazards in the production, long-term use and disposal of PER products. PER is also a relative specialty product, whose primary use is a foam product marketed for children's toys and yoga mats. It's not a silver bullet to replace PVC wholesale.

Thank you for you answer. So I was correct in this being a green-washing of a slightly less bad PVC? I would be interested in your opinion of Thermoplastic elastomer (TPE) as a PVC replacement. Would that require posting another question?

I would go ahead and post another quation on that. TPE isn't any one material, but a class of copolymers and polymer blends. ABS is a leading replacement for PVC in home plumbing as the trend against BPA-containing plastics continues.

Not ot be confused with French PER which is Poly Ethylene Reticulated water pipe industry standard, because progress means that french people have plastic house pipes : PER est le sigle du polyéthylène réticulé haute densité (PEX pour les Anglo-Saxons).

PVC is to my knowledge not giving off significants amounts of its monomer. Inappropriately burnt, it gives off PCDD's (polychlorinated dibenozdioxines).


text:
What is the definition of organic compounds?
According to the definition of organic compounds on Wikipedia:

An organic compound is any member of a large class of gaseous, liquid, or solid chemical compounds whose molecules contain carbon.

Carbonates, simple oxides of carbon, and cyanide molecules do have carbon, but they are not regarded as organic.  The same Wikipedia article on organic compounds reads:

The modern meaning of "organic compound" is any one of them that contains a significant amount of carbon.

The current main definition for organic compounds doesn't fit for carbonates, simple oxides of carbon, steel, etc. So, what should be the redefined definition for organic compounds?

The next sentence in the Wikipedia page you cite is “The distinction between "organic" and "inorganic" carbon compounds, while "useful in organizing the vast subject of chemistry... is somewhat arbitrary”

So, what should be the redefined definition for organic compounds?

We should not define organic compounds. As F'x pointed out in the comments, the distinction between organic and inorganic is largely arbitrary and mostly based on a distortion of historical precedent.
Rather, we should classify compounds by the type(s) of bonding interaction present in the substance. By this criterion, we have several classes of compounds:

Molecular (or covalent) compounds - the particles that make up these compounds are molecules held together by covalent bonds. This class of compound includes traditionally inorganic $\ce{CO2}$ and organic $\ce{CH4}$ carbon compounds. Covalent compounds also the majority of organometallic compounds, so we do not need a separate category for these. Other traditionally "inorganic" compounds, like glass, also belong in this category.
Ionic compounds - the particles that make up these compounds are largely ions held together by Coulombic forces. For the sake of simplicity most people would include in this class of compounds those ionic compounds where one or both ions is a small molecular ion held together by covalent bonds $($e.g. the $\ce{NH4^+}$ ion in $\ce{NH4Cl}$ or the  $\ce{CO3^{2-}}$ ion in $\ce{Na2CO3})$.
Mixed covalent/ionic compounds - those in which there are some ionic functional groups, but mostly covalent bonds, like methyl orange. Many folks might list these substances in Category 1. This category would include most coordination compounds including such things as metal-organic frameworks and probably many composite and/or ceramic materials.
Metallic compounds - those substances with extremely delocalized bonding, typical of metals. Alloys, including steel, belong in this category.

Under this system, what we traditionally study as organic chemistry becomes redefined as the study of the structure, properties, and reactions of covalent compounds (Classes 1 and 3), especially the nature of the covalent bond and how it can be formed and broken selectively. What you might learn about the structure, properties, and reactions of  organic compounds can easily be extended to the structure, properties, and reactions of molecular inorganic compounds. For example, we do not need two different branches of chemistry to study benzene and borazine.

The term "organic" in chemistry historically means "requiring a living organism for its creation", as proposed by the vitalism theory. The modern definition differs and can vary, as the answers to posts on this site demonstrate. It might help to quote IUPAC's definition, curiously found under the definition of "functional group":

Organic compounds are thought of as consisting of a relatively unreactive backbone, for example a chain of $\ce{sp3}$ hybridized carbon atoms, and one or several functional groups. The functional group is an atom, or a group of atoms that has similar chemical properties whenever it occurs in different compounds. It defines the characteristic physical and chemical properties of families of organic compounds.

This fits the definition provided by Ben Norris in this post, but disagrees with that by tomij, which requires the presence of atoms of covalently bonded carbon in the compound, although that definition is common.
References
(1) IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.
(2) PAC, 1994, 66, 1077. (Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)) on page 1116.


text:
How to calculate the heat capacity of a calorimeter based on a neutralisation reaction?
I've looked at the other similarly worded questions but I can't seem to make heads or tails out of them.
So here's my case. I need to find the heat capacity of the calorimeter (in Joule/degree Celsius), and I've been provided with the following information:
$$\ce{NaOH (aq) + HCl (aq) -> NaCl (aq) + H2O}$$

$\Delta T = 11.5\ ^\circ \ \text{C}$
$\Delta_r H = -58.3 \text{ kJ/mol}$
$s_{p,\ce{NaCl}} = 3.91 \text{ J/(g}\cdot \ ^\circ \text{C)}$
$d_{1\ \text{M NaCl}}=1.0037\text{ g/mL}$
$V= 100 \text{ mL}$

So far I've calculated the mass of $\ce{NaCl}$:
$$m = d\cdot V = (1.037 \text{ g/mL})(100 \text{ mL}) = 103.7 \text{ g}$$
Then I calculated the heat capacity of $\ce{NaCl}$...at least I think that's what I did. I'm not even sure if I was supposed to do this:

$q = m\cdot s_{p,\ce{NaCl}}\cdot \Delta T$
$q = (103.7 \text{ g})(3.91 \text{J/(g} \cdot\  ^\circ \text{C}))(11.5 \ ^\circ \text{ C})$
$q = 4662.8 \text{ J}$

Now I'm confused as to the next step, and I don't know how to use $\Delta_r H$ in this problem.
Any help would be immensely appreciated.

Edit: Below I've added more things I've tried in my attempt to find the answer.
I saw this equation in a textbook so I gave it a shot:

$q = - \Delta H\cdot n_{\text{limiting reactant}}$
$q = (- 58,300 \text{ J/mol})(1.62 \text{ mol HCl})$
$q = 94,446 \text{ J}$

But I don't know how/if this fits anywhere in the problem.
Another equation I tried:

$q = C_P\Delta T$
$-58,300 \text{ J}= C (11.5 \ ^\circ \text{C})$
$C = 58,300 \text{ J} / 11.5 \ ^\circ \text{C}= 5069.5 \text{ J/}\ ^\circ \text{C}$

Which is at least the correct unit of the heat capacity I'm supposed to find.
Finally, I tried subtracting the heat capacity I found for $\ce{NaCl} \ (4662.8 \text{ J})$ from the supposed heat capacity of the system $(5069.5 \text{ J})$, I get $406.76 \text{ J}$
I'm kind of at a loss...

You are mixing totally different quantities, be much more careful when expressing variables, what they really mean. E.g. the first calculation m = ... It is not mass of NaCl. Maybe of the solution. Even then, the numbers do not match the input block.
Additionally, there is not enough information provided. You need concentration of the species, at least of the product.

First, you have to find out the concentrations so you can calculate the amount of heat released. Since the density is given for a 1 M solution, I will assume this concentration.
Next, the mass of our solution:
$$ m = \rho \times V = 103.7~\mathrm{g}$$
Now we have to calculate the amount of energy that gets released through the reaction, but viewed from the environment and not the reaction (that's the reason for the negative sign before $q$):
$$ -q = \Delta_\text{r}H \times n = -58.3 \times 0.1 ~\mathrm{kJ\, mol^{-1}\, mol} = -5.83~\mathrm{kJ}$$
The total heat capacity of your system is given as
$$ c_\text{tot} = c_\text{s, sol}\times m + c_\text{cal}$$
where $c_\text{s, sol}$ is the specific heat capacity of your solution and $c_\text{cal}$ is the heat capacity of the calorimeter.
Now we know all the quantities needed for the following equation:
$$ q = c_\text{tot}\Delta T$$
We expand:
$$ q =( c_\text{s, sol}\times m + c_\text{cal}) \times \Delta T$$
Solve for $c_\text{cal}$:
$$ c_\text{cal} = \frac{q-c_\text{s, sol}m\Delta T}{\Delta T} = \frac{5.83~\mathrm{kJ} - 3.91~\mathrm{J\, g^{-1}\, K^{-1}} \times 103.7~\mathrm{g} \times 11.5~\mathrm{K}}{11.5~\mathrm{K}} = 101~\mathrm{J\, K^{-1}}$$
And now you know the heat capacity of the calorimeter.


text:
How to estimate pH?
How you you calculate/estimate the pH of a solution with 3 components all of different pHs?
Is there a paper describing this?
For example:
0.1M Buffer PCTP at pH 9.5
1.6M Sodium Dihydrogen Phosphate at pH 3.39
0.2M Sodium Acetate at pH 9.0255
The measured pH is ~5.5.
If i take a weighted average i.e (0.1 * 9.5 + 1.6*3.39 + 0.2 * 9.0255)/ 1.9 the pH is estimated at 4.311.
I assume the buffer "drags" up the pH but how to estimate this?

What have you tried to solve this problem? Like, writing the definition of pH, expressing the equilibrium constants, ...?

Could you describe more of what you want to do? An example of the problem you are trying to solve would be very helpful. For example, do you want to be able to answer questions of the type "What is the pH of a solution that is 0.0234 M in HCl, 0.212 M in acetic acid and 0.0216 M in potassium acetate?"

Sorry for the delay, I've amended the question.

As I stated in my previous comments: take equilibrium constants of all the chemicals inside, calculate the equilibrium concentration of $\ce{H^+}$ and from that easily pH. There is no way around it.

You can't just take an average of pHs- supposing it was a simple dissolution question rather than one with chemical reactions. pH=-log(H). H=10^(-pH)  First convert all the pH values to concentrations of hydrogen ions and then take an average of those values and then convert to pH- that will at least get you close the correct value.


text:
Does hydrogen peroxide exhibit intramolecular hydrogen bonding interactions?
We know that intermolecular hydrogen bonding is seen in hydrogen peroxide $(\ce{H2O2})$. But is intramolecular H-bonding, i.e. hydrogen bond between an oxygen atom and a hydrogen atom attached to the adjacent oxygen atom possible? How strong will it be, compared to intermolecular H-bonding and why?

If there is intramolecular h-bonding occuring (which, off the cuff, I strongly doubt), my guess is that it would be very weak with respect to intermolecular h-bonding due to the angle, A(OOH), being far from optimal in a normal h-bond.  I'll see what an NBO analysis says.  It could give us a clue if there is any sort of overlap that could give rise to a hydrogen-bond like intramolecular interaction.

Preamble:
I will conveniently avoid entering into a debate as to what a hydrogen bond is, given the numerous theories which attempt to do so from various perspectives.
I will, however, attempt to use the relatively cheap and easy natural-bond orbital (NBO) analysis method to examine the possible existence of an intramolecular hydrogen bond (HBond) in the ground-state, global minimum configuration of H$_2$O$_2$.  In doing so, we will assume that the mere existence of an $n\rightarrow\sigma^{*}$ overlap indicates that an intramolecular HBond is indeed present (a commonly accepted metric).  For comparison, I will look at the planar form of H$_2$O$_2$ (containing a trans-type bond configuration) in order to minimize any overlap between the oxygen lone pair and the corresponding $\sigma^{*}$ orbital.  I note here that the NBO analysis is purely a qualitative method, and caution should be used when attempting to make quantitative comparisons with this method.
Methods:
The H$_2$O$_2$ molecule was constructed in $C_2$ symmetry (Structure A). A full optimization and corresponding harmonic vibrational frequency computation was employed at the MP2/heavy-aug-cc-pVTZ level of theory, where non-hydrogen (i.e. heavy) atoms are augmented with diffuse functions.  The basis set is denoted henceforth as haTZ.  The magnitudes of the residual Cartesian gradient components of the optimized structure were less than $4.0\times10^{-5}$ $E_{\textrm{h}}$ $a_0^{-1}$.  The frozen-core approximation was invoked for all computations.
A subsequent NBO analysis of the fully optimized H$_2$O$_2$ molecule was performed at the MP2/haTZ level of theory using the SCF density.
This procedure was also applied to a H$_2$O$_2$ molecule constructed in $C_{2h}$ symmetry (Structure B).
All computations were performed with the G09 (Rev. D01) software package.
Results:
Structure A is a minimum on the MP2 potential energy surface with zero imaginary modes of vibration whereas structure B is a transition state with one imaginary mode of vibration.
The NBO analysis of structure A reveals two $n\rightarrow\sigma^{*}$ interactions (where $n$ is an electron lone pair on oxygen and $\sigma^{*}$ is the $\sigma$ antibonding orbital of the neighboring O$-$H entity.  The existence of exactly two of these interactions is due to the symmetry of the molecule, and only one will be reported here for simplicity.
The following figure illustrates the overlap of a lone electron pair on O with the adjacent $\sigma^{*}$ orbital in structure A.  The occupation number for this antibond, 0.00351, indicates a very small overlap.  The corresponding second-order perturbation energy E$^{(2)}$(donor$\rightarrow$acceptor) is a modest 1.74 kcal mol$^{-1}$.  We note that this is merely an estimated measure of the strength of the interaction.

Interestingly, upon rotation of the torsion angle, $\tau$(H-O-O-H), to 180$^\circ$ (structure B), the $n\rightarrow\sigma^{*}$ interaction disappears.  The intramolecular hydrogen bonding  may play an important role in stabilizing the lowest-energy conformer in H$_2$O$_2$ (structure A) as the transition state, structure B, does not contain this interaction.
Comparing the strength of an intramolecular HBond to an intermolecular HBond is not straightforward.  Intermolecular interactions are frequenly and efficiently computed by using the supermolecular approach by taking difference of the electronic energy of a complex and the energies of each non-interacting body which make up the complex.  Intramolecular interactions cannot be determined in this way without breaking bonds, thereby skewing the resulting number.  To facilitate comparison to the intramolecular HBond in H$_2$O$_2$, an NBO analysis was performed on the water dimer (H$_2$O)$_2$ (in $C_S$ symmetry) due to the presence of a strong, intermolecular hydrogen bond.   The corresponding E$^{(2)}$ for an $n\rightarrow\sigma^{*}$ interaction is a much larger 6.82 kcal mol$^{-1}$, nearly 5 kcal mol$^{-1}$ stronger than that seen in H$_2$O$_2$
Conclusions:
An NBO analysis of the lowest energy conformer H$_2$O$_2$ indicates the possibility of the presence of weak, intramolecular hydrogen bonds.  These intramolecular hydrogen bond-like interactions could play a role in stabilizing the H$_2$O$_2$ complex.
The strength of the interaction is much smaller than that computed for the water dimer by about 5 kcal mol$^{-1}$.
(Note: I strongly recommend performing a Bader analysis on H$_2$O$_2$ to compare with the NBO analysis before reaching any solid conclusions.)

ADDENDUM:
I have tried to avoid calling this intramolecular interaction a hydrogen bond.  Based on the metric imposed in the preamble, it has a property of a hydrogen bond, however, the strength of this intramolecular interaction is so weak, it really does not reflect a true hydrogen bond (e.g. A typical hydrogen bond is about 10 kcal mol$^{-1}$ and a weak hydrogen bond is usually considered to be on the order of 2-3 kcal mol$^{-1}$.  If we take the NBO results at face value, the interaction is just too weak to be considered as a hydrogen bond.  Coupled with the fact that the geometry itself does not facilitate reasonable HBonding configurations.
Upon reflection of these issues, I conclude that there is, in fact, no intramolecular hydrogen bonding in H$_2$O$_2$.

I went through your thorough answer,and though i could not understand much of the middle part,I understood that there cannot be an intramolecular hydrogen bond.How did you do this analysis?Is it from a research paper?

I was just thinking about this and water,not referring to experimental data,can you help me guess whether the strength of(intermolecular) hydrogen bond in $H_2O_2$ is more/less than in $H_2O$?I think the presence of two oxygens may make the hydrogen atom more electrophilic,hence making a stronger H-bond.Am I right?

@scienceauror I just decided to do this off the cuff as a mini-project.  I wouldn't dream of lifting this kind of work from a paper without citing it.  It was fun and I learned something.  Also, when you are talking about intermolecular HBonding in H$_2$O$_2$, you are referring to something like a dimer now? If so, it would be trivial to figure out a reasonable HBond strength.

So,is there a method of doing such an NBO analysis?This was brilliant.
No,I'm not suggesting its existence as a dimer.Weak intermolecular hydrogen bonds are known to have contributed into some physical and chemical properties of compounds.Let's consider the case of boiling points,where the strength of attractions between the molecules determines the ease in separating them.Do you think strength of H-bond in $H_2O_2$ will be stronger than water,due to electrostatic attraction of electron clouds by the Oxygen atoms,making H-atoms more positive in character?

@scienceauror I went back and looked at the R(HO) bond lengths in H$_2$O$_2$ and H$_2$O.  They were 0.97 Ang and 0.96 Ang respectively.  From this alone I would say that the OH bond in water is stronger than in hydrogen peroxide.  This sort of makes sense as the neighboring oxygen in H2O2 is 'much larger' than a hydrogen atom (in water), thereby 'pushing' electron density away from the adjacent oxygen, causing a lengthened OH bond.

The effect you are describing could be Hyperconjugation?

My molecular models set has led me to believe that hydrogen peroxide can conform to form 2 bent hydrogen bonds and one bent covalent bond between the $\ce{OH}$ groups.
This means that there can be a triple bond between the 2 oxygen atoms. The triple bent bond (i.e., 3 banana bonds) of which 2 are of the hydrogen bond type and one is the Linus Pauling bent-covalent bond type, like in ethylene ($\ce{CH2CH2}$), tau bonds, banana bonds, bent bonds, and indirect angular bonds.

You're wrong - hydrogen bonds can't be bent. Even if they could they wouldn't be able to offset ring strain.

Can you provide any evidence for these claims? They seem pretty unlikely to me.

The normal direct covalent bond between the 2 OH groups becomes bent due to 1 repulsion between the non-bonding electron pairs, 1 on each oxygen atom and 2 attractions of the 2 hydrogen atoms for the 2 Non-bonding electron pairs. These attractions (2) and 1 repulsion are all in the same direction causing the direct bond to become bent, now forming a triple bent bonds. This lends greater stability to the gaseous form of H2 O2so in the conformation where

The normal direct covalent bond between the 2 OH groups becomes bent due to 1 repulsion between the non-bonding electron pairs, 1 on each oxygen atom and 2 attractions of the 2 hydrogen atoms for the 2 Non-bonding electron pairs. These attractions (2 bent hydrogen bonds and 1 repulsion ( non-bonding pair repulsion) are all in the same direction  causing the direct bond to become bent, now forming a set of triple bonds, all bent. This lends to far greater stability to the gaseous form of H2O2. In the liquid, H2O2 forms intermolecular hydrogen bonding with itself and with H2O & would predominate

You will need a molecular model set that can form bent bonds of the covalent type (Linus Pauling type bent bond) and to  show the bent hydrogen bonds.  My models are unique and can easily illustrate non-bonding electron pairs and their interactions with their surroundings & can easily illustrate interactions to form bent bonds. I have no way to show a photo of the model  here but being creative you can draw this molecule with pencil & paper, if you follow these directions above

With your molecular model set, build a traditional model of HO-OH, then rotate to a conformer that shows the attractions of the H atoms for the adjacent non-bonding e- pairs and also shows the e-pair, e-pair repulsion. The forces involved in this conformer will produce the triple bent bonding, simply and easily for gaseous H2O2, a di-radical. A hydrated oxygen di-radical Atom has the formula OH2O forming from H2O vapor and an oxygen atom from Ozone (O3). which gives a hexagonal structure for Ozone & held together by 3 bent bonds or banana bonds or indirect covalent bonds, angular bonds....

With your molecular model set, build a traditional model of HO-OH, then rotate to a conformer that shows the attractions of the H atoms for the adjacent non-bonding e- pairs and also shows the e-pair, e-pair repulsion. The forces involved in this conformer will produce the triple bent bonding, simply and easily for gaseous H2O2, a di-radical. A hydrated oxygen di-radical Atom has the formula OH2O forming from H2O vapor and an oxygen atom from Ozone (O3). which gives a hexagonal structure for Ozone & held together by 3 bent bonds or banana bonds or indirect covalent bonds, angular bonds...

There is no, zero ring strain in my proposed molecule as you inferred. This may all be new chemistry for you. What are you Science & chemistry background and credential(s)

H2O2 is not a di-radical, my error in above comment. Please respond with your comment to my explanations. What is your chemistry experience? Are you a student or a PHD?


text:
What is the solid crud that appears inside my kettle?
Almost every share-house I've lived in has had some nasty looking solids collecting inside.  I always regret looking inside the kettle because a cup of tea never tastes quite as good after you've seen this:

What is that stuff?  What's the physical process that causes it to collect and why is it usually concentrated around the heating element?  Is it related to calcium in tap water, or is there any trick to prevent it appearing in your kettle?  Is it dangerous to ingest and should be removed, or is it harmless?

You're wouldn't be living with Rick, Mike, Neil and Vyvyan Basterd would you? Just be thankful - the last kettle had a breakdown and exploded!

As noted, it is limescale, which is typically white-ish (or if there are iron present, orange-ish). It's curious that it's yellow - could just be the camera... just wanted to offer that a natural way to clean the limescale (in addition to vinegar, which smells and might adversely affect your tea if you don't fully rinse it) is to use lemon or lime juice, which are also acidic in nature but would be a lovely addition to your tea. :-)

This is limescale and you have it because your tap water is rich in minerals (mainly, calcium, as you note). You could remove these minerals with an ion exchanger, but you wouldn't to it with a tap water since these minerals give it a better taste. Limescale is not dangerous to ingest, but certainly not pleasant either (and I see on the photo that there is a slot for a mesh that should filter scales). Thick layer of limescale is also bad for the heating element – it hinders heat transfer to water and leads to overheating.
It is a good idea to remove limescale once in a while. You can dissolve limescale with any weak acid, like vinegar, but vinegar smells, so I suggest buying a descaling agent.


text:
Reacting calcium with warm water and magnesium oxide with carbon dioxide, what are the products?
So I have several descriptions of reactions and I have to give the net ionic equation. Are these just the equations or are they the proper net ionic ones?
Solid calcium is added to warm water.
$$\ce{Ca_{(s)} + 2 H_{2}O_{(l)} -> Ca^{2+}_{(aq)} + 2 OH_{(aq)}^- + H2_{(g)}}$$
Powdered magnesium oxide is added to a container of carbon dioxide gas.
$$\ce{MgO_{(s)} + CO2_{(g)} -> MgCO3_{(s)}}$$
Did I do those right, or am I on the wrong track?

Purpose of ionic equation is to describe ions in (aqueous) solution, without the need to bother about details.
The first equation looks reasonable, you could possibly end up with $\ce{Ca(OH)_2}$, which is not very water soluble, but it's IMHO the type of detail you should not bother with.
The second one is more difficult, though. At first, there is no solution to speak about, therefore the request for ionic equation itself is invalid. The reaction written is correct, as long as it describes possibility. If it would really happen is question of reaction conditions, therefore totally outside the scope of homework. As such, I'd say, correct.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Can we have an adiabatic process in a sealed metal container?
Let us consider a non-isothermal process inside a metal container. If we somehow are able to make the temperatures on the inside and outside the system remain same throughout the process, can it be called an adiabatic process?

Adiabatic means that energy is transferred between the system and the surroundings only as work and not as heat.
Let's look at your process. You have a non-isothermal process, so $\Delta T_{sys} \ne 0$ for the process. An adiabatic process does not need to be isothermal, so this condition is okay. What makes your process non-adiabatic is the condition that constant temperature is somehow maintained (likely through a constant temperature bath which serves as the surroundings). This requires heat exchange between the system inside the metal container and the surroundings outside the container.
If you remove the constraint that the overall temperature is held constant, then you can have an adiabatic process.


text:
Why is Density Functional Theory notoriously bad at describing oxygen molecules?
I read in several papers that DFT is notoriously bad at describing free oxygen molecules but in none of them an explanation or reference concerning the problem has been provided. I guess this has something to do with its triplet ground state but I haven't come across a good explanation yet. Could you help me out?
Update: If someone wants to know the papers I'm refering to, here are the links and the passages that are most relevant (both groups use gradient-corrected (GGA) density functionals, so maybe this isn't a general problem of DFT but only of GGA- or LDA-functionals):

Rossmeisl et al. have devised a method for calculating the Gibbs Free Energy of the intermediate in the oxygen evolution reaction. The intermediates are treated with DFT but in order to calculate the Free Energy of the final products, including desorbed oxygen molecules, they use tabulated values.

The free energy change of total
  reaction: $\ce{H2O -> 1/2 O2 + H2}$ is fixed at the experimentally found value of 2.46 eV per water molecule. This is done in order to avoid calculations of $\ce{O2}$, since this molecule has a complicated electronic structure, which is not described accurately with DFT.

There are other groups studying water splitting that use the experimental value too, instead of calculating it with DFT.
The paper by Bloechl includes calculations of the binding energies of a hydrogen molecule, which are in rather good agreement with experiment (4.338 eV (DFT) vs. 4.448 eV (exp.)), and of an oxygen molecule, where the DFT value is off by 0.795 eV.

The binding energy of an oxygen molecule is calculated to be 5.912 eV, including the zero-point vibartion energy of 0.110 eV, which is comparable to the atomization energy of 5.906 eV, obtained with the Becke-Perdew-Wang gradient-corrected density functional. At 5.116 eV, the experimental binding energy is substantially smaller (by 0.795 eV) than the theoretical prediction.

@Philipp So, off the cuff, DFT in general notoriously bad at describing open-shell systems.  Molecular oxygen may have an even number of electrons but the ground state of O$_2$ is a triplet state where two electrons are unpaired.  It would be my guess that perhaps a problem lies with the exchange portion of a DFT method.  Its tough to say (given the speedy development of functionals these days) if any functional exists that would properly characterize O$_2$.  However, many commonly used functionals are expected to fail in this type of situation.

The answer is most certainly the Self interaction error of an electron with itself in the exchange term.
This would probably be more helpful than what I can write here, as I have little time to make this answer more complete.
A. Droghetti PRB 2008
Also have a look at
Klupfel2012

Thanks for providing the links and the keyword "Self interaction error".

No problem, if you are looking for keywords, see also Self-Interaction Correction (SIC). I think that the first serious work on this (with relation to what you are talking about) was here: http://dx.doi.org/10.1063/1.476859

I would make sure to compare the correct endpoints of the "reaction". The molecule should be a triplet (i.e. spin-polarized) and the separated atoms should be spin-polarized and Hund-compliant. Similarly, the expt energy itself should be the atomization energy of the triplet molecule into spin-polarized atoms.
That said, a typical value for the binding energy of the triplet (without the zero point vibration quantum, which reduces it) is about 4.95 eV/atom in GGA compared to spin-unpolarized atoms.
On the SIC, the first serious SIC work was actually Phys Rev B 23, 5048 (1981). A good practical SIC, unfortunately used less than it deserves, is Phys Rev B 84, 195127 (2011), also reviewed in a previous version in Eur. Phys. J. B 71, 139 (2009).

You can add links to the references you used to appreciate them.


text:
In what order should the compounds containing Ru and Os, or Mo and W be named?

According to IUPAC recommendations for naming of inorganic compounds, the more electropositive element is written first with a suitable prefix, and the more electronegative element is written at the last with a suffix -de and a prefix. For example, $\ce{H2O}$ (Water) can be called dihydrogen monooxide. [this is an modified statement extracted from the wiki encyclopedia-IUPAC nomenclature of inorganic chemistry]
  In the above example, $\ce{H}$ (Pauling scale = 2.20 or Allen scale = 2.300) is less electronegative than $\ce{O}$ (Pauling scale = 3.44 or Allen scale = 3.610). So, $\ce{H}$ is placed first, followed by $\ce{O}$.

If suppose we were needed to name the compounds containing

$\ce{Ru}$ (Pauling scale = 2.2) and $\ce{Os}$ (Pauling scale = 2.2)
$\ce{Mo}$ (Allen scale = 1.47) and $\ce{W}$ (Allen scale = 1.47)

, in what order should we write the elements?
If we use pauling scale as referance, we would face difficulty in naming $\ce{Ru}$ and $\ce{Os}$, if we use allen scale, we would be in trouble with respect to the elements like $\ce{Mo}$ and $\ce{W}$, etc, as we can see that elements have same electronegativity value. I don't know whether the compounds with the above mentioned elements exist or not, but if we suppose, assume that, they exist, in what order should they be named?

Electronegativity hasn't been very well defined for the $d$ and $f$ blocks, so I think it may not be a good idea to take its importance in nomenclature too strictly. Also substances with more than one metal element are usually alloys rather than molecules or salts, and only in rather rare occasions do they come without any other elements that have a clear electronegativity difference.

@NicolauSakerNeto yes, but… don't underestimate the power of IUPAC :)  even if it's not likely to happen, there is a rule to govern it…

@F'x Haha, I knew there was something weird about IUPAC basing standards on such a tricky property, but didn't consider they could very well define their own scale!

When IUPAC rules refer to electronegativity (and electropositivity), they don't mean actual electronegativity measured by one or another scale. IUPAC decided, for nomenclature's sake, on a formal ordering of electronegativity, which goes like this:

This is Table VI, on page 260 of IUPAC's very own 2005 Red Book, aka Nomenclature of inorganic chemistry. The cases in which it is used are detailed in IR-2.15.3.1 (page 42).
The explanation of the changes from earlier IUPAC recommendations is in IR-1.6.3 of the Red Book:

The element sequence of Table VI
  In Nomenclature of Inorganic Chemistry, IUPAC Recommendations 1990 (Ref. 11), the position of oxygen in certain element sequences was treated as an exception. Such exceptions have been removed and the element sequence of Table VI is now strictly adhered to. In particular, oxygen is treated as the electropositive component relative to any halogen for constructing compositional names (Section IR-5.2) and corresponding formulae (Section IR-4.4.3) for binary compounds. This results in, for example, the formula O2Cl and the name dioxygen chloride rather than the formula ClO2 and the name chlorine dioxide.
In Ref. 11, the formulae for intermetallic compounds were also subject to an exceptional rule although no guidance was given for naming such compounds, and the term ‘intermetallic compound’ was not defined. The problem is to define the term ‘metal’. Therefore, no attempt is now made to make a separate prescription for either the formulae or the names of intermetallic compounds. It is stressed, however, that the present recommendations allow some flexibility regarding formulae and compositional names of ternary, quaternary, etc. compounds. Several ordering principles are often equally acceptable (see Sections IR-4.4.2 and IR-4.4.3).
The element sequence of Table VI is also adhered to when ordering central atoms in polynuclear compounds for the purpose of constructing additive names (see Section IR-1.6.6).

For more on this topic, you've got to go read (and upvote) this excellent answer by Manishearth, which is where I first learnt that useful and intriguing nugget of IUPAC idiosyncrasy!


text:
How toxic is disilanol?
Silicon has been known to form analogues of organic compounds. For example, silicon with 4 hydrogen atoms forms silane as opposed to methane. This continues up the alkane list, with disilane for ethane, and so forth.
Silanes also have with hydroxyl groups to form "silanol." Thus, disilanol is the silicon analogue of ethanol. ($\ce{H3Si-SiH2OH}$)
How toxic is disilanol?

Difficult to say. After a fair amount of Googling, I was unable to find any reliable sources on the compound you ask about, $\ce{H3Si-SiH2OH}$. From the Wikipedia article on silanols, it appears that the simple silanols are not stable, but no reference is provided.
It stands to reason that if $\ce{H3SiOH}$ is not stable, there is a good chance that $\ce{H3Si-SiH2OH}$ is also not very stable. Disiloxane $\ce{H3Si-O-SiH3}$ is much more stable, and silanols likely convert into siloxanes.
Good knowledge about the health hazards of a substance requires thorough investigation, which is not possible when the substance is not stable.
Alkyl silanols, like trimethylsilanol are stable and used for a variety of applications. Acutley, trimethylsilanol(pdf) is a volatile liquid that irritates skin, eyes, and mucus membranes (e.g. the lungs). The language on the MSDS is "may be harmful if..." This language suggests that you want to avoid contact with it but that some exposure will not cause serious harm. Note the msds for ethanol(pdf) has much scarier language.


text:
Amount of substance produced in chemical reaction
If $15.0\ \mathrm{mol}$ of nitrogen are reacted with $30.0\ \mathrm{mol}$ of hydrogen, how much ammonia will be produced?
$$\ce{N2 + 3H2 -> 2NH3}$$
What I’ve tried is the following:
$$\dfrac{n(\ce{N2})}{n(\ce{H2})}=\dfrac{1}{3}=\dfrac{15.0\ \mathrm{mol}}{x}$$
$$\dfrac{n(\ce{N2})}{n(\ce{H2})}=\dfrac{1}{3}=\dfrac{x}{30\ \mathrm{mol}}$$
I’m not sure how to proceed from here. Any help would be appreciated.

In the reaction:
$$\ce{N_2 + 3H_2 \to 2NH_3}$$
If 15 moles of $\ce{N_2}$ is used along with 30 moles of $\ce{H_2}$, then $\ce{H_2}$ will be the limiting reagent (Reactant which is consumed completely). This is because 15 moles of $\ce{N_2}$ will need 45 moles of $\ce{H_2}$ ($15 \times 3$, by the balanced reaction) for reacting completely, but as there is only 30 moles available hence it will react in a fixed proportion, hence you will have to assume the reacting moles of $\ce{N_2}$ as $x$. By this the equation goes,
$$\frac{\ce{N_2}}{\ce{H_2}}=\frac{1}{3}=\frac{x}{30}$$
From this
$$x=\frac{30}{3}=10$$
Hence from 15 moles of $\ce{N_2}$ only 10 moles will react, hence the amount of $\ce{NH_3}$ formed will be 20 moles ($10 \times 2$, by the balanced reaction).


text:
Overlap between computer science and chemistry?
This isn't particularly a question about chemistry itself, but more-so the prospect of chemistry.
What kind of opportunities arise when you overlap chemistry and computer science? I'm a computer science major, but prior to going to university I had (and still have) a deep passion for chemistry. My highest marks were always in chemistry, but I decided to go down the compsci route because likewise it is a passion of mine. An opportunity has presented itself to double major in compsci and chemistry and I'm curious to know whether it's worth it.
My knowledge of chemistry is fairly barebones (a limitation of highschool education), but as I said, it was my strongest subject and for some reason it just clicks with me. I'd often spent sleepless nights perusing chemistry textbooks and online periodicals because it fascinates me to no end.
I'm not looking for greater job prospects (I already work in both compsci as a programmer and chemistry in an electronic cigarette company), I'm more-so looking for a more fulfilling pursuit of knowledge. While irrelevant, my university recently opened a new building dedicated primarily to chemistry, so facilities are recent and faculty is adept and education is modern. I'd very much like to take advantage of the opportunity, but I'm just unsure if a university-level chemistry program will scratch that itch for me.

Chemistry job marker is shrinking. It is saturated with chemists. Companies merge, hence massive lay offs and hiring freezes. A lot of chemists (me included) flipped to software development. There are a lot of devs with background in chemistry.

Try to get a phone in chemistry related field. You wouldn't pass grep filter because there are no jobs.

Have a look at list of computational chemistry software. It gives you overview, how the computers are used in chemistry. Many of them fall more into the chemical physics field, some into the chemoinformatics, to name just few. So definitely there are lot of places to contribute to use of computers in chemistry. If you are able to mix the software engineering and the science itself in a good way, possibilities are immense.
As for the last question - almost whatever field of chemistry you choose, you can contribute to the programs used within it. There is no such thing like completed scientific code. And university level is a necessary prerequisite to do reasonable science. Just concentrate on the models and mathematics behind the scenes and everything will follow.

(1) Gaussian  (2) ChemOffice/ChemSketch (3) Scifinder. Not much. And there are millions of chemists competing for positions in these companies.

I'm late to the party and everyone else here has suggested how computer science experience can contribue to chemistry. I'm going to do a bit of that, but also point out how chemistry experience could potentially contribute to some (very select) CS opportunities.
I find myself in a similar position to you. I recently declared a double major in Chemistry and Applied Math-Comp.Sci emphasis because I enjoyed both subjects. I was initially planning to double in pure Comp. Sci., but our CS department is full of horrible lecturers and the math is full of amazing lecturers so I kinda got shoehorned into the other.
Computational chemistry is one of many fields in which CS and chem will overlap these days. It's not obvious, but there are many, many others, as people who are good with computers and science are rarer than they should be. Many labs, even those that are not computational in nature, rely on some basic modeling. I currently do some work with a nanoparticles group, and the PI was recently complaining that most of his grad students couldn't program anything to save their lives.
Failing that, our field could always use some good software devs. Perhaps I'm a little spoiled, but some our so-called "cutting edge" software suites have to be partially recomplied every time they are launched and suffer from spontaneously having communications modules crash, necessitating reboots of several computers in order to get things up and running again. It's pretty obvious to me that a lot of the stuff was designed by a software developer and not by a practicing scientist--such as the million button presses needed to load data and work with it, and the general recalcitrance of such software to spit out various data types.
If you don't want to work on that problem, we're still struggling with the issue of how to share code and models/data. Everything is moving to computers, and research is no exception. Researchers are still trying to figure out how to share files easily, store them, and publish them (more than one study has gone to press as a letter without any of the code attached to it).
Ultimately, I don't know what you want. Based off your last few paragraphs, this probably isn't what you were looking for, but since you opened with a question about the overlap between two fields, and not how one could contribute to the other, I'm leaving these here as ideas about how several fields can potentially interact.
P.S. For an interesting example, Philip Guo's PhD thesis revolved around the question of how the same code could give the same results when run on different systems, in spite of different environments. The question led him to create a tool (I believe Python) which allows simulations to run in a safely sandboxed (and thus identical) environement without loss of performance. http://www.pgbovine.net/PhD-memoir.htm

+1 for "very select" aka "there are opportunities, but those are rare"

Computational chemistry is a huge field. The 2013 Nobel Prize in chemistry was awarded to researchers who laid the foundations for modern computational chemistry to advance forward, and is essentially a commentary by the Chemistry Nobel Prize committee on the importance of computational chemistry. Computer models are particularly critical for developing an understanding of systems that are hard (or impossible) to access experimentally.
This essentially involves creating mathematical models for how molecules and atoms behave, writing programs to solve them, and running them on (usually massive) computers. Molecular dynamics is a common activity, and involves simulating a handful of molecules that follow a handful of rules (how do they collide, how do they spin, how do they attract or repel one another, etc.). The more computing power you have, the more molecules you can simulate, and so the better your extrapolations to larger scales are. Simulations that are run on supercomputers can provide a direct link between the microscopic (molecules) and the macroscopic (fluid properties, like viscosity or conductivity). Writing programs that will run (and run efficiently) on supercomputers involves a lot of moving parts, most of which are pure computer science.
If you have a computer science background and you enter a chemistry degree program, or go to work at a chemical company where you're working with chemists, you'll have an edge because of your computational skills, and will have the intuition to think about problems in a computational way - the value of which can't be overstated. As mentioned above, computer models are used often in chemistry, so programming fluency is important. On the other hand, you'll have plenty of catching up to do on the math and physics, which is a different kind of intuition you may not have that others will.
National laboratories and universities do a significant amount of computational chemistry work, less so in industry R&D (caused by, among other reasons, lack of access to enough computing resources and difficulty of obtaining immediately applicable results). Pharmaceutical companies are the 800 pound gorillas of industrial computational chemistry. If you're looking for work in industry, though, consider chemical engineering. Chemical engineering is another huge field with an equally huge number of opportunities (read: low-hanging fruit) for someone with a computational bent.

In short:

quantum chemistry, if you like academic chemistry,
chemoinformatics /bioinformatics, if you like more industrial chemistry, pharma,
simulations in nanotechnology, if you do not afraid of a little physics, and like interdisciplinar fields

In long:
Quantum chemistry / molecular simulations is nice and interesting, but there are two main possible cons (depending on your taste):

QC is generally very academy oriented. Though it is used in industry,
too, QC people are generally stick to academia or small scientific
software companies. Just google job advertisements to have a rough
idea, before you make any big decision.
You may fell be late from the party. QC has many-many implementations, most of them is in
ancient FORTRAN dialects. There are even more unimplemented
algorithms, however you may find difficult to start a completely new
project from scratch, and you may find painful spagetthicoding in
FORTRAN if you join others.

You should note that in that respect molecular simulation people (molecular mechanics, MM based MD etc) are very similar to quantum chemistry folks.
Chemoinformatics is a different field, and it generally requires different kind of math (much more machine learning / statistics oriented):

Chemoinformatics is much less popular in academia, and the academic community is much smaller, on the other hand much more popular in industry. If you look for "computational chemistry" jobs, bigger part of the advertisements will be some flavour of chemoinfo.
Generally it requires large (reliable) databases which can limit your research to certain fields. However, there are some innovative use of cheminfo, e.g. done by Grzybowski group at Notheastern digitalizing chemical synthetic databases.
The field has an almost incestous relationship with pharma, so if it not your main interest, you may find most topics boring.

Combined fields: there is always interesting blends of  theory and calculation (and experiment), see e.g. the work of Christofer Wilmer about design of MOFs.
Simulations for nanotechnology should be similar to quantumchemistry, however it is an emerging field with slightly different culture, much less established codebases and much more diversity of theories. It is often a field where Physicists, Engineers and Chemists collide. It also mean you should have a better idea about what you do. It is essentially the Wild West.
In my experience, nanotech experimental groups are also much more willing to collaborate with computational / theoretical people than classical chemists and they have much more intimate relationship with machines and quantitative data than e.g. organic chemists.
One word of caution, though:
Scientific computational guys are smart and everything in many ways, but the majority of them are HORRIBLE programmers. By horrible programmers I mean people who never heard about version control, do not test their code properly, and willing to write totally undocumented, unreadable code in languages that most programmer have goosebumps. It is a field where even visualisation software are written in FORTRAN.
This can be a huge pro or a huge con, depending on your situation.

Unfortunately I chuckled a little on your last paragraph, maybe I should have cried. The lack of documentation - even in commercial code - can drive you insane. Just one example which was recently posed here on the network:  Definition of the B3LYP functional in common QC programs

Do study Python (faster to learn) or Java (better paid) if you are a chemist.
Do NOT study chemistry if you are a dev.
Universities pump tones of chemists to the job market. Same rate as in 1960s. But there are no jobs because in 1960s you need a lot of chemists to run a factory. Now, thanks to automation, you just need one.
Before every province in Europe would have a factory producing chemicals. Europe moved out dirty production processes to India/China because of strict environmental laws in Europe.
If you study chemistry now - change your major. Srsly.

Well, I think you didn't read whole question and it isn't exactly answering it.

I actually agree with this answer and don't think it's an off-topic. This might sound sad, but this is true and also partially addresses the question IMO.

@Mithoron I did. It is pointless to learn chemistry for a software dev because it takes 10 years (chemistry labs included) to become good in it. If you do so you find yourself competing with a lot of devs with background in chemistry. Chemistry is like archery nowadays. Adds no value to your  dev resume. Sad, but true.

Well, OP said "I'm not looking for greater job prospects" and that's about the prospects...

If we ignore "greater job perspectives" point of view then the question is "primarily opinion based" and is much like "I am a cook - should I study astronomy? This is for fun, not career related". Well, why not? But after studying it for years (which you have to do in a University with a lab - thus spend time and money) you end up with zero gain in value. And say "why no one told me that chemistry is useless nowadays??" If time and money are unlimited, then, sure, go ahead and study whatever.

Gordon Moore got his PhD in chemistry. The overlap is actually in making the boards and the electronics. What materials and what properties something has and why you would use it in a computer.

I'd argue that this isn't really computer science--the actual laying of silicon is usually studied in disciplines like electrical engineering.

Software and material science are fully  decoupled in electronics. Chemist needs to know how what is a field-effect transistor and how to measure conductivity. His insight in programming doesn't give him any benefits because this transistor works the same way for all programming languages. Likewise, coder can't care less about hardware part.


text:
Calculating the equilibrium constants for parallel reactions
So, I am attempting to calculate the equilibrium constant for two separate reactions.
$$\ce{A + B<=>C + D}$$
$$\ce{A + B <=> C + E}$$
I have figured out the final pressures at equilibrium for $\ce{A}$, $\ce{B}$, $\ce{C}$, $\ce{D}$, and $\ce{E}$.
I was wondering if the equilibrium constant is calculated as normal using just the final pressures and molar ratios in the equations, or if the final pressures have to be adjusted somehow to compensate for the fact that the second reaction is also consuming the same products and producing one of the same reactants.

You can combine the two chemical reactions into a single global reaction scheme by adding them:
\begin{align}
\ce{A + B&<=>C + D}\tag{1}\\
\ce{A + B&<=>C + E}\tag{2} \\ \hline
\ce{2A + 2B &<=> 2C + D + E}   \tag{Global}
\end{align}
and the equilibrium constant for this overall reaction scheme is equal to the products of the two equilibrium constants:
\begin{align}
K_\mathrm{eq,global} &= K_\mathrm{eq,1} K_\mathrm{eq,2} \\
&= \frac{\ce{[C]^2[D][E]} } {\ce{[A]^2[B]^2}}
\end{align}
(where $\ce{[i]}$ denotes either molar concentration or partial pressure of component i). You do not need to adjust the final total pressure or mole amounts.