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Please explain why dioxygen difluoride is so dangerous
I just read this article that mentioned that dioxygen difluoride is very dangerous. The terms it uses are "awful", "violently hideous" and "deeply alarming".
But I couldn't get a handle on exactly what it would do that warrants those descriptions.

Isn't that sufficiently explained in the linked In the Pipeline blog post?

@MadScientist No, does unstable mean flammable or explosive? If it's explosive, how are people being poisoned by it? If it's explosive, give a comparable substance to compare it to: like gasoline, TNT or antimatter?

"Unstable" does not mean "flammable" or "explosive". Unstable means "reacts with everything it can get its grubby little valence electrons on in interesting, messy, violent, or uncontrollable ways." Unstable can also mean "tends to decompose quickly so as it cannot be isolated or stored." Gasoline is flammable, but it is not unstable - otherwise filling stations would randomly go up as the gasoline reacted violently to things like the storage tanks it sits in.

As far as being able to make it, there are a lot of chemical agents that are very useful in the industry but that do not tolerate exposure to "ambient environmental conditions", specifically room temperature and exposure to normal air. Most of these are "air-sensitive"; they rapidly turn into something else when exposed to the oxygen and/or water vapor in the air, and they're handled by using syringes and pressure from inert gases to move them from place to place in the lab equipment. Others simply have to be kept reasonably cold to reduce their energy and thus their tendency to decompose.

$\ce {O_2F_2}$ doesn't spontaneously combust. It is a supporter of combustion, which means that it's basically a better version of oxygen when it comes to supporting fires.
Basically, when placed in contact with $\ce{O_2F_2}$, other materials spontaneously combust. Here's an analogy: Substances such as nitroglycerin and TNT are like a person with a short temper. They are easy to combust. Substances like $\ce{ClF3}$ and $\ce{O_2F_2}$, on the other hand, are like a person with a knack of making others lose their temper. Mild-mannered on his own, but he can make people around him combust with relative ease.
It still can be prepared, though, because it becomes a lot more mellow at low temperatures. Also, it won't act on stuff which is already a good oxidizing agent (oxygen, ozone, fluorine, sulfuric acid for example), or at least it won't act so vigorously. Coupling these two together, one can probably prepare it in a low temperature environment inside a container lined with a relatively inert material.

Update:
(h/t to KeithS for pointing this out)
Note that $\ce{O2F2}$ has problems on its own as well. It decomposes when on its own to $\ce{OF2}$. Generally, when $\ce{O2F2}$ is oxidising something, it is really the $\ce{OF2}$ which does this (though $\ce{O2F2}$ is strong enough to do the same as well).
After burning something, $\ce{OF2}$ leaves behind some dangerous residues as well, like $\ce{HF}$ (which will eat through most of the material that was saved from burning)
(There are more details in @KeithS' answer)

No, it doesn't spontaneously combust. It spontaneously decomposes, and that's worse, because no matter how pure and well-isolated your sample is, all you need is just a little too much thermal energy.

@KeithS: Hmm, good point. I've only addressed the making-others-combust, I forgot that it still has dangers on it's own. I'll write up a bit about that :)

XKCD's source material is an article from the blog of one of the funnier chemists I've read, Dr. Derek Lowe. The chemical in question made his "Things I Won't Work With" list, and the article is found here.
Dioxygen difluoride, $\ce{O_2F_2}$, sometimes evocatively labeled according to its atomic arrangement $\ce{FOOF}$, is first and foremost a vigorous oxidizer; in plain English, it starts roaring fires with almost anything it touches, even with things that simply will not burn in any other circumstances, like sand, concrete, brick, iron, copper, even asbestos and fiberglass. Both the oxygen and the fluorine are pretty highly electronegative (fluorine is the most electronegative element in the table, and oxygen's #2), and so they will readily react with anything even slightly more willing to donate an electron; carbon, silicon, metals, alkalis, most other gases, etc. When that happens, the high-energy bonds of this molecule (it can only be formed at 700*C, and that thermal energy remains "locked" in the structure of the molecule) are rearranged into structures that require less energy, and the excess energy is released. The resulting rapid, aggressive, highly exothermic reaction of dioxygen difluoride with just about any other material would not be described as anything less than an explosion.
Second, even in the absence of something to oxidize, the oxygen would much rather exist in its "native" diatomic gas state. Again, there's a lot of energy tied up in the molecular bonds of this chemical, and the lower the energy required to form a bond, the more stable the molecule. Therefore, even at extremely cold temperatures (-160*C) in complete isolation, dioxygen difluoride decomposes to oxygen difluoride ($\ce{OF_2}$) and $\ce{O_2}$ gas. At room temperature, this decomposition is uncontrollable (no-one's gotten it anywhere close to RT without it coming apart), and even at extremely cold temperatures, the heat produced by slow decomposition can cause what's called a "thermal runaway"; the heat produced by one molecule decomposing is absorbed by nearby molecules causing them to decompose, releasing more heat, and so on. When things rapidly decompose into more stable gaseous components, the event is normally accompanied by a loud noise and shrapnel, formed from the remains of whatever had been holding the substance and anything else nearby.
Of the products formed by the self-decomposition of FOOF, oxygen gas is a pretty strong oxidizer, but it's the least of our worries here. The other product, oxygen difluoride, is almost as strong an oxidizer as dioxygen difluoride, which is only stronger because it's oxygen difluoride plus more oxygen. The "shared" electron pairs forming the bonds between the oxygen and each fluorine atom are much more closely attracted to the fluorine, as opposed to a structure like water where the oxygen is the more attractive. This gives the oxygen a formally positive oxidation state of +2.
By its very nature, oxygen yearns to flip that sign and achieve its "natural" -2 ox state by bonding with something less electronegative (which, as previously mentioned, is almost everything else in the periodic table besides fluorine). However, unlike the other oxygen atom in the FOOF molecule which flew the coop, this second oxygen can't escape, because the two fluorine atoms won't let it go. The fluorines prefer to bond to it rather than to each other... until something else, anything else, is introduced that's even more attractive.
$\ce{OF_2}$, like its FOOF parent, reacts very strongly and exothermically with almost anything, especially water. It will literally burn ice, rapidly; that's one of the reactions guaranteed to produce a powerful explosion. The most stable ultimate products of that reaction are hydrogen fluoride gas and more oxygen, and when you call HF gas a "stable" product of any reaction you are speaking in very relative terms; a release of HF gas into the air is one of those "drop everything and run" types of industrial accidents. HF gas dissolves in water, including the mucus of your lungs, to produce hydrofluoric acid, a persistent, toxic, corrosive acid that will shred just about any organic chemical it comes across from hemoglobin to proteins to DNA, and you really want to keep those kinds of things intact.
Basically, there are a lot of things in an inorganic chemistry lab that have a Medusa-like ability to kill at a glance, but fluorine deserves special notice, in part because of the number of notable chemists who were poisoned or blown up in the early heady days of isolating it. When you combine it with other elements on the right hand side of the periodic table in "lightweight" combinations (between two and about ten atoms per molecule), your day is about to get reeeeealy interesting; see the work of A.G. Streng, who was tasked not only with producing large quantities of dioxygen difluoride, but also with reacting it with things that normal chemists would keep very far away from it (and in fact "normal chemists" stay very far away from FOOF itself for the obvious reasons).
Fluorine only starts becoming more stable and easier to work with as an ionic salt, such as sodium or calcium fluoride (in which form it naturally occurs as the mineral fluorite; still poisonous in sufficient doses but much less of an explosion/reactivity hazard), or when you start throwing in elements of the carbon group (the carbon-fluorine bond is among the strongest known to chemistry, and fluorocarbons have a variety of uses as propellants, refrigerants and polymers, including Teflon), or heavier chalcogens like sulfur (sulfur hexafluoride, $\ce{SF_6}$, has virtually no reaction chemistry and is favored as a dielectric and to blanket air-sensitive materials, though other sulfur-fluoride compounds like sulfur tetrafluoride and disulfur decafluoride are highly toxic).

It is a very strong oxidizer. That means it can make a lot of materials "burn", i. e. undergo a reaction (oxidation) that releases energy, often in violent ways, because there's so much energy released in such a short time that it has trouble dissipating in any other way than exploding with vigour. This includes a lot of materials that you wouldn't normally consider burnable. Derek Lowe has another article ("Sand won't save you this time", about a similarly hideous substance that's also a fluorine compound), which has some very illustrative examples.
It contains fluorine, which forms many highly toxic compounds that you don't want to have flying around hot and steaming. Even on itself (as $F_2$), it is a gas that's far from harmless.

If you like firsthand stories about compounds like this, and some insight on why people might want to experiment with them at all (although imho, the real reason is they are simply lunatics), I can recommend the book "Ignition!" about rocket fuel research. It's out of print, but PDFs float around the web.


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What determines a substance's energy of activation with air?
Ice will melt when heat is applied; paper will catch fire. In trying to figure out why — what the difference is between things that melt and things that catch — I found "Burn, Char, Melt" by Roberto Gregorius, who says it depends on

whether the substance easily forms compounds with some oxidant [usually oxygen]. This term "easily" can be taken to mean that the energy required to start the reaction (called the energy of activation) is low enough so that the reaction starts (as in striking a match), or the compounds produced in the burning are low enough in energy that there is a tendency to form it. Of these two, energy of activation is usually the controlling factor in deciding if a reaction will take place.

All right, so what determines a substance's energy of activation with air? (Wikipedia isn't helping.) In case it isn't obvious, I'm uneducated in chemistry, and an answer tailored accordingly, to the extent possible, would be ideal.

Energy of activation is pretty much an experimental thing. You can try to explain obtained results, but predicting exact values without numerical analysis is not really possible.
Basically, given a compound to react with air, you first devise a reaction mechanism -- how the reaction proceeds. Now, using some rules, you determine the highest energy transition state. Finally, you try to approximate the energy difference between this and the reactants by comparing with known quantities.
About the rules that are used to compare energies and energy differences: There are many of them, and out of the scope of this answer (indeed, one would be hard pressed to fit them all on a single bookshelf). Basically, one uses general rules that compare stability. For example, a positively-charged carbon is more stable when it has more alkyl groups attached to it (the reason usually given for this is hyperconjugation). A carbon conjugated with a double bond is considered even more stable, due to resonance.

Regarding paper, paper is made of molecular fibres which are effectively infinitely long. To melt paper, one would have to disentangle these (as molecules need to be much farther apart and with much less attraction in liquid form), and even then one would be left with a very, very viscous liquid (which is not too different from a solid). So paper would be extremely hard to melt, if not impossible.

source: Stack Exchange

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Polar group strong dipole effect molecular axis
There are some fundamental charactaristics of liquid crystal. One of them is that all liquid crystals are organic compounds and at one side of their long molecular foundation there exists a polar group such as -CN, -NO2, -NH2, -OR etc.
A strong dipole effect in molecules is created by the existance of polar group; so the molecular axies are parallel.
Actually I don't understand the characteristic perfectly because of having confusion at several terms.
My questions are :

What are the polar polar groups?
What is the strong dipole effect?
What is molecular axis?

I especially want to know  about molecular axis with an extant explanation.

Could you perhaps include the source of where you got this?

Molecular axis is the 'long direction' for a molecule that is roughly cylindrical or rectangular flat.  the rest of the terms you explained in your question.

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

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

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

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

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How are samples prepared for analysis with ICP-OES? Can it analyse gases?
In ICP-OES (Wikipedia page), you introduce a sample into a very hot plasma and then run a spectral analysis on the resulting light emissions. Liquid samples are sprayed into the flame. From the Wikipedia article:

A peristaltic pump delivers an aqueous or organic sample into an analytical nebulizer where it is changed into mist and introduced directly inside the plasma flame.

From that I can imagine that you would probably dissolve solid samples (the technique is widely used in metallurgy and mineralogy) in something suitable to obtain a liquid sample. If that's correct, I'm wondering how the feeding apparatus is constructed to withstand possibly aggresive solvents, like nitric acid or aqua regia. The injection tube is apparently made from fused silica, I guess that's a probable material for the feed as well.
And how about gases? Are they analysed with ICP-OES as well? Are they simply pumped into the injection tube? For example, can we just introduce a stream of air into the plasma and get the emission spectrum for 78% nitrogen, 21% oxygen (I guess the argon will be hidden because the plasma gas is argon)? I guess since every sample will be beyond vaporized anyway by the intense heat, it can't make much difference, but are gas samples too "thin", i. e. do they contain too little atoms per volume for useful analysis?

From that I can imagine that you would probably dissolve solid samples in something suitable to obtain a liquid sample. Under normal circumstances, that is absolutely correct. As mentioned on the answers, acids are not a problem - in fact, concentrated acids are usually used to "digest" organic samples in sample preparation. Then the samples are diluted to be suitable for the ICP. There's a whole science behind sample preparation, though. Each sample and each analyte require very specific sample preparation methods in order to be properly detected in equipments such as an ICP-OES.

About the measurement of solids
Injections tubes are indeed often made out of fused quartz. Dealing with aggressive solvents is actually not that much of an issue because of the extremely low amounts of material that can be detected with ICP-OES. Therefore, you can simply dilute your aggressive solvent before use. You can look for example at this research note where they dissolve a sample in $\ce{HCl}$ and $\ce{HNO3}$ and then dilute it before use.
About the measurement of gases
Yes, it is possible to measure gases directly, but it is not recommended and some adaptations to the normal procedure are needed. In this research paper they investigated aerosol analysis by ICP. The main issue was keeping sampling lines at elevated temperatures to allow proper calibration. A secondary problem arose because their target analyte, metallic compound in flue gases, which are volatile at (or close to) room temperature, are typically highly unstable or explosive. Therefore, they recommend dissolving areosols in a liquid solution of water and some organic solvent (1-methoxy-2-propanol in their case) and use this as feed.
For the measurement of 'normal' gases it seems that the problems they encounter will not occur, because you don't need elevated temperatures in your calibration. So in principle that means that you can indeed measure air. It should be noted that in some machines a shear flow of dried air or nitrogen is used to cut the plasma (see wikipedia) which will disallow measurement of air, as this Perkin Elmer instrument (overview table in slide 7 of this presentation). Some other instruments, like the one in the research paper I referenced, do not use such a shear flow and will thus be able to detect air.
Gas samples are not at all 'too thin', because the normal nebulized feed that goes into the plasma is similarly dilute.

:) That first research note incidentally is about the exact application I am interested in, specifically, the analysis of steel alloys.


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Why do chalcogens (Group VI) stink so badly?
For more fun with Dr. Derek Lowe, see this for a primer: Things I Won't Work With - Carbon Diselenide. The short of it is that we carbon-based life forms generally like oxygen. However, move just one row down in the group, to sulfur, and things start going horribly wrong. Sulfur, and the thiol/mercaptan analogs of otherwise relatively benign hydrocarbons and alcohols, gives us the thrilling eaudeur of rotten eggs, burning tires, skunks, farts, etc.
Move one row further, to selenium, and abandon all hope ye who enter here; the tamest common example of selenium in our daily lives is garlic, an instantly recognizable scent, especially when your body starts excreting the selenium-containing essential oil back out of your pores. It's supposed to be the last word in mosquito repellent; down here in Texas the little biters apparently have a taste for Italian. Selenium analogs of thiols often defy chemists' descriptions of their olfactory power; one of Dr. Lowe's other selenium-based posts, on selenophenol (the selenium analog of thiophenol, which is bad enough), puts it thusly: "Imagine 6 skunks wrapped in rubber innertubes and the whole thing is set ablaze; that might approach the metaphysical stench of this material".
This poses the rather interesting question; why do things change so suddenly with this group? I'm willing to believe that all of them have similar chemical properties, and we as humans and other oxygen-respiring life forms differentiate oxygen from the rest of the group because oxygen's good for us while the others are bad. But, that borders on the simplistic, and poses more questions, like why we work so much better on oxygen than sulfur (there are a few bacterial species hanging out near deep-sea vents that just can't get enough of those sulfur gases, so obviously there's sulfur-based "organic" metabolism just as there is oxygen-based). I guess what I'm looking for is an explanation of why two chalcogens, both obeying the octet rule as Periods 2 and 3 tend to, are treated so differently by our sensibilities.

I'm not confident enough in this supposition to make it an answer, but I suspect that it's probably because sulfur-containing compounds are common and volatile products of the fungal and bacterial decomposition of organic matter (namely, sulfur-containing amino acids). This is likely why amines also smell bad (cadaverine and putrescine, for example, being common products of biological decay), as well as many carboxylic acid (butyric acid, for example, is produced when certain fats break down). Being averse to these odors is therefore probably an evolved and advantageous genetic trait.

I think that there are two reasons: the sulfur is quite common in biochemistry and at the same time a lot of sulfur compounds are toxic. The bio-degradation of sulfur compounds mostly leads to some sulfides which very strongly bind transition metals, i.e. they can destroy/block Metalloenzymes. If you add some millions years of evolution it's a v.good feature to smell them (that would also work for small phosphines and amines). Your chances for food-poisoning are lower, chance for survival higher (Long Live "The selfish gene").

They stink because the nasal receptors activated by those compounds trigger nerve pathways that activate neurons corresponding to the perception of stench. The question is phrased as if the elements themselves were responsible for stench, when really it is the nasal receptor cells/proteins and neuronal pathways that are to blame. This questions belongs on biology.se unless the question is rephrased to inquire about the mechanism of protein receptor binding to said elements.

This sidesteps the actual question; oxygen doesn't have a repulsive smell, but most sulfur and selenium compounds do. The reaction chemistry of these three, though not identical, is very similar (hence their positions in a single column of the Table). A biologist would pretty much give Kris_R's answer, but I'm looking for something more systematic and chemistry-based that would explain how a receptor could signal strongly on sulfur or selenium but not oxygen.

What you wish to know is which protein odor receptor(s) detects those elements respectively and why they respond more strongly to one than another...


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Balancing a chemical equation
I'm in chemistry and I still have a hard time comprehending how to balance a chemical equation.
What are the main steps and general things to know?
thank you.

OK, here goes a semester of college chemistry in one post...
First off, remember the law of conservation of matter; in chemical reactions, matter is neither created nor destroyed. Therefore, every atom of every element on the left side of the arrow has to appear on the right side, with all their protons and neutrons intact. No strictly chemical reaction will cause a carbon to become a nitrogen by gaining a proton.
What can change in a reaction is the number of electrons each atom "owns". For that, remember the concept of valence shells; the number of electrons in the outermost "shell" of its orbitals. Most elements do not have a stable valence shell, and so will want to react in ways that increase their stability. For elements not in the "transition metal" groups, this is pretty straightforward; elements to the left of the transition metals generally want to lose electrons to empty their valence shell (acquiring the stable configuration of the next lightest noble gas), while elements to the right of the transition metals want to share or gain electrons to fill their valence shell, acquiring the configuration of the next heaviest noble gas. The transition metals are "stable" in various valence configurations based on their more complex d-orbital valence structure (the s and p orbitals exposed by the first two and last six columns of the table are more predictable).
When atoms outright gain or lose electrons, they become ions, and gain a charge based on the difference between the protons and electrons in their configuration. When they gain them, they become negative anions, and then they lose them, positive cations. Cations are then attracted to anions, and these charges must balance; a +2 cation such as Mg2+ or Ca2+ requires one -2 anion (carbonate, sulfate, hydrogen-phosphate) or two -1 anions (chloride, fluoride, bicarbonate).
The alkali metals, in the 1 and 2 groups, usually want to lose their one or two valence electrons; the heavier the alkali metal, the more the atom wants to lose the electrons, with cesium being the most reactive alkali metal that has a non-radioactive isotope. The elements in the halogen group (fluorine, chlorine, bromine, iodine) like to gain electrons, especially from the alkalis. The lighter the compound, the more it wants to gain an electron, with fluorine being the most aggressive electron receptor in the table (to the point where it will try to take electrons from other elements that don't want to give them up, like other halogens and noble gases).
Columns between the halogens and transition metals (from left to right, the boron group or triels, the carbon group or tetragens, nitrogen group or pnictogens, and oxygen group or chalcogens) are more of the "sharing" type; they'll take electrons outright from the alkalis, and if they really have to they'll give them up to a halogen like fluorine or chlorine, but they prefer a "give-and-take" relationship, and form shared-orbital "covalent bonds" with most other elements of these groups and with the transition metals. How strong these bonds are depend on what elements are being bonded and how many electrons each is sharing with its neighbor. A double or triple bond is very strong, while certain other combinations are strong for other reasons (bonds between carbon and the "Big 3" highly electronegative elements, namely oxygen, fluorine and chlorine, are commonly very strong). You can compute a balanced bond structure using valence electrons, much like calculating ionic charges; oxygen needs two valence electrons and so it will form no more than two bonds with other elements. When two atoms in these columns get together, expect to see a few multiple bonds; oxygen and nitrogen love forming double and triple bonds when they can, and carbon will oblige with these two and a few others, but typically prefers single bonds with other carbons and most other nonmetals (allowing it to form those long chains that create the backbone of life).
When determining what the reaction is, keep in mind that there has to be a driving force that causes an equilibrium shift. This can be a number of things, but basically, the reactants have to want to react and produce a product more than they want to stay where they are, and also more than any other component of their environment wants them not to react. Compounds generally want to react when the bonds of the newly-formed compounds are of lower energy and therefore more stable; such reactions are exothermic. Compounds can be forced to react when additional energy is added to the system (usually in the form of heat, sometimes with electricity to get the electrons jumping).
Compounds also generally want to react when they're attracted so strongly that nothing can keep them apart, so learn your solubility rules. Sodium chloride and aluminum bromide will both sit in solution together quite happily as dissociated ions; you can call some of it sodium bromide and some of it aluminum chloride, but because all four possible products are soluble and no electrons change hands, the solution is already in equilibrium and so there's no reaction. However, calcium chloride and sodium carbonate will react because one of the possible cation/anion combinations, calcium carbonate, is insoluble in water; these two will meet, bond, and precipitate out, leaving sodium chloride in the water.

Basically, balanced chemical reactions must 1) have the same number of each element on each side, and 2) have the same net charge on each side. Consider the reaction $H_2 + O_2 → H_2O$. It is unbalanced because there are two oxygen atoms on the left side but only one on the right. If we assign variable coefficients to each reactant and product, $a H_2 + b O_2 → c H_2O$, we can set up a system of equations by multiplying the variables by the number of atoms in each species. For oxygen, $2b = c$ and for hydrogen, $2a = 2c$. While the system has infinitely many solutions, we take the lowest integer solution. Thus, the balanced equation is $2 H_2 + O_2 → 2 H_2O$.
Chemical reactions involving charges are a bit more complicated. A good explanation can be found this tutorial.

ok that makes sense. my professor rushes over everything too quick to pick up everything. thanks.

From a paper I wrote on the algebraic method for balancing chemical equations:
Instead of the inspection method, shown in most chemistry texts, a more general method of balancing chemical reaction equations is the algebraic method, proposed by the Russian academician, V. A. Kistiakovskij, in 1919. Let’s illustrate the method using the following chemical reaction, which is a complete oxidation (or burning) of pentane fuel in pure oxygen to give carbon dioxide and water as the only products:
C5H12 + O2 = CO2 + H2O
The idea is to write down the answer, only using algebraic letters instead of numbers, and then solve the system of equations resulting to obtain the numerical values of the coefficients:
aC5H12 + bO2 = cCO2 + dH2O
It turns out that most of the time, there is one more variable (the algebraic coefficients we set) than equations (different atom balances), and in these cases, we can set a = 1 and solve the equations using this starting value. This works because as all good chemists know, the compounds in a chemical equation react and form in ratio amounts, always (the mole concept).
The next step in the process is to set up the system of equations to solve, which are just balances of atoms of each element on the left and right sides of the original equation (required according to the law of conservation of mass).
In each case, the number of atoms of a particular element is equal to the coefficient in front of the compound containing that element multiplied by the subscript on the element in that compound, added to any other instances of the element in other compounds on either side of the equal sign in the original equation.
For the example, we have balances on carbon (C), hydrogen (H) and oxygen (O):
C: 5a = c
H: 12a = 2d
O: 2b = 2c + d
Substituting a = 1, we have:
5 = c (so c = 5)
12 = 2d (so d = 6)
2b = 2•5 + 6 = 16 (so b = 8)
Substituting the values obtained from the coefficients back into the original equation gives:
C5H12 + 8O2 = 5CO2 + 6H2O
The last step is to do a quick check to make sure our algebra was done correctly:
C: 5 = 5 (check)
H: 12 = 6•2 (check)
O: 8•2 = 5•2 + 6 (check)
This method can be employed for the vast majority of chemical reactions studied in high school and even beginning college chemistry, in both the inorganic and organic worlds. More importantly, it can be automated, and so it is used as the algorithm that does the calculating for on-line and off-line, computer-based chemical equation balancers. But it is simple enough that for most chemical reactions studied in high school, it can be used by students with some knowledge of algebra, given a pencil and paper, and for more difficult reactions, can be completed faster than the inspection method.
It should be noted that fractional coefficients are often obtained using the substitution
a = 1 shown above. Since the convention is that coefficients of chemical equations are whole numbers, to get the final answer it is often required to multiply the set of coefficients through by the lowest common multiple (LCM or LCD) of all the denominators of the fractional values.
An example used in several papers in the Journal of Chemical Education from the 1930s considers the famous experiment of putting a copper penny in a flask of nitric acid, which dissolves the penny, producing copper nitrate, while giving off nitric oxide (which quickly oxidizes in air to the brown gas nitrogen dioxide):
Cu + HNO3 = NO + Cu(NO3)2 + H2O
Using the algebraic method, we write down the answer using algebraic letters as coefficients:
aCu + bHNO3 = cNO + dCu(NO3)2 + eH2O
Then do the atomic balances on each element (I always go from left to right):
Cu (copper): a = d
H (hydrogen): b = 2e
N (nitrogen): b = c + 2d (remember that the subscripts occurring after parentheses are distributed into the elements inside like exponents in algebra)
O (oxygen): 3b = c + 6d + e
Now set a = 1 and solve:
1 = d (so d = 1)
Substituting b = 2e and d = 1 into the nitrogen and oxygen balances gives:
5e = c + 6 (oxygen balance)
 – 2e = c + 2 (nitrogen balance)
   3e = 0 + 4 (subtracting the bottom equation from the top one)
So e = 4/3, and therefore b = 8/3 (from the hydrogen balance) and c = b – 2d (from the nitrogen balance) or c = 8/3 – 2 = 2/3. The coefficients are therefore:
a = 1
b = 8/3
c = 2/3
d = 1
e = 4/3
The LCM of the denominators of the fractions is 3, so we multiply through by 3, giving:
a = 3
b = 8
c = 2
d = 3
e = 4
The balanced equation is:
3Cu + 8HNO3 = 2NO + 3Cu(NO3)2 + 4H2O
Our quick check shows:
Cu: 3 =3 (check)
H: 8 = 4•2 (check)
N: 8 = 2 + 2•3 (check)
O: 8•3 = 2 + 6•3 + 4 (check)

You mention a paper you wrote - care to provide the reference?

So far published only in this blog. Used (by me) extensively in tutoring and teaching sessions. Reference for the method is from a company selling software based on this method. Other references were from the Journal of Chemical Education, where the method was published previously, in the 1930s.

Balancing a chemical equations is fundamentals of chemistry. There is one recommended way to balance the equations is that you balance the Oxygen in the equation first, do not use fractions only natural numbers are allowed.


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How can I make a graphene supercapacitor?
I have heard that a super capacitor can be made with a dvd burner. What materials do I need to to get and what software for my DVD burner do I need to make this amazing device?

Could you provide a link to where you heard/read about this? It would help use help you.

http://io9.com/5987086/meet-the-scientific-accident-that-could-change-the-world

Chemistry World UK have made this YouTube Clip that has a professor from UCLA explaining what is needed and how to go about it - even though the video is about 2.5 minutes, but it is a good demonstration.  The associated article is DVD player burns graphene to disc
Additionally, according to this JensLabs page DIY Graphene, it states that the following readily available materials are needed:

Graphite oxide
Light scribe capable DVD burner
Light scribe DVD
Substrate
Sonicator - the article suggests that a ultrasonic cleaner would suffice
Pipette
Ion-porous separator
Electrolyte


text:
How do you change the cis-trans isomerism of compounds?
I would like to know a general procedure for the synthesising cis isomers from their trans counterparts and vice versa.

In general, any setup that gives enough energy to the compound can isomerize it. This can be done by heating it, irradiating it with visible or UV light, or by using a catalyst (reducing the energy required for transition rather than supplying more energy). Excitation of electrons from the $\pi$ to the $\pi^*$ (bonding orbital to anti-bonding orbital) allow free rotation about the $\sigma$ bond. Relaxation back to the $\pi$ orbital locks the configuration. The specific procedures depend on the compound of interest.
FYI, photoisomerization (light-induced isomerization) is the principle on which sight is based.

By the way does photoisomerisation only induce the transition between cis and trans isomers? i.e. can dextrorotatory and levorotatory swapping occur due to photoisomerisation?

@BrentonHorne Changing from cis to trans may change the molecule from + to -, but it will not change the R/S configuration of any stereocenter. The +/- property does not depend on the absolute configuration of the stereocenters in any useful way.

As a particular example of what buckminst is talking about, one common undergraduate organic chemistry laboratory experiment is the one where a tube full of dimethyl maleate is isomerized to dimethyl fumarate, with the help of sunlight and a teeny amount of bromine. Since solid dimethyl fumarate has low solubility in liquid dimethyl maleate, this makes for a nice visual demonstration. See e.g. this and this.


text:
Tetrahedral or Square Planar
How can one predict whether a given complex ion will be square planar or tetrahedral when its coordination number is 4 using crystal field theory ?
Is it possible to theoretically predict this?

This is a really quick and dirty method since I don't know much about the calculations involved in crystal field theory. This method is based on some general observations I've made.
First of all, if your complex metal ion is not in group 8 ($\ce{Ni}$, $\ce{Pd}$, $\ce{Pt}$) or it doesn't have a $d^8$ configuration (like $\ce{Rh(I)}$ or $\ce{Au(III)}$), it is usually tetrahedral (Assuming you already know that the coordination number is 4).
If your metal ion is in group 8 or has a $d^8$ configuration, look at the crystal field splitting diagram. Square planar complexes have a four tiered diagram (i.e. four different sets of orbitals with different energies).

If it has a two tiered crystal field splitting diagram then it is tetrahedral.

But this assumes you have the crystal field splitting diagram of the complex.
As to how you obtain these diagrams (the calculations involved), I don't know exactly how it's done for specific molecules.
Like I mentioned before, this is just a very basic way to distinguish between the two geometries. I'm sure the calculations involved to obtain these diagrams would be really complicated but would make more sense in some cases if you're looking for the actual reasons behind why some complexes are square planar while others are tetrahedral.
EDIT:
Predicting the structure of a complex depends on how much info you're given. If you only have the number of ligands in the complex or the molecular formula of the complex, then you can predict the structure based on this list here.
Then again, steric factors and electronic factors are also important in determining the structure and your prediction would be more accurate if you know what ligand(s) you're dealing with (for steric effects) and the electronic configuration and oxidation state of the central metal ion.
I don't know much about how exactly these factors influence the structure but in general, you only have to be worried about these if:

the ligands are bulky or have some other special features
certain special electronic configurations (I've read something about certain electronic configurations causing distortions in the structure especially in copper(II) complexes)

This is a good start. For the first row transition metals, you also need to know whether the electron configuration lends itself to high spin or low spin. The second and third row metals are always low spin.

Supposedly , I am given a complex ion of the first period of transition elements and asked to predict its geometry , then how to do it ? Is there a way , to do this ? I am not being given experimental data or anything

@BenNorris Thanks for mentioning that. I completely overlooked that point.

Seems there's a typo... Ni, Pd, Pt are in group 10.

There are all sorts of approximations available, but the short of it is that in general one cannot predict square planar vs tetrahedral without either doing experiments or doing some very detailed calculations.
Put another way, ligand field theories and the like are after the fact justifications of observations rather than predictions of observations.

Our teacher told us this trick to tell if complex is going to be square planar.
We are considering the fact that the coordination no. is 4.
If the metal has a $\ce{d^7}$, $\ce{d^8}$ or $\ce{d^9}$ configuration along with a strong field ligand or $\ce{d^4}$ with weak field ligand then complex will be square planar otherwise tetrahedral.
One exception i have come across for this is $\ce{[PtCl4]^2-}$ which is a square planar. You can read the explanation for it here.

How can pairing happen in $\ce{d^9}$ configuration?

A method I use requires the memorization of the spectrochemical series, which arranges the ligands in increasing order of ligand strength. A strong ligand will almost always be able to cause pairing of electrons, producing a low spin, inner orbital complex, which in case of the coordination number 4, may correspond to the $dsp^2$ or $d^3s$ hybridization.
For instance,
$\ce{Ni^{+2}}$ produces a square planar complex with $\ce{CN-}$, as cyanide is a strong ligand. Nickel carbonyl, on the other hand, exists as a tetrahedral complex because nickel's oxidation is 0.
You shall be able to use this method most of the time, but in certain cases, there are some exceptions. For instance, Cobaltic ions behave differently than other $3d$ metal ions, and form a low spin complex with $\ce{H2O}$.
Good Luck!


text:
What is Molar Conductivity of an electrolytic solution
Please explain the physical meaning it carries ? And how is it that it can increase with dilution ?
And why is it called Molar ?
And what is equivalent conductivity ?

Some of the terms you're asking about can be easily found on the net. Did you try researching first?

I will answer your question about the definition of molar conductivity and why it increases with decreasing concentration.
As Jerry pointed out, you can probably learn what you need about equivalent conductivity using a Google search.
The molar conductivity of an electrolyte solution is the conductivity of the solution divided by the molar concentration of the dissolved electrolyte.
Conductivity ($\kappa$, in $\text{S/cm}$) is proportional to the square root of the concentration of ions ($\sqrt{[\text{ions}]}$, withy $\ce{[ions]}$ in $\text{mol/L}$). In the following equation, A is an empirical proportionality constant with units needed to convert from $\text{M}^{1/2}$ to $\text{S/cm}$, namely $\frac{\text{S/cm}}{\text{M}^{1/2}}$
$$\kappa = A\sqrt{\ce{[ions]}}$$
The ion concentration is related to the electrolyte concentration ($C$) by the van't Hoff factor ($i$).
$$[\ce{ions}]=iC$$
$$\therefore \ \kappa = A\sqrt{iC}$$
The molar conductivity $\kappa_M$ is the conductivity divided by the molar concentration of electrolyte ($C$).
$$\kappa_M=\frac{\kappa}{C}=\frac{A\sqrt{iC}}{C}=A\frac{\sqrt{i}}{\sqrt{C}}$$
As $C$ decreases, so does $\sqrt{C}$, and thus $\frac{1}{\sqrt{C}}$ increases, while $A$ remains constant. The van't Hoff factor changes a little with concentration, but not dramatically. ($A$ probably also varies a little with concentration. It certainly varies with temperature.)
For example, if you have a series of solutions of sodium chloride, concentration decreases faster than conductivity, so conductivity divided by concentration increases. Note that the conductivity data are fabricated with an arbitrary value of $A$. I could not find the specific value of $A$ for sodium chloride. The van't Hoff Factors come from the "Colligative Properties" Chapter of Chemistry: Atoms First by Burdge and Overby, McGraw Hill, 2012.

Conductivity: The degree to which a specified material conducts electricity is called conductivity.
Molar conductivity: Conductivity of a solution on dilution. Conductance and conductivity both decrease but molar conductivity increases.
Conductivity is directly proportional to the concentration.

Molar conductivity of an electrolyte is defined as the conductivity of a volume of solution containing one mole of electrolyte.

You didn't answer the other questions the OP asked.


text:
Salt Hydrolysis
Referring to this, when "hydrolysis of a salt" is its reaction with water:
How is it that both the cation and anion of the salt are reacting in an equal amount with water?
To state this specifically in reference to the link provided, how is it that there is a unique $h$ (degree of hydrolysis)?

I feel a bit icky about the site you linked. And why did you delete your previous question to ask it again in another account? Doesn't all of the $CH_3COONH_4$ get hydrolysed? remember that all ammonium salts are soluble in water.

I don't see much value in that particular method of viewing this phenomena.

@Lighthart: I dunno... the textbook way(seen it multiple times) of explaining hydrolysis of weak acid-weak base salts assumes that the degree of hydrolysis is the same for both ions. It was a question that occurred to me and many of my friends when I was taught this. IIRC the answer has to do with the dissociation of water limiting the degrees of hydrolysis. So it seems like a legit question to me :) If you feel like this should be closed, just let me know with a more detailed explanation.

The question is legit, I just never found much educational value in this analysis.  I always preferred to look at this in the context of solvolysis and free energy.  Tim Toady.

I think @ManishEarth has it right.  The actual phenomenon involves not only the hydrolysis of the salt, but the autoionization of water.  To keep things simple the autoionization of water in this case is often omitted in elementary textbooks.
In reality the hydrolysis of the anion and the cation are not normally identical and the resulting solution is slightly acidic or basic, depending on the nature of the ions.
Nevertheless, for each such reaction there is a definite amount of hydrolysis (which you call $h$ in the question).  This constant will vary somewhat with temperature.


text:
Does tolerance to alcohol affect blood/breath alcohol levels?
When blood / breath alcohol levels are taken to determine how intoxicated a person is, does it depend on their tolerance ?
i.e. If you had two people of identical age, weight, fitness, etc.
person A, a non drinker, drank 1 measure of alcohol over a period and felt quite "drunk"
while person B, a frequent drinker drank the same measure of alcohol over the same period, but because of their built up tolerance felt and acted considerably more sober.
would they give the same blood, or breath alcohol readings, despite the effects on the person being different ?

Sorry if this is a very poorly formatted question from a chemistry point of view, i know absolutely nothing, nor have interacted with any of the sciences since school.
I was also not sure whether this comes under chemistry or biology, if it needs changing to the biology stack exchange site please advise

Typical roadside 'breath-testing' measures metabolites produced in the metabolism of alcohol (primarily, they measure ethanal, oxidised ethanol). The level measured level of intoxication therefore definitely differs from person to person, and even day-to-day for everyone. Some people have higher metabolic rates, and will metabolise the alochol faster. Some people have greater amounts of alcohol dehydrogenase (enzyme that breaks down alcohol) than others. It follows, that a larger person will likely be able to consume more alcohol compared to a smaller person (compare the concentration of 1 g of salt in a small glass, compared to a large glass)
However, there isn't really any fudge room for arguing against the tests. And most police-forces will back up breath testing with far more rigorous blood tests.

This doesn't answer the question. Will built up tolerance affect measure and how?

authors: w̢in̡͢g͘̕ed̨p̢͟a͞nt̡͜͝he̸r̴


text:
Why do transition elements make colored compounds?
Why do transition metals element  make colored compounds both in solid form and in solution? Is it related to their electrons?

it is due to d-d transition and charge transfer spectra

You are absolutely correct, it all about the metal's electrons and also about their d orbitals.
Transition elements are usually characterised by having d orbitals. Now when the metal is not bonded to anything else, these d orbitals are degenerate, meaning that they all have the same energy level.
However when the metal starts bonding with other ligands, this changes. Due to the different symmetries of the d orbitals and the inductive effects of the ligands on the electrons, the d orbitals split apart and become non-degenerate (have different energy levels).
This forms the basis of Crystal Field Theory. How these d orbitals split depend on the geometry  of the compound that is formed. For example if an octahedral metal complex is formed, the energy of the d orbitals will look like this:

As you can see, previously the d orbitals were of the same energy, but now 2 of the orbitals are higher in energy. Now what does this have to do with its colour?
Well, electrons are able to absorb certain frequencies of electromagnetic radiation to get promoted to higher energy orbitals. These frequencies have a certain energy which correspond to the energy difference between different orbitals. Now most substances are only able to absorb frequencies of radiation which are outside the visible light spectrum, for example they might be able to absorb radiation which has a frequency of $300$GhZ (that is infrared radiation). This means that it reflects all other types of radiation, including the full spectrum of visible light. So our eyes see a mixture of all the colours; red, green, blue, violet, etc. This is seen as white (this is why several organic compounds are white).
However transition metals are special in that the energy difference between the non-degenerate d orbitals correspond to the energy of radiation of the visible light spectrum. This means that when we look at the metal complex, we don't see the entire visible light spectrum, but only a part of it.
So for example, if the electrons in an octahedral metal complex are able to absorb green light and get promoted from the $d_{yz}$ orbital to the $d_{z^2}$ orbital, the compound will reflect all other colours except for green. Therefore by using the colour wheel, we can find the complementary colour of green which will be the colour of the compound, which is magneta.

This explains why not all transition metal complexes are colourful. For example copper sulfate is a bright blue compound, however zinc sulfate on the hand is a white compound despite being a transition metal. The reason behind this is because zinc's d orbitals are completely filled up with electrons, meaning that it is not possible for any electron to make a d-> d transition as they are all filled up. Hence you might sometimes see zinc referred as not being a transition metal.

The crystal field theory is deprecated, but what you describe is also valid for the ligand field theory, which should be used instead now.

Then how come KMnO4 has color (a very deep purple)? The manganese would have an oxidation state of 7-, which means the atom would no longer have occupied d orbitals at all.

A broader question: How does crystal field theory apply to transition metals that are part of polyatomic ions?

Won't the electrons emit the light in the same frequency as it absorbed when it comes back to the ground state?

Nanoputian, the complementary colour explanation works, but the comment above asks a good question. Is it that the electron excitation-deexcitation method of explaining this just doesn't work?

@GovindBalajiS https://chemistry.stackexchange.com/a/99638/108785
I hope this answer by some other user answers your query.

The partially full d-orbitals in transition metals have energy splittings that happen to lie in the visible range.  Depending on the arrangement of substituents (known as ligands) that attach to them, the electron energies split according to crystal field theory.  Similar splitting in the s or p orbitals produce gaps in the ultraviolet, and any visible light goes right through, so we don't see any color.  In transition metals, however, visible light excites the electrons from a lower d orbital to a higher one and only letting some light through.

A perhaps deeper explanation for the first sentence in your answer can be found in this article. It seems that the fact the 3d orbitals are anomalously compact explains why many electronic transitions in fourth period d-block metal complexes are in the visible region.

Why can't electron transition take place between fully filled split d orbitals?

Because of the Pauli exclusion principle that says that two electrons can't be in the same place at the same time. For the transition to take place, an electron has to move from a lower split level to a higher split level. If there is no empty spot, no transition can take place.

How we are sure that the color doesn't come from the transition between p (filled) to s (empty) orbital?

Transition elements have partially filled d orbitals. We also know that when electrons jump from one orbital to another light is emitted due to which the compounds of transition elements seem to be colored compounds.
Note: This can also happen in some organic compounds though in this case, it is p orbitals and not d orbitals.

In case of transition metals ($3d$) why we can't have a transition from $4s$ to $5s$ where would be in the visible range?

Firstly, in complexes usually the 4s electrons are removed to form the central positive charged ion. Secondly, the energy gap between the 4s and 5s orbitals is extremely high that visible light photons does not have enough energy to overcome. Such transition would requre photons of at least ultraviolet light. And we are not supposed to see ultraviolet light. In short, it is theoretically possible for a 4s-5s transfer, but we cannot see it.

Colored compounds of transition elements are associated with partially filled (n-1)d orbitals. Tthe transition metal ions containing unpaired d-electrons undergoes an electronic transition from one d-orbital to another. During this d-d transition process, the electrons absorb certain energy from the radiation and emit the remainder of energy as colored light. The color of ion is complementary of the color absorbed by it. hence, colored ion is formed due to d-d transition which falls in the visible region for all transition elements.


text:
Strength of intramolecular vs intermolecular hydrogen bonds
Why are intramolecular hydrogen bonds weaker than intermolecular hydrogen bonds?

The premise of the question isn't true and OP didn't even mention why he would think so...

@Mithoron I actually like the question because of its flaws and brevity. I could even see it in textbooks of a country presented as fact… (if you catch my drift).

@Martin-マーチン The books of that country don't state this as a fact, but many teachers keep making such nonsense statements.

I don't think there is much of a difference between the strengths of intramolecular or intermolecular hydrogen bonds.
However the strength of interactions between molecules depends on the type of hydrogen bond.
Consider a typical example: ortho-Nitrophenol has a lower melting and boiling point than para-Nitrophenol.

The interactions between molecules are much stronger when there are intermolecular hydrogen bonds (as in para-Nitrophenol) because the bonds are formed between molecules. This results in more stability and higher melting and boiling points (which might appear to be because of stronger bonds) because the molecules prefer to be closer to each other.
If intramolecular hydrogen bonds are present (as in the case of ortho-Nitrophenol) the interactions within individual molecules are greater and so there is less attraction between molecules. This causes less stability and lower melting and boiling points because the molecules are not really attracted to each other much and not because the hydrogen bonds are weaker.

My textbook said that. And also, can we say that due to "strain", intramolecular hydrogen bond is weaker ?

Intermolecular: More hydrogen bonds between molecules indeed increase stability, but it leads to higher melting/ boiling points, opposite to what you have stated. Intramolecular: Hydrogen bonds within a molecule lead to lesser interactions between molecules, hence the melting/ boiling points decrease.

Second picture is quite wrong.

@Martin-マーチン Do hydrogen bond within a molecule necessarily lead to fewer interactions between molecules? If so, could you please explain why?

"I don't think there is much of a difference between the strengths of intramolecular or intermolecular hydrogen bonds".Thats nonsense!The strength of the covalent bond is much much stronger than the intermolecular force between 2 H2.If that was the case H2 would never be formed at room temperature.

@Mithoron Why is the second picture wrong?

@Wolgwang First one isn't great too, but hydrogen bonding, like in the second one, simply isn't possible.

It is mainly due to kinetic motion in molecules
The kinetic energy of molecules follows the Boltzmann distribution. Therefore, at room temperature molecules are constantly moving.
Intermolecular interaction, such as intermolecular hydrogen bonding only take place for a short time when two molecules are in each other's vicinity. Shortly after, the molecules separate and the interaction does no longer exist.
Intramolecular interactions on the other hand are much stronger since the two atoms forming the hydrogen bond remain in each others proximity due to structure of the molecule. In other words, they exhibit strain.

Typically hydrogen bonds are between a hydrogen atom attached to an electronegative atom, $\ce{-XH}$, and another electronegative atom, $\ce{-Y}$:
$$\ce{R^1\bond{-}XH\bond{...}Y-R^2}$$
It is mostly an electrostatic interaction, but there are orbital interactions, too. While the first is independent of direction, for the latter that is actually the crucial part.
In first order approximation the bonding in $\ce{-XH}$ is a σ-orbital with a corresponding σ*-orbital. These have their "densities" opposite of each other through hydrogen, hence their angle is $180°$. The other electronegative atom donates electron density into the σ*-orbital.
Therefore the best overlap is with a linear coordination around hydrogen.
However, orbital contribution is small even for strong hydrogen bonds.
Intermolecular hydrogen bonds typically achieve this arrangement more easily, while intramolecular hydrogen bonds are typically more bent.
You may find correlation with that. Correlation is not causation. There are many more factors at play, which must all be carefully considered.
If you take away any rule of thumb here, it should be: Strong hydrogen bonds are closer to a linear coordination around hydrogen.

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


text:
How was the quantum three-body problem solved?
What method was used to solve the quantum three-body problem?  I heard it was solved and was not sure about the status of the problem.

It never was solved, analytically (which means an exact mathematical solution)
It was solved approximately, using perturbation theories.  In the case of the helium atom, an estimate of the energy electron-electron repulsion is added to the calculation.  The appropriateness of the perturbation added is based on the comparison of the result to experimentally determined values.

Three-body special cases can have closed solutions.  For $\ce{H2+}$, the heavy nuclei are assumed to be stationary while the electron travels. Confocal ellipsoidal coordinates then describe the system to first order, followed by perturbation treatments because the nuclei do move. As stated, there are no general analytical forms for the three-body problem overall.


text:
Molecular orbital theory & predicting the stability of a molecule?
A question in my general chemistry textbook gives me a list of formulas for molecules and ions. It then asks me to determine, using molecular orbital theory, whether or not each would exist in a relatively stable form.
How can I tell this? The chapter covers molecular shapes, valence bond theory, and molecular orbital theory which all builds on what was taught in the previous chapter (mainly on Lewis structures.) I can't tell what concepts to use in telling what will be stable and what won't.

I can't tell which concept I should use it depends on the molecules and what you've seen so far on molecular orbital theory. Are they simple, inorganic molecules or more complex organic molecules? Could you provide a list of said molecules?

You need to mix the orbitals, populate them with the electrons and see if you have net bonding.
Eg:  H + H two 1s orbitals mix to form sigma and sigma*.  Two electrons total, both occupy the sigma orbital, two more electrons in bonding than antibonding orbitals, the compound is stable.

Eg:  He + He; same mixing as above.  Four electrons, two in the sigma, two in the sigma*.  Since there are as many bonding electrons as as antibonding, there is no net bond.  He2 is not possible.

Eg:  He + H; same mixing as above.  Three electrons, two in sigma, one in sigma*.  One more electron in bonding than antibonding.  He-H forms a very weak bond.  Please note the diagram is for He2+ but the He-H is very similar

Eg: Li + H; Li has 1s + 2s, while H has 1s.  This mix to form a sigma orbital from H1s+Li2s, a sigma* orbital and H1s-Li2s, and a non bonding orbital from Li1s (lower in energy than the sigma).  Four total electrons.  Nonbonding sigma is occupied, and then the sigma orbital is occupied.  Net effect: Li-H forms a stable bond.

You might have noticed that the first three diagrams had identical structure, even if their energy levels were different.  This is a property of quantum mechanical symmetry, and solving the problem only requires knowing a few diagram types and filling the electrons in.  Consider the second row.  The diagram is as :

You simply count the valence electrons and fill in the diagram.  Then you subtract the number of bonding orbital electrons from antibonding, and if you have more, you will have a bond.
A couple of webpages are very good for further explanation:
Source of the diagrams:
http://www.chem1.com/acad/webtext/chembond/cb08.html
and
http://www.meta-synthesis.com/webbook/39_diatomics/diatomics.html

Is there a particular reason why the same graphic is included in this post three times?

The diagram is topologically the same.  Also I wrote this three years ago when I was newer to site.  The post itself explains.


text:
Why is heating increasing the solubility of sodium hydroxide?
\begin{align}
\ce{NaOH &-> Na+ + OH-} & \Delta H^\circ &= \pu{-44.51 kJ/mol}
\end{align}
The dissolution of sodium hydroxide in water is an exothermic process, and so, according to Le Chatelier’s principle, cooling the container should shift the reaction to the right. Shouldn’t this mean that cooling the container will increase the solubility?
According to a solubility chart for sodium hydroxide, heating the container will increase the solubility; why is this the case?

The question to the Newton - Ask a Scientist program by Danna C Griffiths (to be found via the Internet Archive) has an answer for it.
What I understood from the explanation on that site:
Solubility is defined as the concentration of the solute in a saturated solution. So when considering the increase in solubility with temperature, you have to check the enthalpy of solution of $\ce{NaOH}$ in a saturated solution of $\ce{NaOH}$.
When you dissolve $\ce{NaOH}$ in pure water, the process is highly exothermic, but as the concentration of $\ce{NaOH}$ increases, the process becomes less exothermic and eventually near the saturation point it becomes endothermic (it becomes less favourable for more $\ce{NaOH}$ to dissolve).
So according to Le Chatelier’s Principle, although the dissolution process is overall exothermic, since solubility is determined only at the saturation point, the solubility increases with increase in temperature.

This reaction has a positive change of entropy (ΔS > 0). Hence, from the equation ΔG = ΔH - TΔS, we find that its Gibbs energy will decrease with an  increase of the temperature. As a results, the solubility of NaOH increases when we increase the temperature (heating the container), since we have ΔG = -RTlnK.

No, it's the sign of the enthalpy, not the entropy, that determines the temperature-dependence of the equilibrium constant.   See:  https://en.wikipedia.org/wiki/Van_'t_Hoff_equation

According to
$$\Delta G = -RT \ln K$$
K is dependent on Temperature.
Le Chatelier's principle is explicitly a mass balance issue, and thus is not affected by temperature, although that distinction may be pedantic.

No. The Free Energy is a description of Le Chaetleier's principle.  However temperature is not a 'product' or 'reactant'  and so cannot 'shift' the equilibrium in any way traditionally related to the application of Le Chateliers's principle.

No, raising temp does affect the Keq (see Haber process, or NO2 -> N2O4). Plus, change in entropy in this reaction is negative not positive (-11 for OH + 59 for Na - 64 for solid NaOH = -16 J/K*mol).

T does affect Keq, per DG=-RT ln K above.  However, this is not leChatelier's principle.  Also, the entropies you've quoted are not relevant; you need the solvated ions.

It is part of Le Chatelier's principle... (see here) The entropies above are of the solvated ions Na+ and OH- (see here) Thus, the change in entropy of the reaction is negative.

Le Chateliers principle is explicitly mass balance.  Temperature is irrelevant in that context.  Using products minus reactant for entropy, 59 - 11 (products) - (-64) reactants = 112, a positive  entropy gain, based on your numbers.

It's not -64, it's +64. (-11 + 59) - (64) = -16 J/K*mol

I'll edit above.  Those numbers are unexpected.


text:
Hot and cold air interaction in a sealed environment
What would happen if air heated to 1200 degree's Fahrenheit in a sealed environment was injected with colder air?
No specific pressure defined yet just a general idea of potential interaction between the two.
Edit
My apologies, I meant within a sealed environment like an engine cylinder. I know that the hot air would naturally expand and push a piston but how would the injection of cold air alter this, if at all?
For example purposes we can assume the cylinder is iron while the piston is aluminum which is common design today.

Maybe this question should be moved to Physics SE.

@kaliaden: A note: While this is on topic on Phys.SE, it is not off topic here (gas laws/thermodynamics is fine here) -- migrations of such posts are only done if the OP requests or if they don't get answers in a few days :)

In order to work out this problem, several parameter values are required to define the initial state, preferably

amount $n_1$, volume $V_1$, or mass $m_1$ of the hot air in the cylinder
initial temperature $T_1$ of the hot air
initial pressure $p_1$ of the hot air
amount $n_2$, volume $V_2$, or mass $m_2$ of the injected cold air
initial temperature $T_2$ of the cold air
initial pressure $p_2$ of the cold air

Unfortunately, only the initial temperature of the hot air is defined as $T_1 = 1200\ ^\circ\mathrm F$, which is $T_1 \approx 649\ ^\circ\mathrm C \approx 922\ \mathrm K$ in proper units (the use of the unit degree Fahrenheit is deprecated). This leaves us with five unknown variables, which is certainly too many for a vivid analytical description.
Nevertheless, we may investigate this problem by guessing reasonable parameter values as follows. By way of example, we arbitrarily assume that equal masses of hot and cold air $(m_1 = m_2 = 1\ \mathrm{kg})$ are mixed. The assumed temperature of the cold air sample is equal to room temperature, i.e. $T_2 = 68\ ^\circ\mathrm F = 20\ ^\circ\mathrm C \approx 293\ \mathrm K$, whereas the temperature of the hot air sample is given as $T_1 = 1200\ ^\circ\mathrm F \approx 649\ ^\circ\mathrm C \approx 922\ \mathrm K$. The assumed pressures of the air samples is equal to normal atmospheric pressure, i.e. $p_1 = p_2 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$. These conditions define the initial state and also allow to determine other interesting properties using tabulated reference values for air, e.g. density $\rho$, Volume $V$, specific internal energy $u$, and internal energy $U$.
1. Hot air sample

temperature $T_1 = 1200\ ^\circ\mathrm F \approx 649\ ^\circ\mathrm C \approx 922\ \mathrm K$
pressure $p_1 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$
mass $m_1 = 1\ \mathrm{kg}$
density $\rho_1 = 0.3827\ \mathrm{kg/m^3}$
volume $V_1 = \frac{m_1}{\rho_1} = 2.613\ \mathrm{m^3}$
specific internal energy $u_1 = 819.4\ \mathrm{kJ/kg}$
internal energy $U_1 = u_1 \cdot m_1 = 819.4\ \mathrm{kJ}$

2. Cold air sample

temperature $T_2 = 68\ ^\circ\mathrm F = 20\ ^\circ\mathrm C \approx 293\ \mathrm K$
pressure $p_2 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$
mass $m_2 = 1\ \mathrm{kg}$
density $\rho_2 = 1.2046\ \mathrm{kg/m^3}$
volume $V_2 = \frac{m_2}{\rho_2} = 0.830\ \mathrm{m^3}$
specific internal energy $u_2 = 335.3\ \mathrm{kJ/kg}$
internal energy $U_2 = u_2 \cdot m_2 = 335.3\ \mathrm{kJ}$

3. Mixture at constant volume
The hot air sample is confined by a piston in a cylinder. The cold air sample is added, thus allowing the two air samples to mix. The total mass $m_3$ is given by
$$\begin{align}
m_3 &= m_1 + m_2 \\
&= 1\ \mathrm{kg} + 1\ \mathrm{kg} \\
&= 2\ \mathrm{kg}
\end{align}$$
For simplicity’s sake, we ignore any work $W$ performed by or on the system (e.g. the work that is required to inject the cold air into the cylinder). Furthermore, we assume that the process is adiabatic, i.e. no heat $Q$ is transferred into or out of the system. Therefore, in accordance with the first law of thermodynamics, the total internal energy $U_3$ is given by
$$\begin{align}
U_3 &= U_1 + U_2 \\
&= 819.4\ \mathrm{kJ} + 335.3\ \mathrm{kJ} \\
&= 1154.7\ \mathrm{kJ}
\end{align}$$
and the new specific internal energy $u_3$ is
$$\begin{align}
u_3 &= \frac{U_3}{m_3} \\
&= \frac{1154.7\ \mathrm{kJ}}{2\ \mathrm{kg}} \\
&= 577.3\ \mathrm{kJ/kg}
\end{align}$$
At constant volume $V$, i.e.
$$\begin{align}
V_3 &= V_1 \\
&= 2.613\ \mathrm{m^3}
\end{align}$$
the new density is
$$\begin{align}
\rho_3 &= \frac{m_3}{V_3} \\
&= \frac{2\ \mathrm{kg}}{2.613\ \mathrm{m^3}} \\
&= 0.7654\ \mathrm{kg/m^3}
\end{align}$$
The new temperature $T_3$ and pressure $p_3$ can be determined from the density $\rho_3$ and the specific enthalpy $h_3$ using tabulated reference values:
$$\begin{align}
T_3 &= 658\ ^\circ\mathrm F \approx 348\ ^\circ\mathrm C \approx 621\ \mathrm K \\
p_3 &= 1.37\ \mathrm{bar} = 0.137\ \mathrm{MPa}
\end{align}$$
4. Mixture at constant pressure (ignoring work)
Allowing the system to expand to normal atmospheric pressure by moving the piston, the new pressure $p_4$ is
$$p_4 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$$
We know that the mass $m$ of the air in the closed system does not change, i.e.
$$\begin{align}
m_4 &= m_3 \\
&= 2\ \mathrm{kg}
\end{align}$$
and we still (incorrectly) assume that no work $W$ is performed by or on the system and that no heat $Q$ is transferred into or out of the system. Therefore,
$$\begin{align}
U_4 &= U_3 \\
&= 1154.7\ \mathrm{kJ}
\end{align}$$
and thus
$$\begin{align}
u_4 &= u_3 \\
&= 577.3\ \mathrm{kJ/kg}
\end{align}$$
The new density $\rho_4$ can be determined from the pressure $p_4$ and the specific internal energy $u_4$ using tabulated reference values:
$$\rho_4 = 0.5681\ \mathrm{kg/m^3}$$
Thus, the new volume $V_4$ is
$$\begin{align}
V_4 &= \frac{m_4}{\rho_4} \\
&= \frac{2\ \mathrm{kg}}{0.5681\ \mathrm{kg/m^3}} \\
&= 3.521\ \mathrm{m^3}
\end{align}$$
The volume change $\Delta V$ during expansion is
$$\begin{align}
\Delta V &= V_4 - V_3 \\
&= 3.521\ \mathrm{m^3} - 2.613\ \mathrm{m^3} \\
&= 0.908\ \mathrm{m^3}
\end{align}$$
The corresponding pressure-volume work $W$ can be calculated assuming an external pressure of $p_\mathrm{ex} = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$.
$$\begin{align}
W &= p_\mathrm{ex} \cdot \Delta V \\
&= 101\,325\ \mathrm{Pa} \times 0.908\ \mathrm{m^3} \\
&= 92.0\ \mathrm{kJ}
\end{align}$$
The amount of energy transferred to the surroundings as work $W$ significantly reduces the internal energy $U$ of the system. We ignored this fact when we assumed that $W_4 = 0$ and $U_4 = U_3$. We can eliminate this error by calculating a new state as follows.
5. Mixture at constant pressure (including work)
If the internal energy is reduced (due to energy transfer to the surroundings as work) to
$$U_5 = 1087.8\ \mathrm{kJ}$$
and thus the specific internal energy is reduced to
$$\begin{align}
u_5 &= \frac{U_5}{m_5} \\
&= \frac{1087.8\ \mathrm{kJ}}{2\ \mathrm{kg}} \\
&= 543.9\ \mathrm{kJ/kg}
\end{align}$$
the new density $\rho_5$ after expansion to normal atmospheric pressure
$$p_5 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$$
can be determined from the pressure $p_5$ and the specific internal energy $u_4$ using tabulated reference values as
$$\rho_5 = 0.6111\ \mathrm{kg/m^3}$$
Thus, the new volume $V_5$ is
$$\begin{align}
V_5 &= \frac{m_5}{\rho_5} \\
&= \frac{2\ \mathrm{kg}}{0.6111\ \mathrm{kg/m^3}} \\
&= 3.273\ \mathrm{m^3}
\end{align}$$
the volume change $\Delta V$ during expansion is
$$\begin{align}
\Delta V &= V_5 - V_3 \\
&= 3.273\ \mathrm{m^3} - 2.613\ \mathrm{m^3} \\
&= 0.660\ \mathrm{m^3}
\end{align}$$
and the corresponding pressure-volume work $W_5$ is
$$\begin{align}
W_5 &= p_\mathrm{ex} \cdot \Delta V \\
&= 101\,325\ \mathrm{Pa} \times 0.660\ \mathrm{m^3} \\
&= 66.9\ \mathrm{kJ}
\end{align}$$
We find that
$$\begin{align}
U_5 &= U_3 - W_5 \\
1087.8\ \mathrm{kJ} &= 1154.7\ \mathrm{kJ} - 66.9\ \mathrm{kJ} \\
\end{align}$$
This result is in agreement with the first law of thermodynamics, which states that the change in the internal energy $U$ of a closed system is equal to the amount of heat $Q$ supplied to the system (which is $Q = 0$ in this case), minus the amount of work $W$ done by the system on its surroundings.
$$\small
\begin{array}{llllll}
\hline
\text{Quantity} & \text{Symbol} & \text{Unit} & \text{State 1} & \text{State 2} & \text{State 3} & \text{State 4} & \text{State 5} \\
\hline
\text{Temperature} & T & ^\circ\mathrm F & 1200 & 68 & 658 & 658 & 580 \\
& & ^\circ\mathrm C & 649 & 20 & 348 & 348 & 304 \\
& & \mathrm K & 922 & 293 & 621 & 621 & 577 \\
\text{Pressure} & p & \mathrm{bar} & 1.01325 & 1.01325 & 1.37 & 1.01325 & 1.01325 \\
& & \mathrm{MPa} & 0.101325 & 0.101325 & 0.137 & 0.101325 & 0.101325 \\
\text{Mass} & m & \mathrm{kg} & 1 & 1 & 2 & 2 & 2 \\
\text{Density} & \rho & \mathrm{kg/m^3} & 0.3827 & 1.2046 & 0.7654 & 0.5681 & 0.6111 \\
\text{Volume} & V & \mathrm{m^3} & 2.613 & 0.830 & 2.613 & 3.521 & 3.273 \\
\text{Specific internal energy} & u & \mathrm{kJ/kg} & 819.4 & 335.3 & 577.3 & 577.3 & 543.9 \\
\text{Internal energy} & U & \mathrm{kJ} & 819.4 & 335.3 & 1154.7 & 1154.7 & 1087.8 \\
\text{Work} & W & \mathrm{kJ} & \text{n/a} & \text{n/a} & 0 & 0 & 66.9 \\
\hline
\end{array}$$

If the volume is constant (sealed container that is not a rubber ballon, for example), the overall temperature would cool.  And the net pressure would be a weighted average of the two pressures if the volume was constant.  Without actual amounts it cannot be determined if the pressure would be higher or lower than the original.
If the container were a rubber balloon, it is not possible to determine the final volume (even as bigger or smaller) without information about relative amounts and the temperature of the 'colder air'.

So this is a closed system energy balance then with the heat transfer rate between the hot and cold air is a function of internal energy change only? An eventual equilibrium temperature may be calculated by $Q_\text{lost by hot air} = Q_\text{gained by cold air}$?


text:
Does SiO₂ have polar covalent bonds or coordinate covalent bonds?
First off, I know $\ce{SiO2}$ is a network solid. I am looking at its intramolecular forces between the $\ce{Si}$ and the $\ce{O}$ molecule. On the one hand, it seems to me that this should be a polar covalent bond, but on the other, both electrons are given to the $\ce O $ by $\ce{Si}$ which makes it coordinate covalent.

As long as you think of molecules, you are dead wrong. There are no molecules.

I'd never heard the term network solid before, thanks! It would certainly help in an answer to this question.

The bonds are polar covalent.
'Given' is a strange term in this context and generally only of relevance for ionic compounds.
Coordinate covalent bonds imply the donation of two electrons to a bond with another atom.  In the case of $\ce{ SiO2 }$, The structure is actually tetrahedral, with an oxygen between and two silicon atoms, and where each oxygen forms two bonds and each silicon forms four bonds.  In these bonds, each of the four valence electrons for silicon is shared singly with a single electron from one of the oxygens.  Thus, in each bond, one of the electrons is from silicon and one from oxygen, which is why we don't normally consider this a coordinate covalent or dative bond.
This is one of those examples where the empirical formula, $\ce{ SiO2 }$, is highly misleading.

The $\ce{Si-O}$ bond has to be polar due to the difference in electronegativity of these atoms. However, due to the tetrahedral arrangement of these bonds, the net dipole moment should be zero.


text:
Measuring Thermodynamic Stability
We can predict whether a reaction can occur spontaneously by calculating the change in Gibbs' free energy.
I was just wondering whether there is any general measure of the thermodynamic stability of any substance?
Can I look at a random substance and using a table of thermodynamic values predict whether this substance is stable or not within a range of given conditions (pressure, temperature, etc.).

Thermodynamics isn't quite that general.  It will tell you about the possibility of going from one well-defined state to another well-defined state, but as has been said, there is no way to do it in general.  What you are asking is akin to "is it hard to get there from here?" without specifying where both here and there are.
So you need to specify enough variables such as temperature or composition to unambiguously determine both the initial and final states.  How many variables do you need?  The Gibbs phase rule provides the answer.  Once that has been done, your question can be answered.

I might not be sure on this, but the instantaneous partial molar gibbs free energy change at a particular P and T (chemical potential) shows the chemical equivalent of potential energy and thus the graph of it versus the conditions gives us an idea of the "driving force" of the reaction which is maximum if the rate of change of chemical potential is maximum and hence the substance under consideration has the greatest tendency to react or rather to move from one phase to another, i.e equivalent to say that it is less stable.
Partial molar gibbs free energy change is nothing but the change in gibbs free energy per a small  change in phase i.e for a small part of reactants to be converted into products in a reaction under consideration. If this is high, slope of the graph of G versus Moles is high the reaction has a greater tendency to move forward
EDIT:-
well unfortunately there is no know-it-all thermodynamic function ( atleast that i know of!) because, similar to the variations in stability caused by different pressure and temperature, number of moles and other possible phases in the system also affect the stability. Hence you can only talk about the resistance of of a substance to a particular reaction or a change of phase (in the thermodynamic sense).
 For example, methane is quite stable in presence of intense sunlight alone, but if fluorine or chlorine is introduced into the system under consideration, there is explosive conversion of methane into methyl halides and thus the stability is out of question. therefore you would have to compute the potential changes for a mixture with all properties defined before you can comment on its stability.
 But as is given in most texts, an almost improper way of understanding stability, is through seeing how negative the heat of formation is but it only gives details as to how stable it is in comparison only to the reactants.


text:
Deodorizing using baking powder instead of baking soda
I am an engineer and I understand the difference between baking power and baking soda as baking powder is baking soda (alkali) with tartaric acid (acidic) and an inert starch (why? Is this a catalyst in the presence of humidity and temperature).
So why use baking soda to deodorize a carpet and not baking powder? Is it because baking powder has both an acid and base and thus one would neutralize the other and would render the baking powder useless for adsorbing smells?
My carpet would be at a temperature of about $25~\mathrm{^\circ C}$. I am guessing this temperature and a relative humidity of $50{-}60~\%$  will not catalyze a violent reaction with either baking powder or baking soda?
Any input from chemists would be useful!
I don't want to use either on my synthetic carpet in my rented house to experiment before I get enough input from people with know-how!

It's the bicarbonate ion that is doing the work, but in baking powder, it will react with the tartaric acid, so you'll have a significantly lower amount of bicarbonate to do the cleaning.

Is the reaction of the bicarbonate with the tartaric acid an exothermic one at room temperature? Should I be worried about using baking pwd. on my carpet (notwithstanding it's lower "cleaning prowess" as compared to baking soda)?

The action on the carpet is unlikely to differ.  The BP may turn out to be slightly less effective.  The starch is likely a free flowing agent and not of much concern unless you wet it and it leaves a white residue. that is hard to vacuum out.

Baking soda is an effective deodorant because of its basicity. Many foul-smelling compounds are acidic. Neutralizing the compound gives it an ionic character which reduces the vapor pressure (meaning it is less available for the nose to smell it) and makes it much more water soluble (it can be washed away). Odorous compounds like fatty acids (butyric acid is BO) and thiols  have much higher $pK_a$s (~4.8 and 7-10, respectively) than tartaric acid ($pK_a$ ~3) means that they would remain un-neutralized with baking powder. It would not be an effective  deodorizer.


text:
What defines the specific enthalpy of atomization?
1 mol of $\ce{F2}$ is turned into $\ce{2F}$ (atoms).
Is this the specific enthalpy of atomization, or two times that?

ChemGuide says that 'The standard atomisation enthalpy is the enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state.'
For example,
\begin{array}{rc}
\ce{1/2 Cl2 (g) -> Cl (g)} & \Delta H_{a}^{0} = +122~\mathrm{kJ/mol} \\
\ce{1/2 Br2 (l) -> Br (g)} & \Delta H_{a}^{0} = +112~\mathrm{kJ/mol} \\
\ce{Na (s) -> Na (g)} & \Delta H_{a}^{0} = +107~\mathrm{kJ/mol}
\end{array}

Enthalpy of atomisation is energy required to break 1 mole of E-E bonds.
Eg: $\ce{1 F2 -> 2 F}$ .............................$\Delta H_{atomisation}$


text:
Is it possible for an acid and base to react to form an even stronger base or acid?
Is there a combination of an acid and a base that when mixed results in a solution that is more basic or more acidic than the initial base or acid, respectively?
If so, do such combinations have a name?

This is strongly dependent on what you mean by "acid" and "base."

Technically, you could say this occurs for the formation of Fluoroantimonic acid $\ce{H2FSbF6}$, one of the strongest known acids. It can formed by reacting $\ce{HF}\text{ and }\ce{SbF5}$: $$\ce{2 HF ⇌ H2F+ + F−}$$
$$\ce{SbF5 + F− → SbF6-}$$ One could argue that the acidic character comes from a combination of the instability of the acidic $\ce{H2F+}$ and the very weak base character of $\ce{SbF6-}$

I'd argue that satisfies the question.

$\ce{6HF + Al(OH)3 (s) -> H3[AlF6] + 3H2O}$
$\ce{6HCN + Fe(OH)2 (s) -> H4[Fe(CN)6] + 2 H2O}$
$\ce{TlOH + HOCl + H2O -> Tl(OH)3(s) + HCl}$

May be one of these will work. Even if they don't, you should get the idea where to search.

If you are talking about mixing acid or base with acid or base then the other answers have summarised it correctly; We can mix two acids to make a super strong acid. (Often Brønsted acid + Lewis Acid).
However if you are asking about mixing acid and base, that will never lead to stronger acid or base. This is because in the genetral reaction:
$$\ce{\text{Strong Acid}+\text{Strong Base} <=>> \text{Weaker Acid} + \text{Weaker Base}}$$
The equilibrium always lies towards the direction of weaker acid and base.

I think that's the crux here.  Are there any exceptions to this rule, as seen in e.g. Tyberius's answer.  I've read of others with e.g. cocrystals forming instead of salts, but that's out of the league of my idle curiosity.

There are no exceptions. Terbius is mixing two acids. I will provide more info later.

When you add a Bronsted-Lowry acid to a Bronsted-Lowry base, you always get a weaker Bronsted-Lowry acid and a weaker Bronsted-Lowry base.  But you could possibly get a stronger Lewis acid or base.  One example of such a reaction involves two chemicals, ordinarily both "acids" , that you probably already have in your laboratory.
What we call "nitric acid" looks like a base in sulfuric acid solvent, reacting as follows:
$2 \ce {H2SO4}+(\ce {O2N}-\ce {OH}) \rightarrow \ce {NO2+}+\ce {H3O+}+2 \ce {HSO4-} $
The nitryl ion, $\ce {NO2+}$, is a powerful Lewis acid or electrophile that does its thing, especially with aromatic substrates.  This is the essence of organic nitration.

NO, adding acid in base is addition of $\ce{H+}$ ions in $\ce{OH-}$ ions which will certainly reduce the concentration of $\ce{OH-}$ ions in solution and solution will become less basic.
As,
$$\mathrm{pH} = 14 - \mathrm{pOH} = 14 + \log_{10}[\ce{OH-}]$$
EDIT:(after further question in comment)
Whatsoever, you say addition of $\ce{H+}$ would be followed by a decrease in concentration of $\ce{OH-}$ due to release of water. In the case of amphoteric oxides, they behave depending on the nature of the medium; in basic pH it will act as an acid. A polyprotic acid will neutralise the base faster and even in case or organic acids the process will be slow (being weak acids).

Thanks. I have doubts that this is true in all cases; it would be helpful if you provided evidence to support your claim. I suspect there must be cases involving amphoteric substances, polyprotic acids or organic compounds whose products interact. I am sure there must be some example involving ammonia. :)


text:
Solid hydrophilic transparant materials that don't dissolve in water
I am looking for hydrophilic transparant materials which are solid at room temperature and that don't dissolve in water. I need these for an experiment in which I condense droplets on a plate made out of/or coated with the material
Just thinking `chemically' I imagine that any polymer with a lot of hydroxyl groups will be hydrophilic (like Polyvinylalcohol), but PVA is water soluble. Polyethylenetereftalate (PET) is reasonable but still has a contact angle around 70 degrees (common polymer contact angles)
Does anybody know transparant materials with contact angles closer to 0 that don't dissolve in water? An additional `demand' is that the material does not break easily (like glass)

Does it have to be a polymer? For example, some very hydrophilic (but insoluble) minerals might have small contact angles: quartz and limestone, for example.

Not necessarily polymer, but preferably yes. The reason is twofold: the material should not break easily (like glass) and it should be transparant

If you do not need chemical stability, cellulose esters like nitrocellulose may work. If you find sheet of mica big enough, it may work too.

Several commercially available titanium dioxide coatings are available, commonly used to kill bacteria or aid UV stability (I think). When exposed to light, the titanium dioxide has a contact angle of almost 0 degrees.
You should be able to deposit a thin layer on something like a PVA or polycarbonate sheet. The commercial coatings are somewhat expensive, though, as far as I can tell, and the common process for doing it yourself involves heating the surface in a mid-temperature flame for 10-15 minutes which excludes pretty much all plastics I know of that would otherwise be suitable.

Great suggestion! And this could be made transparent by just coating a sufficiently thin layer I guess?

@michielm: There are at least commercial coatings designed for windows, so I'm assuming so.


text:
In acidic solution, what does ClO₃⁻ reduce into?
In acidic solution, what does $\ce{ClO3-}$ reduce into and, most importantly, why? It's $$\ce{6e- +ClO_{3}^- +6H^+ -> Cl^- + 3 H_2O}.$$
Now, with the gift of foresight, why was the equation for its oxidation not instead $$\ce {2e^- +ClO_{3}^- +2H^+ \rightarrow Cl^- + O_2 + H_2O}?$$
Obviously it balances and charge is conserved. However, the latter equation would seem more likely: less charges have to be pulled apart (so less work has to be done to move these charges), and the bond energy of bonds in a water molecule is almost twice that in an oxygen molecule?
Where is there a flaw in my logic and does there exist a rough set of rules to determine how things will split up under oxidation or reduction?

The chlorine atom in $ClO_3^-$ is actually being reduced in this reaction since it gains electrons and goes from a +5 to a -1 oxidation state.

and the bond energy ... in a water molecule is almost twice that in an oxygen molecule
This indicates that the first equation is preferred.

The first equation is correct because $\ce{ClO3-}$ is reduced to $\ce{Cl-}$ and oxygen maintains its -II oxidation state in both educts and products. As chlorine changes its oxidation state from +V in chlorate to -I in chloride, $6 \ \mathrm{e^-}$ are required for this reduction.
In your second equation, chlorate is reduced to chloride as well; however, only $2 \ \mathrm{e^-}$ are explicitly supplied. The remaining electrons would have to come from the concurrent oxidation of oxygen from -II in $\ce{ClO3-}$ to 0 in $\ce{O2}$, and this is unlikely to happen, as oxygen is more electronegative than chlorine.


text:
How to find the molecular weight of a given gas based on its density?

The density of a gas $\ce{X}$ is $10$ times that of hydrogen. In that case, what is the molecular weight of gas $\ce{X}$?

Well what I've done up to now is this:
The molecular mass of hydrogen is $M(\ce{H2})=2$, the density of hydrogen is $\rho(\ce{H2})= \pu{0.089 kg/m3}$.
Assuming that equal volumes $V$ of both gases are taken, the mass of hydrogen $m(\ce{H2})= (0.089 \cdot v) \pu{kg}$ and the mass of the unknown gas $m(\ce{X}) = (0.89 \cdot v) \pu{kg}$.
Although this hasn't really gotten me anywhere. How will it be done?

Indeed you can use Avogadro's law here, but I guess it is useful to give a little bit more insight in Avogadro's Law. What you can do is apply the ideal gas law:
$$p V = n R T $$
Here $n$ is the amount of substance of gas, which you can also write as:
$$ n=\frac{m}{M},$$
where $m$ is the total mass of the gas and $M$ is the molecular mass.
You can rewrite the ideal gas law now to:
$$p M = \frac{m}{V} R T,$$
where you should recognize that $\frac{m}{V}=\rho$.
The gas constant $R$ is (as the name shows) a constant so you can write
\begin{align}
  R &=\frac{p M}{\rho T} &
  \rightarrow            &&
  \frac{p_1 M_1}{\rho_1 T_1}&=\frac{p_2 M_2}{\rho_2 T_2}
\end{align}
And therefore, for equal pressure and temperature you will find that
$$\frac{M_1}{\rho_1}=\frac{M_2}{\rho_2},$$
which means that if the density of gas 2 (the unknown gas $\ce{X}$) is $10$ times as high, the molecular mass should be as well.

What is interesting to add is that this assumes the ideal gas law to be applicable. If, for example, your gases would be at high pressure (e.g. $\pu{200 bar}$) this is typically not a good assumption so the molecular mass that you would find for a $10$ times as dense gas will not be equal to $10$ times the molecular mass of hydrogen.

Assuming equal volume $V$, equal temperature $T$, and equal pressure conditions $p$, we can apply Avogadro's Law:
 An equal volume of two gasses contain an equal number of molecules at the same temperature and pressure.
Now, you can proceed:
\begin{align}
  \rho &= \frac{\text{Mass}}{\text{Volume}}\\
       &= \frac{n\times M (\text{molar  mass})}{\text{Volume}}\\
\implies\frac{\rho_1}{\rho_2}&=\frac{M_1}{M_2}\\
\implies M_{\ce{X}}&=10\times M_{\ce{H2}}
\end{align}


text:
How does sodium bicarbonate act as a deodorizing agent?
I was reading this other question on the site: Deodorizing using baking powder instead of baking soda.
Apparently, baking soda is used quite often as a cleaning and deodorizing agent. From what I know, it's cleaning effects are because it can denature proteins and saponify oils (convert oils to soap) since it is a weak base.
But when you take a look the Wikipedia page on aroma compounds, the only compounds that could possibly be affected by a weakly basic substance are esters (which would undergo hydrolysis). Most of the compounds are polyunsaturated aldehydes, ketones and alcohols which are not affected by (weak) bases.

Note: I assume here that compounds which cause bad odors would have a similar structure to compounds that are fragrant since they both have to activate whatever receptors there are in the nose. Also I have read that usually the same compound can cause both a bad and a good smell depending on the concentration of the compound, such as indole.

So I ask: Does $\ce{NaHCO3}$ really have any deodorizing effect and if so, what is the mechanism (reactions involved) behind it. Also would it be possible to achieve the same effect with any random alkaline substance ($\ce{NaOH}$ maybe, although not very practical).

Body odor is coming from bacterial fermentation products, mostly organic acids. Baking soda is a mild killer of bacteria, and also forms slat with acids. Acids are easier to evaporate when neutral, therefore they are smelly, however in their deprotonated form are ionic, therefore their vapor pressure is very low.

The key factor is that baking soda is amphoteric. This means that it can react both with strong acids and with strong bases which are typically smelly compounds (sour milk, rotten fish, fatty acids in sweat).
The explanation on the third page of article by Lamb [1] gives a nice summary:

The chemical explanation of why bicarbonate of soda should act as a deodorant
  is of interest. Sweat (3) has an acid reaction, $\ce{pH}$ of $5.2\text{–}6.75$ which is due to the fatty acids or acid phosphates of sodium and potassium which it contains. It also contains $\ce{NaCl}$, $\ce{KCl}$, alkaline salts, organic acids and urea. Minute amounts of unstable fats and oils are secreted in sweat and hydrolysed to their corresponding fatty acids and glycerin. Free fatty acids may be secreted. Many of these acids are volatile and have a characteristic disagreeable odor. These include butyric acid, formic acid, caproic acid and valerie acid. The following are among the possible explanations of the manner in which sodium bicarbonate acts as a deodorant in the axilla.

It forms a sodium salt with butyric, caproic and valerie acids. These sodium salts are comparatively mild in odor in comparison to the volatile "rancid-smelling" fatty acids.
By changing the $\ce{pH}$ of the axillae, the bacteria which decompose the fats are unable to act. However, as a result of this the staphylococcic flora which prefer higher alkalinity might thrive and axillary furunculosis might result.

To give an example. The reaction of baking soda with butyric acid looks like this:
$$\ce{C3H7COOH + NaHCO3 -> C3H7COONa +H2O + CO2} $$
while butyric acid is extremely smelly (an intense vomit smell), the salt hardly smells at all.
Given this explanation, I think $\ce{NaOH}$ should obviously also work well for deodorizing in case the smell is caused by acids. The problem would be that smells caused by basic compounds are left untouched.
References

Lamb, J. H. Sodium Bicarbonate: An Excellent Deodorant. Journal of Investigative Dermatology 1946, 7 (3), 131–133. https://doi.org/10.1038/jid.1946.13

I would think the volatility of $\ce{CO2}$ would also be important, since loss of $\ce{H2CO3}$ will drive the neutralization reaction even further away from the stinky organic acids, per le Chatelier.