text:
Formation of peracetic acid from acetic acid and hydrogen peroxide and its stability in their presence
I've been trying to find out as much as I can about peracetic acid, especially regarding its use as a sanitizer. In the Wikipedia entry it notes that

Peracetic acid is always sold in solution as a mixture with acetic acid and hydrogen peroxide to maintain its stability.

This is definitely born out by commercial examples: Loeffler Lerasept PAA (a sanitizer I've used before) lists 31% hydrogen peroxide, 17% peracetic acid, 16% acetic acid and 1% phosphonic acid among its ingredients.
But, my understanding was that mixing acetic acid and hydrogen peroxide makes peracetic acid. From the same Wikipedia article:

It forms upon treatment of acetic acid with hydrogen peroxide.

So my question is: what exactly keeps the excess hydrogen peroxide from reacting with the excess acetic acid to simply form more peracetic acid? Does the phosphonic acid inhibit this reaction? Is there something else not listed that does this? Or is it just some bit of chemistry I'm not aware of?

phosphonic acid (HEDP) is added as a stabilizer. Binding to metal ions which might de-activate hydrogen peroxide.

The formation of peracetic acid from acetic acid and hydrogen peroxide is an equilibrium reaction and so in order for the peracetic acid to remain at a constant concentration as desired, acetic acid and hydrogen peroxide must also be present so that the reaction can be at equilibrium. $$\ce{CH3COOH + H2O2 <=> CH3COOOH + H2O}$$

what exactly keeps the excess hydrogen peroxide from reacting with the excess acetic acid to simply form more peracetic acid?

Nothing. The hydrogen peroxide reacts with the acetic acid to form peracetic acid and water but the peracetic acid and water also react to reform hydrogen peroxide and acetic acid. At equilibrium the rates of these reactions are equal and so the concentrations of the species do not change. At room temperature the equilibrium constant is somewhere around 2.5 (see references) which indicates that significant amounts of reactants and products are present, supporting the reasoning above.
The paper Chin J Proc Eng 2008 February, 8 (1), 35–41 has some data on the values of the equilibrium constant and the variation in concentration of peracetic acid formed.
Another value for the equilibrium constant on page vii of Unis, Melod (2010) Peroxide reactions of environmental relevance in aqueous
solution. Doctoral thesis, Northumbria University.

OK that makes sense. My understanding of chemistry is pretty basic, and I always seem to forget the whole equilibrium thing (which seems pretty vital, really). Would it be more accurate to say they simply "exist" together in solution, rather than being "sold together"?

But, my understanding was that mixing acetic acid and hydrogen peroxide makes peracetic acid

Exactly! But within the mixture, peracetic acid reacts back to hydrogen peroxide and acetic acid. With other words, the components are in an equilibrium.
\[
\ce{CH3COOH + H2O2 <=> CH3CO3H + H2O}
\]
For disinfection purposes, these mixtures are sold as made. This is safe (and cheap) for both the manufacturer and the consumer. The fairly stable and safe to use, while concentrated solutions of peracetic acid tend to decompose - violently.
Phosphonic acid with a (first) pKa around two is the strongest acid within the mix. One might speculate that it is added to adjust the pH, keep more acetic acid protonated and therefore shift the reaction a bit more towards the peracetic acid.

Hallo Klaus, This "phosphonic" is a typo I assume. Phosphoric acid is a traditional additive to stabilize peroxides. (Presumably by masking iron impurities)

Hallo Georg, that really makes much more sense! Admittedly, I was struggling a bit to find an "explanation" for the phosphonic acid ;)

I don't think it is a typo. See this MSDS, which contains both phosphoric and phosphonic acid. I think they use it as a metal chelator for acid cleaning products.

@FranklinPCombs I'm not convinced, since that's a peracetic acid free mixture. Georg's explanation is very reasonable. Phosphoric acid is stable towards oxidation, strong enough to adjust the pH and cheap as dirt: Perfect for the manufacturer.

OK I believe you. Thanks for the answer. Both being equally helpful, I had to accept the one that came in first.


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How to store distilled water without introducing impurities?
How can I store distilled water without introducing impurities?  I'm going to be doing so electrical studies of pure water and ice and I need a way of keeping the water pure during the experiments.  The water needs to remain relatively pure (I'm not yet entirely sure how relative yet), over at least the course of $1$ day.  What is a good container that won't be leeching any salts or other chemicals into the water?

You can't leach silicic acids from teflon bottles :)

Distilled water I know of is sold in PE containers.

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Is it possible to make a drug that liquefies heart plaque to treat heart disease?
Is it possible to make a drug that liquefies heart plaque to treat heart disease without damaging other parts of the body?  If so, would the liquefied plaque be eliminated as regular fluid is through the kidneys?  I believe that if this is possible, it would definitely be worthwhile to make as it could save many lives!
I heard there are drugs that can dissolve a blood clot.
I also heard of a technique being developed in Israel that dissolves heart plaque with an ultrasonic probe.
It looks like there are some efforts already in this area, but it appears they halt rather than dissolve the plaque:
http://heartdisease.about.com/cs/coronarydisease/a/plaquebuster.htm

statins inhibit coQ10 and vitamin K2 that leads to calcification of blood vessels and impairing heart muscles statins are not 100% safe as your doctor tells you... statins are not good to remove built up plaque in blood arteries, vitamin k2 would be better option if ur LDL is not high. https://pubmed.ncbi.nlm.nih.gov/25655639/

What you have written is not a proper answer to the question, it is a comment on statins.

Plaques are continually being removed from arteries by natural mechanisms within the body.  Statin use, when combined with aggressive dietary changes, can slow down the rate of plaque deposition to the point that the rate of plaque removal is actually higher than the rate of plaque deposition, and, as a consequence, plaque thinning can be observed.
Reducing plaques faster (plaque liquefication) has some potential, life threatening drawbacks.

We've all seen how ice cubes, when frozen together into a clump and then placed in a water bath, dissolve.  Some ice cubes break off from the main structure and float away to dissolve separately.  The same thing can happen with a large plaque deposit.  If a piece of the plaque breaks off during the liquefication process and continues to travel through the blood stream it can lead to downstream blockages, particularly in the brain (stroke).  Rapid dissolution of plaques would increase the risk for such adverse events.
While plaques form more or less everywhere as relatively uniform deposits, the real concern is areas where plaques have grown to large size and begin to block arteries.  It has been found that these larger deposits are initiated by events related to damage of the arterial wall at that specific location. The damage leads to infection (that's why highly sensitive C-reactive protein [hs-crp] measurements are routinely made in conjunction with cholesterol measurements).  The body mobilizes the immune system to fight the infection; microphages and white blood cells are directed to these infection sites.  Once this happens deposition of other materials (lipids, calcium, etc.) start at the infection site and abnormal plaque growth results.  Because of the 1) initial damage, 2) the infection and 3) infection fighting, the arterial wall becomes abnormally thinned at the site.  To some degree, the plaque actually serves to stabilize the thin wall preventing wall rupture and leaks.  Rapid removal of the plaque would increase the chances for the wall to rupture,  producing adverse events and increasing mortality.

So while it is likely that drugs could be designed to remove the plaque extremely fast (e.g. liquefication), the risks might far outweigh the benefits.

would the liquefied plaque be eliminated as regular fluid is through
  the kidneys?

A major component of plaque is lipoprotein.  Lipoprotein is transported by blood as insoluble particles of varying density and size.  Lipoprotein itself is an assembled complex of proteins and fats.  The fats are various esters of glycerol including esterified cholesterol.  The lipoprotein transports cholesterol and fats, via the bloodstream to cells where these compounds are needed for biosynthesis. After the lipoprotein has released the transported molecules it returns to the liver where it is "reloaded" and sent out again. Lipoproteins are not eliminated (unless damaged) from the body, but are rather recycled.
If a large amount of lipoprotein were suddenly released into the body, I'm sure some would be eliminated through the kidneys.  Certainly the lipoprotein receptors would sense this event and stop synthesis of new lipoprotein until systemic balance was restored. it seems likely that adverse events like kidney damage, gallstones, bile duct clogging, etc. would occur.
Since statins were first released onto the market, drug companies have been considering where to go next.  I have to believe that plaque liquefication was among the options considered.  While the reduction of plaque build up (statins) have their share of problems, I would guess that these problems are minor compared to the possible problems that can result from liquefication (discussed above).  Indeed, with the recent announcements (see here for example) of monoclonal antibodies achieving plaque reductions superior to statins, it seems that drug companies are continuing to focus on avoiding the problem in the first place, rather than fixing things (liquefication) down the road.

@Ron, thank you for your detailed answer!  It sounds like this might actually be not such a good idea, unless there is a way to mitigate the risk of thinned artery walls rupturing, and to mitigate the risk of the plaque breaking up and traveling to other parts of the body.  Any ideas here?

@Jonathan  I've been thinking about that and haven't come up with anything.  I'll post back if I do.

A little late, but maybe you could have a minimally invasive surgery to put a filter downstream to catch anything too large to avoid clots elsewhere. Maybe even a filter that catches most LDL to help mitigate the large release. I don't know of anything else as physically large as LDL is in blood, maybe a simple particle size filter would work.

A good starting point to read about drugs that can potentially remove plaque is The prevention and regression of atherosclerotic plaques: emerging treatments Vasc Health Risk Manag. 2012; 8: 549–561.
Yes, there is is evidence that drugs can remove plaque, but the word "liquify" is probably not the best description.  Also, the removed plaque would not be especially excreted by the kidneys.  The main carbon compound in urine is urea, which has 2 nitrogen atoms per carbon atom, and the plaque doesn't have that much nitrogen.  If the plaque is excreted, it would be after metabolism and the carbon in the plaque would mainly be excreted as CO2.
So, going back to the cited article, rather than liquify, the term "regress" is used:

It is now known that plaques are able to regress. Plaque reversal occurs by removal of lipids and necrotic material, endothelial repair, or halt of vascular smooth muscle cell proliferation. Several mechanisms explain this reversal, such as high-density lipoprotein cholesterol (HDL-C) action, destruction of foam cells and macrophages in lymph nodes and restoration of endothelium by neighboring cells or circulating progenitors.

Reference is made to An integrated approach for the mechanisms responsible for atherosclerotic plaque regression. Exp Clin Cardiol. 2011;16(3):77–86 for further information.
Then the article goes through several studies on drugs that may remove plaque.
The drug treatments discussed fall into the three broad catagories of statins, phosphodiesterase inhibitors, and thiazolidinediones and specfically are:

rosuvastatin alone and in combination with ramipril.
atorvastatin alone or in combination with ezetimibe
pitavastatin
cilostazol alone or in combination with probucol
pioglitazone alone or in combination with a statin

This drug would have to attack the cholesterol accretion or help the body do it by raising the HDL levels in relation to the LDL levels.
Statins are commonly used for this.

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Why is Ma*Va=Mb*Vb still valid for weak acids/bases?
I asked my chemistry teacher this today and didn't get an answer. It would seem to me that the equation for determining the morality of the unknown in a titration, MaVa = MbVb, would only work if the acid and the base were both strong.
I always interpreted the equation as a way to say (moles H+)=(moles OH-), but if the acid was weak, the moles of H+ would not simply equal MaVa, because the acid wouldn't fully dissociate, and the same for a base and OH-.
Yet the formula still works for both strong and weak acids and bases. Why?

We normally titrate a weak acid by a strong base, or  weak base by a strong acid, to ensure that the reaction of titration is total (quantitative). We don't titrate weak acid by a weak base (and vice-versa).
I will calculate the equilibrium constant of the titration of  weak acid $\ce{HA}$ by a strong base $\ce{OH-}$:
 $$\ce{HA + OH- <=>A- + H2O}$$ The equilibrium constant is $$K=10^{(\mathrm{p}K_w -\mathrm{p}K_a)}$$ For common weak acids (acetic acid $\mathrm{p}K_a=4.7$ ): $$K>> 10^3$$
So, the equation of titration (neutralization) must be written:$$\ce{HA + OH- ->A- + H2O}$$
At the equivalence point: $$n(\ce{HA})=n(\ce{OH-})$$


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DFT Functional Selection Criteria
I have a very very general question:
In DFT functional selection , mostly people speak about the most recent ones. For example my professor always asks: " which DFT Functional did you select ? "  and if I say B3LYP, he says : " No ! that's too old ! " but if I answer: M06 , he says : " hmm ... sounds promising, that's a modern functional ".
I think it is too naive to select functionals just based on their chronologic sequence. I want to ask if there is any good and reliable criteria for functional selection. For example a criteria that says for an alkene with some specific characteristics, go for M06-L and for an alkane with other characteristics, use B97xxx family and so on..
Is there such a criteria ?   I hope this topic can become a good guideline for future reference. !

DFT functionals usually contain constants, that were fitted for better reproductions of real-world data on some set of specific cases. For example, b3lyp is fitted using small molecules of light elements. Given that, the answer is simple: find source article for the functionals you can use in your package and choose one fitted for case similar enough for yours.

What's up with all that magic? (A chapter formerly known as Introduction)
The hunt for for the holy grail of density functional theory (DFT) has come a long way.[1] Becke states in the introduction of the cited paper:

Density-functional theory (DFT) is a subtle, seductive, provocative business. Its basic premise, that all the intricate motions and pair correlations in a many-electron system are somehow contained in the total electron density alone, is so compelling it can drive one mad.

I really like this description, it points out why we use and need DFT, and as it also points out the flaws, that every computational chemist has to deal with: How can something with such a simple approach be correct?
Something that is often forgotten about DFT is, that in principle it is correct. It's the implementations and approximations, that make it incorrect, but usable. Becke states this in the following quote:

Let us introduce the acronym DFA at this point for “density-functional approximation.” If you attend DFT meetings, you will know that Mel Levy often needs to remind us that DFT is exact. The failures we report at meetings and in papers are not failures of DFT, but failures of DFAs.

I sometimes here that the abbreviation DFT is often used in the wrong context, since we are not talking about the theory itself any more, but about the implementations and approximations of it. One suggestion I heard was, that it should rather be used as density functional technique.
With that in mind I would like to state, that I absolutely agree with the previous answer by user1420303 and it's subsequent comment by Geoff Hutchison. Since you asked for a somewhat more practical approach, I like to offer the advice I usually give to students new in the field.
Old is bad, isn't it?
Some of the functionals are now around for about thirty years. That does not make them bad, maybe even the opposite. It shows, that they are still applicable today, giving reasonable results. One of my personal favourites is the conjunction of Becke 1988 and Perdew 1986, often abbreviated as BP86.[2] It's a pure functional which is available most modern quantum chemical packages.[3] It performs usually well enough for geometries and reasonable well for energies for simple systems, i.e. small organic molecules and reactions.
The magical functional B3LYP was one of the first hybrid functionals, and it was introduced by Gaussian's very own developers.[4] A lot of people were surprised how well it worked and it quickly became one of the most popular functionals of all time. It combines Becke's three parameter functional B3[5] with Lee, Yang and Parr's correlation functional.[6] But why are we surprised it works? The answer is quite simple, it was not fitted to anything. Frish et. al. just reworked the B3PW91 functional to use LYP instead of PW91. As a result, it heavily suffers or benefits from error compensation. Some even go as far as to say: “It is right for the wrong reasons.”[7-9] Is it a bad choice? No. It might not be the best choice, but as long as you know what you are doing and you know it is not failing for your system, it is a reasonable choice.
One functional is enough, is it?
Now that we established, that old functionals are not out of fashion, we should establish something very, very important: One is never enough.
There are a few things, where it is appropriate to do most of the work with one functional, but in these cases the observations have to be validated with other methods. Often it is best to work your way up Jacob's ladder.[10]
How do I start?
It really depends on your system and what you are looking for. You are trying to elucidate a reaction mechanism? Start with something very simple, to gain structures, many structures. Reaction mechanisms are often about the quantity of the different conformers and later about suitable initial structures for transition states. As this can get complex very fast, it's best to keep it simple. Semi-empirical methods and force fields can often shorten a long voyage. Then use something more robust for a first approach to energy barriers. I rely on BP86 for most of the heavy computing. As a modern alternative, another pure density functional, M06-L is quite a good choice, too.[11] Some of the popular quantum chemistry suites let you use density fitting procedures, which allow you to get even more out of the computer. Just to name a few, without any particular order: Gaussian, MolPro, Turbomole.
After you have developed a decent understanding of the various structures you obtained, you would probably want to take it up a notch. Now it really depends on what equipment you have at hand. How much can you afford? Ideally, more is better. At least you should check your results with a pure, a hybrid, and a meta-hybrid functional. But even that can sometimes be a stretch.[12]
If you are doing bonding analysis, elucidation of the electronic structure, conformation analysis, or you want to know more about the spectrum, you should try to use at least five different functionals, which you later also check versus ab initio approaches. Most of the times you do not have the hassle to deal with hundreds of structures, so you should focus of getting the most accurate result. As a starting point I would still use a pure functional, the worst thing that could happen is probably, that is reduces the times of subsequent optimisations. Work your way up Jacob's ladder, do what you can, take it to the max.[13]
But of course, keep in mind, that some functionals were designed for a specific purpose. You can see that in the Minnesota family of functionals. The basic one is M06-L, as previously stated a pure functional, with the sole purpose of giving fast results. M06 is probably the most robust functional in this family. It was designed for a wide range of applications and is best chosen when dealing with transition metals. M06-2X is designed for main group chemistry. It comes with somewhat built in non-covalent interactions and other features. This functional (like most other though) will fail horribly, if you have multi-reference character in your system. The M06-HF functional incorporates 100% Hartree-Fock exchange and was designed to accurately calculate time dependent DFT properties and spectra. It should be a good choice for charge transfer systems. See the original publication for a more detailed description.[14]
Then we have another popular functional: PBE.[15a] In this initial publication an exchange as well as a correlation functional was proposed, both pure density functionals, often used in conjunction.[15b] I don't know much about it's usefulness, since I prefer another quite robust variation of it: PBE0, which is a hybrid functional.[15c,d] Because of its adiabatic connection formula, it is described by the authors as a non-empirical hybrid functional.[15d]
Over the years there have been various developments, some of the are called improvement, but it often boils down to personal taste and applicability. For example, Handy and Cohen reintroduced the concept of left-right correlation into their OPTX functional and subsequently used it in combination with LYP, P86 and P91. Aparently, they work well and are now often used also as a reference for other density functionals. They went on and developed a functional analogous to B3LYP but outperforming it.[16]
But these were obviously not the only attempts. Xu and Goddard III extended the B3LYP scheme to include long range effects. They claim a good description of dipole moments, polarizabilities and accurate excitation energies.[17]
And with the last part in mind, it is also necessary to address long range corrections. Sometimes a system cannot be described accurately without them, sometimes they make the description worse. To name only one, CAM-B3LYP, which uses the coulomb attenuating method.[18] And there are a couple of more, and a couple of more to come, head on over to a similar question: What do short-range and long-range corrections mean in DFT methods?
As you can see, there is no universal choice, it depends on your budget and on the properties you are interested in. There are a couple of theoretical/ computational chemists on this platform. I like BP86 as a quick shot and answer questions relating to MO theory with it, shameless self-promotion: Rationalizing the Planarity of Formamide or Rationalising the order of reactivity of carbonyl compounds towards nucleophiles. And sometimes we have overachievers like LordStryker, that use a whole bunch of methods to make a point: Dipole moment of cis-2-butene.
So I picked a functional, what else?
You still have to pick a basis set. And even here you have to pick one that fits what you need. Since this answer is already way longer than I intended in the first place (Procrastination, yay!), I will keep it short(er).
There are a couple of universally applicable basis sets. The most famous is probably 6-31G*. This is a nice ancient basis set that is often used for its elegance and simplicity. Explaining how it was built is easier, than for other basis sets. I personally prefer the Ahlrichs basis set def2-SVP, as it comes with a pre-defined auxiliary basis set suitable for density fitting (even in Gaussian).[19]
Worth mentioning is the Dunning basis set family cc-pVDZ, cc-pVTZ, ... . They were specifically designed to be used in correlated molecular calculations. They have been reworked and improved after its initial publication, to fit them to modern computational standards.[20]
The range of suitable basis sets is large, most of them are available through the basis set exchange portal for a variety of QC programs.
Sometimes an effective core potential can be used to reduce computational cost and is worth considering.
*Sigh* What else?
When you are done with that, consider dispersion corrections. The easiest way is to pick a functional that has already implemented this, but this is quite dependent on the program of you choice (although the main ones should have this by now, it's not something brand new). However, the standalone DFT-D3 program by Stefan Grimme's group can be obtained from his website.[21]
Still reading? Read more! (A chapter formerly known as Notes and References)

Axel D. Becke, J. Chem. Phys., 2014, 140, 18A301.
(a) A. D. Becke, Phys. Rev. A, 1988, 38, 3098-3100. (b) John P. Perdew, Phys. Rev. B, 1986, 33, 8822-8824.
Unfortunately this functional is not always implemented in the same way, although the differences are pretty small. It basically boils down as to which VWN variation is used in the local spin density approximation term. Also see S. H. Vosko, L. Wilk, and M. Nusair, Can. J. Phys., 1980, 58 (8), 1200-1211.
P. J. Stephens, F. J. Devlin, C. F. Chabalowski, and M. J. Frisch, J. Phys. Chem., 1994, 98 (45), 11623–11627.
Axel D. Becke, J. Chem. Phys., 1993, 93, 5648.
C. Lee, W. Yang, and R. G. Parr, Phys. Rev. B, 1988, 37, 785–789
Unfortunately the B3LYP functional suffers from the same problems that are mentioned in [3].
The failures of B3LYP are known and often well documented. Here are a few recent papers, but there are many, many more. (a) Holger Kruse, Lars Goerigk, and Stefan Grimme, J. Org. Chem., 2012, 77 (23), 10824–10834. (b) Joachim Paier, Martijn Marsman and Georg Kresse, J. Chem. Phys., 2007, 127, 024103. (c) Igor Ying Zhang, Jianming Wu and Xin Xu, Chem. Commun., 2010, 46, 3057-3070. (pdf via researchgate.net)
Just my two cents, that I am hiding in the footnotes: “Pretty please do not make this your first choice.”
John P. Perdew and Karla Schmidt, AIP Conf. Proc., 2001, 577, 1. (pdf via molphys.org)
Yan Zhao and Donald G. Truhlar, J. Chem. Phys., 2006, 125, 194101.
Note, that not always full recomputations of geometries are necessary for all different functionals you apply. Often single point energies can tell you quite much how good your original model performs. Keep the computations to what you can afford.
Don't use an overkill of methods, if you already have five functionals agreeing with each other and possibly with an MP2 calculation, you are pretty much done. What can the use of another five functionals tell you more?
Y. Zhao, N.E. Schultz, and D.G. Truhlar, Theor. Chem. Account, 2008, 120, 215–241.
(a) John P. Perdew, Kieron Burke, and Matthias Ernzerhof, Phys. Rev. Lett., 1996, 77, 3865. (b) The exchange functional was revised in Matthias Ernzerhof and John P. Perdew, J. Chem. Phys., 1998, 109, 3313. (c) Carlo Adamo and Vincenzo Barone, J. Chem. Phys., 1999, 110, 6158. (d) Kieron Burke, Matthias Ernzerhof, and John P. Perdew, Chem. Phys. Lett., 1997, 265, 115-120.
(a) N. C. Handy and A. J. Cohen, Mol. Phys., 2001, 99, 403-12. (b) A. J. Cohen and N. C. Handy, Mol. Phys., 2001, 99 607-15.
X. Xu and W. A. Goddard III, Proc. Natl. Acad. Sci. USA, 2004, 101, 2673-77.
T. Yanai, D. P. Tew, and N. C. Handy, Chem. Phys. Lett., 2004, 393, 51-57.
(a) Florian Weigend and Reinhart Ahlrichs, Phys. Chem. Chem. Phys., 2005, 7, 3297-3305. (b) Florian Weigend, Phys. Chem. Chem. Phys., 2006, 8, 1057-1065.
The point where the use of more basis functions does not effect the calculation. For the correlation consistent basis sets, see a comment by Ernest R. Davidson, Chem. Phys. Rev., 1996, 260, 514-518 and references therein. Also see Thom H. Dunning Jr, J. Chem. Phys., 1989, 90, 1007 as the original source.
DFT-D3 Website; Stefan Grimme, Jens Antony, Stephan Ehrlich, and Helge Krieg,
J. Chem. Phys., 2010, 132, 154104.
Have fun and good luck!

@Martin Very informative answer. I also prefere Ahlrichs def2 family :) . As a personal opinion, I would always try to avoid the Dunning basis set when feasible. Only use them for special cases where auxiliary basis sets are not available for other basis set families.
"Still reading? Read more!"

Beautiful review, but PBE is pure functional, not a hybrid one. It was made into made into a hybrid PBE0 by Adamo.

@Wildcat Thank you very much, for the compliment and correction. Indeed I was already thinking about PBE0, I hope the edit clarified it now.

Point number 22 is very important :)
Thanks for the answer @Martin-マーチン !

There's now also a thorough article from Grimme et. al on this: https://doi.org/10.26434/chemrxiv-2022-n304h

There are many criteria that you can follow to choose a functional, in no special order:

Availability: Do you have access to a software package in which the functional is implemented? If yes, are analytic derivatives also implemented? Do you need extra dispersion corrections? Are they available?
Computational cost: Functionals in different steps of "Jacob's ladder" (pure, hybrids, double-hybrid) scale in different way with system size (N^3,N^4,N^5).
Accuracy of the functional for the property that you want to calculate.

Of course it is not a good idea to select a functional based in a chronological sequence. It is true that newer functionals tend to be more accurate that old ones. BUT, it is not always the case, there are very new functionals that have high theoretical interest, but lacks of outstanding performance.
¿What about B3LYP? It is true that the 'magical functional' is not the best performer today, but it is widely accepted, no one will complain about its usage. For most applications, it is very good for geometry optimization in my opinion.
There is not a simple recipe, the best you can do is just compare the results of many functionals in many test sets. Look at the test set designed to test the properties that you need to predict.
In short, till today, for almost any application you will be ok if just read the results of this paper:
Lars Goerigk and Stefan Grimme, Phys. Chem. Chem. Phys., 2011, 13, 6670-6688. (DOI: 10.1039/C0CP02984J)
There are too few functionals out there in which you can be interested

+10. Excellent response. I think the one caveat I'd add is for "accuracy" since different functionals are designed for different properties. So if you want an accurate functional for intermolecular interactions, it might be different than one for transition state barriers.

This answer really helps but I am looking for something more structured and practical  ( as what is in here:   http://en.wikipedia.org/wiki/Minnesota_Functionals )   that's why I decided to put a bounty .

I agree with what you said, except the part with the B3LYP functional, I would always complain about that, but that is because the magical functional is also the lazy functional.

@Martin, Although we may complain, I think that most referee won't. Would you also complain about the B3LYP usage for geometry optimizations? If so, What functional do you suggest for that purpose?

@Aug Did you take a look to the tables on the paper and in its supplementary information? There you can get objective and quantitative information of the performance of a large list of functionals in test sets designed to test a specific property. Descriptions based on design can be not reliable when you test it in real world. I suggest, for example, see the tables of the M-12 functionals paper.

@user1420303 Is the study looking at minima?  If so, is B3LYP yielding actual minima on the PES?  Is this data reported? As a referee I would be demanding that sort of information.  Is B3LYP doing a poor job describing the set of systems as minima? If so then another functional is needed.  B3LYP opts in this case would not be suitable for publication... but perhaps only preliminary optimizations for more rigorous ones later.

I have been a busy bee and obsessively compiled my two cents, not particularly to answer your comment, that was just a side effect.|| @LordStryker In what cases don't you need another functional ;)

@Martin Going by what I see published in something like "Journal of Medicinal Chemistry", some people think that you don't need anything other than Hartree-Fock.

@user1420303 in Colorado library we had not this item !  I ordered it via inter-library system and should have it tomorrow. I am sure that is a great paper.

@user1420303 Just a question about "Jacob's Ladder ". What is the computational cost of Minnesota Series ( especially M06 ) ? When I try a molecule with M06, it seem it takes forever to finish !

@LordStryker: Equilibrium B3LYP geometries for a large basis set are normally good enough to make post ab initio calculation (for example Gn and Wn methods, the later with sub chemical precision for equilibrium energies). Even if it can fail, more reliable alternatives are too expensive (In my experience, B3LYP is more reliable for geometries than MP2)

@Aug , M06 and M06-2X are hybrid functionals, they should have a scaling of N^4 with system size (the same as B3LYP), BUT, that doesn't mean that the computation time for a given system is the same for M06 and B3LYP. Roughly speaking it means that if you double the system size the computation time for one iteration will be 2^4 longer for both functionals.

@Aug This scaling is VERY important because it determinate if the method can be used with larger systems/bases. It is not so much important if it take 2 or 3 hours, but for larger systems it can change from 1 week to a century.

In addition to the good answers that have already been given, it is important to note that for DFT to be properly implemented, you must calibrate the functional(s) to whatever it is you are examining.  While this is idealistic, it is far from common practice.
Calibrating is very tedious and time intensive than just running with a single functional.  For instance, when examining systems with large dispersion components to say, an intermolecular interaction, a slew of  13 DFT functionals (and corresponding variants) were tested to find qualitative agreement against a variety of computed properties (nature of the stationary point, binding energy, geometry, etc.) and compared against the MP2 wavefunction method (see: http://pubs.acs.org/doi/abs/10.1021/jp502588h).  Some functionals were clearly underperforming while quite a few functionals were in very good (qualitative) agreement amongst each other as well as MP2.  This is one approach you can use to validate your choice of functional.
You have to be careful about promoting one functional over another based on this type of analysis alone if experimental results are not available (or some previous calibration study) as was the case in this paper.  Experimental values were available for formaldehyde dimer but not for any of the sulfur containing systems.  The computational methods were validated by producing numbers that agree with experiment and extended to the sulfur containing systems.  But this does not mean that you will get the same performance on these novel systems.  In this case, more rigorous examinations are required and the authors of this paper state that future work include implementing the CCSD(T) level of theory to verify not only the MP2 results but also the functionals used.

Short answer: No.
Long answer: There are too many different ways to categorize the myriad conceivable chemical species, and too many properties of potential interest, for such a 'lookup database' of functionals to be feasible.
Numerous theoretical chemists are actively engaged in benchmarking of functionals for their performance in predicting particular properties of particular sets of related compounds; searching the literature for benchmark studies pertinent to the system/property you wish to examine is likely the best course of action.

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What does this property of transition metals mean?
what does it mean:
"Transition metals either have incompletely filled d subshells or readily give rise to cations that have incompletely filled d subshells.
I don't understand the portion starting from " or readily give rise to..."

Take the electronic configuration of copper (atomic number $29$) $\ce{[Ar] 3d^10 4s^1}$. According to the first portion of the definition, copper doesn't have incompletely filled subshells. But, copper readily gives rise to cation $\ce{Cu^{2+}}$ that has the electronic configuration $\ce{[Ar] 3d^9 4s^0}$ with incompletely filled d subshells. So, copper is considered as a transition metal.
Now, you can guess why zinc (atomic number $30$) is not considered as a transition metal.


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sp² hybridized orbital
I do not understand why and how an atom with $\ce{sp^2}$ hybridization has only one p orbital??
For example in pyridine why is the lone pair of nitrogen not counted as a pi electron.

The picture below from the UC Davis ChemWiki summarizes why that lone pair is not available. It also explains how pyridine can function as a base.

The lone pair in the $sp^2$ orbital is perpendicular to the $p$ orbitals and $\pi$ bonds in the molecule. the electron pair from the nitrogen that participates in aromaticity is the one in the $\ce{C=N}$ bond.
To answer your first question about hybridization, we have a conservation law that guarantees that when orbitals mix (for any reason), we need to get the same number of orbitals out as we put in. The law is the Conservation of Angular Momentum, which does apply in quantum systems.
Nitrogen, in its valence shell, has four orbitals: $2s,2p_x,2p_y,2p_z$. The $sp^2$ hybridization is achieved by mixing two $p$ orbitals and one $s$ orbital. Because of the conversation of angular momentum, four orbitals in means four orbitals out. Thus, the $p$ orbital that was not mixed is still there (and is the only one still there).
$$[s,p,p],p\implies[sp^2,sp^,sp^2],p$$
We cannot have two $p$ orbitals at an $sp^2$ hybridized atom because then we would have 5 orbitals (or we would not have $sp^2$).

Now, let's create a general rule for aromatic heterocycles.

Look at pyrrole (also from ChemWiki);

In the structure of pyrrole, the $\ce{N-H}$ bond uses the $sp^2$ orbital and the lone pair is in the $p$ orbital. In general, if there is a double bond drawn to the nitrogen atom (like in pyridine), then the $\pi$ bond uses the $p$ orbital and the lone pair is in the $sp^2$ orbital and does not count toward aromaticity. If there is no double bond drawn to the nitrogen atom, usually there is a $\ce{N-H}$ bond (or some other bond to the nitrogen). This $\ce{N-H}$ bond uses the $sp^2$ orbital and the lone pair uses the $p$ orbital and counts toward aromaticity.

Good answer.But I have a doubt.If an acid is added to pyridine,will the aromaticity be preserved or destroyed as in case of pyrrole ?

Pyridine remains aromatic when protonated. The lone pair that enables pyridine to be a base is orthogonal to the pi system.

Actually pyrrole gets C-protonated IIRC

@Mithoron - it likely does, but either way, aromaticity is lost.


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Amount of Energy Needed for Boiling a Litre of Water
What is the amount of energy (J or Wh) needed to boil one litre of tap-water in a regular kettle starting at room temperature (20°C)?
The calculation itself assuming ideal and standard prerequisites would be interesting as well as factoring in the non-purity of tap-water and a typical kettle performance factor.

You want to know the enthalpy of vaporization.

It is simple. Density of water is $1\;\frac{\text{g}}{\text{ml}}$ as we have 1 kg water.
Now according to the laws of calorimetry the amount of heat required

$Q = c \cdot m \cdot \left( {100{\text{ }^\circ\text{C}} - T} \right) + \Delta {h_{\text{vap}}} \cdot m$

Here

$c$ = specific heat capacity of water, which is 1 cal/(g °C)
$m$ = mass of water; i.e. 1 kg
$T$ = the room temperature in °C
$\Delta {h_{\text{vap}}}$ = specific enthalpy of vaporization of water; 540 cal/g

So you will get the amount of energy needed (in cal) to boil a litre of water
To get answer in Joule multiply the answer in cal by 4.184 J/cal. And be careful with units!

The answer assumes total vaporization of the water.
If you only want to heat the water to the boiling point e.g. to make some tea, then the second term should be Δhvap⋅m1, where m1  stands for the water that got vaporized in the process. Typically 0<m1<m.


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What is the hybridization of oxygen in furan?
I fail to understand why the hybridization of oxygen in furan is $\ce{sp^2}$. I think that since there are 2 bond pairs and two lone pairs so it should be $\ce{sp^3}$ hybridized?
Is it to do with the aromaticity of furan?

Furan is aromatic because one of the lone pairs of electrons on the oxygen atom is delocalized into the ring, creating a $4n+2$ Hückel aromatic system.
The steric number is sometimes incorrectly described as the sum of the number of sigma bonds and the number of lone pairs. However, it is more accurately given by the number of sigma bonds plus the number of localised lone pairs.


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Thermochemistry: forms of energy & pressure
I've been in a few classes now that teach about energy, and I feel they're always so bad. There is always such a rush to start using equations to get problems solved, and I often feel I don't have an intuitive grasp on energy at all. I guess this question is also about pressure.
Questions:

Does pressure decrease with time? If pressure is caused by the force of gas molecules hitting the inside surface of a container, don't they transfer energy and eventually slow down enough to totally reduce pressure?
Along the same lines, energy can be released or absorbed by a reaction as work or heat. If the only change that happens is through work (in something that is totally insulated - where gas is formed and pressure is created), after the pressure is created, doesn't it require continuous energy to maintain pressure against a constant force of 1 atm (or whatever it is)? Heat dissipates to all things, and that makes sense as everything gets colder in time. The same does not seem true for pressure.

Thanks in advance.
Noobsauce

Does pressure decrease with time? If pressure is caused by the force of gas molecules hitting the inside surface of a container, don't they transfer energy and eventually slow down enough to totally reduce pressure?

No. As you know, an aerosol can remains pressurised indefinitely until it is used. The molecules do not slow down if no energy transfer occurs, because there are no smaller bodies to transfer their energy to.
At the start of a game of snooker/pool/billiards (whatever you play in your part of the world) the balls are arranged in a triangle at the bottom of the table. You place the white ball at the top of the table and hit it into the other balls. When the collision occurs, the white ball shares its energy (normally unevenly) with the other balls. What we see is an increase in disorder. Even if we neglect friction, once this has occured it is very unlikely (impossible even) that all the other balls would stop and the white ball would have all the energy again. The way the energy of the white ball is spread between the other balls is a macroscopic display of how kinetic energy is dissipated by friction.
You will learn later on that while energy must be conserved, disorder in a closed system can only stay the same or increase; it can never decrease. Disorder is quantified by entropy but the quantification is beyond the scope of your question.
The balls on the table will soon stop moving, due to friction. What happen is that the kinetic energy is shared around in a more disordered way amongst the molecules of the balls and table. The energy is still there (the balls and table are very very slightly warmer) but it is now distributed in random movements of the molecules so that the balls are no longer moving.
Once it gets to this level, the molecules just keep vibrating (or in the case of a gas, colliding.) There is no friction in these interactions, because there is nothing smaller than a molecule for energy to be dissipated into.
(well actually there is one thing: the molecules could shake themselves free of the solid they are in and become a liquid, shake themselves free of the liquid they are in and become a gas, or even shake themselves apart and become smaller molecules or atoms. But breaking these bonds or interactions can only occur with additional heat absorption, or by cooling of the material.)

after the pressure is created, doesn't it require continuous energy to maintain pressure against a constant force of 1 atm (or whatever it is)?

"work" (which is a transfer of mechanical energy) only occurs when something is moved against a force. For example, when you put your finger over the end of a bicycle pump and squeeze the plunger, your squeezing arm does work. If you release the plunger, it moves outward again, releasing the stored energy.
The fact that you find it difficult to maintain the force on the plunger indefinitely does not mean that any work is being done. Your body is not designed to maintain a force with perfect efficiency; energy is expended within your body to maintain that muscle contracted. But if you had a latch on the bicycle pump, you could close that and maintain the pressure indefinitely.
Similarly, when you lie down on an airbed, you increase the pressure inside it. You have no expenditure of energy while you are lying on the bed, yet you maintain the pressure. Similarly the aerosol can maintains its internal pressure with no expenditure of energy.

1.Does pressure decrease with time?

The ideal gas law tells us that
$$PV=nRT$$
or
$$\mathrm{P=\frac{[nRT]}{[V]}}$$
If the temperature, number of moles or volume does not change, then the pressure will not change.

If pressure is caused by the force of gas molecules hitting the inside
  surface of a container, don't they transfer energy and eventually slow
  down enough to totally reduce pressure?

Yes, that's correct.  If energy is transferred out of the system, then the temperature will fall and the pressure will decrease.  But if everything is already at the same temperature, there will be no energy flow in or out and the pressure will remain constant.

after the pressure is created, doesn't it require continuous energy to
  maintain pressure against a constant force

No, there is no requirement that energy must be supplied to maintain the pressure.  Again, if the gas, the vessel and the surroundings are all at the same temperature the pressure will not change.  However, if the entire system is not at thermal equilibrium, then heat will flow so as to put things at equilibrium and the temperature will change.


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Pt[BrCl(PR₃)₂] cis, trans isomer nmr - Which is cis and which is trans?
The two square planar isomers of $\ce{Pt[BrCl(PR_{3})2]}$ have different $\ce{^{31}P}$-NMR spectra. One shows single group of lines, the other shows two distinct resonances, each one being similar to the single resonance region of the first one. Which isomer is cis and which is trans?

Look at the two structures drawn below

In the trans-isomer each phosphorous atom is in the exact same chemical environment.  Each one has a chlorine on one side, a bromine on the other side and a phosphorous diagonally across.  Therefore only one signal is observed for the phosphorous.
In the cis-isomer, each phosphorous is in a different chemical environment.  One phosphorous has a chlorine diagonally across from it, the other has a bromine diagonally across from it.  In this case we observe two phosphorous signals, one for each of the two different phosphorous environments.

Chemical shifts and multiplicity are always controlled by:
Who sees what?
If two centres have the same environment/neighbours or are related by a symmetry operation, they will have identical shifts.
Here, the phosphor atoms in the trans isomer both "see" the same and appear as a single set of lines.

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Non-halogen intermediate between alkane and alcohol under basic conditions?
Using organic chemistry I have failed to find a method that doesn't involve halogens (gasseous especially).
Please ignore the enzyme mono-oxygenase, which adds $\ce{O2}$ directly to alkanes to form alcohols.
[edit] Regarding selectivity the reaction should preferably stop at the alcohol stage.
The most promising method I have found involves Boranes, but only works for terminal alkanes. The termial organoborane can then be converted to an alcohol.
I found it using a google search inspired by the answers below: "photochemistry of alkanes."

A haloalkane is not really an alkane. What starting materials can you use? It is not clear from the question.

Conversion between alkane and alcohol can be done in various ways. Btw I don't think this q. is unclear.

I think that you're asking if there is a way to convert an alkane to an alcohol without using halogens in the process.
Alkanes are relatively unreactive and so high energy species (like free radicals) are most often used to carry out reactions on alkanes. The  reaction of alkanes with halogens that you've mentioned is an example of this free radical process.
Other high energy molecules that can be used to convert alkanes to alcohols include ozone and (photo-) chemically generated singlet oxygen.  Just like with the halogens, these reactive molecules show little selectivity, so if primary, secondary and tertiary hydrogens co-exist in your alkane, then a mixture of primary, secondary and tertiary alcohols result.  However in molecules such as cyclohexane, where there is only one type of hydrogen, somewhat useful results can be obtained.  I say "somewhat useful" because ozone and singlet oxygen can oxidize the resultant alcohol further to a carbonyl.  Here is a reference to an abstract on the use of ozone to functionalize alkanes.
Reagents such as potassium permanganate and chromic acid also oxidize alkanes by a free radical mechanism to produce alcohols.  Again the issues of selectivity and further oxidation remain.
Here is a link to a nice review entitled "Alkane Hydroxylation." It covers a wide variety of reagents.  Most of the molecules depicted contain functional groups which can help improve selectivity, but there are a number of hydrocarbon examples as well.  Perhaps you'll find an example that is similar to your molecule of interest.

Nice answer. You definitely have stronger oxidants than Kalus! I chose Klaus' answer because his method is more specific in that it is capable of stopping at the alcohol stage.

@JoeHobbit  I don't think so.  Wikipedia says, "The free radicals generated by this process then engage in secondary reactions. For example, the hydroxyl is a powerful, non-selective oxidant. Oxidation of an organic compound by Fenton's reagent is rapid and exothermic and results in the oxidation of contaminants to primarily carbon dioxide and water."

Maybe if a protecting group were added to the solution that would latch onto the alcohols produced before further degradation.

That's a possibility, something that reacts rapidly with alcohols (TosCl?).  Have you had a chance to look through the "Alkane Hydroxylation" link I added?  I was surprised by how selective some of the pure hydrocarbon reactions were, and that conditions could be tuned to optimize alcohol yield.

The use of Platinum Chlorides to make termial alcohols is  great!

I will giver preference to methods using zero halogens, but this comes fairly close.

ron has already pointed out that the functionalization of alkanes often involves radicals.
A classical process is the Fenton reaction, typically using $\ce{FeSO4}$ and $\ce{H2O2}$.
Here, the reactive species is the hydroxyl radical, $\color{\red}{\ce{HO\cdot}}$, which is generated via:
\[\ce{Fe^{2+} + H2O2 -> Fe^{3+} + OH- + \color{\red}{HO\cdot}}\]
In a variant, the photo-Fenton reaction, the process is facilitated by UV/VIS irradiation.
Irradiation in the visible range enables photoreduction of $\ce{Fe^{3+}}$ to $\ce{Fe^{2+}}$, furnishing more hydroxyl radicals:
\[\ce{Fe^{3+} + H2O ->[h\nu] Fe^{2+} + \color{\red}{HO\cdot} + H+}\]
Upon irradiation at wavelength below 400 nm, direct excitation (and homolytical cleavage) of hydrogen peroxide provides an additional source for the reactive species:
\[\ce{H2O2 ->[h\nu] 2\color{\red}{HO\cdot}}\]
However, due to the lack of selectivity, this process is rarely used in synthesis and rather applied for the opposite: degradation of organic matter in wastewater treatment.

Precisely what I was looking for! Would the photo-Fento reaction would be at all hindered by a moderately alkaline environment? pH=9

Admittedly, I've never seen Fenton ot photo-Fenton reactions under alkaline conditions. The solution of iron(II)sulfate itself is quite acidic, btw. And again, this isn't a really useful method other than for the degradation of organic matter, such as in waste water treatment (reduction of TOC and degradation of chlorophenol in paper industry).

I wonder if a dilute solution could achieve a reasonable yeild of alcohol.


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Hemoglobin equilibrium - effect of increased carbon dioxide and lactic acid
The binding of oxygen by haemoglobin giving oxy-haemoglobin is partially regulated by the concentration of $\ce{H3O+}$ and dissolved in $\ce{CO2}$ in blood
$$\ce{HbO2 + H3O+ +CO2<=>H+.Hb.CO2 + O2 + H2O}$$
If there is production of lactic acid and $\ce{CO2}$ in a muscle during excretion then is more oxygen released or is more $\ce{HbO2}$ formed?

Is this homework? What research have you done? Where exactly are you stuck?

See im a bit confused with this...i wanted to know if my guess is correct...is it oxygen because if we add more CO2 equilibrium should shift to right..i just wanted someone to verify...no its not my homework...just practising problems..


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Solubility determination of Ca(OH)₂ in water
In one problem it's asking me to find the solubility of $\ce{Ca(OH)2}$ in water. In that problem it assumes the starting concentration $\ce{OH}$ to be zero. But for other problems like $\ce{BaSO4}$ in $0.1~M~~ \ce{Na2SO4}$, we account for the $\ce{SO4}$ in the $\ce{NaSO4}$. So in the first problem, why don't we worry about the $\ce{OH}$ from the self ionization of water? Is it just that extremely low of an amount?

I don't get your point... It's like saying why we don't assume friction when computing the energy we spend to get a glass of water from the fridge. Never have we solved problems that talked about the true conditions in real life.

Yes, compared to the amount of $\ce{SO4}$ contributed by 0.1 M $\ce{Na2SO4}$, the amount of $\ce{OH^{-}}$ contributed by water is exceedingly small.

The self ionization of $\ce{OH}$ is $10^{-7}$. If the concentration of $\ce{Ca(OH)2}$ is greater by a few magnitudes, then the self ionization should not matter much. In fact, $\ce{Ca(OH)2}$ is considered soluble, so if not in miniscule amounts, the self ionization should not matter much


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Please explain the following graph for a quantum mechanical harmonic oscillator

Graphs such as the above keep coming up when talking about harmonic oscillators in a quantum mechanical sense. However, I simply cannot make sense of them. What does each line represent why are they waves and what is the parabola? Also, I see sometimes graphs such as:
These seem to follow on from the initial graph. It's something to do with the recovery of classical behavior but I don't understand the first graph so stand no chance of getting the second ones at high n - perhaps you could explain the relation between the first and second graph please? Thanks

Background
Let's start by considering the case of 2 hydrogen atoms separated by a large distance.  If they happen to be on a trajectory that brings them closer together they can interact with one another.  If they get close enough together they can form a bond and create a hydrogen molecule.  This process is pictured in the following graph.

image source
This "potential well" is often referred to as a "Morse potential".  Out on the right edge of the curve (say 5 angstroms), we have our 2 separated hydrogen atoms.  As the distance between them lessens, stabilization occurs, they can fall into the well and form a hydrogen molecule.
The red lines in the Morse potential represent different vibrational states of the hydrogen molecule.  Initially, the hydrogen atom is formed in one of the higher vibrational states.  It can either absorb more energy through collisions and go back to 2 dissociated atoms or it can give away some excess energy through collisions and fall to a lower vibrational level.
The classical view of the hydrogen atom is two weights connected by a spring.  The more energy in the system, the wider the oscillation of the spring.  With lower energy the oscillations are smaller.  The energy of the spring system (y-axis) as a function of separation (x-axis) is given by Hooke's Law as
$$E(x)={1\over2}kx^2$$
This is really an equation that describes a parabola, so the lower part of a Morse potential can be accurately described by a parabola.  Your top drawing represents the lower part of the Morse potential diagram.
While we have used hydrogen as our model in the above explanation, the same reasoning applies to any bond formation (and subsequently bond stretch) process.  For example, we could describe a $\ce{C-H}$ bond stretch in a completely analogous manner.
Your Questions

What does each line represent

Different vibrational levels, as explained above

why are they waves

Each energy state (vibrational level) is described by its own wavefunction ($\Psi_n$).  The square of the wavefunction (${\Psi_{n}^2}$) describes the probability of finding the electron at a certain point on the x-coordinate for that specific vibrational energy level. From the solution to the wavefunction we find that as we go to higher and higher energy levels, the number of nodes (a point where the wavefunction = 0) in our wave continually increases.  The square of the wavefunction produces positive values (the "upside-down" parabolas). We always have the same number of nodes in the wavefunction and the probability distribution.  For example, in the lowest vibrational energy level (n=0) there is no node in either the wavefunction or the probability distribution.  In the n=1 vibrational level there is one node in the wavefunction and probability distribution.

what is the parabola?

Explained in the background section


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Ionize Hydrogen and Oxygen Gas in Electrolytic Cell?
As is well known, water can be electrolyzed into oxygen and hydrogen gasses. However, all redox reactions can be run in reverse. If I were to electrolyze water in 2 closed containers (in order to trap the gasses) and i were to run the cell in reverse using a battery, would the hydrogen and oxygen gasses turn back into ions and, therefore, water? If so how would this work on a physical level? And what path do the ions take to get to each other?

Since the monatomic hydrogen and oxygen have combined, respectively, in H2 and O2 gases, those molecular bonds must be broken in order to get them to recombine as water and produce electricity. That is precisely what a fuel cell does. You can lease a hydrogen fuel cell powered Hyundai Tucson, if you're close to H2 filling stations, and Iceland is looking into setting up a hydrothermal-powered H2 fuel station system.

What if you do not have a catalyst though unlike a fuel cell?

As mentioned, there is activation energy which must be exceeded. At high temperatures, H2 and O2 will react directly; see http://en.wikipedia.org/wiki/Solid_oxide_fuel_cell.


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How does a catalyst affect the rate equation?
If I determine the order of the reactants in a reaction without using a catalyst, and then use a catalyst, will the order of the reactants then be different?

Yes. It is possible for a catalyst to change the order of a reaction since it shows different path to the reaction i.e. reaction occurs through different mechanism. In the new path, rate determining step (slow step) may involve different number of reactant molecules.
For example, in case of reaction catalyzed on metal surfaces, most of the times, the rate does not depend on the concentration of the reactants at higher concentration since now it is surface area that governs the rate.


text:
Calculating Vapor Pressure from Delta G/ Kp

For the process, $\ce{CH3OH(l) -> CH3OH(g)}$
  $\Delta G^\circ = 4.30\ \mathrm{kJ/mol}$ at 25 °C. What is the vapor pressure
  of $\ce{CH3OH(l)}$ at 25 °C in mmHg?
  (A) 0.176 mmHg
  (B) 14.0 mmHg
  (C) 134 mmHg
  (D) 759 mmHg

This question is from the 2012 National Chemistry Olympiad (US).
Attempt at solution:
$\Delta G^\circ = -RT\ln K$
$4.3\times10^3\ \mathrm{J/mol}=-8.314\ \mathrm{J/(mol\cdot K)}\cdot 298\ \mathrm{K} \cdot\ln K$
thus
$0.1763\ \text{(unitless)} = K$
Now, I’m supposed to know that $K$ is in units of atm and convert to mmHg accordingly. How would it know that this unitless quantity is atm and not pascal or mmHg?

http://chemistry.stackexchange.com/questions/5811/unit-of-the-equilibrium-constant-contradiction-of-bridgmans-theorem This question has an explanation to a similar question like yours

This topic came up the other day but in regards to the evaporation of water.  You might find my answer to that question helpful.
Here's how I would solve the problem.

Explicitly note standard state used and assumptions.  I will assume:

Methanol vapor is an ideal gas
Standard states for all species are 1 bar of partial pressure 25 °C.  (This assumption matters a lot.  The meaning of the $^\circ$ symbol in $\Delta G^\circ$ can't be understood without it.
The free energy of methanol liquid does not vary with pressure, i.e. $G_{liq} = G_{liq}^\circ$ at all pressures.  I'm fairly confident that this assumption is equivalent to assuming methanol liquid is an incompressible fluid.

Note the equilibrium condition.  At equilibrium,

$$\Delta G = 0 = G_{vap} - G_{liq}$$

Substitute the pressure-dependent free energy of an ideal gas

$$ 0 = G_{vap} - G_{liq} = \left [G_{vap}^\circ + RT \ln{\frac{p}{p^\circ}}\right ] - G_{liq}$$

Use the assumption of incompressibility of liquid methanol that we made above

$$0 = \left [G_{vap}^\circ + RT \ln{\frac{p}{p^\circ}}\right ] - G_{liq}^\circ$$

Rearrange and substitute

$$0 = \left [G_{vap}^\circ - G_{liq}^\circ \right ] + RT \ln{\frac{p}{p^\circ}}$$
$$0 = \Delta G^\circ  + RT \ln{\frac{p}{p^\circ}}$$
$$0 = 4.3 \frac{\textrm{kJ}}{\textrm{mol}} + RT \ln \frac{p}{p^\circ}$$

Solve for $p$

$$\ln{\frac{p}{p^\circ}} = \frac{-4.3 \frac{\textrm{kJ}}{\textrm{mol}}}{RT} $$
$$p = p^\circ e^{\frac{-4300 \textrm{ J/mol}}{RT}}$$
$$p = p^\circ e^{\frac{-4300}{8.314\cdot 298}}=  p^\circ e^{-1.73} = 0.176~p^\circ  $$

Figure out what $p^\circ$ is.  Just like it does for $\Delta G$, adding a $^\circ$ indicates that we are referring to a standard state.  Thus $p^\circ$ refers to the partial pressure of methanol vapor at its standard state.  This is almost always defined as one atmosphere (or in more modern times, one bar, which is nearly but not exactly the same thing) of pressure. Assuming that the standard state is the usual one, then $p^\circ$ is one bar, which is equal to 750 mmHg.  This is what we have assumed in step 1.  If that is the case, then $p = 0.176~ p^\circ=0.176\cdot750~\textrm{mmHg}=134~\textrm{mmHg}$.

How would it know that this unitless quantity is atm and not pascal or mmHg?

Hopefully my answer above helps you answer your question.  It has to do with the common use of $\circ$ to indicate a standard state of one bar of partial pressure for all gases.

Thanks for your very comprehensive answer! I adopted a somewhat less rigorous but easy to understand explanation, in that it is p over p standard, or 1 atm (or bar) but you help show how that is so

Seems familiar from my USNCO days. If you assume methanol is ideal, you can just use fugacity = 1.

I'm not sure why you went through all the trouble of #2-5 instead of just using the equation ∆G° = -RT ln K.


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Why is 2-methylpropene less in energy than its alkene counterparts?
Other isomers include 1-butene, cis 2-butene, and trans 2-butene. Why would 2-methylpropene be less in energy if there is more steric hindrance?

There are two things we need to understand before we can answer the question.
1) More highly substituted double bonds are generally more stable than less substituted double bonds.
This is because the $\ce{sp^3}$ hybridized carbon in the alkyl group is electron donating towards the $\ce{sp^2}$  hybridized carbon in the double bond. Electron density likes to flow from low to high s-character orbitals since the more s-character in an orbital, the lower the energy of the electrons in that orbital.  As the electrons flow from the $\ce{sp^3}$ carbon to the $\ce{sp^2}$ carbon the energy of the electrons (and therefore the energy of the molecule) is lowered.
A "shorthand" way to illustrate this electron flow is by drawing hyperconjugated resonance structures like the following.

Consequently 2-butene (where we can draw 6 hyperconjugated resonance structures) is more stable than 1-butene (where we can only draw 2 hyperconjugated resonance structures).
2) Carbanion stability decreases in the order primary > secondary > tertiary
As we've just shown above, alkyl groups are electron releasing.  If we have a negative charge on a carbon (a carbanion), the last thing we want is to have more alkyl groups attached to that carbon releasing more electron density onto an already negative carbon atom.
Now we can return to your question and try to understand why 2-methylpropene is more stable than trans-2-butene.   Carbon-hydrogen hyperconjugation is at the center of the answer.  Let's compare some of the hyperconjugated resonance structures that can be drawn for these two molecules.

Remembering that a) more highly substituted double bonds are more stable and b) that primary carbanions are more stable than secondary carbanions, we can compare hyperconjugated resonance structures A and B. Resonance structure A contains a disubstituted double (more stable) and a primary carbanion (more stable), while resonance structure B contains a monosubstituted double bond (less stable) and a secondary carbanion (less stable). Therefore, resonance structure A should be more stable than resonance structure B.  Since resonance structure A is more stable than resonance structure B, 2-methylpropene will be more stable than trans-2-butene.
Finally, we note that cis-2-butene is less stable that trans-2-butene for steric reasons.  This allows us to place these $\ce{C_4}$ isomers in the following order of stability:
2-methylpropene is more stable than trans-2-butene which is more stable than cis-2-butene, which is more stable than 1-butene. If we examine the heats of combustion for these isomers, we see that our order of stability is confirmed.

image source

Why do you say that a primary carbanion is more stable than a secondary carbanion?

@Cyclopropane As Point 2 above notes, alkyl groups are electron releasing, hence attaching them to a carbanion (a carbon already bearing a negative charge) further destabilizes the ion; 3 alkyl groups more than 2, more than 1.


text:
How to find molecular charge of fragments using quantum calculations?
Assume in a software like Avogadro, I put molecules A and B in a specific distance and I know they will involve in a charge transfer reaction ( as Marcus Theory outer sphere transfer ). Then I run a calculation package as Gaussian or GAMESS.
Now I want to see if any charge has moved from A to B.  One approach is to measure their charges and see if they are partially charged.
The problem is that I don't know how to find that molecular partial charge in a selected part of the system in such softwares (like I select a part of the system on the screen and see its charge )
One approach I found is in Avogadro: View>properties > atom properties and find the charge of each single atom and then add the selected atoms together , but it is very very time consuming in larger systems. I am pretty sure there must be a more effective method.

I do not know if that is possible. I would make a script for the output file if there are too many atoms. Also, be careful about the meaning of the results.

The way to do obtain the charge of a fragment in a molecule is quiet similar to How to calculate molecular volume from the wavefunction?.

The way to obtain the wave function from GAMESS is with the AIMPAC keyword. Look at page 2-18 of the GAMESS manual
As just repost the answer given over there and just editing the correct words to this post:
The way to determine the charge of a fragment in a molecule is by doing a quantum calculation to obtain the wave function and, then, integrate it.
You can obtain the charge of your desire fragment in the molecule by following these steps:

Do a quantum calculation with the desire method and a very good wave function.
In the input file, you have to specify to the software of your preference to print out the wave function file (in gaussian, you can do it with output=wfn, more directions: Directions to print wave function from gaussian).
After the quantum calculation is done, you have to integrate the wave function with a specialized software (see at external links of QTAIM software).
After integrating the wave function, you will get the integrated values of several properties, one of them is the atomic charge of each atom. (See page 16 of The Quantum Theory of Atoms in Molecules).
Finally, sum up all the atomic charges of the fragment you want to know.

It was a great suggestion. The only problem is that I am in Colorado and we have not Gaussian on our supercomputers. I have to use GAMESS and don't know how to set wavefunction request in the input file. Do you have experience on it ?

I exactly how to print thw wfn file from gamess, nope... I don't know how properly you can do it. Beside, I can complement you with the manual of gamess, a discussion where some got the wfn from gamess and, if you ind the way to obtain wfn files from gamess, a script to convert to the properly wfn structure to be used with the software previously mentioned

:D I found the way to do it from GAMESS :D

In the manual I shared you, look at page 2-18
The keyword is AIMPAC and look the other page (wfn from gamess) to see how to obtain the properly wfn file ... I'm going to update my answer

Thanks . I think this is a perfect answer and brings up my other question:  How to extract the wavefuction from GAMESS.  As I think that is a separate question, I made another thread on it here:http://chemistry.stackexchange.com/questions/28855/how-to-extract-wavefunction-from-gamess-output

you may want to have a look at it as you almost have the answer for that question also:  http://chemistry.stackexchange.com/questions/28855/how-to-extract-wavefunction-from-gamess-output

That is good, I'm going to check it. btw, a naive question ... Did not I get the bounty price in the volume related question? heeh ... just like a pampered child :D

Thank you for the answer but I think you got it ! didn't it show on your reputations ?

Thank you to you too and here we are to help each other through the knowledge we have :D

Your question boils down to "where are the electrons" in a quantum chemical system?
The problem, is if there's any level of interaction between the two fragments, you don't know. The charge will be delocalized across the entire A+B complex.
Put it another way.. Let's say you have a bunch of atoms and you run a calculation and the overall charge on the molecule is +1. Where's the extra positive charge?
Well, you might make some guesses, but it's probably delocalized across multiple atoms. The same thing happens with intermolecular interactions, but to a lesser degree.
There are two approaches:

Constrained DFT: This is a method where you force the charge on a particular fragment to be the way you want. You can then compare this semi-localized charge with the distributed (unconstrained) calculation and compare.
Fragment MO: You can diagonalize the wavefunction onto multiple fragments that you define.

I don't use GAMESS much, but a quick search implies that FMO is supported in GAMESS.


text:
Why is nitric acid a stronger oxidising agent than sulfuric acid?
We were taught that nitric acid can oxidize $\ce{Cl-}$ to $\ce{Cl2}$, but sulfuric acid cannot. Is this due to its smaller size, or structure, or what?

Does this answer your question? Why is HNO3 a stronger oxidising agent than H3PO4?

The nitrate ion is a strong oxidant whereas the sulfate ion is a milder oxidant that only shows oxidizing properties in high concentrations. The oxidizing atom in question is the oxygen atom in the acid, not the hydrogen ion. This is because hydrogen ion is not very electrophilic, which is why non-oxidizing acids cannot dissolve copper: even copper has a higher affinity to electrons than hydrogen.
In a nitrate ion, the molecular structure is inherently not very stable, as the nitrogen atom has to bear a positive charge to maintain the molecular structure. Besides, nitrogen is in itself a very electronegative element, so that makes the nitrate ion even more unstable, hence its oxygen atom readily reacts with other atoms or molecules that oxygen can oxidize, including chloride ions:
$\ce{Cl- + O -> OCl-}$
$\ce{H+ + OCl- <=> HOCl}$
$\ce{H+ + Cl- <=> HCl}$
$\ce{HCl + HOCl <=> H2O + Cl2 }$

Oxygen in these acids is in lowest ox. state so it can oxidate anything.

@Mithron isn't the formal oxidation state of oxygen here -2? Otherwise are you talking about partial charge?

Sorry I meant it cannot because it's -2.

@Mithoron not entirely. Formal oxidation state serves as no more than a reference for determining stability of compounds, while the reality isn't too accurately reflected by this. It is indeed true that the nitrogen atom takes a much greater oxidizing role than oxygen here, the reaction still takes place through the oxygen atom as if oxygen is the oxidant here, and the stability of whatever product is formed is determined by oxygen's oxidative power and not nitrogen's.

The acids $\ce{HNO3,H2SO4}$ , and the ions $\ce{NO3-,SO4^2-}$ are involved in redox reactions . standard reduction potentials of these ions are
$\ce{(1) SO4^2- +4H+ +2e-  -> H2SO3 + H2O (E^0= -0.93V)}$ and
$\ce{(2) NO3- +2H+ +e- ->NO2  + H2O (E^0= +0.83V)}$; Reduction potentials reveal that $\ce{NO3-}$ is more prone to be reduced i.e. nitrate ion is stronger oxidizing agent;  The possible reasons for this behavior are:-
(1) $\ce{SO4^2-}$ is tetrahedral and sulfur is comparatively blocked to be attacked  moreover, charged oxygens in sulfate repel the incoming $\ce{Cl-}$/ electrons, while, $\ce{NO3-}$ is planar & nitrogen is more open to be attacked& less repelled.
(2) N of $\ce{HNO3}$ is more electronegative than S of $\ce{H2SO4}$  for the reasons that N  stays positively charged (formal charge in the lewis structure) & $\mathrm{sp^2}$ hybridized while S stays with no formal charge & $\mathrm{sp^3}$ hybridized. These factors can account for nitric acid's enhanced oxidizing nature. Hope this answer is helpful.


text:
Does nitric acid release hydrogen only when reacting with Mg and Mn?
$\ce{HNO3}$ always reduces its anion to give nitrogen containing compounds ( $\ce{NO2}$, $\ce{NO}$, and even $\ce{N2O}$) when reacting with metals, but with $\ce{Mg}$ and $\ce{Mn}$ it releases $\ce{H2}$. Is there a specific reason for this variation?

Might help you : http://en.wikipedia.org/wiki/Nitric_acid#Reactions_with_metals

Actually, it usually doesn't. It produces all types of products ranging from $\ce{NO2}$ to $\ce{NH4NO3}$ with exact products depending on concentration and conditions.
The reason for it is high oxidation power of nitric acid, exceeding that of hydrogen cations, so nitric acid reduces first. As result, to produce some nitrogen, the oxidation power must be brought below that of hydrogen cations. I'm not aware if it is actually possible, but possibly high dilution may help.
As for Mg, it reacts with hot water, so it's unclear if the reaction of $\ce{Mg}$ with diluted nitric acid counts. Not aware about $\ce{Mn}$ though.

"Produce different products with exact products" what did you mean? You meant with the same reactants produce different kind of products like NO2 to NH4NO3 depends on conditions and concentration?

"Oxidation of nitric acid exceeding that of hydrogen cations" you meant oxidation of NO3 - ions exceeding that of hydrogen cations?

@Lifeforbetter Yes and yes.

If you form "NH4NO3" with magnesium, then yes you get hydrogen because, independent of any reaction with nitric acid, magnesium displaces hydrogen from aqueous ammonium salts.  Thus NH4NO3 and hydrogen should be considered as being formed together.

It's because of the activity of hydrogen ions in comparison to the metals.
Certain metals like $\ce{Mg}$, say, will be oxidized by the hydrogen ions. Note that $\ce{Mg^2+}$ has a reduction potential of -2.37V, which is less than $\ce{H+}$'s reduction potential of 0V (not accounting for overvoltage and what not). Hence, $\ce{H2}$ gas is produced spontaneously.
However, metals like $\ce{Cu}$ (which is the example given by your Wikipedia link) and $\ce{Ag}$ have a reduction potential greater than 0V. This means that H+ will no longer be able to oxidize these metals. Instead, the $\ce{NO3-}$ ion will serve as the oxidizing agent.

Because reaction of Metal+Acid= Salt + hydrogen
$$\ce{Mg +2HNO3 -> Mg(NO3)2 +H2}$$
$\ce{Mg}$ has a charge of +2 while nitric acid ($\ce{HNO3}$) has a charge of -1
Metals always react with acids to form salt and hydrogen gas

thats just the point! Nitric reacts with ALL metals to release NO, NO2,N2O BUT releases H2 with Mn and Mg. Why the VARIATION?

It also releases hydrogen with other metals not just with Mn or Mg

actually, the wikipedia reference suggested by binary geek agrees w/ my facts, but doesn't supply a reason.


text:
Resonance structures of some aromatic compounds
Do compounds like tropolone, furan, and thiophene (which are aromatic) show resonance?

If yes then what are the resonance structures that can be drawn? Also, do all aromatic compounds show resonance?

I think aromaticity by definition implies resonance

Binary is right. A prerequisite for aromaticity is cyclic conjugation. Conjugation in itself allows resonance.
Although do not confuse resonance with aromaticity. Resonance is a necessary, but not sufficient condition for aromaticity.

Resonance structures can be drawn for all aromatic compounds.  Tropolone, furan and thiophene are considered to be aromatic compounds because

we can draw resonance structures
and they are cyclic compounds with 4n+2 (n=1) pi electrons

Here are the resonance structures we can draw for tropolone, the positive charge can be delocalized to every carbon in the ring.

Just like in benzene, where instead of drawing 2 resonance structures (top of following figure) we usually just draw the one resonance structure shown at the bottom of the figure;

instead of drawing all 8 resonance structures for tropolone, we usually represent them by just drawing the one resonance structure shown at the bottom of the tropolone figure.
and here are the resonance structures for furan

and for thiophene, they are the same as furan, just replace the oxygen with a sulfur atom in the resonance structures.


text:
Why is p-methoxyphenol more acidic than p-methylphenol?
Why is p-methoxyphenol more acidic than p-methylphenol?
According to me, inductive effect really doesn’t matter at para positions and p-methoxyphenol should be less acidic due to the high mesomeric effect of the $\ce{-OCH3}$ group in comparison to the weak hyperconjugative effect of the $\ce{-CH3}$ group.
p-Methoxyphenol – $\mathrm{p}K_\mathrm{a} = 10.1$
p-Methylphenol – $\mathrm{p}K_\mathrm{a} = 10.3$

Related: Acidity of substituted phenols

First off, it is a very small difference in $\mathrm{p}K_\mathrm{a}$ that you are asking about.  In fact, I would say that the two compounds have similar acidity.  It is often difficult to explain such small differences, but what follows is a general approach that can be used to probe the question and to try and answer questions related to equilibria.
In this question we are comparing the effect of two different substituents on an equilibrium.  To perform this comparison we must look at the effect of the substituent on both the starting material (the substituted phenol) and the product (the substituted phenoxide anion).  If the substituent stabilizes the phenol it will tend to shift the equilibrium towards the phenol side making the phenol less acidic; if the substituent stabilizes the phenoxide it will tend to shift the equilibrium to the phenoxide side making the phenol more acidic.
From your other question we see that the p-methoxy group destabilizes the phenoxide anion to a greater extent than the p-methyl group.  Based on this alone, we would say the substituent effect on the phenoxide side of the equilibrium should be such that the p-methoxy substituent pushes the equilibrium more strongly to the phenol side than the p-methyl group will.  This would tend to make p-methoxyphenol less acidic than p-methylphenol.
Looking at the phenol side  of the equilibrium, both the methoxy and the methyl groups are electron releasing through resonance, with the methoxy having a much stronger effect.  The phenolic $\ce{OH}$ group is also strongly electron releasing through resonance. Perhaps having two strongly electron releasing groups (hydroxyl and methoxy) attached to the aromatic ring causes some destabilization due to electron repulsion associated with the high electron density in the ring.  Certainly such an effect would be greater with methoxy than methyl.  If true, then we could argue that the p-methoxy substituent also destabilizes the starting phenol (more then the methyl group) and this would tend to push the equilibrium to the product side making p-methoxyphenol more acidic than p-methylphenol.
If the substituent effects on the phenol and the phenoxide are comparable, then they would cancel and we might expect p-methoxyphenol and p-methylphenol to have comparable acidities and if the substituent effects cancel we would also expect both phenols to have $\mathrm{p}K_\mathrm{a}$'s similar to phenol ($\mathrm{p}K_\mathrm{a}=10$), which they do.
The approach used here is sound.  I think the explanation of the substituent effects on the phenoxide are also sound.  The explanation of the substituent effects on the phenol side is (admittedly) weak.  Nonetheless, this explanation does correctly explain why p-methoxyphenol and p-methylphenol have comparable acidities and why these acidities are close to the acidity of phenol.

This may be a bit beyond the scope of the question, but can you include a couple more examples where there are conformational changes to increase stability?

While cycloheptatriene exists in a non-planar conformation converting it to the tropylium carbocation causes it to adopt a planar conformation in order to enhance resonance interaction.  Cyclooctatetraene exists in a tub conformation, if it were planar, resonance would destabilize the molecule (it would be antiaromatic). On the other hand, cyclooctatetraene dianion adopts a planar conformation as this produces a resonance stabilized 4n+2 system.

Due to the -I effect of OCH3 group the conjugate base of para methoxyphenol is more stable than para methylphenol. Due to this, para methoxyphenol is more acidic in nature then para methylphenol.


text:
How were the absolute masses of atoms calculated before the mass spectrometers?
I understand that true (non-relative) masses of atoms are calculated with a mass spectrometer. using this relationship: Centripetal Force = Force due to magnetic field(B)
But I was wondering how masses of atoms were calculated before the mass spectrometers were invented.
(Please don't get this mixed up with relative atomic mass. That's not what I'm talking about.)

Absolute mass can be determined electrochemically, once the coulomb is defined in terms of numbers of electrons, as through the Millikan oil-drop experiment, to be  $6.24×10^{18}$ electrons per coulomb.
See Determination of Avogadro's Number and Absolute  Determination  of  the  Electrochemical  Equivalent and  the  Atomic  Weight  of Zinc.
One thing that caused confusion in the determination of atomic mass is natural isotopic separation, e.g. sulfur derived from condensation from vulcanism vs. from mineral compounds, so that there were slightly different experimental results depending on source.


text:
Cannizarro reaction of chloral
When chloral undergoes reaction with a concentrated alkali, will it undergo Cannizarro reaction to form the corresponding alcohol and acidic salt, or should it form chloroform instead by departure of trichloro carbanion instead of the hydride ion?
I understand that the trichlorocarbanion is stabilized by the $-I$ effect of chlorine, so why not form that instead?

In the Cannizarro reaction an aldehyde with no alpha-hydrogens is treated with $\ce{OH^-}$.  Through a series of equilibria the aldehyde goes on to disproportionate into the corresponding carboxylate and alcohol as shown in the following mechanism.

Again, the only requirement for the Cannizzaro reaction to occur is that the aldehyde not have any alpha-hydrogens.  Formaldehyde and benzaldehyde are two very common aldehydes that undergo the reaction when treated with $\ce{OH^-}$.  Chloral ($\ce{CCl3CHO}$), having no alpha-hydrogens also undergoes the Cannizzaro reaction (reference)
Edit: Response to OP's comment
A key part of the Cannizzaro mechanism is shown in the second line of the above mechanism.  Here a hydrogen is transferred in a concerted step to the second molecule of aldehyde.  Free $\ce{H^-}$ is not generated in the process, rather it is a smooth, concerted transfer of a hydrogen atom.
[Note: This has been proven experimentally, when the reaction is run in $\ce{D2O}$, no deuterium winds up attached to the alcohol carbon.  This shows that free $\ce{H^-}$ was never involved, but rather a concerted transfer of the aldeyhdic hydrogen takes place].
Thanks to Mithoron for noting that, with chloral, the haloform reaction can apparently compete with the Cannizarro reaction.  Presumably, the pathway followed depends upon the reaction conditions.

Your reference says on the same page that it does undergo haloform reaction...

@Mithoron Does the percentage ratio of haloform and Canizzaro reaction depend on the concentration of base?

@Mithoron  On page O-339, where abouts on the page?

Penultimate chain of reactions starting with ethanol.

If chloral undergoes Cannizzaro reaction then the Haloform reaction with acetaldehyde or Chloral should fail.
I think Cannizzaro with Choral is not possible as CCl3- is a good leaving group rather than hydride(H-) ion.
Ref:Cannizzaro reaction mechanism


text:
What does the peak separation of carbon dioxide IR rotational-vibrational spectrum corresponds to and why?

Spectrum of $\ce{CO2}$ at $2349\space \mathrm{cm^{-1}}$   region
For some other molecules (for example $\ce{CO}$) the peak separation corresponds to $2B$ (where $B$ is the rotational constant wavenumber)
However for $\ce{CO2}$ when I calculated the peak separation and compared it to literature value, it seems approximately to equal to $4B$? Any idea why is this?
Also, what is the explanation for $\ce{CO2}$ spectrum being asymmetric? and why does the $P$ branch have larger peak separations than the $R$ branch?

(In this answer I am using the symbol $\require{begingroup} \begingroup \newcommand{\tb}[0]{\tilde{B}} \tb$ to represent a wavenumber instead of $B$, which I choose to define as an energy – not that we will be using it.)

Why is the peak separation $4\tb$ instead of $2\tb$?

This has to do with nuclear spin statistics. Because oxygen-16 is a boson (spin 0), the wavefunction of the entire molecule (which comprises translational, rotational, vibrational, electronic, and nuclear spin components) must be symmetric upon exchange of the two oxygen-16 atoms.
The translational component is symmetric, as usual. The electronic ground state is ${}^1\Sigma_\mathrm{g}$, which is totally symmetric. The nuclear spin component must be symmetric by virtue of the fact that both oxygen atoms are spin 0; this means that there is only one possible spin state that they can be in (cf. hydrogen-1 has a spin of 1/2 and two possible spin states). Consequently, exchange of the oxygen atoms leads to no change in the nuclear spin wavefunction.
As a result, this means that the vibrational component times the rotational component must be symmetric; or in other words, the vibrational and rotational components must have the same parity.
Since the peak structure at $\pu{2349 cm-1}$ arises due to the antisymmetric stretch, the vibrational ground state $v = 0$ is symmetric with respect to exchange of oxygen, and must therefore be paired with rotational levels with even $J$, which are likewise symmetric. In a similar manner, the vibrational excited state $v = 1$ is antisymmetric, and must be paired with odd $J$ levels.
In $\ce{CO}$ where there are no nuclear spin statistics at work, the following transitions are all allowed in the $R$-branch ($\Delta J = +1$):
$$\begin{align}
(v,J) = (0,0) &\to (1,1) & \text{line 1}\\
(0,1) &\to (1,2) & \text{line 2}\\
(0,2) &\to (1,3) & \text{line 3}\\
(0,3) &\to (1,4) & \text{line 4}\\
\end{align}$$
and so on. However, in $\ce{CO2}$ the state $(v,J) = (0,1)$ is not permissible, and hence line 2 does not appear in the spectrum. Similarly, neither does line 4, or line 6, and so on; the separation between adjacent lines is therefore doubled.

Why are the peak separations in the $P$-branch larger than in the $R$-branch?

This is because the rotational constant $\tb_0$ in the vibrational ground state is different from the rotational constant $\tb_1$ in the first vibrational excited state. The rotational energy levels (in wavenumber units) are therefore by:
$$\tilde{G}(v,J) = v\omega_0 + \tb_vJ(J+1)$$
where $\omega_0$ is the wavenumber of the pure vibrational transition (i.e. the band origin). For the transition $(v,J) = (0,J) \to (1,J-1)$ in the $P$-branch, the associated energy is therefore
$$\tilde{P}(J) = \omega_0 + \tb_1J(J-1) - \tb_0J(J+1)$$
The separation between adjacent lines in the $\ce{CO2}$ spectrum is then
$$\begin{align}
\tilde{P}(J) - \tilde{P}(J+2) &= \tb_1J(J-1) - \tb_0J(J+1) - \tb_1(J+1)(J+2) + \tb_0(J+2)(J+3) \\
&= 4J(\tb_0 - \tb_1) + 6\tb_0 - 2\tb_1
\end{align}$$
You can easily verify that if $\tb_0 = \tb_1 = \tb$, then this simplifies directly to $4\tb$, as expected.
However, because $\tb_0 \neq \tb_1$, the gap between adjacent lines acquires a $J$-dependency, because the term $(\tb_0 - \tb_1)$ is nonzero. In the case of the asymmetric stretch, it turns out that $\tb_0 > \tb_1$ (can you figure out why? It's best explained using the moment of inertia $I$, since $\tb = \hbar/8\pi cI$). Therefore, as $J$ increases (i.e. as we move farther left along the spectrum), the separation between lines increases.
An exactly analogous calculation will explain why the lines in the $R$-branch tend to "bunch":
$$\tilde{R}(J+2) - \tilde{R}(J) = 4J(\tb_1 - \tb_0) + 6\tb_1 - 2\tb_0$$
$\endgroup$


text:
Reaction of hydrogen peroxide and catalase
All Google searches have simply returned more info on catalase. I'm looking for a catalyst that isn't found inside living organisms which can break down $\ce{H2O2}$. Looking for any resources to look at, or just examples of other catalysts.
Side question: Can catalase be used for anything other than the decomposition of  $\ce{H2O2}$?

Manganese (IV) oxide, iron (III) chloride, and lead (IV) oxide are common examples for a catalyst that isn't found inside living organisms and which can break down $\ce{H2O2}$. Potassium iodide can also be used as a catalyst, along with other transition metal compounds, of metals such as silver or platinum.
This didn't take much effort to find so I'm sure you can find more examples easily.
As for catalase, it seems to just decompose $\ce{H2O2}$, but this is useful in many different areas. For example, decomposing $\ce{H2O2}$ in contact lens cleaning solutions, preventing foods from being oxidised in wrappers (in conjunction with glucose oxidase, another enzyme), and removing $\ce{H2O2}$ from milk before cheese is made. The decomposition reaction is also used as a test for catalase in bacteria.
Sources: General use - National Center of Biotechnology Education - and more uses (including the cheese example) - Worthington Enzyme Manual

The OP should be aware of the difference between a catalyst and and enzyme. This might help in any Google search.  Catalase is an enzyme as it is a protein found in living organisms.


text:
Why do E2 reactions (dehydrohalogenation) only occur when the hydrogen and leaving group are in the same plane?
E2 reactions (dehydrohalogenation) only occur when the hydrogen and leaving group are in the same plane.
I can't think of a reason that they would need to be in the same plane, so what is the reason for this strange phenomenon?

When the $\ce{C-X}$ and $\ce{C-H}$ $\ce{sp^3}$bonds break, the orbitals left behind on the two carbons will rehybidize into p orbitals and form the pi bond in the forming double bond.  For these two p orbitals to form a pi bond they must be in the same plane (or very close).
This is illustrated in the following figure.  This is called the anti-periplanar conformation (the soon-to-be p orbitals are in the same plane and arranged ~180° with respect to each other) and is the preferred geometry for E2 elimination.

image source
Elimination can also occur from the syn-periplanar conformation (the soon-to-be p orbitals are in the same plane and arranged ~0° with respect to each other) as shown below, but this conformation is slightly less favored.

This is why in molecules where the $\ce{C-X}$ and $\ce{C-H}$ bonds are locked in a conformation (perhaps by an inflexible ring system) where they are oriented 90° apart, E2 elimination becomes very difficult, if not impossible.

Why is $180^\circ$ orientation of p orbitals is preferred to their $0^\circ$ orientation with respect to each other?

@Abcd 1) electronic factors - imagine how the original sp3 orbitals transform to p-orbitals as the hydrogen and LG move further away and note that there is more\better  overlap (on both the top and bottom lobes) between the forming p-orbitals in the 180° case as compared to the 0° case  2) steric factors - will favor the 180° elimination particularly if "B" and LG are large.

@ron If syn-periplanar conformation just has higher activation energy this doesn't mean that in a reacting sample there is no possibility that two molecules will react with that conformation, right?

@adosar Correct.  As I noted in my answer, "Elimination can also occur from the syn-periplanar conformation...but this conformation is slightly less favored".


text:
What forces hold ink on paper?
On regular paper, ink is absorbed, while on photopaper ink stays at the surface and dries for a longer time. What forces act in these two different cases?
(see also related questions about pencil marks and colorants)

Hmm, there are two types of inks: Inks used for printing and inks used for writing. Which one do you mean? (More precisely, do you wonder the how of what you've written, or printed?)

@MARamezani, I meant printed, but I would appreciate an answer that covers both.

There's actually two questions being asked here:

What is holding ink on paper?

and

Why do some inks take longer to dry?

Ink sticks to paper because it has a high affinity for the medium it's being applied to. To group things into simple categories we could say, there are hydrophobic inks (non-polar inks that don't like water) and hydrophilic inks (polar inks that do like water). As with the inks, surfaces can also fall into this category. When ink is thrown onto a surface that it doesn't like, it aggregates, clumping together to avoid all interaction with the surface.
A simple illustration showing why this is unfavorable can be seen with water interacting with lipids. In the picture below, water is forming ordered cages around the hydrophobic part. These cages cause a decrease in entropy, and by the 2nd law of thermodynamics, this is unfavorable. You can think of the ink as the water, and the hydrocarbon chain of the lipid as the surface.

When inks do favor their substrate (surface) they will attach via h-bonding, london dispersion forces, and/or other inter-molecular forces.
As for the second question, the answer relies more on what the ink is made out of. According to the wiki on inks:

Ink can be a complex medium, composed of solvents, pigments, dyes, resins, lubricants, solubilizers, surfactants, particulate matter, fluoresces, and other materials.

Generally, the ink, which has become mobile in its solvent, will remain mobile until the solvent evaporates. So you'll find that the dry time is dependent upon the rate of evaporation. That's why the dry times can be longer for some oil-based pens.