text:
X-Ray Diffraction: Phase Identification Strategy
I am having some trouble with qualitative phase identification using x-ray diffraction (XRD). Here follows some basic information about the experiment:
Instrument: PANalytical X'Pert$^3$ Powder
Software: PANalytical Highscore Plus
Database: PAN ICSD
Sample: respirable fraction aerosol samples from ferro- and silica manganese plants.
The goal: which manganese phases are present, and which dominate?
Grasping the software is not what I am concerned about. I lack an analytical strategy. I know how to determine the background, but I am not sure when the background is appropriately determined. Too high? Too low? Does this strongly affect the phase identification?
Then I do a peak search, and 50 peaks are detected. Should I focus just on the major peaks? Are the minor ones important too? When I do a search-match analysis, I get 400 suggested phases, where 20 phases share the highest score. Then another 20 phases share the highest score minus 1. I lack criteria needed to accept a phase as present, and discard other phases as not present.
I am not able to find good literature on this subject. I did find an article called "Hints on Phase Identification Using Powder X-ray Diffraction", which certainly raised a lot of potential problems and procedures, but which did not necessarily solve these problems or explain why the procedures would help. Does anyone know of any books or other literature that could help me?
My problem boils down to lack of "strategy" or "analytical procedure", and maybe someone here could offer some advice. I hope my question is "answerable", and not too broad; I know there is a "specific-and-precise-question" policy in here.

I don't have much information on your particular system: are they nanocrystals?  Are you expecting just one or two phases?  Dozens of phases? Do you have a lot of amorphous material present?
I have one publication you can look at which addresses a rather complicated sample with amorphous, nanocrystalline, and crystalline objects.  It is (hopefully) overkill for your sample, but it will give you ideas on how to partition the problem (and some of the references may help too):
Gainsforth, Z., Brenker, F. E., Simionovici, A. S., Schmitz, S., Burghammer, M., Butterworth, A. L., et al. (2013). Stardust Interstellar Preliminary Examination VIII: Identification of crystalline material in two interstellar candidates. Meteoritics and Planetary Science, 49(9), 1645–1665. doi:10.1111/maps.12148
You may also find it helpful to play with simple fitting softwares such as CrystalMaker/CrystalDiffract where you can try out different mixtures and fits.  Sometimes the peak-finding softwares are a little simplistic with a large number of false positives.  If you try manually combining the possible phases, you can probably rule many of them out and simplify your list.
A few tips:
1) Look for sets of peaks with FWHM that are similar.  This works with nanocrystalline materials.  If you have a nanocrystal with a larger crystal, then the larger crystal will have sharp peaks and the nanocrystal will have broad peaks.  Then you can select which peaks apply to Phase A and Phase B and simplify your search space drastically.
2) If Peak A is present in your experimental data, and you get a "hit" for phase X, but Phase X also predicts Peak B but you don't see it in your experimental data, then you can rule Phase X out.  You may need to understand the characteristics of the phase in a little detail because you can have situations where one atom substitutes for another and you get a change in peak intensities or positions.
3) Do you know the chemical composition of your sample?  If one of your "found" phases is a Yttrium based superconductor, and you're pretty sure you don't have Yttrium, then you can drop that phase.
4) Try looking at it with an electron microscope (SEM or TEM).  This will provide basic constraints as to whether you have many phases, or just a couple, and how big the crystals are.  This is a lot of work, but maybe less work than spending the next month chasing false positives.  TEM can also give you diffraction on any phase, no matter how small, and will be pretty unambiguous.
5) You can also provide constraints using techniques such as XANES, NMR, FTIR, etc., depending on what kind of sample you have.  Which of these are useful is also dependent on the resources you have present at your university/company/facility.
Hope that helps!

Thank you for your help. I have just done an element analysis using ICP-OES, which certainly will help. SEM analysis is also planned, along with ion chromatography. I will keep in mind your points; they are very helpful! I have realized that experience is alpha and omega in phase ID!

Glad to help!  I wish you success with the analysis.


text:
Difference between S and I in the context of magnetic resonance experiments
In the lecture our professor told us that nuclei with $S>1/2$ have a quadrupole and are therefore detected with NQR instead of NMR.
The majority of all elements have nuclei with $S>1/2$.
I wondered why he used $S$ (electron spin) instead of $I$ (nuclear spin) in the slides?

The nomenclature you mean has nothing to do with the electron spin S, it is something completely different. In heteronuclear experiments with two spins it is common to call them $I$ and $S$, with $I$ being the more sensitive nucleus. They're still both spin 1/2 nuclei.
$I$ is one spin, and $S$ is a different spin in this nomenclature, with a scalar coupling between them.
This nomenclature is usually introduced with the INEPT transfer and experiments like HSQC or HMQC.

I don't think this is why the lecturer is using S instead of I. I think it is being used simply as an abbreviation for the term spin. Although I is the correct symbol for spin, a lot of NQR papers tend to refer to Spin-1 (or Spin-3/2 etc) nuclei. I think it is simply an extension of that terminology.

Mad Scientist has provided a nice answer to your question about S and I.  I'd like to comment on another aspect of your post.

our professor told us that nuclei with S>1/2 have a quadrupole and
  are therefore detected with NQR instead of NMR

"Instead of nmr" might be too strong.  Most nuclei with a nuclear spin >1/2 can also be observed by nmr.  Some of these nuclei may have a low relatively sensitivity to the nmr experiment, or due to nuclear quadrupole relaxation might produce signals where information has been lost, in which case NQR might be preferred.  While there are many cases where the NQR experiment would be more informative than the NMR experiment, there are also many cases where NMR would work just fine.
An interesting example related to proton-nmr compares the following 3 compounds, pentadeuterioacetone, N-methylaniline and chloroform.

Deuterium, nitrogen and chlorine are quadrupolar nuclei. Deuterium and nitrogen-14 (the major isotope of nitrogen) both have a nuclear spin of 1, while chlorine-35 and -37 (the 2 predominant isotopes of chlorine) have a nuclear spin of 3/2. If one examines the proton-nmr of pentadeuterioacetone, one can observe the coupling between the deuterium and hydrogen nuclei.  By the same token, if one examined the deuterium-nmr of this compound, one would observe the same H-D coupling.  If one examines the proton-nmr of N-methylaniline, the coupling between nitrogen and the various protons is washed out, broadened signals where the coupling information has been lost are observed.  In the case of chloroform, the proton-nmr shows only a sharp singlet.
Why the change in coupling to a proton as we change the coupled nucleus from D to N to Cl?  The situation is very similar to physical exchange phenomena.  If a proton is attached to a nucleus, then coupling can be observed between that proton and adjacent protons.  If the proton is exchanging with the environment at a rate similar to the timescale of the nmr experiment, then the coupling will broaden.  If the proton is exchanging very rapidly compared to the timescale of the nmr experiment, then the proton is effectively decoupled from the system and a sharp singlet will be observed.
In the case of the 3 compounds discussed above, instead of physical exchange, the proton is being relaxed (decoupled) at varying rates by the attached quadrupolar nuclei.  The rate of quadrupolar relaxation caused by a coupled deuterium atom is slow on the nmr timescale, so all coupling is preserved and observed.  The quadrupolar relaxation rate of nitrogen-14 is comparable to the timescale of the nmr experiment and so the coupling is beginning to wash out.  The chlorine nuclei undergo rapid quadrupolar relaxation on the nmr timescale and are effectively decoupled from other nuclei.

"Instead of nmr" might be too strong. Utterly absurd actually.
The principal advantage of NQR over NMR is that it can relies on energy level differences induced by electric fields, rather than magnetic fields. This makes it a far more portable technique, and is sometimes referred to as zero-field spectroscopy. It has some nice real-time applications (explosives detection, crude oil evaluation), but there is a clear lack of prominence in peer-reviewed literature compared to NMR studies of quadrupole nuclei. I wonder if said lecturer has a research interest in NQR spectroscopy.....

Thanks to Mad Scientist's answer, I understand the topic better.

An electron can has a spin of $s=1/2$ (lower case s).
The total spin quantum number $S=1/2$ (upper case S) is the sum of the single spins.

Although protons are nuclei, they are elementary particles (which have usually a spin of $s=1/2$). Protons have only a spin $s=1/2$ and therefor $S=1/2.$

In contrast to the mentioned two cases of elementary particles examples an atom (or ion with at least one electron; a proton is also an ion) have a spin quantum number $s$ and a angular momentum quantum number $l.$ So the two quantum numbers sum up to the nuclear spin which is usually determined by $I$ as the sum over all $s$ plus the sum over all $l.$ Very often we find as well instead of $I$ the letter $J = L + S$ in spectroscopy.

I also agree with long's comment

I don't think this is why the lecturer is using S instead of I. I think it is being used simply as an abbreviation for the term spin. Although I is the correct symbol for spin, a lot of NQR papers tend to refer to Spin-1 (or Spin-3/2 etc) nuclei. I think it is simply an extension of that terminology.

and I think that the lecturer wrote $S$ as an abbreviation of spin without distinguishing $I$ und $S.$

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

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

sort: votes


text:
How to find the percentage contribution of a compound via gravimetric analysis?

You have a sample of silver alloy and would like to know how much silver is in it. Dissolve the silver sample in nitric acid. Filter to remove any insoluble material. Add an excess of $\ce{KCl}$ solution, which causes the formation of silver chloride precipitate. Filter, dry, and weigh the precipitate. What do you estimate the percentage of silver in the original sample is if its mass was $5.2~\mathrm{g}$, and the mass of silver chloride was $6.4~\mathrm{g}$?

The atomic mass of silver is $M(\ce{Ag})= 107.87~\mathrm{\frac{g}{mol}}$. That of chlorine is $M(\ce{Cl}) = 35.45~\mathrm{\frac{g}{mol}}$.
My answer is $81\%$, because $5.2/6.4 \cdot 100 = 81$. However, I am quite unsure.

\begin{align}\ce{
3Ag + 4HNO3 &-> 3AgNO3 + NO + 2H2O\\
AgNO3 + KCl &-> AgCl + KNO3\\
}\end{align}
Moles of $\ce{AgCl} = 6.4/143.5 = 0.0446~\mathrm{mol}$
Mass of $\ce{Ag} = 0.0446 \cdot 108 = 4.816~\mathrm{g}$
Percentage of $\ce{Ag} = 4.816/5.2 \cdot 100=92.62\%$

I have updated your post with chemistry markup. If you want to know more, please have a look here and here. Please do not use markup in the title field, see here for details.

@Martin that wa typo, division was right

I noticed that :)

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Does potassium nitrate solution degrade?
I have created a solution of $\ce{KNO3}$ and table sugar in order to create fuse from hemp rope. I wet the rope in the solution and let it dry - I repeat this process several times.
It works and produces very slowly burning rope which is easy to set off. The problem is that after 24 hours, my solution started to stink. Something between garlic, onion and vinegar. I have a bad feeling about that.
Does my solution degrade, or is this stink caused by some impurities (the solution isn't very pure)?

I'm afraid bacterias came for your sugar ;)

I wonder why did I write sugar... I'm pretty sure I was just dissolving KNO3 alone.

As mentioned in the comments, a likely culprit is microbial contamination.  In a $\ce{KNO3}$ solution, you are providing plenty of water, a nitrogen source (nitrate), and even a high-energy, respiratory means for growth (nitrate respiration).
Living things also need (i) food, (ii) trace amounts of sulfur, (iii) and even smaller trace amounts of phosphorus.  But if you are soaking hemp in this solution, enough leachable S and P-containing impurities are probably extracted into the solution to provide these.  (Or also low levels of S and P may already be present in your $\ce{KNO3}$, or water.)  The food could be organic compounds leached from the hemp.  Or (less likely) the microbial contaminants might be capable of photosynthesis, in which case their food is atmospheric $\ce{CO2}%edit$.

Should I understand it so that $KNO_3$ actually supports life, in this case?

If you are referring to my comment about nitrate respiration, then yes, for some organisms it supports like in the same way that oxygen does.  Oxygen doesn't help you live on its own, you need to combine it with food to make energy.  Certain bacteria can combine nitrate with food (e.g. glucose) to make lots of energy in a similar way.  http://en.wikipedia.org/wiki/Anaerobic_respiration

I just saw the substantial edit you made to the question.  If you weren't using sugar, carbon containing compounds that can act as food for microbes could also be leached out of the hemp.

I am sorry for the edit. It was mentioned in comments, I just forgot to add it. Now the things that confuses me that potassium nitrate is used as conversant for foods. Something that supports bacteria isn't very conserving. Please do not become offended because I have doubts about the answer.

Tomas, you raise a good point.  There are bacteria that can live off of nitrate, but they usually do so only at low concentrations.  If you are using very high concentrations, it is a bit of a mystery where the odor is coming from.  If you soak your hemp in plain water for 24 hrs, does it generate the same smell?


text:
Outermost electronic configuration of f block elements
Why is it $(n-2)\mathrm{f}^{1-14}(n-1)\mathrm{d}^{0-1}n\mathrm{s}^2$.
The only exception I thought to this formula was Thorium $\ce{5f^{0} 6d^{2} 7s^{2}}$.
So I feel it should be $(n-2)\mathrm{f}^{1-14}(n-1)\mathrm{d}^{0-2}n\mathrm{s}^2$.

As you said, if you look at the electron configurations of Lanthanides and Actinides, Thorium seems to be the only exception from this empirical rule. You could expand the rule to also include the case of Thorium but I would say it is more useful to have the rule describe the majority of cases and remember $\ce{Th}$ as an exception.


text:
Can we determine apriori the volume of solvent in an electrolyte
Suppose I would like to create a one molar solution of NaCl in water. The molar concentration is defined as
$c_i=\frac{n_i}{V} $
$n_i=1$ in this case. I need to take 58.5 grams of NaCl (atomic weight of Na=23 & Cl=35.5) and dissolve it in water until the total solvent volume is 1 litre. The total volume of the water added is less than 1 litre. But what exactly is that amount of water and how can I determine it apriori from thermodynamics?

This will require some reading: http://en.wikipedia.org/wiki/Partial_molar_property. In particular, you're interested in partial molar volumes.

You can calculate it using partial molar volumes.
The partial molar volume is defined as the differential change in volume with respect to change in mole fraction, and is determined experimentally. From the definition of partial molar properties, you can calculate the amount of one component needed to achieve a certain volume, if you know all of the partial molar volumes and pure component molar volumes.
There may be some complications to that for this system because you are talking about mixing a solid-phase ionic compound with water, instead of two liquids directly.
In practice, it is much easier to just use a volumetric flask so that you know the total volume is 1 L. If you measure the amount of water you actually used you will find that it is very close to 1 L.

Thank you. I was just curious and maybe it will be useful to automate for industrial level production.

Your best bet for automation would be to use a batch process that had a tank water-level sensor to shut off water flow when it reaches the correct volume. You could then measure out the right amount of salt by mass (also automated), dump it into the mixing vessel, and add water until it reaches the shutoff sensor. It might be possible to set up a continuous process instead of batch, but I don't think it would be worth the effort unless you needed very large amounts of continuously supplied solution.


text:
How does sulbutiamine work in the same way as thiamine, even if it doesn't have a thiazole ring?
So sulbutiamine is said to be a thiamine substitute.

Yet - it doesn't contain the thiazole ring that's so important for many of thiamine's functions (highlighted in red).
How does it function as a thiamine substitute?

To get started, here are the structures of some compounds that are relevant.
Thiamine

Sulbutiamine

Allithiamine

Looking at these structures we see that you are of course correct, sulbutiamine does not contain a 5-membered thiazole ring like thiamine.
Thiamine (vitamin B1) was an early focus of interest because sailors at sea for long periods of time developed beriberi due to a nutritional thiamine deficiency.  Since those early days, thiamine substitiutes were of medicinal interest.  One of the early substitutes was allithiamine.  It is a natural product that can be extracted from garlic.
You can see the structural similarities between allithiamine and sulbutiamine.  Sulbutiamine has the advantage that both sides of the disulfide (S-S) bond contain the same active ingredient, so one would expect increased pharmacological properties from sulbutiamine.  The advantage that both allithiamine and sulbutiamine possess over thiamine is that they are more soluble in lipids (the positively charged thiazole ring in thiamine makes it very soluble in water, but not so soluble in lipids).  This enhanced solubility in lipids means that they can cross the blood-brain barrier much more effectively than thiamine, this improves their bioavalabilty.
With that background we can now answer your question, "how does sulbutiamine work in the same way as thiamine."
Once in the brain, sulbutiamine can be easily metabolized to thiamine.  Look at the structure of sulbutiamine and here is what happens:

reductively break the disulfide bond and form two identical mercaptans
$$\ce{RS-SR ->C[{[H]}]\   2 ~RS-H}$$
protonate the acetyl (that's on the nitrogen that will become part of our thiazole ring) oxygen to make it more attractive to nucleophilic substitution
Let the mercaptan sulfur nucleophilically attack the acetyl carbonyl carbon
eliminate water and form the 5-membered thiazole ring
and we now have thiamine

But a much larger dosage in the brain than had we simply ingested thiamine.


text:
Quantitative relationship between corrosiveness of acid and other properties
Is the corrosive nature of acid relative to some substance quantifiable like acid dissociation constant ($K_\mathrm{a}$)? If yes, what can be used to describe the corrosive nature of an acid?
For example, it is known that citric acid is far less corrosive than, say, sulfuric acid. But, can it be quantified? Also, if corrosiveness can be quantified, then can the caustic nature of bases be quantified too?

In addition to the acid dissociation constant, some other factors that add to the corrosiveness of an acid are its ability to act as an oxidizing agent and its affinity for water (hence, ability to dehydrate material). Concentrated sulfuric acid has both these abilities in addition to simply being a strong acid. I do not know if these factors have been quantified, but they could be.
Mixing chromic oxide ($\ce{CrO3}$) with sulfuric acid makes it an even better oxidizing agent, especially when hot - so-called - "cleaning solution" is great in the lab for removing residues of tar from your glassware, but it is rather dangerously corrosive and has to be handled very carefully.


text:
What actually is the Wavefunction?
I am aware that the square of the Wavefunction gives the probability density of finding an electron at a particular point in space. I have also heard that it's a complex number but since it's a function I am unsure as to how that could be the case (perhaps someone could please clear that up for me as a sub-question). Moreover, I have seen it described as the amplitude of something but I literally have no idea what is meant by that. However, the crux of the question is what is the Wavefunction itself (i.e not what the square of it is). I have read in the book "Why Chemical Reactions Happen" that it is essentially synonymous with the word/concept: orbital. I have also seen in various places w(x,y,x) or w(r,theta,thi) [where "w" represents the sign for a wavefunction] so, from what I can gather it is a function in three dimensions that represents the shape of a particular orbital where the function of (x,y,z) or (r,theta,thi) would, of course, be different for the s,p,d and f orbitals. However, in the book "Why Chemical Reactions Happen" it goes onto say that wavefunctions of different atoms interfere with each other to form molecular orbitals and the atomic orbitals somehow have a phase?! For my idea of a wavefunction to be correct, the idea of interference of these wavefunctions must be an idea that was thought of to try and rationalize the process of the formation of molecular bonding and anti-bonding orbitals from atomic ones to make it easier to understand and conceptualize. If I am right about the wavefunction being a purely mathematical representation of the space where an electron is likely to reside then the molecular wavefunctions are just as curious in terms of their shape and suchlike as the shapes of the atomic orbitals they came from (i.e nobody really knows why they are that shape but when atoms or molecules are arranged in a particular way, orbitals "just are" that shape). Also the idea of phase seemed to bring my argument down but then I thought that perhaps phase is simply defined as being opposite relative to another part of an orbital if it is the other side of an angular node (I suspect this is wrong but could someone please shed some light on this?)
Going back to when I said - "the molecular wavefunctions are just as curious in terms of their shape and suchlike as the shapes of the atomic orbitals they came from" - I thought this because I found it bizarre that empty orbitals could "interfere" with an orbital from another atom. For example one p orbital is empty yet all three interact with the four hydrogen s orbitals. Unless, the electrons can jump between the the degenerate p orbitals at such a frequency that all can be considered occupied I find it incomprehensible that just a portion of space can interfere with an orbital. As a result, I came to the conclusion that I quoted at the start of the paragraph. Basically meaning that when atoms come together to form molecules the electrons find the lowest possible energy spaces to reside; or in other words, a molecuar orbital / molecular wavefunction forms which is equally as curious as the atomic wavefunctions with their strange shapes.
Any help with this concept would be greatly appreciated - I am off to Oxford university this October and Molecular Orbital Theory is a small part of the suggested reading list before arrival into the first year. Also, on that note, please refrain from heavily mathematical answers because I simply won't understand as I have only done A-level Maths and A-level Chemistry. Thanks.

Too many (mostly trivial) questions. Either try to separate the and ask one-by-one, or better go and read some book on introductory quantum mechanics.

Or, maybe, you are trying to learn this just "too soon". I mean, you are just entering the university, where they surely will teach you quantum mechanics. Maybe not the first year, maybe not even the second year, but they will teach you the subject. So, do not jump the gun.

MO theory can be studied with little math, but at least basics of functional analisys and vector operations are very,very useful.

Let me see if I can get at some of your questions. As mentioned above, it's much easier when you ask individual specific questions.
One problem with books on introductory quantum mechanics is that, put simply, the language of quantum mechanics is math. Specifically, most people use the Schrödinger equation which involves second derivatives and differential equations.
When I teach quantum chemistry to undergraduates, there are typically problems due to the mathematical nature.
There are multiple "interpretations" of quantum mechanics. That gets into philosophy. Wikipedia has some nicely-written descriptions.
As you mention, the most widely used interpretation (the Copenhagen interpretation) of the wave function centers on the square of the wave function, or rather $|\psi^*\psi|$ as a probability density. As you mention, the wave function could be imaginary or complex, so this notation indicates a mathematical way of getting a real number for a probability density.
You ask about the shape and why atomic orbitals have particular shapes.
One thing that's amazing about quantum mechanics is that we put in some relatively simple equations to describe the motion of an electron in a hydrogen atom, and the solutions are exactly these atomic orbitals (s, p, d, f, etc.).
So I disagree that "no one knows" why atomic orbitals have particular shapes. The answer is that they are the mathematical solution to the Schrödinger equation. That is, we want to know the energies of the system and there are only certain quantized solutions.
And yes, the whole notion of chemistry is that electrons (and hence orbitals) on different atoms interact. Otherwise, there is no bonding.
I'll leave you with this. Without the math, it can be exceptionally hard to understand quantum mechanics. When you do understand the math, I think the results are quite beautiful. The shapes of orbitals aren't arbitrary. This is actually just a reflection of the mathematical nature of quantum mechanics.

The problem is that many programs teach the required math necessary for quantum over the course of too many semesters.  There are way too many credit hours to take in order to gain the appropriate math background by the time quantum is supposed to be taken.  This either discourages students and they switch majors or they enter quantum having to learn the math as they go leading to the problematic disconnect.

@LordStryker Indeed! We recently created a "math for chemistry majors" course, since students entered quantum without differential equations. Hopefully this concept will spread.

And at my school the professor says "orbitals aren't real" and QM is bull**** and QM-inclined faculty have been "popping too many math pills."

@LordStryker, well, I think the problem is not that a lot of math is required, but that students are discouraged by this. And that is their own problem. Modern chemistry is quantum chemistry, quantum chemistry is the application of quantum mechanics to chemical systems, quantum mechanics is a physical theory that requieres a lot of math. So, if you want to become a chemist, you have to learn a lot of math. Period.

@GeoffHutchison, calculus is not enough. They will certainly help to grasp the basic concepts of wave mechanics, but wave mechanics is not the whole story...

@Wildcat I believe that a lot of math is required for quantum.  From personal experience, there was debate in the department about requiring yet another math class because students going into quantum was having problems understanding the material.  A closer look revealed that by requiring another math class would have pushed quantum to a senior semester class just so all the math could be covered before hand.  Its atrocious.  Maybe a focused line of math classes strictly for chemistry students would fix the problem.  I don't know.

@GeoffHutchison, sounds great! Chemistry had changed completely in the last century, it, in a sense, finally emerged as a scientific discipline and not just a random collection of recipes. Some even argue that chemistry is over (the laws that govern chemical species are known), but in my opinion it is only beginning: we finally know how to do chemistry on a solid theoretical ground. The only problem is, we need much more powerful computers. :D

@Wildcat I've heard chemistry be defined as simply "condensed phase physics" by very notable people.  But the very fact that life operates in the condensed phase under biological conditions gives chemistry enough credence to be its own, thriving entity.

@Geoff Hutchinson, we also have a maths for chem course, to help students with weaker backgrounds. We supplement this with pre-lab problems in lab classes and in tutorials. Also we use Maple (a computer algebra programme) to help students get over technical maths details and allow them to focus on understanding the problem at hand. It also helps us to teach some coding which is essential for prospective postgrads to know.

@Wildcat Although I'm not a synthetic chemist, its easy to knock them for the recipes they use. But that you be unfair, for the body of knowledge built up by a systematic  & logical  approach to synthesising molecules has lead to  a wide & general understanding of how to predict the synthesis of millions of  new compounds.  No other approach has yet been anywhere as successful.

I know, I would better write a comment, but suddenly comments are limitted in size, so I will write my thoughts (rather than actual answer) here.

Also, on that note, please refrain from heavily mathematical answers
  because I simply won't understand as I have only done A-level Maths
  and A-level Chemistry.

That is the root of your problems. But do not be afraid, you are not alone! As I pointed out in comment to Geoff Hutchison answer, many chemistry students (and let us be honest here, not only students) are strongly reluctant to the following two simple very well established and no longer questioned facts:

In order to understand chemistry you have to understand quantum chemistry. In other words, at the very fundamental level, there is no any other chemistry except for quantum chemistry.
In order to understand quantum chemistry you have to understand quantum mechanics which requires a lot of math.

So, the right way to proceed is to learn some required math first (at least calculus), then physics (at least classical and quantum mechanics in the form of wave mechanics), and only then quantum chemistry. Thus, my advice is: do not put the cart before the horse! If you have some free time, learn some math.

Mathematician's answer
Wavefunction is quantum mechanical conterpart to trajectory in classical mechanics. Unfortunately, since we interact with the world in a way exposing to us mostly classical newtonian physics, there is no way to find pleasant, easily understandable everyday analogy. However, you may consider atomic orbitals as building blocks for 'trajectory' of electrons. They don't have to be filled to be used as building blocks.
To truly understand the role wavefunction plays in quantum mechanics, at least rudimentary understanding of quantum mechanics is required.  In includes understanding of what is

coordinate space
matrix operator
eigenvalues and eigenvectors
Knowledge of at least one example of basis in some function space, say, fourier series.

To understand MO theory, a very rudimentary understanding of Hartree-Fock formalism may help greatly.
AFAIK, wikipedia provides basic knowledge on the topics I mentioned. It is OK to have very vague understanding of concepts I mentioned until 3rd year in university. There is no need to have deep understanding of quantum mechanics to successfully apply MO theory.

This is physicist's answer. The mathematician's would be that the wave function is an element of the corresponding $L^2$ space which represent the state space of a quantum system, i.e. it is an equivalence class of functions that are  square (quadratically) integrable on a particular domain.

Short answer:
Wavefunctions are functions that represent the physical state of certain quantum systems.
Extended answer:

Ordinary wavefunctions are complex functions defined in a Hilbert Space (HS), corresponding to pure quantum states. There is not a one-to-one correspondence between pure states and wavefunctions. For
instance if $\Psi$ is a valid wavefunction in the energy basis (with
Hamiltonian operator $\hat{H}$), then the wavefunction $\Psi´ =
e^{i\phi}\Psi$ represents the same state because
       $$E = \int d^3x \> (\Psi´)^{*} \hat{H} \Psi´ = \int d^3x \> \Psi^{*}    \hat{H} \Psi$$
Wavefunctions are used in the wavefunction formulation of quantum
mechanics, which is only one of many available formulations. There
are other formulations of quantum mechanics, where wavefunctions are
used for different purposes and no longer represent a quantum state,
and there are formulations where wavefunctions are not used at all.
You can find a basic review of nine formulations of quantum mechanics
in this article and a new formulation of quantum mechanics
without wavefunctions in this article.
In the position representation, wavefunctions are expressed like
$\Psi = \Psi(x,t)$, and the square of the wavefunction is often
physically interpreted as the probability density of finding the
particle at a given position. This is correct for wavefunctions that
satisfy the Schrödinger equation, but it is not for Klein-Gordon and
Dirac equations. I.e. the square of a Dirac wavefunction does not
give the probability density of finding the particle at a given
position.
Wavefunctions do not describe mixed quantum states. Mixed quantum
states are outside the scope of the wavefunction formulation of
quantum mechanics. For instance, an atom in a molecule does not have
wavefunction associated to its quantum state, because atoms in
molecules are open systems and quantum correlations with neighbor
atoms prohibit the description of the properties of the atom with
wavefunctions. A heat bath at non-zero temperature does not have
wavefunction describing its quantum state either. Mixed quantum
states are properly described by other formulations of quantum
mechanics.
Ordinary wavefunctions cannot be used to rigorously describe instable
quantum states. Ordinary wavefunctions --sometimes named "Dirac Kets"
in the abstract representation of wavefunctions-- can be applied to
stable particles, but not to decaying particles or resonances.
There are attempts to extend the ordinary wavefunction formulation to
instable systems. Those generalized states are sometimes named "Gamow
kets" and are defined in a Rigged Hilbert Space (RHS). You can find
this formalism in this book.
From the above, it looks like if wavefunctions are a kind of analog
of the phase space densities $\rho$ used in classical mechanics. This
is more than a mere formal analogy and we can develop a phase-space
formulation of quantum mechanics where wavefunctions are replaced by
quantum phase-space $\rho$ and operators are replaced by phase-space
functions, for instance the energy of a quantum system in the
phase-space formulation is computed as
$$E = \int d^3x \int d^3p \> H \rho$$
which looks as a classical average, but is a proper quantum average.
Alternatively one can formulate classical mechanics in terms of
wavefunctions, which satisfy the classical equation
$$ i\hbar \frac{\partial \psi_\mathrm{class}}{\partial t} = \left[
   -\hbar^2\frac{\nabla^2}{2m} + V + \hbar^2\frac{\nabla^2 \sqrt{\psi_\mathrm{class}^{*}\psi_\mathrm{class}}}{2m
   \sqrt{\psi_\mathrm{class}^{*}\psi_\mathrm{class}}} \right]
   \psi_\mathrm{class}$$
sometimes named the classical Schrödinger equation in the literature.
Note that the last term between brackets is absent in quantum
mechanics. This classical equation is nonlinear and this nonlinearity
kills the superposition principle and other aspects associated to
quantum mechanics.

I think that @permeakra's last comments are very important, which is that you don't have to have a deep knowledge of quantum mechanics to be able to apply it and also that a lot of the maths detail can be left until your last undergraduate year.
One other point not mentioned above is that any area has its own often obscure language and notation. Quantum is one of the worst in this respect, e.g. basis sets, orbitals, wavefunctions, operators, hamiltonians, eigenvalues, eigenfunctions, bra's, ket's, stationary states and so on. Often textbooks (and lecturers) assume that one is already familiar with these concepts and if you are not (the general case) this jargon just gets in the way when you start to learn.
As a chemist you have to understand quantum mechanics primarily to help you understand spectroscopy, (vibrational, rotation, & most importantly nmr) as well as bonding.
But to the chemist, QM is a tool we need to use, and a fairly simple one. (If you think QM is hard, consider making your own nmr machine or even a laser from scratch). Although you will need to know how to solve the Schrodinger equation, all the cases you meet as an undergraduate have been solved long ago, with well know solutions; textbook are full of these. It is, however, essential to understand the physical assumptions and limitations that lead the solution, but not necessarily all the nitty-gritty.
What is also fundamentally important is to understand how to formulate a problem, and to identify unphysical solutions you may come across.  That way you begin to think like a scientist by clearly focusing on the problem and do not get distracted/overwhelmed by any maths. Believe that solving equations is, at the end of the day, a simple skill!


text:
Method of allylic and benzylic oxidation
I am told that $\ce{MnO_2}$ oxidizes allylic and benzylic alcohols. However, what is the mechanism of this oxidation? All the professor had to offer me was that the metal ion, manganese 2+, chelates the substrate; there is some sort of interaction between a pi bond, the metal ion, and the electrophilic carbon atom bearing the hydroxyl group. No further explanation was given, and a Google search hasn't unearthed anything of use. What, therefore, is the mechanism of this oxidation? I would like to know more about the mechanism to better understand why $\ce{MnO_2}$ only touches allylic and benzylic alcohols rather than any vanilla alcohol.

This is one of those heterogenous reactions occurring on a surface which are  mechanistically messy. As expected for a reaction that occurs on the surface of a reagent, factors such as surface area and reagent preparation (freshness, surface contamination) are significant variables.  A free radical mechanism is generally accepted and a reasonable mechanism is posted below.

image source
When an allylic alcohol with stereochemistry (cis, trans) around the double bond is oxidized, the stereochemistry around the double is preserved.

image source
This suggests that we are not really dealing with a free, long-lived radical intermediate, but rather a free radical held in close association with the manganese ion.

Hmm, it works for me.  Let me put the url here again in case anyone else has a problem.  http://www.citab.utad.pt/refbase/files/1890_JI-DONGLOU2011.pdf

The link does not work

@SchwarzKugelblitz I've added the mechanism directly to the answer.


text:
Ring Strain and C=O Stretching Frequency
Why does $\ce{C=O}$ stretching frequency increase with ring strain?
Why does conjugation decrease $\ce{C=O}$ ring strain?
I am told that cyclopropanone exhibits a higher $\ce{C=O}$ stretching frequency than does cyclohexanone. I know that higher stretching frequencies are correlated with a shorter, stiffer, and stronger bond. What I can't see is why would the $\ce{C=O}$ bond have more of these characteristics in a strained ring system than in a less strained ring system.
Are there hyperconjugative effects going on?
Are there other rationalizations that we can make? I.e. would smaller $\ce{C-C-C}$ angles suggest higher $p$-character in the $\ce{C-C}$ bonds and therefore higher $s$-character in the $\ce{C=O}$ bond?

Why does C=O stretching frequency increase with ring strain?

The carbonyl carbon prefers to be (more or less) $\ce{sp^2}$ hybridized.  Imagine taking the two other bonds attached to the carbonyl carbon and bending them towards each other so that the R-C-R angle is forced from 120 to 90 degrees (the same thing as incorporating these bonds into a smaller and smaller ring).  As we do this, the hybridization of the orbitals the carbonyl carbon uses to form these bonds changes from $\ce{sp^2}$ to $\ce{p}$.  As these bonds use more p-character, more s-character remains for use in the C-O sigma bond.  This would strengthen the CO bond and raise the energy required to stretch it (increase its ir stretching frequency).
An alternate explanation can be made based on resonance structures.  Look back at the acylium ion resonance structure used in our earlier acid chloride discussion (resonance structure IV).  As we strain the carbonyl by placing it in a small ring, the bonds between the carbonyl carbon and the adjacent carbons become weaker.  This means that resonance structure IV, with a strong $\ce{C#O}$ triple bond, becomes more important.  Here, in the present case, we would say that resonance structure IV involves carbon-carbon hyperconjugation.

Why does conjugation decrease C=O ring strain?

Do you mean, "why does conjugation to the carbonyl decrease the ir stretch frequency"?  Look at the resonance structure on the right side below for the conjugated ketone, cyclohexenone.  It's inclusion leads to a weakening of the carbonyl pi bond.


text:
Chemical bond of adsorbed atoms
When oxygen is adsorbed on iron metallic surface, does it create $\ce{Fe-O}$ bonds like the one in iron oxide? Can it be considered as a monolayer of iron oxide on the surface?
If not, how is the atomic bond of adsorbed oxygen and iron? How much is it strong? and what is its difference with normal chemical bond in iron oxide molecule?


text:
Is a single carbon atom stable?
It is well known that single atom of oxygen is not stable, and it forms $\ce{O2}$ molecule. But elements like carbon form a network of repeated bonds. As answered in another question, last atoms in such network are unstable and need to create some bonds.
Is it possible to have a (free-standing) single atom of carbon? Stable in the form of a stable powder like graphite or diamond.
If not, what is the minimum size of a carbon particle (diamond or graphite network)?
Further Clarification: I meant normal chemical stability to have a powder material at room temperature. I do not mean highly reactive carbon, quickly turning into a compound.

You seem to have several questions here.  I've answered your question title, but you also ask 1) about carbon specifically and 2) about how much carbon it takes to make a stable molecule.

Ultimately, stability has to be determined with respect to something. Helium gas is stable with respect to chemical reactions, but unstable with respect to nuclear fusion (even at 0 K, thanks to quantum tunneling; tremendous pressure and temperature just speed it up). But then all matter as we know it in the Universe is unstable with respect to subatomic particles. All atoms are expected to suffer proton decay (and consequently neutron decay too) over sufficiently long timescales (at least around $~10^{35}\ yr$, but not more than $~10^{200}\ yr$).

@NicolauSakerNeto, yes, but it is Chemistry site, so you can safely assume that "stable" means "chemically stable".

There are a couple interesting platonic hydrocarbons which are related to buckyballs: http://en.wikipedia.org/wiki/Platonic_hydrocarbon

Is it possible to have a (free-standing) single atom of carbon?

It's definitely possible, but such an atom would be highly unstable and likely would have a short lifetime. An example of this happening is if you blasted some carbon material with a laser. Here's a really cool paper where researchers blasted carbon nanotubes and graphene with lasers and observed single-atom chains (not an individual atom in this case, but the same principle applies) of carbon exhibiting some very interesting electronic properties:
http://www.pa.msu.edu/cmp/csc/eprint/DT086.pdf

If not, what is the minimum size of a carbon particle (diamond or
  graphite network)?

Buckyballs (60 carbon atoms) come to mind.
Tangentially related....
A few years ago, a team of Australian scientists fabricated a single-atom transistor (the basic building block of integrated circuits). The atom was Phosphorus. Granted, the atom was not free-standing, but still very impressive. Such a device is extremely difficult to fabricate, and requires cryogenic temperatures, so it's doubtful that this would ever become a viable technology.
http://www.nature.com/nnano/journal/v7/n4/full/nnano.2012.21.html

The smallest member of the Fullerene family is $\ce{C20}$. While the Buckminsterfullerene is the most common of course.

Is it possible to have a (free-standing) single atom of carbon?

Yes, it is just the matter of having appropriate physical conditions. So, for instance, atomic carbon exist in the gas phase above 3,642 °C (according to Wikipedia).

Well, the OP asked about room temperature, before he dies.

@PeterSchneider, initially there was no restriction on temperature.

A single atom of helium is stable in nearly all circumstances.  Stability isn't a binary thing, though; whether an atom will react to form a molecule depends on what other atoms are available to react with (for instance, xenon is generally non-reactive, but will form molecules with fluorine), as well as temperature (as Wildcat mentions) and pressure.


text:
How to find the molecular formula for a hydrocarbon compound from known combustion products?

What is the molecular formula of a compound containing only carbon and hydrogen if combustion of $1.05~\mathrm{g}$ of the compound produces $3.30~\mathrm{g}~\ce{CO2}$ and $1.35~\mathrm{g}~\ce{H2O}$ and its molar mass is about $70~\mathrm{g}$?

Here is my work:
$$
\begin{array}{cccccc}
                  & \ce{C_{a}H_{b}} & \ce{->} & \ce{CO2} & + & \ce{H2O} \\
\text{masses (g)} & 1.05            &         & 3.30     &   & 1.35
\end{array}
$$
\begin{align*}
\ce{CO2} &\rightarrow \ce{C} \\
44~\mathrm{g} &\rightarrow 12~\mathrm{g} \\
3.30~\mathrm{g} &\rightarrow x
\end{align*}
$$
x = 0.9~\mathrm{g},~\text{moles of C} = \frac{0.9}{12} = 0.075
$$
\begin{align*}
\ce{H2O} &\rightarrow \ce{2H} \\
18~\mathrm{g} &\rightarrow 2~\mathrm{g} \\
1.35~\mathrm{g} &\rightarrow y
\end{align*}
$$
y = 0.15~\mathrm{g},~\text{moles of H} = \frac{0.15}{1} = 0.15
$$
$$
\text{empirical formula}~\ce{C_{0.075/0.075}H_{0.15/0.075} -> CH2}
$$
$$
\frac{70}{14} = 5
$$
$$
\text{molecular formula is}~\ce{C5H10}
$$

The molar mass of any compound would not be 70 g, although it might be 70 g/mol.

What actually is the question here?

$$\ce{C_{$a$}H_{$b$} + $\left(a+\frac b4\right)$O2 -> $a$CO2 + $\frac{b}{2}$ H2O}$$
Suppose you had $n$ moles of hydrocarbon, then we have $a\cdot n$ moles of $\ce{CO2}$ and $\frac{b}{2}\cdot n$ moles of $\ce{H2O}$ dividing their moles we'll get $2\frac{a}{b}$:
$$2\frac ab=\frac{330/44}{135/18}=\frac{7.5}{7.5}=1\implies \frac ab=\frac12$$
So the empirical formula is $\ce{CH2}$
Now for actual formula $\frac{70}{14}=5$
Yes the formula is $\ce{C5H10}$.


text:
Concerted hydride and alkyl shifts
I noticed while looking through a couple of books (specifically, Wade, Carey, and Vollhardt/Schore) that the mechanism of concerted hydride and alkyl shifts is always shown with alcohol leaving groups.  I understand that these are difficult, equilibrium-driven processes.  But, I would assume they can happen with other leaving groups, such as a halogen, right?  Is it just the matter of fact that dehydrations are usually done at elevated temperatures while SN2 is not?  If that's all it is, though, we would expect to see rearranged primary E2 products, which Carey mentions does happen, but the other books remain silent on.

The SN2 reaction involves an attacking group and a leaving group moving in concert. A pentacoordinate transition state is involved.  Typically, there is no rearrangement in an SN2 reaction.  Similarly for an E2, as a base removes a proton. a leaving group is ejected in concert to create an olefin, no rearrangement products are observed.
It is the SN1 and E1 reactions that proceed stepwise through distinct carbocation intermediates.  As the carbocation is being formed in these reactions, a hydride or alkyl group can migrate to produce a more stable cation.
The leaving group can be things other than a hydroxyl group, for example chloro, bromo, or iodo groups make good leaving groups.  Back in the early days of physical-organic chemistry many alcohols were used in reactions simply because they were readily available as natural products.  Alcohols could also be readily converted into acetates,  tosylates, brosylates, or other sulfonate esters, which were even better leaving groups and allowed for reactions to proceed under even milder conditions.

I appreciate the quick response, but this post does not answer the question.  I am specifically asking about concerted hydride and alkyl shifts.  Rearranged primary products have been seen that obviously do not come from the rearrangement of a carbocation.  The hydride or alkyl shift is concerted with the the leaving group leaving giving a 2/3 carbocation.  For example, according to Carey p. 207, 1-butanol treated with sulfuric acid at 140*C will give primarily a mix of cis/trans 2-butene (88% yield).  My question is whether there is a scientific reason why this is always shown with alcohols.


text:
How do you solve a net ionic equation without a balanced equation?
In class, my professor went over a method for finding the net ionic equation without determining the balanced equation (e.g., if finding the balanced equation was too difficult for us students). It involved writing the ions of each of the solutions being mixed and determining which ions would be spectator ions.
For example, finding the net ionic equation for the following mixture of solutions:
$\ce{NaC_2H_3O_3 + BaCl_2 +NH_4NO_3 + K_2CrO_4}$
Unfortunately, I don't quite grasp the process. How does one determine which ions and compounds will remain in the net ionic equation?

This sort of question is really all about solubility. What you're really asking when you're trying to get a net ionic equation is What combinations of the ions are insoluble? Net ionic equations involve only ions that end up as precipitate on the right side of the equation. If you do enough of these problems, you get a feel for the general solubility trends of common ions, but there should be a table in your textbook or have a look here.
For this specific example, salts with alkali metal and ammonium cations are soluble. Nitrate and acetate anion salts are soluble. Therefore, all of those end up as spectator ions and we're left with just $\ce{Ba^2+}$ and $\ce{CrO_4^2-}$. Chromate salts other than those of alkali metal cations or ammonia are insoluble (rule 8 in the link), so the net ionic equation is:
$$\ce{Ba^2+_{(aq)} + CrO_4^{2-}_{(aq)} -> BaCrO_{4(s)}}$$
In this case, it's already balanced, but that may not necessarily be true.


text:
Bent's rule for Sulfur trioxide and Sulfur dioxide
How is Bent's rule applied on $\ce{SO3}$ and $\ce{SO2}$?
Bond length of $\ce{SO3}$ < $\ce{SO2}$. Is this a direct consequence of applying the Bent's rule?

I really don't see why you would need to apply Bent's rule.

Here are structures of dominant resonance structure (only one of 3 and 2 differencing only by rotation) for sulfur tri- and dioxide. Despite common misconception, d-orbitals apaprantely play very insignificant role in chemistry of p-block elements, so the structures include formal charges.

As can be seen, the average order of sulfur-oxigen bond lowers with move from di- to trioxide from $1.5$ to $1.(3)$. However, average formal charges grows from $+1$ to $+2$ on sulfur and $-0.5$ to $-0.(6)$,  on oxygen, thus greatly increasing ionic bonding (roughly from $1*0.5 = 0.5$ to $2*0.(6)=1.(3)$) Generally, ionic bonding greatly outperforms covalent bonding, and here charges grown considerably so bond shortening may be attributed to that. The point, however, is, that in this particular structure more than one effect is in play, and their effects outperform whatever small effects are considered when Bent's rule is applied.

But how would you apply Bent's rule here?

@shauryagupta I wouldn't. There are more significant factors in effect here.

There are several factors which play a role in determining the bond length in a molecule.

Here, in $\ce{SO3}$, we have no lone pairs on the sulfur atom whereas, in $\ce{SO2}$, the central atom has a pair of lone electrons.
According to Bent's rule, the lone pairs tend to occupy the orbital with more s-character which in turn decreases the s-character in the $\ce{S-O}$ bonds and increases the p-character. Also, since there's an increment in p-character, there's an increment in bond length and thus S-O bond length in $\ce{SO2}$ > $\ce{SO3}$.
I hope this might get you on clear grounds of how Bent's rule can be applied in this case.


text:
Hydrogen and deuterium exchange
Is this a valid mechanism? I think it is. It is just a simple acid base mechanism.
Could I increase yield though by using a stronger base than heavy water? Perhaps heavy sodium hydroxide? Or is chloroform sufficiently acidic?

The mechanism is sound, but be certain, that there will be an equilibrium and a statistical distribution of the involved species.

It would help to know more about the reaction conditions. I would not expect this reaction would happen in plain water at room temperature, but maybe my intuitions are wrong.

CCl3- is unstable and almost always forms :CCl2/Carbene

If you are interested in making deutero-chloroform, there are better ways to do it.
From a purely mechanistic standpoint, chloroform isn't a very strong acid ($\ce{pKa \sim 15}$) so $\ce{D2O}$ is unlikely to deprotonate it at a useful rate.  Further, chloroform isn't very soluble in water making your desired reaction even less likely (but you could engineer your way around this problem by using a phase transfer catalyst).
Your suggestion to use a stronger base is reasonable, but the problem you run into here is that a different reaction is brought into play.  Specifically, once a stronger base like potassium t-butoxide (in t-butanol) is used, dichlorocarbene formation occurs.  Once the trichloromethylcarbanion is formed, it quickly ejects $\ce{Cl^-}$ to form dichlorocarbene.  Dichlorocarbene is an extremely energetic species and would quickly react with whatever else is around - perhaps the t-butanol to produce dichloromethyl-t-butyl ether.

Deprotonation at a useful rate or a useful extent?

I'd say the rate of deprotonation (or reaction) would be extremely low and therefore the extent of deprotonation (or reaction) would be extremely low.


text:
How does life choose it's nominal temperature preferences?
If all life relies on water, and water freezes at 32F and boils at 212F.
Then why isn't the nominal temperature for life 122F?
122F will literally kill some animals, including us.
Do our biochemical reactions need "cold" (most people prefer ~72F) for some reason?

A large part of life's 'operating temperature range' is directly linked to the effect of the sun on the Earth and Earth's distance from the sun.  A range of temperatures exist on Earth and evolution will take over from there.  Seems like a good question for SE:Biology.

I don't think there is any reason, $a priori$, for the preferred temperature to be exactly half way between freezing and boiling. The thermodynamics of stability for things like liposomes, proteins, and DNA are not tied to water directly that I can tell.

generally, there are bacteria, capable of operating at -5 C (and survive absolute zero) and there are some, thrieving at 100 C

I think this question has a lot to do with enzymes, which are proteins that act as biological catalysts. Enzymes have a specific temperature range, too cold and they function too slowly, but if it's too hot they get 'denatured' and cease to function.
This doesn't answer your question - because the point remains: how is this temperature optimum determined?
I was just commenting this to add specificity.

Here's a link to a brief Scientific American article that presents the results from a study of this question.  Basically, for every $\ce{1^{o}C}$ increase in body temperature, 6% fewer fungal species can survive - so the hotter, the better (the longer you're likely to survive).  The downside of "hot" is the increased metabolic rate required, which translates into more time spent consuming food.  If you were to plot the effect of these two competing effects on longevity versus body temperature you would get something like an inverted (upside-down) parabola.  The study found a maximum longevity at 98.1 F, not far off from the observed average body temperature of 98.6 F.


text:
Why does urine smell like ammonia?
Apparently, water or $\ce{H2O}$ combines in some way with nitrogen (from the air or elsewhere?) to form ammonia, or  $\ce{NH3}$. Where/how in the body is this $\ce{NH3}$ (or whatever nitrate) produced before it is excreted with urine?

When excess proteins in the body are broken down, urea is formed (nitrogen containing compound). This is excreted in urine and in sweat.

All animals (including humans) consume significant amount of proteins. Proteins consist of aminoacids, which include amine nitrogen. This nitrogen traces back to mineral fertilizers humans use in agriculture. As proteins in our bodies are destroyed, the waste is utilized, formine urea (carbamide), so ammonia didn't cause too much trouble. However, urea is slowly hydrolized in water back into ammonia, so old urine stinks of (among other things) ammonia.

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


text:
Why N2O is a "laughing gas"?
It is (mostly) used as an "oxidizer" in motors and engines, I understand that much. The oxygen "oxidizes" with other chemicals, while the nitrogen returns to the air.
But $\ce{N2O}$ also produces "euphoria" in human beings, works as a mild anesthetic, and causes people to laugh. Why does this happen?

This appears to be more of a question for SE:Biology

In contrast to $\ce{O2}$ (or $\ce{CO}$), $\ce{N2O}$ does not bind to hemoglobin and it is not metabolized in the human body. It is taken up through the lungs, gets in the blood and leaves the body the same way being exhaled.
According to a 2004 study by P. Nagele in PNAS (DOI), $\ce{N2O}$ acts as an antagonist for the  $N$-methyl-$D$-aspartate (NMDA) receptor. Other infamous antagonists are drugs like ketamine or phencyclidine.
Further information on NDMA receptor antagonists can be found here.


text:
How many grams of acetic acid are present in 2.80 L of this vinegar?

A vinegar sample is found to have a density of $1.006 \mathrm{\frac{g}{mL}}$ and to contain $8.7~\%$ acetic acid by mass. How many grams of acetic acid are present in $\pu{2.80L}$ of this vinegar?

I multiplied: $$\left(\frac{1.006~\mathrm{g}}{\mathrm{ml}}\right)\cdot \left(\frac{1000~\mathrm{ml}}{1~\mathrm{L}}\right)\cdot 2.80~\mathrm{L} = 2816.8~\mathrm{g}$$
I then converted $8.7~\%$ into $0.087$ and multiplied it into $2816.8$, to get the answer $245~\mathrm{g}$.
Am I correct?

Please add what you have attempted towards solving the problem into the body of your question.  For more information, see the site's homework policy for how to ask homework questions. Thanks!

(you've done this very well in your other questions so far, so if you could do the same with this one, that would be great)

Can you tell me if I am on the right track?

Yes, I get the same answer as you.

@JohnSnow I appreciate you working hard in editing the questions with MathJax. I would appreciate it even more, if you would have a look at this meta post, specifically the section about units and variables.

Your math and stoichiometry are correct.
$2.80L*(\frac{1000 ml}{1 L})*(\frac{1.006 g}{ ml})*8.7$%$=245g \checkmark$


text:
How is the dichromate anion generated from potassium dichromate?
How does $\ce{K2Cr2O7}$ become $\ce{Cr2O7^2-}$ when dissolved in water?

Questions based on related concept: How to determine which ions will pair to form a compound in an aqueous solution? and Acid-base reaction vs. salt dissolving reaction

You don't seem to be overly familiar with the concepts of salts (made of cations and anions), ionic bonds and solvation of salts in water.
Lets's walk through some examples in which a solid material made from cations and anions is dissolved in water in such a way that the anions and cations are each surrounded and stabilized by water molecules (i.e. they are in an aqueous state, denoted as aq):
$$\begin{align}
\ce{NaCl (s) &->[H2O] Na+ (aq) + Cl- (aq)} \\
\ce{MgSO4 (s) &->[H2O] Mg^2+ (aq) + SO4^2- (aq)} \\
\ce{Na2SO4 (s) &->[H2O] 2Na+ (aq) + SO4^2- (aq)} \\
\ce{K3PO4 (s) &->[H2O] 3K+ (aq) + PO4^3- (aq)} \\
\ce{Mg3(PO4)2 (s) &->[H2O] 3Mg^2+ (aq) + 2PO4^3- (aq)}
\end{align}$$
As a general rule for the solvation of the salts, we can derive:
$$\ce{M_mX_n (s) ->[H2O] mM^n+ (aq) + nX^m- (aq)}$$
Applying this to potassium dichromate, we recognise that the cation is $\ce{K+}$ and the anion is $\ce{Cr2O7^2-}$. Therefore, it is the dissolution of potassium dichromate in water which generates the dichromate anion:
$$\ce{K2Cr2O7 (s) ->[H2O] 2K+ (aq) + Cr2O7^2- (aq)}$$
As for the physical reason behind why the salts dissolve, you may find a more in-depth discussion here: Why do salts such as NaCl dissolve?


text:
Derivation of the relation between temperature and pressure for an irreversible adiabatic expansion
I have a formula for finding the final temperature in an irreversible adiabatic expansion:
$$T_2=\left[\frac{C_V+\frac{P_2}{P_1}}{C_p}\right]T_1,$$
where $T_1$ is the initial temperature.
$C_V$ and $C_p$ are the molar heat capacities at constant volume and constant pressure, respectively.
I am interested in how this formula is derived.

What have you yourself tried to derive this equation? This site requires you to show some effort otherwise your question might be in danger of being closed. As for the topic at hand: Where have you gotten this equation from? The equation, as it stands, cannot be correct, because if you make a dimensional analysis you see that $C_v$ and $\frac{p_2}{p_1}$ don't have the same unit. However, I've seen a similar equation that contains a term $\frac{p_2}{p_1}R$ instead of $\frac{p_2}{p_1}$.

I have not the slightest idea from where has this equation come. It has been given in one of the rather incomplete solutions. Maybe it is wrong. However, even if it were $\frac{p_2}{p_1}R$ instead of just $\frac{p_2}{p_1}$, how is it arrived at? Or is it necessary to have one more parameter to determine the final temperature?

When you first posted this question I was very busy, but now I've found the time to look at it again and can give you an answer:
As I already mentioned in one of my comments your equation is slightly incorrect. It should actually be
\begin{align}
  T_2=\left[\frac{c_{V} + R\frac{p_2}{p_1}}{c_{p}}\right]T_1
\end{align}
Now, I will come to the derivation:
You should always start out by identifying what you already know about the system. You know the expansion will be performed adiabatically, so there can be not heat exchange with the environment, i.e.
\begin{align}
  \delta Q = 0 \ .
\end{align}
This means that from the first law of thermodynamics you get:
\begin{align}
  \mathrm{d} U = \delta Q + \delta W = \delta W
\end{align}
where $U$ is the internal energy and $W$ is the work.
Furthermore, you know that the expansion process is irreversible. If the only work performed is the pressure-volume work, where the system has to expand against a constant external pressure, which is $p_2$ in your equation, then this work is given by:
\begin{align}
  \delta W = - p_{2} \mathrm{d} V \qquad \text{where:} \qquad p_{2} = \text{const.} \neq f(V)
\end{align}
So, together with the first law of thermodynamics this yields:
\begin{align}
  \mathrm{d} U = - p_{2} \mathrm{d} V \ .  \tag{1}
\end{align}
Now, you need to assume that the expanding gas is an ideal gas. For ideal gases you know that
\begin{align}
  \mathrm{d} U &= \underbrace{\frac{\partial U}{\partial T}}_{= \, C_{V}} \mathrm{d} T + \underbrace{\frac{\partial U}{\partial V}}_{= \, \Pi \, = \, 0} \mathrm{d} V \\
  \mathrm{d} U &= C_{V} \mathrm{d} T \tag{2}
\end{align}
where $T$ is the temperature, $C_{V}$ is the heat capacity at constant volume, and $\Pi$ is the internal pressure, which is equal to zero for an ideal gas.
Other relations that hold for ideal gases are the ideal gas law
\begin{align}
  p V = n R T
\end{align}
where $n$ is the amount of the gas and $R$ is the universal gas constant,
and
\begin{align}
  c_{p} - c_{V} = R  \tag{3}
\end{align}
where $c_{p} = \frac{C_{p}}{n}$ and $c_{V} = \frac{C_{V}}{n}$ are the molar heat capacities at constant pressure and constant volume, respectively.
Now, you just have to combine equations $(1)$ and $(2)$:
\begin{align}
  - p_{2} \mathrm{d} V = C_{V} \mathrm{d} T \ ,
\end{align}
integrate the result (under the assumption that $C_{V}$ is temperature independent)
\begin{align}
  - p_{2} \int^{V_2}_{V_1} \mathrm{d} V &= C_{V} \int^{T_2}_{T_1} \mathrm{d} T \\
  - p_{2} \left( V_2 - V_1 \right) &= C_{V} \left( T_2 - T_1 \right) \ ,
\end{align}
use the ideal gas law $V = \frac{n R T}{p}$ to substitute $p$ for $V$
\begin{align}
  - n R p_{2} \left( \frac{T_2}{p_2} - \frac{T_1}{p_1} \right) &= C_{V} \left( T_2 - T_1 \right) \\
  - n R \left( T_2 - \frac{p_2}{p_1} T_1 \right) &= C_{V} \left( T_2 - T_1 \right) \ ,
\end{align}
introduce the molar heat capacity via $C_{V} = n c_{V}$
\begin{align}
  - R \left( T_2 - \frac{p_2}{p_1} T_1 \right) &= c_{V} \left( T_2 - T_1 \right) \ ,
\end{align}
and separate $T_1$ and $T_2$
\begin{align}
  T_1 \left( c_{V} + \frac{p_2}{p_1} R \right) &= T_2 \left( c_{V} + R \right) \ .
\end{align}
Finally, you can use equation $(3)$ to get the desired result:
\begin{align}
  T_1 \left( c_{V} + \frac{p_2}{p_1} R \right) &= T_2 \bigl(\underbrace{c_{V} + R}_{= \, c_{p}} \bigr) \\
  T_1 \left( c_{V} + \frac{p_2}{p_1} R \right) &= T_2 c_{p} \\
  \Rightarrow \ T_2 &=\left[\frac{c_{V} + R\frac{p_2}{p_1}}{c_p}\right]T_1 \ .
\end{align}

Hi there, you can use \tag{...} at the end of a line to make enumerations of equations. It's a bit easier, saves some coding and does not mess up the alignment. I have edited that in, hope you don't mind ;) Excellent answer!

@Martin Thanks for making me aware of \tag{...}. That's indeed much better than my manual approach. It never crossed my mind that there is a proper mathjax function for this kind of thing, but it's about time for me to be dragged kicking and screaming into the century of the fruit bat :) I will look over my old questions and replace my old label constructs with \tag. Up until now I usually avoided labelling equations because I thought it had to be done manually. Again, thank you very much.

No problem at all. Not long ago, only one tag was available per environment, so I had to do it also manually. But apparently they fixed that. Have a look at my sandbox on meta

why $p_2=const.$

@eaxdpiotnyeantial In a reversible expansion you change the volume in infinitesimal steps from the initial to the final value and after each step you give the system time to equilibrate, i.e. the external pressure is equal to the inner pressure of the gas for all intermediate volume steps. So the external pressure is a function of the volume $p_{2} = f(V)$. In an irreversible expansion you change the volume abruptly from the initial to the final value, so the inner pressure of the gas is not equal to the external pressure during the process (only at the end after equilibrium is reached)...

... The external pressure is constant and not a function of the volume, i.e. $p_{2} \neq f(V)$.

$$\color{purple}{X-axis: Pressure},\color{green}{Y-axis: Volume}$$

Use conditions for reversible adiabatic process:
$$P_1^{1-\gamma}T_1^\gamma=P_2^{1-\gamma}T_2^\gamma|\quad P_1V_1^\gamma=P_2V_2^\gamma|\quad T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}|\quad\gamma=\frac {C_p}{C_v}$$
and the combined gas law:
$$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}=\frac{P_2V_3}{T_3}$$

The problem with this approach is that you can't solve for the final temperature because you don't know the final volume $V_{3}$.

@Philipp so there exists no definite formula for temperature as tere can ennumerous adiabatic cycles

Well, there is a way to treat the irreversible process but that entails some different conditions and equations compared to the chain of reversible processes you are suggesting. But you are right in so far that a treatment of the chain of reversible reactions would always be ambiguous as long as you don't know $V_3$ beforehand.


text:
Metal Compounds that bond covalently
I would like to ask if anyone has a list or knows which covalent compounds have metals in them. For example, beryllium and aluminium are both metals but they bond covalently with chlorine to form beryllium chloride and aluminium chloride respectively. Other than these 2 compounds, I do not know any other and thus would like to ask this question.

Hmm. I see some good answers below. There are entire classes of organometallic (i.e., covalent metal-carbon) compounds. Are you looking for other metal-chlorine covalent bonds? If you can be more specific in what you want to know, that would really help.

Metal-Chlorine Covalent Bonds would be good.

Silver sulfide has a large covalent component.

There is no sharp line between ionic, metallic or covalent bonds.
Most of transition metal oxides and sulfates have strong covalent characteristics. For example, the whole reason of using ligand field theory instead of crystal field theory is that the ionic description (CFT) breaks down, and the covalent effects are important in all but the simplest cases.
If you want to take even more covalent molecules, look for organometallic compounds, metal carbonyls, metallocenes, etc.

Some very clear evidence of a highly covalent metal compound is found when some metals are extensively fluorinated. For example, tungsten hexafluoride is a gas in ambient conditions (one of the densest, in fact), which completely exposes its covalent nature. Even though it's a metal fluoride with large electronegativity differences, the high oxidation number of the tungsten atom allows it to strongly attract electrons in the fluorine atoms, precluding complete charge separation into discrete ions and the formation of an ionic lattice.

There are examples of metal compounds that are regarded by a majority of chemists as covalent in organometallic chemistry. The problem is that not everyone regards some of these compounds as covalent, because the border between ionic and covalent isn't so strict and it doesn't matter.
Alkyllithium compounds and Grignard reagents ($\ce{R-Li}, \ce{RMgX}$) are considered to be covalent, and they are represented like this $\ce{CH_3CH_2-Li, CH_3-Mg^+I^-}$ (no +,- signs above the carbon and lithium and only the magnesium-halogen bond in Grignard reagents is ionic), while alkyl-sodium and potassium compounds are considered to be ionic (like sodium-acetylide).
An Organoaluminium example of a covalent compound is $\ce{Al(C_2H_5)_3}$.
Organocopper exhibits covalent bonds between copper and carbon, like in $\ce{(CH_3)_2CuLi}$.
There are many more examples in organometallic chemistry.
The d-block metals have significant polarizability and thus the bonds they participate in can have a significant covalent character like in $\ce{RuO_4}$, $\ce{Hg(CH_3COO)_2}$, $\ce{Pb(CH_3COO)_4}$, $\ce{CuI}$, etc.
Mercury(I) exists in solution as $\ce{Hg_2^{2+}}$ where the two mercury ions are covalently bonded.

No list is really complete without Grignard reagents, one of the best known examples.

There is no sharp border between ionic and covalent bond. Generally, single-atom anions are not stable enough to justify full transfer of an electron from cation, so simple binary compounds usually have significant covalent character, even if their crystal structure suggests otherwise. For example, $\ce{CsCl}$ said to have 25% covalent character. On the opposite end, heteroatomic bonds usually have some ionic character. Given that, we don't have clear groups of ionic and covalent metal compounds, but rather a spectrum of them.

$\ce{CsClO4}$ would be an almost pure ionic compound
$\ce{CsCl}$ would be mostly ionic
$\ce{AlCl3}$ would be highly polar, but significantly covalent
$\ce{CuS}$ would be almost purely covalent
$\ce{Li2}$ in gas phase is purely covalent

To get a structure with purely covalent bond we have to consider homoatomic bonds with atoms in identical neighborhood, otherwise some amount of polarity and consequently ionic bonding will arise. Such molecules with metal-metal bonds do exist.

Let's start with the most clear (and, admittedly, least chemically significant) example: group one metal diatomic molecules. Such molecules do exist in gas phase and can be detected spectroscopically, though the bond energy is quite low because of low density and relatively insignificant overlapping of the orbitals.
For group two there is a report of a rather complex compound with a Mg-Mg bond
For the third row there are reports of various compounds with the most simple being $\ce{((((CH3)3Si)2CH)2Al)2}$ with a direct Al-Al bond.
There are reports of $\ce{Sn2H6}$, i.e. $\ce{(SnH3)2}$

However, it must be noted that with atoms of the lower-left corner of periodic table homoatomic bonds are very unstable, to the point of impossibility to obtain stable compounds.
Many more examples are known for d-elements. For starters, $\ce{(Mn(CO)5)2}$ is a relatively stable compound with a singular Mn-Mn bond. However, d-elements favor complex structures with unusual delocalized, but clearly covalent bonding, similar to boranes. Such structures are especially common for late transition metals, say, from the Cr-W-Pt-Ni rectangle. An example would be $\ce{Rh6(CO)16}$, adopting structure with $\ce{Rh}$ atoms forming an (almost) perfect octahedron.
Looking from another side, we can look for molecules with metal as a central atom. They are much more common in regular chemistry, though the nature of bonding may be sometimes disputed. For example, copper acetate adopts an unusual paddle-wheel structure, but the copper-oxygen bond may be argued to be ionic. On the other hand, molecules with high d-element oxidation state usually have an effective charge of the central atom far below its formal oxidation state, so such compounds may be considered highly polar, but still covalent in nature. This said, many higher chlorides, oxochlorides and fluorides may be considered covalent, such as $\ce{TiCl4}$, $\ce{VCl4}$, $\ce{CrO2Cl2}$, $\ce{OsO4}$, $\ce{WF6}$, $\ce{AuF5}$.
Looking from third side we can argue to be mostly covalent compounds where metal is bound to radical, that clearly cannot stabilize a negative charge, such as $\ce{CO}$ molecule or $\ce{CH3}$ and some other organic fragments. Such compounds are known for most metals in the periodic table, say, $\ce{LiCH3}$, $\ce{CH3MgCl}$ (the nature of these compounds is more complex than is usually written), $\ce{Al2(CH3)6}$, $\ce{PbEt4}$ (tetraethyllead), and many d-element compounds, such as $\ce{(CH3)2CuLi}$.
Here I mostly avoided a huge can of worms named 'coordination compounds', which includes (mostly) interactions of d-element atoms with different electron pair donors. Some of such compounds may be argued to be mostly covalent, such as most metal carbonyles, like $\ce{Cr(CO)6}$, but some may be argued to be mostly ionic, such as $\ce{[FeF6]^{3-}}$, or ion-dipole, such as $\ce{[Co(NH3)6]^{2+}}$

There is no clear separation between ionic and covalent bonds.
Electronegativity values are useful in determining if a bond is to be classified as nonpolar covalent, polar covalent or ionic.
The rule is that when the electronegativity difference is greater than 2.0 (Electronegativity according to Pauling), the bond is considered ionic.
If the electronegativity difference  is less than 0.5, then the bond is nonpolar covalent.
If the electronegativity difference is between 0.5 and 1.6, the bond is considered polar covalent.
For this reason $\ce{CaO}$ is ionic, while $\ce{Be3N2}$  is covalent.

Covalent bonding can be either two atoms of the same element becoming a molecule (single bond) or atoms from different elements bonding (covalent compound). Examples of atoms becoming a molecule are basically diatomic molecules such as H2, Cl2, I2, etc. Examples of more complex covalent compounds are methane, water, ammonia and carbon dioxide. There are also solid covalent compounds which are mainly allotropes of carbon such as diamond and graphite.

I know this, but what I am trying to say is that covalent bonding normally occurs between non-metals, whether of the same element or not. However, there are exceptions like beryllium chloride and aluminium chloride. Hence, my question is, other than these 2 covalent compounds, which covalent compounds are formed from a covalent bond of a metallic element and a non-metallic one?  Thanks for your help anyway and I apologise if my question was not clear enough.

Well to be honest I've never heard of a covalent bond between a metal and a non-metal, since that would normally be classified as an ionic bond. But now that I've searched about it those are the only two I was able to find.


text:
Finding the pressure equilibrium constant of the reverse reaction

At $1000~\mathrm{K}$, $K_\mathrm p = 1.85$ for the reaction.
  $$\ce{SO2 + \frac12O2 <=> SO3}$$
What is the value of $K_p$ for the reaction
  $$\ce{SO3 <=> SO2 + \frac12O2}$$

Now, since the task gives us $K_\mathrm p$, it seems strange to ask for it again with the equilibrium put the other way around. Does this make sense? Is it a typo? Either that or my understanding is flawed. It would make more sense to ask for $K_\mathrm c$.
The chapter has this equation: $K_\mathrm p = K_\mathrm c(RT)^{\Delta n}$, I think this is relevant given the temperature is available to solve it.
Source: task 15.21 in the 11th edition of Chemistry: The Central Science, and the answer is suppose to be $0.541$.

They are actually just looking for the equilibrium constant for the reverse reaction.
$$K_p(-\xi) = K_p^{-1}(\xi) = \frac{1}{1.85} = 0.541$$


text:
Why do we write NH3?
We've learnt that the electropositive element is written first. Then why is ammonia written as $\ce{NH3}$ ?

Perhaps they don't want anyone to think it is a binary acid.

Ammonia can act as an acid though.

@Dissenter: Many compounds can act as many things in various weird conditions, but in aqueous solutions ammonia doesn't generally act like an acid.

See $\ce{CH4}$ , $\ce{BH3}$ etc This rule is generally applied for hydrogen when it is protic: $\ce{H2O}$, $\ce{HCl}$, IUPAC has a whole sets of rules that nobody cares to use, especially in nomenclature.

The same is the case with phosphine.

Maybe because it is predominantly covalent. Besides, we write other covalent compounds in a similar way. Example - Cl2O7

According to current nomenclature rules, $\ce{H3N}$ would be correct and acceptable.  However some chemical formulas, like $\ce{NH3}$ for ammonia, that were in use long before the rules came out, are still accepted today.

Are the hydrogens in ammonia really more electropositive than the nitrogen?  Combining H+ ions with anions other than OH- yields compounds which behave as acids in aqueous solutions, but NH3 behaves like a base.

Yes, hydrogen is more electropositive than nitrogen. The electronegativity of hydrogen is 2.2, while nitrogen is 3.04, see  http://en.wikipedia.org/wiki/Electronegativities_of_the_elements_(data_page)

I know that in general hydrogen is more elecropositive, but within a covalent structure things sometimes behave in weird ways.  Hydrogen can behave as +1 as in HF or -1 (as in NaH), and nitrogen can be +3 (as in Li3N), -3 (as in NI3), or various other oxidation states.  Is the behavior of ammonia closer to that of a nitride of hydrogen, or a hydride of nitrogen?

PS -- look at http://chemistry.stackexchange.com/a/538/7274 for a table which suggests that for purpose of nomenclature, hydrogen's electronegativity is considered to be between that of the first group V element (nitrogen) and the last group VI element (polonium).  See also http://chemistry.stackexchange.com/a/975/7274 which would imply that compounds of hydrogen that are more acidic than water are written with the hydrogen first, and those which are more basic are written with the other element first.

@supercat $\ce{NaH}$ and $\ce{Li3N}$ are not covalent compounds. And nomenclature does have nothing to do with chemical behaviour. Oxidation states are only a bookkeeping tool.

@supercat how does oxidation state correlate with electronegativity?

@Martin: Those aren't covalant, but I think NH3 is.  My point is that while it's common with some covalant compounds to write them according to oxidation state (one writes H2O, for example, rather than OH2), I don't think it's clear that the hydrogens in NH3 should be viewed as having a +1 charge and the nitrogen as -3.  Oxidation states are a bookkeeping tool, true, but the question is about nomenclature rather than chemical behavior, bookkeeping would seem relevant to that.

The fact that hydrogen is more electropositive than nitrogen is irrelevant to the basic character of ammonia, which is due to the lone pair on nitrogen and not due to hydride ions (as it would be in LiH, for example).

The atomic symbols in a formula may be ordered in various ways. For example, the electronegativity criterion is most often used in binary species (for intermetallic compounds, however, older recommendations prescribed alphabetical ordering rather than by electronegativity).
Section IR-4.4.2.1 of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) explains what is meant by the ordering principles ‘electronegativity’:

IR-4.4.2.1 Electronegativity
If electronegativity is taken as the ordering principle in a formula or a part of a formula, the atomic symbols are cited according to relative electronegativities, the least electronegative element being cited first. For this purpose, Table VI is used as a guide. By convention, the later an element occurs when the table is traversed following the arrows, the more electropositive is the element.

This table is also included in the IUPAC Technical Report Brief guide to the nomenclature of inorganic chemistry. Pure Appl. Chem. 2015, 87(9–10), 1039–1049 as well as in the corresponding four-sided lift-out document, which is available as supplementary material.
Note that hydrogen is placed between the elements of group 16 and the elements of group 15.
Therefore, all the following examples are consistent with the above convention:
$\ce{HF}$
$\ce{HCl}$
$\ce{HBr}$
$\ce{HI}$
$\ce{H2O}$
$\ce{H2S}$
$\ce{NH3}$
$\ce{CH4}$
Note that, according to this convention, the formula for the hydroxide ion should be $\ce{HO-}$ (not $\ce{OH-}$).

Indeed, the hydroxyl radical as most commonly written (at least in the hard-core free radical chemistry community) conforms to this: $\ce{HO}^\bullet$

See also, my answer to this question

I think you quoted the wrong paragraph. That one says, elements that came first in the arrow order are cited first. I think you should rather quote IR-2.15.3.1: "In compositional names of binary compounds and corresponding formulae, the element encountered last when following the arrow in Table VI is represented first in the formula as well as the name."

The table misses seven elements we now have after Rg. Cmon IUPAC, get the update in!

Possibly $\ce{OH^-}$ got grandfathered in. In an ionic compound like sodium hydroxide, $\ce{NaOH}$ reflects the true bonding connections (sodium is bonded to oxygen), so the hydroxide ion appears as $\ce{OH^-}$.

The answer of this question lies in the nomenclature of the late 1700’s when chemical knowledge was primitive.  At the time, most chemists agreed on a class of acrid substances, for example vinegar, called “acids”.  Another class of substances reacted with acids to form the “base” of compounds.
When systematizing these compounds, Lavoisier came up with the word oxygen by combining roots which translated meant “to generate acid” because he thought oxygen was used to form acid.  Acid was also thought to be formed by hydrogen as well.  Most of the early chemical formulas for compounds specified both hydrogen and oxygen as the acid forming component.
It was not discovered until 1832 hydrogen, unlike all of the other known elemental gases, shared a property with metal: it was electro-positive. So in the late 1800’s the nomenclature debate surfaced.  Some of the old nomenclature was kept so the opposite nature of acids and bases was reflected in the nomenclature.  This nomenclature was said to facilitate understanding of molecules combining and breaking apart.
As an example, notice how the nomenclature in the following reaction suggests two molecules with opposite properties are sticking together.
$$\ce{ NH3 + HCl -> NH4Cl }$$
The short answer would be early chemist developing nomenclature did not have modern understanding of hydrogen, so hydrogen often was placed in the same location of formulations as oxygen in similar compounds.
Sources:
Williamson. Remarks on Chemical Nomenclature and Notation. J. Chem. Soc. 1864, 17, 421–432.
Robert Kane, John William Draper, Elements of chemistry, page 152.


text:
para-dichlorobenzene - number of proton NMR signals
I am told that para-dichlorobenzene exhibits two proton NMR signals. This doesn't jibe with me; all the protons on the molecule are equivalent. Here is a book claiming the same thing. Am I wrong, or is the person who told me that para-dichlorobenzene has two signals on its proton NMR wrong?

You are right, p-dichlorobenzene has only one proton nmr signal.

Be very careful with how you interpret this molecule. Although the 1H NMR spectrum appears as a single line, this is actually an AA'A''A''' system, meaning that, although chemically equivalent,  all four protons are magnetically non-equivalent, but have coincidental chemical shift. The coupling between all protons is different; JA-A''' is different to JA-A''' is different to JA'-A'' is different to JA'-A''' and so on.
So yes, the spectrum is as you thought. p-dichlorobenzene IS a single peak in the 1H NMR spectrum.
Below is a set of simulations for a generic para disubstituted benzene system, which is an AA'BB' system. A is chemically equivalent to A', but magnetically non-equivalent. A is chemically AND magnetically non-equivalent to B. Your system is just a special case example of these types of systems, with all protons being chemically equivalent, but magnetically non-equivalent. These are second order systems, which have a lot of complex additional splitting structure in their spectrum. Like all spin systems, as the chemical shift difference becomes less (the peaks move closer together), the tilting, (or tenting or roofing, depending on which text book you read) becomes more pronounced, and the wings on the main peaks become smaller and smaller. Eventually, they collapse into a single peak at very small chemical shift difference.
The fine text which you can't read (sorry about that), gives the coupling constants for the system, which are bog-standard ortho/meta/para couplings: Jortho=7Hz, Jmeta=3Hz; Jpara=1Hz. The only variable in the simulations is the chemical shift of the downfield peak, which slowly moves towards its upfield partner.

P-dichlorobenzene only has 1 type of H (check the symmetry) and thus appears as a singlet in the HNMR.
However, it is not an AA'A"A"' system as another responder implies, it is an A4 system because there is only 1 type of nuclei in the spin system.

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

For clarification, wouldn't you see one peak in a "weak" magnetic field, but  multiplets in a "strong" field?

Wrong. A methoxy group is an A3 system, because all coupling constants are identical. Not here.