text:
Is it true that a pH higher than the pKa of an acid make the H automatically disassociate?
Will the hydrogen atom always disassociate if the pH of the solution is higher than the pKa of the acid?

It's an equilibrium, so no.

Could you elaborate?

The $\text{p}K_{\text{a}}$ of an acid is the $\text{pH}$ where it is exactly half dissociated. At $\text{pH}$s above the $\text{p}K_{\text{a}}$, the acid $\ce{HA}$ mainly exists as $\ce{A^{–}}$ in water; at $\text{pH}$s below the $\text{p}K_{\text{a}}$, it exists mainly as undissociated $\ce{HA}$. This doesn't mean that at $\text{pH}$s higher than the $\text{p}K_{\text{a}}$ suddenly every molecule of the acid dissociates, it means that a greater portion of the initial concentration of $\ce{HA}$ will be dissociated as you go higher in $\text{pH}$. There will always be some (potentially tiny) amount of the undissociated acid left even at very high $\text{pH}$ because it is an equilibrium. For more information on $\text{p}K_{\text{a}}$ see here.

"This doesn't mean that at pHs lower (?) than the pKa suddenly every molecule of the acid dissociates..." – don't you mean "higher"?

@FinlandRules Oh yes, sorry for that. Confused the pH-directions there.

I propose the following  answer:
Let's consider the following equilibrium:
$$ \ce{H}A+\ce{H_2O} \leftrightarrows A^-+\ce{H_3O^+}$$
The equilibrium constant $\ce{K}_A=\dfrac{[A^-][\ce{H_3O^+}]}{[\ce{H}A]}$. So if we take ''$-\log$'' of this equation, we can define the $\ce{pK}_A$ of an acid as: $\ce{pK}_A=\ce{pH}- \log\dfrac{[\ce{A^-}]}{[\ce{H}A]}$.
we rewrite this equation as follows:
$$
\ce{pH}=\ce{pK}_A+\log\dfrac{[A^-]}{[\ce{H}A]}
$$
From this equation, we can deduce that when $\ce{pH}>\ce{pK}_A$, $[A^-]>[\ce{H}A]$,  i.e. the concentration of the conjugated base is bigger than the concentration of the acid. But this doesn't mean that all the acid molecules are dissociated. The bigger the difference between the pH and the $\ce{pK}_A$, the bigger is the percent of dissociated acid molecules.

Try not to base your answer on links/images.


text:
What is the pH of a solution of HCOONa and HNO3?

Calculate the pH of an aqueous solution obtained by mixing
  20 ml of $\ce{HCOONa}$ 0,1 M and 5 ml of $\ce{HNO_3}$ 0.2 M.
  ($K_\mathrm a$ of $\ce{HCOOH}$ $= 10^{-4}$)

My try:
The number of the moles involved are: $n_{\ce{HCOONa}}=0.002~\mathrm{mol}$, $n_{\ce{HNO3}} = 0.001~\mathrm{mol}$.
We have the reaction:
$$\ce{HCOONa + HNO3 -> HCOOH + NaNO3}$$
Where 0.001 moles of $\ce{HNO3}$ si is consumed by reacting with 0.001 moles of $\ce{HCOONa}$ giving 0.001 moles of $\ce{HCOOH}$. Because remains $0.002\ \mathrm{mol}-0.001\ \mathrm{mol}=0.001~\mathrm{mol}$ of $\ce{HCOONa}$, we'll have a buffer solution with the acid $\ce{HCOOH}$.
It'll start the reaction:
$$\ce{HCOOH + H2O <=> HCOO- + H3O+}$$
Where the initial concentration of $\ce{HCOO-}$ is equal to $\ce{HCOONa}$ and is $\frac{0.001~\mathrm{mol}}{(0.02+0.005)~\mathrm l} = 0.04~\mathrm{mol~l^{-1}}$. So we have,
$$[\ce{H3O+}]=K_\mathrm a \frac{[\ce{HCOO-}]}{[\ce{HCOOH}]} \Rightarrow \mathrm{pH} = \mathrm pK_\mathrm a - \log_{10}{\frac{[\ce{HCOO-}]}{[\ce{HCOOH}]}} = \mathrm pK_\mathrm a = 4$$
Is it correct?

Yes. You could also put the molar ratio into the equation for calculating pH (then you didn't need to calculate the concentrations).
And those are not initial concentrations, but equilibrium concentrations.

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text:
Why does lactulose turn dark?
Why does the standard 10gm/15ml lactulose solution turn dark when exposed to heat above 30 degrees Celsius?
This is probably a simple question. It is funny that the warnings/important information on the drug label of lactulose state that this effect is harmless, but then go on to warn not to use if it becomes too dark. How do I know if it is "too dark"? This is obviously very subjective and simply want to know why it becomes dark or "too dark" so it is scientifically understood why it is safe (or unsafe) to use when it is dark.

When milk is heat-processed, some of the lactose will isomerize to lactulose, a disaccharide.  Lactulose

can decompose by the same routes that most sugars decompose.  These are complicated processes with many products and pathways, so sugar decomposition is rarely, if ever, fully understood.

It can hydrolyze to its constituent monosaccharides fructose and galactose.
Lactulose or its released monosaccharides equilibrate with their ring-opened isomers and the alpha-hydroxyl can air oxidize to produce an alpha-dicarbonyl.  These alpha-dicarbonyls are prone to further oxidation to generate formic acid.  Over time, as the solution becomes more acidic from the formic acid build-up, further oxidations and eliminations (dehydration) can take place to produce a mixture of colored products.
In addition to degrading to smaller molecules, sugars (and the dehydration and oxidation products mentioned above) can also oligomerize to produce a variety of viscous, colored products.  Caramelization (see the "Chemistry" section in the link) is a term applied to this chemical polymerization and coloration process.

If you would like to read more about lactulose and its decomposition, take a look at this reference, especially section 7.2.1 starting on page 236.

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text:
Where can I find Equilibrium Constants?
When trying to develop mathematical models to show the concentration in a reaction, where can one find a database or good source of known equilibrium constants if it is not necessarily very convenient to find them empirically?

First, a question: what kinds of reactions are we talking about? Is it a more specialized (in this context) field like biochemistry, or more general chemical reactions? In practice, finding data for the latter is probably simpler than finding data for the former without doing any experimental work.
In general, I would use the information about Gibbs free energy of formation that can be found here (among many other places) for the reactants and products, and then calculate the equilibrium constant from
$$ K = \exp \left(\frac{-\Delta G^\circ_{\mathrm{rx}}}{RT}\right) $$
where $\Delta G_{\mathrm{rx}}$ is the change in Gibbs free energy for the reaction. Beware that if your reaction conditions - e.g. temperature or pressure - is different from standard conditions, this should be taken into account.
Hope this helps a bit!


text:
Can I use Sunflower oil to make soap?
When reading about making glycerine soap, the recipes suggested to melt animal fat and mix it with $\ce{NaOH}$.
Could I use Sunflower oil instead? Or a olive oil for nice flavor? Is heating necessary in that case?

You can use any triglyceride, but the touch, the firmness, the cleaning effectiveness of the resulting soap will be different. Soaps also can go rancid just like their corresponding fat/oil. DIY soapmakers generally mix oil or use specifically selected ones. Coconut oil, castor oil, olive oil, animal fats are generally popular.
Without the intent to advertise any specific site:
http://www.lovinsoap.com/oils-chart/
Olive oil is actually a pretty common component, and you can buy olive oil based soaps even is shops. Flavor? If you do it well, most probably not much remains.
Heating is generally necessarily, saponification is a slow reaction and most home-made soap is not fully reacted (containing lot of NaOH).
Note on safety:
Many people make soap at home, because it is "more natural/organic" just following some youtube videos without knowing much about lab work or chemistry. Using your own blender, doing in your kitchen or using your own kitchenware is a very bad idea. NaOH can make you blind, so do not ignore the "wear googles!" signs. Also just mixing everything up in a highly corrosive mixture contains or not remains of NaOH or produce all kind of unknown byproducts then putting everything on your skin may not the healthiest choice of living.

Can I calculate the proper amounts of oil and sodium hydroxide to get (almost) 100% reaction (no oil or lye). But in fact, my original plan was to make glycerine and nitrate it, so it's not really healthiness I'm looking for.

Sure, you can, but it will be no exact. Glycerine and nitrate.. Not much to do with the topic.

You can use any triglyceride to make soap.  The process is called saponification.  In this process fatty acid molecules are cleaved from glycerol (glycerine).  In order for the reaction to proceed it is heated.  Regardless of the fatty acid source, the soap will probably not taste very good.


text:
How to use signicant figures?
I am confused about significant figures. In a teaching video, the teacher showed this math:
$$ 22.2 + 22.2 + 30.10 + 30.10 = 104.6$$
This is, of course, the actual answer, but in using significant figure count the answer would be $105$, because:
22.2---- 3 significant numbers

22.2---- 3 significant numbers

30.10--- 4 significant numbers

30.10--- 4 significant numbers

+______

104.6

The first number has the fewest significant figures, so we would "round" to 3 significant numbers in our answer. I emailed the course helpline, but they sent a confusing email back about differences in counting the significant numbers whether you are adding and subtracting, or multiplying and dividing. I need help explaining this.

The rule you are applying is for multiplication and division.  For addition and subtraction look at the numbers in the problem.  22.2 is accurate to the tenths place so the answer can be reported to the tenths place regardless of how many sig figs are in the answer.

good answer, but could I maybe see some math to see the difference?

22.2+30.10 = 52.3.  The number 22.2 is accurate to the tenths place.  We do not know what the digit in the hundredths place is.  It could be 0-5.  So, the answer can only be expressed to the tenths place.  The multiplication and division makes less sense.  35X3 = 105.  The lowest number of sig figs in the problem is one, so the answer can only have one sig fig.  We express the answer as 100.


text:
Can ionic and atomic radii be compared?
Can ionic and atomic radii be compared?  For example, this is a common problem in some intro chem courses:
Rank the following from smallest to largest:  sodium ion, magnesium ion, neon, fluoride, oxide.
Any student who has been trained in trends should be able to get the "answer", but if one looks at the actual figures, there are data that show that sodium ion is larger than neon atom.
I am of the opinion that it is comparing apples to oranges and that the problem can be rescued by removing neon from the list.

I don't see how $\ce{Na^+}$ could possibly be larger than $\ce{Ne}$, at any level of theory. I think it's more of a case of atomic sizes being hard to measure (in fact there are several ways to define atomic/ionic radii, and measurements fluctuate within each definition). Are you sure you aren't comparing values from different definitions?

My thought is that they cannot be compared.  Is there a definition of radius that works for both ions and atoms?

See Ionic radius vs Atomic radius might help

A periodic table coupled with simple electrostatic arguments should suffice to provide at least a rough ordering of ions and atoms. For the ions:

We expect POSITIVE ions to be SMALLER than the corresponding atom because there are fewer electrons than protons => an excess of POSITIVE charge => the electron cloud is more attracted to the nucleus on average and becomes more 'concentrated' => $\ce{Na+}$ is 'small' and $\ce{Mg^{2+}}$ is 'very small'.
We expect NEGATIVE ions to be LARGER than the corresponding atom because there are more electrons than protons => an excess of NEGATIVE charge => the electron cloud becomes more diffuse to compensate, lowering the Coulombic interactions => $\ce{F-}$ is 'large', $\ce{O^{2-}}$ is 'very large'.

Even though $\ce{Na+}$ has a lower principal quantum number for its outer shell than $\ce{Mg^{2+}}$, Mg is twice as charged and I believe this can be more important. $\ce{F-}$ and $\ce{O^{2-}}$ have the same principal quantum number for their outer shells, but $\ce{O^{2-}}$ is more charged and accordingly, we expect that $\ce{O^{2-}}$ has a larger radius.
$\ce{Ne}$ carries no unbalanced charge and has comparable valence principal quantum number to the ions, so we expect its radius to fall somewhere between the positive ions (which are smaller) and the negative ions (which are larger). Go over these arguments in your head and try to put the pieces together yourself from here.
(Note: I wouldn't be willing to stake my life on this, so some community validation or corrections would be much appreciated)


text:
How to create the equivalent of luminol for a compound?
How would you go about creating a compound, like luminol, that would visibly react with monomorine I (5-methyl-3-butyloctahydroindolizine) on a household surface (like a kitchen floor)?
It is ok if the compund reacts with other similar compounds as well, as long as they aren't common to a household.
Background:

Luminol is the chemical used by forensic investigators (like those on CSI) to detect trace amounts of blood at crime scenes. The luminol is sprayed over an area and when it comes into contact with blood (specifically, the iron), it exhibits a blue glow.
Monomorine I is an ant trail pheromone (a trail guiding other ants to food).

Objective:
I'd like to be able to apply a compound on a surface and see the trails left by ants.

Probably not easily, especially if the concentrations are really low. Luminol + $\ce{H2O2}$ only works with blood because it's a catalytic reaction, i.e. it doesn't take much iron to create a lot of chemiluminescence. I'm not an organic chemist, but I'm not aware of any sort of similar reaction specific for this class of compounds. I did see this reaction: http://onlinelibrary.wiley.com/doi/10.1002/(SICI)1521-4168(19980501)21:5%3C315::AID-JHRC315%3E3.0.CO;2-V/abstract Unfortunately, it's not open access and I can't share my institutional access, but it's basically a photo-generated ruthenium complex that reacts with tertiary amines and indoles which might be applicable here, though I'm guessing is not sufficiently sensitive. It's not terribly selective so you might run into interference problems and in that paper they use it for HPLC  detection so it may not be bright enough to see with the naked eye. (I have seen similar chemiluminescent compounds before and they're pretty dim, even with the lights out)
There are ways to produce visible changes in response to small concentrations of an analyte: LSPR, ELISA, etc. but they're a lot of work to implement for new analytes and probably too expensive for looking at a large area. Chemical imaging based on mid-IR or Raman spectroscopy may be applicable, though as far as I know, they're not geared for measuring large areas.
It seems that directly visualizing pheromone traces is a non-trivial task. Groups seem to do things like use image analysis (http://www.ploscompbiol.org/article/info:doi/10.1371/journal.pcbi.1002592) to estimate the levels of pheromones. I would like to have seen a comparison of their model with actual pheromone measurements and the way I would propose to do it is to take surface samples after the experiment. If a big sheet of paper was used as the surface, you could cut it up in a grid and extract the pheromone for analysis by LC-MS or whatever (or maybe even used directly as a paperspray emitter for MS). An ambient-ionization mass spec technique like LTP or DART would also probably be a good way to construct a surface map of pheromone concentrations.
Briefly: There are certainly ways of measuring trace amounts of things like pheromones, but a simple CSI-esque naked eye visualization probably isn't easily achievable.


text:
Why don't plus one oxidation state metals form stable complexes with hydroxide ion?
For example, why doesn't AgOH exist as a stable compound?
I read that it rapidly decomposes to silver oxide and water.
$\ce{2AgOH -> Ag2O + H2O}$ $~~~~~~~~~pK_{decomposition}= 2.875$
http://actachemscand.org/pdf/acta_vol_14_p0717-0725.pdf
The above paper gives the pK value for the above equilibrium but doesn't bother to really delve into why silver hydroxide might be so unstable in water.
But why?
Is there repulsion between the silver cation and the proton?

Hard soft acid base theory says silver is soft and hydroxide is hard so they won't form a strong bond.

I know that but why exactly doesn't hard prefer hard?

It does.  The explanation I've seen is hard and hard bond ionically and soft and soft bond with more covalent character.

Oops I meant why doesn't hard prefer soft?

@BrinnBelyea but in Ag2O, Ag+ is soft and O2- is hard, but why do they form a bond?

They will form a bond but it isn't that strong.  In addition, one of the products of the reaction, water has a favorable enthalpy of formation.


text:
Determining the group number of an element from its successive ionization energies

The successive ionization energies of an unknown element X are:
  1st - 9534; 2nd - 11790; 3rd - 13683; 4th - 15309; 5th - 16458.
  To what group number does X belong?

There aren't any significant jumps so that means X is in groups 5-8. After that I don't know.

There are other little jumps to look out for, like the removal of the first electron from  a filled subshell. Consider the ionization energies (in eV) for the third period elements:
$$
\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline
\text{electron}&\ce{Ar}&\ce{Cl}&\ce{S}&\ce{P}&\ce{Si}&\ce{Al}&\ce{Mg}&\ce{Na}\\\hline
\ce{3p}&1521&-&-&-&-&-&-&-\\\hline
\ce{3p}&2666&1251&-&-&-&-&-&-\\\hline
\ce{3p}&3931&2298&1000&-&-&-&-&-\\\hline
\ce{3p}&5771&3822&2252&1012&-&-&-&-\\\hline
\ce{3p}&7238&5159&3357&1907&787&-&-&-\\\hline
\ce{3p}&8781&6542&4556&2914&1577&578&-&-\\\hline
\ce{3s}&11995&9362&7004&4964&3233&1817&738&-\\\hline
\ce{3s}&13842&11018&8496&6374&4356&2745&1451&496\\\hline
\ce{2p}&40760&33604&27107&21267&16091&11577&7733&4562\\\hline
\end{array}
$$
Notice how, in addition to the big jump between $\ce{3s}$ and $\ce{2p}$, there is a small jump between $\ce{3p}$ and $\ce{3s}$ for Ar, Cl, S, P, and Si. If you have five ionization energies, and you don't see the large jump, look for the small jump. If you don't see the small jump, what group is X in?
There is another even smaller jump to look for. Half-filled subshells are also slightly more stable than other configurations (except filled subshells). Thus, it should require a slightly higher increase in ionization energy to remove an electron form any configuration with $\ce{np^3}$ than from $\ce{np^2}$ or $\ce{np^4}$. Let's look at the step wise increase in ionization energy (in other words, how much higher is the ionization energy for the second electron than the ionization energy for the first electron and so on?).
$$
\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline
\text{electron}&\ce{Ar}&\ce{Cl}&\ce{S}&\ce{P}&\ce{Si}&\ce{Al}&\ce{Mg}&\ce{Na}\\\hline
\ce{3p}&-&-&-&-&-&-&-&-\\\hline
\ce{3p}&1145&-&-&-&-&-&-&-\\\hline
\ce{3p}&1265&1047&-&-&-&-&-&-\\\hline
\ce{3p}&1840&1524&1252&-&-&-&-&-\\\hline
\ce{3p}&1467&1337&1105&895&-&-&-&-\\\hline
\ce{3p}&1543&1383&1199&1007&790&-&-&-\\\hline
\ce{3s}&3214&2820&2448&2050&1656&1239&-&-\\\hline
\ce{3s}&1847&1656&1492&1410&1123&928&713&-\\\hline
\ce{2p}&26918&22586&18611&14893&11735&8832&6252&4066\\\hline
\end{array}
$$
You can see that the increase in ionization energy for the third electron to be removed from argon is larger than the increase for the second or fourth electrons. Your numbers don't have this little increase in them at all. Have you considered that X might be a transition metal?

Hi Ben, thanks for your answer! I think you've given a general overview of how to solve such problems. However, I don't see how you have actually solved the OP's question. According to me, I think X is probably a transition metal (because it has no significant jumps). But, I don't know what you were thinking X to be when you wrote the answer, hence I am asking you. What group is X in? Thanks!


text:
Turn lead into gold via radioactive decay
I was looking at the isotope's table and noticed that one of lead's isotopes can actually turn into stable gold through the following mechanism:
$$\ce{_{82}^{197}Pb -> _{81}^{197}Tl -> _{80}^{197}Hg -> _{79}^{197}Au}$$
I know (or at least guess) that such a process must be awfully ineffective. Still, I was wondering:

How could we get a great quantity of $\ce{_{82}^{197}Pb}$?
How long would it take for it to turn into gold? Can that be accelerated?
Once most of it has been turned into gold, how can we extract the gold (there are still remaining $\ce{Pb,Tl,Hg}$ molecules)?
What would be the yield of the process? How much money would it approximately cost compared to how much we win in gold?

Interesting idea, but it has already been done, and not cheaply - read on.

How could we get a great quantity of $\ce{_{82}^{197}Pb}$ ?

There would be two problems with getting a large amount of $\ce{_{82}^{197}Pb}$.  First, the parent nuclide of $\ce{_{82}^{197}Pb}$ is $\ce{_{83}^{197}Bi}$ which is unstable and has a half-life of only 9.33 minutes - so you can't get a large quantity of $\ce{_{82}^{197}Pb}$'s precursor to begin with.  Second, once $\ce{_{82}^{197}Pb}$ is formed, it has a half-life of 8.1 minutes, so it transmutes quickly to $\ce{_{81}^{197}Tl}$.

How long would it take for it to turn into gold?

$\ce{_{82}^{197}Pb}$ half-life = 8.1 minutes
$\ce{_{81}^{197}Tl}$ half-life = 2.84 hours
$\ce{_{80}^{197}Hg}$ half-life = 64.14 hours
After 10 half-lives, ca. 0.1% of the starting material will be left
$$\ce{(1/2)^10 = $0.0009766$}$$
The last step is the slowest by far, so after about 641.4 hours (26.73 days), you should have something around 99.9% pure gold.

Can that be accelerated?

Unlike chemical reactions that can heated, catalyzed, etc., this type of nuclear transformation keeps a set schedule.

Once most of it has been turned into gold, how can we extract the gold
  ? (there are still remaining Pb,Tl,Hg  molecules)

As noted above, you can get whatever purity you desire, just wait.

What would be the yield of the process?

It would be high for the 3 nuclear transformations you listed.  Each of the elements you listed decays directly and only to the daughter isotope you've shown.  However, as noted above, you can't start with $\ce{_{82}^{197}Pb}$, you generate it from $\ce{_{83}^{197}Bi}$, the decay of which adds some impurity along with lead. And then since the bismuth isotope is not long-lived you'd probably start with its precursor, and so on until you find something that has a long enough life that you could assemble a reasonable quantity.
Back around 1980 Glenn Seaborg actually transmuted bismuth to gold, but only a few thousand atoms (see this reference also).

How much money would it approximately cost compared to how much we win
  in gold?

The Wikipedia article I referenced directly above notes, "the expense far exceeds any gain."  There are other ways (fission and fusion) to produce gold, but at least with the methods available today, the cost would be astronomical.

Can't $\ce{_{82}^{197}Pb}$ be obtained by sending particles on normal lead ? - Also, what would the obtained gold look like ? (as it forms particle by particle, would it be some kind of dust provided we have enough molecules ?)

Anything is possible, but... The stable isotopes of lead are 204, 206, 207 and 208, you'd have to knock a lot of protons and neutrons out to get down to 197.

Don't knock off protons otherwise you wouldn't have lead anymore.

Of course if you'd knock three protons off the lead ... cha-ching!

One thing Ron didn't cover is the side effects of all that decay. All that short-half-life stuff will be rather radioactive, to the point where you might not survive to see the results. It's also possible that if you get enough together to be profitable that it will simply vaporize itself. The stable gold should eventually condense so as long as you contain the vapor it might work. Or you'll have a very pretty gold-lined flask.

The stable isotopes of lead are 204, 206, 207 and 208, you'd have to knock a lot of protons and neutrons out to get down to 197.

Shame! Knock a proton off of lead and it turns into thallium. But as that's the point of the exercise just bump off two more and we're done.

Approximately how much energy would be released, and what would the by-products be?

I can’t guess at the cost, but consider this: if it were profitable, many people would do it, and then the value of gold would quickly reduce to a level consistent with the cost of producing it.

I know it's not profitable at all, but I was still wondering how much it would cost with modern methods compared to how much we would gain in gold.

You can't really die from the radiation while the process is happening, if you do it right. So you will need a fat, thick lead box to protect yourself from the alpha, beta and gamma rays/particles. Also, the finally condensed gold will collect on the bottom of the box or on the sides. So technically I calculated that to get about $\pu{450 g}$ (one pound) of 99.9% pure gold, you will need to start with approximately 3 tons of the radioactive bismuth isotope. And to buy that amount, you will need somewhere about $43000.5 \times 10^6$ dollars to do that. So what is the trade-off? It is kind of a bad deal, so it will just be better to mine for gold.

Better make sure the lead in the box isn't pure $\ce{_{82}^{197}Pb}$.  ;-)

The best way I can think of doing it would be to use lead as a cyclotron target, I would need to bombard it with very high energy protons. My intention would be to form Bi-197. By making a proton enter the nucleous of the lead, if the resulting nucleous has a high energy it would emit some neutrons or alpha particles or even some protons.
This process is spallation, this would form an unholy mixture of things with differnet atomic masses. I would expect to form as well as Bi-197 some Bi-205 (radioactive half life 15 days) and Bi-202 which will decay to long lived Pb-202- So the cyclotron target will stay radioactive a long time.
In the mean time the Bi-197 will have undergone a series of positron emissions to form Au-197. I would then need to dissolve the cyclotron target in acid and then use solvent extraction to separate the gold from the mercury, thallium, lead and other things in the target.
I am sure that the electricity needed for the cyclotron, the chemicals needed for the separation and the time of the workers needed will be more valuable than the gold that they foem.

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text:
Why is formic acid considered an acid if it is a reducing agent?
My understanding was that since the acceptance of Lavoisier's revised Easter Memoir of 1778, all acids were understood to be oxidising agents.*
However, formic acid is often listed as a reducing agent (i.e. the opposite of an oxidising agent), despite being an acid.
Please can you reconcile this apparent contradiction for me?
* What is meant by "oxidising" or "oxidising agent" has, of course, changed since then, now referring to electron transfer rather than necessarily involving the element oxygen.

might be easier to describe iron as both reducing agent in Fenton reaction and oxidizer in H-W reaction, just depends on who wants the electrons more.

I realize this is an old post, but since it's risen to the top again, I figured I would mention that the issue isn't with formic acid but, rather, with (modern interpretation of) Lavoisier's memoir. Simply put, formic acid is an acid because of its pKa. Additionally, it is a reducing agent because it oxidizes to $\ce{CO2}$ in the course of reducing other compounds (through hydrogenation).

Formic acid is a carboxylic acid with the formula of $\ce{CH_2O_2}$ and is easily oxidized to carbon dioxide. Other carboxylic acids are not as easy to oxidized.
Added:
If a compound gets oxidized (as formic acid can), then another compound has to be reduced. The compound that gets oxidized is called a reducing agent. The compound that gets reduced is called an oxidizing agent. Most carboxylic acids cannot oxidize other compounds (the acid is not strong enough).
Lavoisier was referring to mineral acids ($\ce{HNO_3, H_2SO_4, }$...) in the text. These acids can oxidized other compounds (such as changing iron into an iron ion).

Thanks, but please could you expand on your answer to address the more fundamental question of how an acid can be a reducing agent (or, if you prefer: why this substance, that is a reducing agent, is still called an acid)?

OK, I'm learning! From Wikipedia: "In organic chemistry, reduction more specifically refers to the addition of hydrogen to a molecule... In organic chemistry, good reducing agents are reagents that deliver H2." This is deeply weird to someone like me who's fairly ignorant of modern organic chemistry, but it explains the first part of your answer; and that in turn makes the next part of it comprehensible. It still, however, does not explain why formic acid is considered an acid.

Formic acid is classified as a carboxylic acid. A carboxylic acid is an organic compound that can release a proton when mixed with water. They are different from minearal acids since not all the protons are released when the compound is mixed in water (only a fraction of the compound is disassociated). Formic acid is easily oxidized to carbon dioxide, whereas most other carboxylic acids require stronger oxidyzing agents. http://en.wikipedia.org/wiki/Carboxylic_acid

Thanks for the explanation of why carboxylic acids are acids and why, therefore, formic acid is an acid. I've accepted your answer :)

A molecule might be an oxidising agent under the electron transfer definition, and simultaneously be a reducing agent under the alternative definition: i.e. that reduction is a loss of oxygen or a gain in hydrogen.
For your specific question, formic acid can be deemed to be an oxidisation agent since it loses a hydrogen during protonation of another molecule. The really silly way to think about this is following deprotonation, formic acid has a higher percentage of oxygen than it did before, i.e. it possesses more oxygen than before.
The case of formic acid being a reducing agent is due to the more modern definition of electron transfer. To see the mechanism of this for clarity, see here. It's clear that the formic acid is donating an electron pair (i.e. itself is being oxidised) and is making the other molecule accept those electrons (i.e. that molecule is reduced). The formic acid is therefore a reducing agent.
So your original issue is due to two different definitions of oxidisation and reduction which still exist to this day, and which are by no means mutually exclusive!
EDIT: Of course, the case of formic acid is more simple because it isn't the acid doing the reducing, it's its conjugate base i.e. formate. So formic acid is an oxidising agent under the proton definition, yet formate is an reducing agent under the electron transfer definition.

Thanks. I followed your link, but it didn't clarify anything for me, not least because it seems to be talking about deprotonated formic acid rather than simply formic acid. I'm still finding this all a bit baffling, so further clarity would be welcomed!

Yes of course! You've got it right (I mentioned it in my edit). I meant in general that it's possible for a molecule to be an oxidising agent and a reducing agent under the two different definitions that exist, but formic acid is more simple: formic acid is the oxidising agent however it's formate that is the reducing agent!


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Why are methanol flames less visible than other flames?
The combustion of methanol tends to generate flames that are less visible than the flames generated by the combustion of other substances?
My guess is that methanol burns quickly as it has a relatively simple structure that provides some of the oxygen needed for combustion.  I am assuming that the flame comes from a mixture of fuel and air that is undergoing reaction.  If the reaction is fast enough, then there will be less flame.  However, methane and hydrogen both have visible flames, much more visible than methanol, leading me to wonder how the one oxygen in methanol makes so much of a difference.

Here's a link to a YouTube video showing a dish of methanol burning next to a dish of ethanol.  The methanol flame is more difficult to see.  Methanol was also widely used as a fuel in the auto racing circuit, until fires that resulted from crashes or fuel filling accidents resulted in "invisible fires" that were difficult to see during daylight, resulting in severe burns to crews and drivers ("you can't fight the fire, if you can't see it", this is one reason why it was replaced by ethanol as a fuel for Indy racing).
Flame color (visibility) is dependent upon the elemental composition of the material being heated and the temperature to which it is heated.  This is because heat will excite electrons in the element(s) to an excited state; when these electrons return to the ground state visible light (flame test) will often be emitted
If we look at the carbon content of methanol and ethanol we find the following weight percents:
methanol - 38% carbon by weight
ethanol - 52% carbon by weight
If we compare the heats of combustion for methanol and ethanol we find:
methanol $\ce{~~}$ -726 kJ/mol (exothermic)
ethanol $\ce{~~}$ -1368 kJ/mol (much more exothermic)
There's just not much carbon there to burn in methanol!  Consequently burning methanol doesn't get that hot and its flame will not be as colored as materials that burn more exothermically and result in higher temperatures.

Am I correct in thinking that carbon burns hotter because of the higher heat of formation of CO2 vs H2O and the higher heat capacity of H2O vs CO2?

"Am I correct in thinking that carbon burns hotter because of the higher heat of formation of CO2 vs. H2O", Yes -394 vs. -242 kJ/mol for CO2 vs. H2O.  I hadn't though about the heat capacities (0.844 vs. 1.93 for CO2 vs. H2O), but that makes sense too.

There's just not much carbon there to burn in methanol ! | while it is true that it is the reason in this case, the way it happens is completely different from what you described.

Perhaps I should elaborate. It is well known and easily provable at home that pale blue flame of propane hand  torch is much hotter, than bright yellow flame of same torch.

@permeakra My explanation was based on 1) lower carbon content leading to 2) lower heat output leading to 3) fewer excited states being produced leading to 4) less coloration of the flame.  I don't understand your comment, specifically which step in my explanation do you disagree with?  Soot is not an issue when adequate oxygen is present.

Your explanation is wrong. Soot is the source of yellow light in the yellow flame you can see in your link. When organic fuel, say methane, is premixed with enough oxygen, say air, it burns pale blue and hot. When it is not, it firstly burns incompletely and colder, producing soot, that soot emit yellowish light and subsequetnly burns completely when mixed with more air.  This is well established elementary school-level knowledge. Ethanol vapors plainly require more air per volume of vapors to burn completely, the percentage of carbon content is irrelevant.

See here, http://youtu.be/FeuMSjRH0nY?t=23s . When the propane torch is ignited, its flame is oxygen deficient and bright. Once the oxygen valve is open, propane is premixed with oxygen, and the flame becomes oxygen rich and pale.

@permeakra The yellow flame in the video is from ethanol, not methanol - read the text below the video!  As your equations indicate, methanol needs only has as much oxygen to burn as ethanol - there is enough oxygen in air for methanol to combust completely as long as the surface area is adequate (as illustrated in my 2 links).  Methanol burns with a pale-blue flame ("invisible" in daylight - see second link) for the reasons I cited. Your answer ("yellow" flame, soot, etc.) addresses ethanol, and has no bearing on methanol combustion.

@ron This comment was about what is wrong in your post, not what is right in my. But if you are going to discuss it, reread it again - my post does contain explanation for pale blue flame, you simply didn't care enough to read it.

+1 here.  Note that as a rule, organic fuels give off a heat of combustion roughly proportional to the oxygen they consume, which in turn is encoded in the number of carbon-carbon and carbon-hydrogen bonds in the molecule.  Ethanol has six such bonds, methanol three, and ethanol has almost twice the heat of combustion as methanol.

The yellow 'flame' is actually an aerosol of black particles heated to temperatures around 1000-1500 Celsium. Depending on temperature, they can emit deep red, orange, yellow and almost white light. So, for yellow flame to occur, black particles must be present. In case of carbon containing fuels the particles are usually carbon particles (soot). Soot is a product of incomplete combustion of carbon containing molecules. Now, let's see the difference between ethanol and methanol, specifically, at amount of oxygen required for combustion of same volume of vapors. Equations of burning:
$\ce{2CH3OH + 3O2 = 2CO2 + 4H2O}$
$\ce{C2H5OH + 3O2 = 2CO2 + 3H2O}$
As it can be seen, same volume of vapors in case of ethanol requires twice as much of air to fully burn, so in case of ethanol incomplete combustion is more common. It is even more common in case of larger molecules, say, higher hydrocarbons, like solid paraffines used in some candles. In case the flame of organic substance has right amount of oxygen (incoming fuel-air mixture contains proper amount of air), it burns with pale flame. It is easily observable in case of hand propane torches, where user regulates amount of air in incoming mixture, depending on amount of air allowed, it is possible to have bright yellow or pale blue flame and everything in between.
The pale blue emission has completely different nature, it is atomic emission of excited radicals and molecules. For example, it is known that $CH$ particle emits a blue light.
Yellow flame may also happen if another source of particles is available and the temperature is right. For example, $\ce{CuO + Al}$ mixture burns explosively, producing bright yellow flash. However, if temperature is too high, bright white flame should be observed. Indeed, magnesium burns with blinding white flame, especially if it is premixed with solid oxidizer (potassium nitrate).
Now, atomic/molecular emission is usually low in common flames. However, even small amounts of sodium in flame makes it bright yellow even if it normally almost colorless. Some other elements may color flame into different colors, ranging from pale violet to deep red. This, however, is a different situation from what you considered.

Under what cases would incomplete combustion of methanol yield soot?  I would expect the combustion to comprise three reactions.  First H3COH->CO+2H2, and then 2CO+O2->2CO2 and 2H2+O2->2H2O.  Given that coal gas is formed by dissociating water to produce hydrogen and carbon monoxide, I would see no reason to expect any elemental carbon to be produced as a temporary decomposition byproduct.

@supercat I never cared to experiment, but guess if a drastically low amount of oxygen is present, it is possible. It is usually easier to achieve with larger flames. The reaction of combustion of ethanol is definitely radical in nature, with many branches. It is likely include such particles as $\ce{.OH , CH2O, .CH2OH, .HCO, CO, HO2, .O.}$ etc., and I'm not ready to discuss it. However, I must note that at low-medium temperatures $\ce{CO}$ is prone to disproportionation into carbon dioxide and elementary carbon.

The spectra  of the yellow flame of burning alcohols is continuous. It is not made of lines.  This means that it is produced by a black body, which is a hot solid substance. It is not produced by a gas. As a consequence, in order for such a flame to be seen, it has to produce from time to time some relatively big particules of solid carbon, containing a huge amount of $\ce{C}$ atoms. Let's suppose that the solid particules are made of $100$ Carbon atoms. To get to such a particule, it is necessary to join the $\ce{C}$ atoms from $100$ methanol molecules. It is not easy. It is much easier to use ethanol to do the same job, as only $50$ molecules are required to "collaborate" to this result. That is why there are more carbon particules in the flame of ethanol. And of course the flame of a higher alcohol, like butanol, will still be more visible, as only 25 molecules has to interact to produce carbon particules. For higher alcohols, the flame produces so much carbon particules that they cannot be burned in the upper part of the flame, and it yields soot.


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Which is the correct way to name pentan-3-one?
I have seen this organic compound named in two different ways.

According to some sources it is written pentan-3-one.. Other sources call it 3-pentanone.
Which one is correct?

Whether you write pentan-3-one or 3-pentanone both are correct.
But according to IUPAC rule pentan-3-one this is correct format of writing the name of organic compound. So I prefer you to write pentan-3-one rather than 3-pentanone.
You should also read this.

Adding to Freddy's answer I would say that, pentan-3-one is preferred overall due to the fact that as molecules start to carry more and more functional groups, it becomes ambiguous of which functional group the number in the front is referring to without knowing the order of priority of naming functional groups.

Some people call it diethylketone.


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Applying the Le Chatelier principle
I am having issues understanding the consequence(s) of a decrease in pressure on a chemical reaction.
I have understood that when we increase the pressure, the side having more number of moles wants its moles to go to the other side where it is more empty(this is how I learnt it). But I fail to understand why if we decrease the pressure by increasing volume the side having less moles goes to the side having more moles?
Also in the case of adding inert gases why does the equilibrium shift in the direction in which larger no of moles of gas are formed?

Why does it want to increase the pressure and the number of moles? Won't it want a balance?

The system doesn't "want" anything.  This is a major misconception with Le Chatelier's principle.
Think about the following reversible reaction, everything in the gas phase:
$$\ce{A + B <=> C}$$
We'll simplify the rate law of the forward reaction to
$$\text{rate} = K_\mathrm{fwd}[\ce{A}][\ce{B}]$$
The reverse rate law is
$$\text{rate} = K_\mathrm{rev}[\ce{C}]$$
If the volume is doubled, what happens to the rate of the forward reaction?  It is quartered.  What happens to the forward reaction?  It is halved.  So, the reaction "shifts to reactants".
If the volume is halved, what happens to the forward rate?  It is quadrupled.  The reverse rate is doubled.  The reaction "shifts to products".
What happens when an inert gas is added?  Since the inert gas appears in neither the forward nor the reverse rate law, nothing happens.
I'm afraid you have learned some misconceptions.
Le Chatelier's principle was developed before much of what is studied in a general chem course today was known.  It has been used to manipulate the outcome of reactions.  All of the predictions made by Le Chatelier's principle can be made by understanding how reactions work at the particle level.  Le Chatelier's Principle is from the macroscopic perspective and leads many students to generate misconceptions.

What happens if we add inert gas at constant pressure. I think there should be 3 cases delta n>=or<0. Please explain to me that also. Thanks

How can the inert gas be added at constant pressure?

Well that's what my books says.

If inert gas is added then to maintain pressure constant,volume is increased. Hence equilibrium will shift in the direction in which larger number of moles of gas are formed

The inert gas has no effect on the equilibrium, hence the word inert.  Ignore the inert gas in any problem.

Also "If the volume is doubled, what happens to the rate of the forward reaction? It is quartered. What happens to the forward reaction? It is halved. So, the reaction "shifts to reactants".
If the volume is halved, what happens to the forward rate? It is quadrupled. The reverse rate is doubled. The reaction "shifts to products"." do you mean to say that for instance in the second case the quadrupling of the forward reaction is dominating the reverse(2times) so it goes in the forward direction?

I think the inert gas has some moles at least which will disturb the equilibrium for sure and increase of volume is needed .

In a reversible reaction both the forward and reverse reactions occur.  At equilibrium the rates are equal.  Le Chatelier's Principle problems in books and classes ask about perturbations to equilibrium. Yes, quadrupling one rate and doubling the other means that the quadrupled reaction's rate is now higher, resulting in net production of (in this case) product.

Bringing time into an equilibrium concept is as big of or bigger of a mistake than taking Le Chatelier's principle at face value... I don't like falling back on equilibrium constants, but that would be more fitting.


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Why does nitrous acid exhibit (unexpectedly) different bond angles?
Data Point 1 - quantum chemical calculations

http://nzc.iap.ac.cn/uploadpdf/Wang_et_al._CSB_2007a.pdf
Key points in paper:

There is some sort of interaction between the hydrogen and the terminal oxygen in the cis isomer; the author leave it up to the reader to decide whether one should consider that a hydrogen bonding interaction or a dipole-dipole interaction. Question - the interaction seems to fit the definition of H-bonding; why raise the possibility of it being a dipole-dipole interaction?
The authors discuss resonance contributors and use bond lengths along with other factors to determine the main contributors for the cis and trans isomers. Their analysis reflects ron's analysis perfectly.
This is perhaps the most significant point and it is also discussed on ron's excellent analysis:

Bent's rule is not explicitly invoked in the paper's analysis but it is implied.

Data Point 2 - Professor's claim - unsourced

Data Point 3 - Another Book's Claim

My questions
1) I have three sets of bond angles for the HONO molecule. The first and the last agree with each other in a general sense: cis-HONO exhibits a larger $\ce{O-N-O}$ bond angle than does the trans conformer. The middle picture - my professor's claim - contradicts the other two. Cis-HONO exhibits a smaller $\ce{O-N-O}$ bond angle than trans-HONO.
Who's right? From the replies I have gleaned that hydrogen bonding is likely to have a small to non-existent role in HONO given that hydrogen-bonding might actually increase (pseudo) ring strain.
I also think that my professor's argument regarding free-rotation along a single bond axis is a bit simplistic at best; in isolation, yes, single bonds can rotate, but so can double bonds (you just have to expend enough energy to break the pi-bond). Plus, there are barriers to free rotation of single bonds. You can't rotate single bonds willy nilly all the time. And according to the quantum mechanical calculation, the cis isomer exhibits a slightly higher energy than the trans isomer.
1a) Over what maximum distance can hydrogen bonding still exert a notable force? I'm curious. With this information I might try to do some trig using the QM data and see if the H and the O in the cis-conformer even come close enough for there to be hydrogen bonding.
2) Can cis and trans terminology be used to describe HONO? Usually these two terms are used with respect to identical atoms or functional groups. How about anti and syn? Or eclipsed and non-eclipsed?
3) Tautomerization. Why can't the proton be attached to the N? What's the barrier to tautomerization?
4) Ionic resonance structures - are these valid in the aqueous phase? Seems that if we have a lone proton hanging around in water it would be immediately solvated and taken away from the rest of the molecule ... this was the issue the prof raised. An obvious counter to his issue is that resonance structures don't exist discreetly; the ionic resonance structure is just a contributor - a partial representation of the true bonding nature in nitrous acid. It doesn't actually suggest that the proton is separated from the rest of the molecule. Dunno why he raised that issue ...
5) Authors of one paper also mention that nitric acid exhibits a similar widening of the cis isomer's ONO bond angle ... what might the two different bond angles for nitric acid be? Hell, does nitric acid even have cis/trans isomers? I don't think so ...
Regarding intramolecular hydrogen bonding in HONO
The below paper is based on the premise that hydrogen bonding exists within nitrous acid.

The influence of intramolecular hydrogen bonding on the
  photodissociation of cis-HONO is probed via measurements on the OH
  fragment ejected by specific [...] stretching (ν2) vibrational levels
  of the Ã state. Due to hydrogen bonding between the H atom and the
  terminal oxygen, the ν2 motion is coupled to the in-plane HON bend ν3.
  Since the latter evolves into fragment rotation, the rotational energy
  and anisotropy of the OH product increases with the number of ν2
  quanta. By contrast, ν2 in trans HONO is a relatively isolated  [...]
  vibration and thus does not influence the OH photofragment's
  properties.

Follow up? http://chemistry.stackexchange.com/q/10091/4945 I cannot see that there is an answer there, but it is the same paper linked. Maybe it would be good removing one of these questions.

I cited that paper in the OP and I have read it @Martin. Good catch anyway!

I was thinking about the question title and especially the use of "unexpectedly" in light of the following.  From previous questions I know that you know that, if two things cannot be interconverted by a symmetry operation, then they must be different.  So right off the bat we know that the two conformers must have different bond angles and lengths.  What's left to discuss is "which direction" and "how much."  For me, the question was, "why does cis HNO2 have a larger ONO angle than its trans counterpart?"

Also @ron the first source seems to be a calculation rather than an actual measurement; how reliable are calculations vs measurements? I put "unexpectedly" because I've been under my professor's impression for a good few months, and now I find calculations and books stating things to the contrary. On the other hand these contradictory claims seem to be calculations rather than actual measurements, and the guy hates calculations.

The reference in my answer provides experimental data.  Maybe you could ask a TA in the class to get a copy of the paper, so we can see what the author's say about this "cis" effect.  Maybe your professor would like a copy too.

@Dissenter "5) Authors of one paper also mention that nitric acid exhibits a similar widening of the cis isomer's ONO bond angle ... what might the two different bond angles for nitric acid be?"  The ONO angle where the terminal O is cis or trans to the terminal H.  "Cis ONO ~ 115.88 degrees, trans ONO ~ 113.85 degrees, see http://en.wikipedia.org/wiki/Nitric_acid

@Dissenter, I'm definitely not certain, but it may be worth considering that negative hyperconjugation is occurring between the nitrogen lone pair and the $\sigma^{*}_{OH}$ antibonding orbital (analogous to what's purported to occur in molecules such as (Z)-1,2-difluoroethene, halogenated diazenes, etc.) in the cis rotamer. I think that would be consistent with the widening bond angles, shortening of the $\ce{N-O}$ bond, and lengthening of the $\ce{O-H}$ bond given in the QM calculations you cite.

@GregE. I'm curious as to how, in general, one knows whether something is going to populate an atom's bonding or antibonding orbital. Sometimes it's obvious - i.e. nucleophilic attack in an SN2 mechanism - the nucleophile is going for that sigma star orbital of the leaving group. But in other cases, how exactly does one know? This might be an elementary question but likely speaks to my relative unfamiliarity with MO theory, due in part to the age of my instructors thus far ... none have really emphasized MO theory.

Perhaps the fact we're populating the sigma star orbital is supposed to come intuitively; we're populating the sigma star orbital; the oxygen already has a full octet and no formal charge, so it's "happy," and any perturbations from this configuration would likely make it "unhappy."

@Dissenter, it's usually impossible to know with any degree of confidence unless one performs calculations. Qualitatively, it's probably reasonable to assume some degree of stabilizing interaction occurs if the orbitals in question are sufficiently close in energy, and that might be the case here. I have to stress that without doing calculations, I'm only speculating.

@GregE. I meant how did you come up with the hypothesis that the sigma star orbital was being populated.

@Dissenter, it was a combination of (a) knowing that similar interactions have been proposed for other molecules, and (b) my suspicion that the nitrogen lone pair and $\sigma^{*}_{OH}$ wouldn't be too grossly dissimilar in energy.

Awesome, I'll look into the other molecules you mentioned. I was actually curious why no one brought up a molecular orbital perspective yet!

These are very small effects that we are discussing, so who really knows what's behind them.  The experimental measurement data was reported in, "Microwave spectrum, structure, dipole moment and quadrupole coupling constants of cis and trans nitrous acids, A. Peter Cox, Alan H. Brittain and David J. Finnigan. Trans. Faraday Soc., 1971,67, 2179-2194, DOI:  10.1039/TF9716702179."  The abstract states, "The structural parameters are markedly different in the two isomers and have been discussed, together with the torsional barrier and relative stability of the isomers, in terms of a cis interaction."  If anyone has access to the paper, perhaps they can let us know what the authors have to say about the nature of this "cis interaction."
Below are some of the resonance structures that can be drawn for the two isomers.  First note that the cis-

isomer has the lone pairs on the same side of the molecule (lone pair - lone pair interactions), presumably responsible for the cis isomer being about 1/2 kcal/mol less stable.
If we examine the (hyperconjugative) resonance structures III and IV in the cis isomer, we might expect some stabilizing, through-space electrostatic attraction of the opposite charges between the positively-charged proton and the negatively-charged, terminal oxygen.
For arguments sake, let's assume that the trans isomer is adequately described by resonance structure I, while the cis isomer is best described by resonance structures III-IV.  In this case we would expect the following structural differences between the cis and trans isomers:

A longer O-H bond in the cis isomer due to the larger O-H separation in cis resonance structures III and IV, than in trans I (this is analogous to longer C-H bonds present on the carbon next to a carbocation due to hyperconjugation)
A shorter internal N-O bond in the cis isomer due to resonance structure III which has a double bond in this position
A longer terminal N-O bond in the cis isomer due resonance structure III which has a single bond in this position
An increased H-O-N bond angle in the cis isomer because the structures cis III and IV involve roughly $\ce{sp^2}$ hybridized O-H oxygen, whereas in resonance structure trans I the O-H oxygen is roughly $\ce{sp^3}$ hybridized (actually shifted towards an even more constrained angle due to the effect of the lone pair); as we shift hybridization of the O-H oxygen from $\ce{sp^3}$ to $\ce{sp^2}$, we expect the H-O-N angle to widen.
An increased O-N-O bond angle in the cis isomer due to Bent's rule (a tendency to direct hybrids of increased s-character towards more electropositive substituents); a partial negative charge resides on both oxygen atoms in cis III and IV, whereas trans-1 has no  negative charge on oxygen; therefor, the oxygens in cis III and IV are more electropositive (less electronegative) and the nitrogen will direct hybrid orbitals higher in s-character towards them as compared to trans I; higher s character in the bond means a larger O-N-O bond angle in cis III and IV compared to trans I

So perhaps very different contributions from the possible resonance structures to the real descriptions of cis- and trans-nitrous acid can account for the observed structural differences between these two isomers.
Side Notes: 1) hydrogen bonding is unlikely due to the attendant ring strain as noted by Ben Norris, 2) the oxygen-oxygen interaction should be electrostatically attractive.

Why would the O-O interaction be electrostatically attractive @ron?

Does this have to do with the formal charges and the resonance structures? How can you be sure that the formal charges correspond with the actual, partial charges? I guess through deciding which resonance structures are major/minor?

Also I edited my OP with what my professor told me; I'd be interested on your take on that @ron

Also I don't understand resonance structure 3 for your cis-isomer ... are the formal charges adding up to 0? I'm perhaps missing something. I think a lone pair is missing.

Your comment #1) in resonance structure 2 the oxygens have opposite charges, this should lead to some oxygen-oxygen attraction in the real molecule; #2) yes, it's all about resonance structures and which ones best describe the real molecule; #3) my first thought, before I looked at the experimental bond angles, was that cis should have a smaller ONO angle due to N-O lone pair repulsions, your prof's angles are in accord with that rationale but don't seem to match the experimental data; #4) yes, the formal charges should add to zero, I didn't show all lone pairs.

Why would resonance contributions differ for cis and trans isomers?

They have to, everything has to be different between cis and trans isomers, rotamers, or anything else where the two entities in question cannot be superimposed.  If they can't be superimposed by a symmetry operation, then they are different.  That means different in all aspects, including electron distribution.  As I mentioned earlier, they must be different, all that's left for us to discuss is, by how much and in what direction.

what's the rationale behind which resonance structures are major/ minor for each isomer?

After studying molecules for a while you develop a feeling for which isomer might be more stable, the same applies here.  There are general rules (see the sections entitled "Major and Minor Resonance Contributors" and "Resonance Contributor Preference Rule" at http://www.chem.ucla.edu/harding/notes/notes_14C_resonance.pdf

That paper says the more covalent bonds the better, therefore why night the ionic resonance forms be preferred?

They're probably not the major contributors (but they probably do contribute).  That's why I said "for arguments sake"; it was just easier to explore (and communicate) their effect on the real structure by focusing on them alone.

So this is probably a hard problem to solve a priori? As in assigning the bond angles to the isomers?

If you mean explaining why the bond angles are what they are, then yes, the differences are so small, it seems like it would be difficult to say "this is why" with confidence.  On the other hand, the authors in the paper I cited seem to indicate that they understand what's going on.

I mean if we were given the information that there were two diff. bond angles for HONO and told to assign bond angles to the cis/trans isomers, is that something that's just about impossible to do correctly a priori? It seems we'd have to consider potential H-bonding and resonance structures, and even then we'd still have to hazard a guess as to which resonance structures are major for each isomer. So probably difficult to impossible a priori?

Also @ron I'm gonna work on getting access to that paper.

Yes, that would be difficult to do.  Any university library should be able to access that article.

I found a print copy in my university library; I will either go tonight or tomorrow and get to the bottom of this issue :D

your analysis not only reflected what was discussed in the paper but went beyond the paper in terms of depth. Excellent mind-reading, ron!

I edited my OP to reflect the fact that H-bonding is not only a possibility in HONO but also something that has been raised in both your "cis effect" paper and in a separate paper which actually probes the nature of the H-bonding in HONO.

What potential stabilization might be derived from an intramolecular hydrogen bond in cis-HONO, is counteracted by strain generated by the bad geometry. In order for the H and the distal O to get close enough to have a bonding interaction, cis-HONO starts to suffer from van der Waals repulsion between the two oxygen atoms. Note that cis-HONO is higher in energy than trans-HONO on your table.
If the hydrogen bond existed, cis-HONO would be a pseudo-four-member ring. For cycloalkanes, four-member rings have almost as much ring strain (110 kJ/mol) as three-member rings (120 kJ/mol). The origin of this ring strain is much more likely van der Waals repulsion than any sort of "angle strain" due to "improper" bond angles. Meanwhile,  $\ce{O-H-O}$ hydrogen bonds average at 21 kJ/mol. The ring strain that would result from hydrogen bonding in cis-HONO would destabilize the molecule more than any stabilization from a hydrogen bond.

The four membered ring would not explain the widening of the ONO bond angle, rather a narrowing.

The fact that the four-member ring destabilizes the molecule when the hydrogen bond is present explains why the hydrogen bond does not overcome the van der Waals repulsion (which is the reason for the bond angle widening).

Where would there be van der Waals repulsion? The intramolecular hydrogen bond can never explain the widening. I also believe that there would be an atractive force acting between the oxygens, destabilising the nitrogen oxygen bonds. The concept of ring strain is a derived concept from hybridisation, which is also only a mathematical concept. The derived numbers are qualitatively arbitrary. The true reason for the widening must lie somewhere else. We are nonetheless talking half a kcal difference, which is nothing at room temperature, also 3 degree is nothing in terms of vibration.

@Martin Heats of combustion and heats of hydrogenation also support the concept of ring strain.

@ron They only support the relative instability towards hydrogenation or combustion products. There is no way to absolutely measure ring strain, as there is no way to prove hybridisation. Numbers assigned to ring strain are always derived from reactions and I would rather look a six membered rings to be extraordinarily stable, than four membered rings being especially strained. But that is just my opinion and that does rarely matter here. I believe the concept of ring strain will be an ongoing debate till the world ends...

@Martin what in your opinion expands the ONO bond angle?


text:
Can a solid and liquid be miscible?
To be specific, what I mean is: is the following situation possible? At given temperature T, there are two substances, X and Y, such that pure X is a solid, pure Y is a liquid, and X and Y mix to form a single phase in all proportions.
The closest I could find was this phase diagram for silver and gold (two very similar metals, I suppose):
http://www.crct.polymtl.ca/fact/phase_diagram.php?file=Ag-Au.jpg&dir=SGTE
But it looks like there is still a narrow region in which there are two phases.

theoretically speaking, a polymeric compound on itself is solid (well, kind of), but some polymers can dissolve in chosen solvents, providing full spectrum from (kind of) solid trough viscous liquid to normal liquid. Good luck finding concrete example with more traditional chemistry.

It's not necessarily a strict definition of a single intermingling of phases, however the closest thing I can think of corresponding to this in reality would be interstitial metal hydrides.
These materials are being investigated as hydrogen storage materials, so essentially the hydrogen molecules interact with the metal to form hydrides, "$\ce{H-}$", which recombine under different conditions to reform $\ce{H2}$ as required. A review of this technique is given here.
It would be interesting, to answer your question more directly, if liquid hydrogen could be used instead of hydrogen gas (this requires temperatures of <33K). I imagine this process would give exactly the same result, albeit would happen more slowly than the gas phase.


text:
Writing half-reaction for the oxidation of iron (III) hydroxide
Here is the unbalanced equation.
$$\ce{Fe(OH)2_{(s)} + O2_{(g)} -> Fe(OH)3_{(s)}}$$
Therefore, one half-reaction is
$$\ce{Fe(OH)2_{(s)} + O2_{(g)} -> Fe(OH)3_{(s)}}$$
and the other half-reaction is
$$\ce{O2_{(g)} -> 2H2O_{(l)}}$$
My question, how are we justified in making the latter half-reaction?

Oxygen should not be in both half reactions. You'll get yourself in trouble.
Try:
$$\ce{Fe(OH)2 ->Fe(OH)3}$$
$$\ce{O2 -> H2O}$$
We are justified in making the latter half reaction because balancing the first half reaction will likely require water added to the reactant side to help balance the $\ce{H}$ and $\ce{O}$ atoms.
However, an equally appropriate second half reaction might be
$$\ce{O2 -> OH-}$$
noting that there is an extra hydroxide in the products of your overall redox reaction.

I would simplify the oxidation reaction further, ignoring the hydroxide ions which do not really participate:
$\ce{Fe^{2+}->Fe^{3+}}$
The reduction of oxygen makes the extra hydroxide in the iron(III) hydroxide product, but it needs water to provide the (protic) hydrogen, thus:
$\ce{O2 + H2O->OH^-}$
You can balance the half-reactions and then the (net ionic) full reaction from there.

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


text:
What element or compound could be used to make the ocean more basic?
The ocean is becoming acidic. This is not a natural evolutionary change but happens through human negative effect on the environment. What element or compound could be used to create a more base ocean? How much of it would be needed? How could it be distributed?

Calcium carbonate has been used to remediate acid rain damage in lakes and ponds.  The scale needed in the ocean might make this unviable.

It is possible to get a coarse estimate for the amount of material needed to directly neutralize all the acid coming from anthropogenic $\ce{CO_2}$ absorbed by the ocean. It turns out that using the material with the highest ability to neutralize acid per mass (lithium metal), we'd need to throw ~100 billion metric tons of the pure substance into the ocean. Compare this with the entirety of the iron mining industry, which currently produces about 4 billion metric tons of iron/steel per year. There is no way we can access the amount of matter required for a simple neutralization reaction.

The ocean is becoming more acidic due to the dissolution of excess $\ce{CO2}$ from anthropogenic activities to form carbonic acid, $\ce{H2CO3}$, which even though mildly acidic is decreasing the pH of the ocean as you rightly point out.
So if we're trying to reduce the effect of this $\ce{CO2}$ on the ocean, we might consider removing it via some sink. One of these theoretical methods could be to react a base, such as $\ce{CaCO3}$,  with this carbonic acid in a conventional neutralisation. In symbols:
$$\ce{H2CO3 + CaCO3 -> Ca(HCO_3)2}$$
This could work, and we could derive everything we wanted to know about how much base to use from information such as the amount of residual $\ce{CO2}$ left over from emissions after photosynthesis has been taken into account (and other sinks), and then assuming that all other $\ce{CO2}$ will be taken up by the sea. From here we can calculate the amount of $\ce{H2CO3}$ present in seawater from the equilibrium constant between $\ce{CO2}$ and $\ce{H2CO3}$.
But we won't do that!
Notwithstanding these nonsense assumptions which could grossly overpredict/underpredict the amount of base required, neutralisation reactions don't deal with the excess carbonate effectively. What you really want is a permanent sink (a sink so stable its contents are practically constant through time) at the bottom of the sea. See the long-term carbon cycle for more information on this.
So your question becomes: how do we get the $\ce{CO2}$ dissolved in the ocean to the deep-ocean sediments? And how do we do it 'permanently'?
An interesting paper came out in the Journal of Global Biogeochemical Cycles in 1990 trying to address the question about why phytoplankton in the Antarctic don't use the myriad nutrients in the Southern Ocean to photosynthesise. One might expect that it could be temperature or sunlight which limit photosynthesis in the Antarctic, but this actually isn't the case. The limiting factor in the Antarctic ocean is actually a lack of iron.
It has since be proposed to increase the amount of iron in the ocean (simply by adding it ourselves) which would promote the photosynthesis of phytoplankton in such locations. When these organisms die, they sink and take their biomass with them which contributes to the ocean sediments.
Both neutralisation and iron supplementation may work to offset ocean acidification, however we really don't know the full consequences of these bioremediation efforts—there are simply too many variables involved.

I corrected your equation and improved the coding. You can enclose the whole reaction equation in $$\ce{...}$$. This way the spacing is better and you safe a lot of typing...

Cheers Martin, still pretty new to the formatting. Will need to write my thesis in this so I'm bound to improve.

Have a look at the documentation of the mhchem package on CTAN.


text:
Which one, Mulliken charge distribution and NBO, is more reliable?
Sometimes the Mulliken and NBO turn out to be so different that I can't decide which one I can trust. I've heard that Mulliken is inaccurate, but is NBO always accurate?
And should I use Gaussian or other software to calculate charge distribution? Are there methods more accurate than Mulliken or NBO?
And charge distribution doesn't represent charge density, right?

There is no proper, strict and agreed definition of atom border, that's the real problem. Gaussian includes option to fit charges on atoms (assumed for the sake of the procedure as points) so they produced electrostatic field as accurately fitting once calculated from the electronic density distribution as possible, and maybe (I haven't used Gaussian for a long time, I'm not sure) dipoles too.

NBO is more accurate than Mulliken in very many ways.  However, NBO is not always accurate.  Generally speaking, you would never want to use Mulliken charges for any sort of production level work unless its been calibrated.

As I do not want to drag this out in the comments, I decided to give a brief answer. Let me first cite the very helpful comments to keep the for (relative) eternity and add my thoughts at the end.

There is no proper, strict and agreed definition of atom border, that's the real problem. Gaussian includes option to fit charges on atoms (assumed for the sake of the procedure as points) so they produced electrostatic field as accurately fitting once calculated from the electronic density distribution as possible, and maybe (I haven't used Gaussian for a long time, I'm not sure) dipoles too. –  permeakra

NBO is more accurate than Mulliken in very many ways. However, NBO is not always accurate. Generally speaking, you would never want to use Mulliken charges for any sort of production level work unless its been calibrated. –  LordStryker

Mulliken is of course the cheapest and fastest way to compute charges. However, this method tends to give qualitative results at best. The reason for this is obvious. It will divide the canonical orbitals equally amongst the participating atoms. There is no polarisation whatsoever. Another problem is, that they are very basis set dependent. And that being said, the worst part is, that the description becomes worse by increasing the basis set.
NBO charges are much more reliable, since the operate on the electron density instead. Localised natural atomic orbitals can be used to describe the computed electron density. Polarisation of bonds is therefore considered. This method obviously only works very well for systems, that can actually be (quite well) described by a Lewis structure. However, they usually give a nice qualitative picture and are quite robust when it comes to increasing the basis set.
There are a couple of more population analysis tools, that can be used and should be considered when it is necessary to derive some conclusions from charges.
Quite common is also the Hirshfeld charges approach, which divides the electron density into spherical basins around the nucleus. Maybe interesting to read is "Are the Hirshfeld and Mulliken population analysis schemes consistent with chemical intuition?" (I can't tell, I do not have access.)
There are also several schemes to obtain partial charges from the electrostatic potential.
Probably the most rigorous approach to atomic partial charges is the Quantum Theory of Atoms in Molecules  (QTAIM). Here the Laplacian of the electron density is used to describe the partitioning in electronic basins. The best feature of this analysis is, that it can be also used for experimentally obtained electron densities. However, as everything, this has also major drawbacks. One of the most important is the computational effort. Unfortunately the integration scheme is not always completely robust when it comes to some particular molecules. However, it is almost completely basis set independent and results could be cross referenced with experimental data.
Since permeakra's initial statement holds true, especially for the agreed part, the best way to deal with partial charges is to be careful and maybe check with different methods. In gaussian there are several schemes implemented, most commonly, Mulliken, Hirshfeld and NBO (3.1) and some others, see the population keyword for more detail.

I've read your answer and the links you provided. Now I see that Mulliken is expendable. Thanks. However, after I look at NBO result, I suddenly feel it's no use. Usually, charge distribution is used to know the dipole moment. But we can know the dipole moment directly from the Gaussian output. I can't think of other uses of charge distribution.

@Ath There are certainly uses for partial charges and they often provide more insight into chemical bonding. The dipole moment is just a vector, but it does not give you the polarity of the molecule. Atomic charges may become important when working on chemical reactions - it can be as useful as oxidation states might have been before that. The NBO concept is a very important one as it provides a lot of chemical insight.

The Mulliken analysis is the most common population analysis method,
  it is also one of the worst and is used only because it is one of the
  oldest and simplest. Table (picture) shows an example of the variation
  in partial charge with respect to basis set for the Mulliken analysis.
  Note that the basis sets which would be best for energy calculations
  are some of the worst for population analysis. This is why the
  Mulliken analysis is not used, it is highly basis set dependent.

Source: http://www.huntresearchgroup.org.uk/teaching/teaching_comp_chem_year4/L7_bonding.pdf


text:
Why is ΔE described as Q-W and Q+W in different contexts?
Depending on the textbooks looked at, the energy change is described as either Q+W or the signage is changed to Q-W. Which is correct? Is it solely context dependent? Could anyone explain the background?

The difference in sign in the two versions of the first law of thermodynamics is to handle the two ways in which work can be defined.
The work done (assuming only pressure-volume work) can be defined as
$$w=P\Delta V$$
This is the definition often used in in scenarios when we care about the fate of the work. This definition is about the surroundings. Thus, when the system expands, it does positive work on the surroundings, which is a good thing. Think about an internal combustion engine. When the reaction occurs in a cylinder, gas is produced causing the volume to expand and move the piston, turning the drive shaft, etc. Since we care about the work after it leaves the cylinder, it is positive because the surroundings are gaining energy.
The work done by a system can also be defined as:
$$w=-P\Delta V$$
We use this definition often in chemistry and in general when we care more about how much energy is left in the system than we care about what the energy that left the system is doing. This definition is about the system. In the case of the engine above, energy has left the system, so the work done is negative to reflect the fact that the system has less energy than it used to.
The sign convention in the definition of the change in internal energy (and I prefer $\Delta U$ to $\Delta E$) reflects that we can define work done with respect to the system or the surroundings. The sign in front of $w$ guarantees that we always get
$$\Delta U=q-P\Delta V$$
With respect to the surroundings:
$$w=P\Delta V$$
$$\Delta U = q-w=q-P\Delta V$$
With respect to the system:
$$w=-P\Delta V$$
$$\Delta U = q+w=q-P\Delta V$$
A better question might be

Why then do we not change the sign of $q$ when we switch reference frames from system to surroundings?

Thanks @Ben Norris, and what would be your thoughts on your closing remark? :)

My guess is that $q=C\Delta T$ and it is much easier to determine the heat capacity of the system than that of the surroundings, thus $q$ is always defined for the system.


text:
Reduce chlorine gas
I have a simple electrolysis setup which uses aqueous sodium chloride as electrolyte. The problem is that, the setup makes tons of chlorine gas along with hydrogen (I only want hydrogen). I didn't even realize it until I recognized that the odor smells like that of ozone. I am using a power supply of 24V @ 6A. Since, it's a pain to buy a new power supply, changing the supply isn't an option. Also, I have experimented with other electrolytes like aqueous sodium bicarbonate but they are far slower than the brine solution. So, the only other option is to consume the chlorine gas. I am not sure what to use here. I don't have a lot of reducing agents on hand. Any ideas?

If anyone is curious as to how I know what ozone smells like, I have experimented with lots of high voltage. It makes your throat feel dry. Chlorine does that too..

plainly keep your cell at temperature well above 50+C. In such conditions sodium chloride is oxidized to chlorate and later perchlorate. Another option is to try and use something else, like sodium hydroxide or magnesium sulfate.

The issue is that oxidation of chloride to chlorine has a lower overpotential than oxidation of water to oxygen. You will always have to have an oxidation reaction going on while reducing water to hydrogen. The one with the lowest standard potential + overpotential will win. The reason the overpotential exists is that electrolysis is pushing the system away from equilibrium, so we need to provide additional electromotive force to get that to happen. The value of the overpotential is related to the activation energy of the redox event.
For example, with a graphite electrode, the overpotential for the oxidation of water is +0.95 V, while the overpotential for the oxidation of chloride is +0.12 V. When you compare the standard reduction potentials for these two reactions, we can see that chlorine is produced more readily (smaller overall voltage needed).
product    Red. Pot.    Overpot.     Total Pot.
O2         +1.30 V      +0.95 V      +2.25 V
Cl2        +1.36 V      +0.12 V      +1.48 V

To avoid chloride production, you need to get ride of chloride. Switch to sodium hydroxide. You have plenty of voltage in your power supply to overcome the overpotential of water/hydroxide.


text:
In cracking, why does C form double bonds?
I've been reading about cracking and I've found various images and animations that show a process like this:

But I haven't been able to find out the sequence of events to get to this, and why this would occur.
I understand that cracking involves thermal decomposition, where a CC bond is cut. But that's as far as I can truthfully say I understand. To get this reaction below, you need:
(1) Two cuts. One cut would be between two Cs, and the other cut between a C and a H. Why does a H have to be cut off as well? Is this example I found actually one of many random and messy ones that just happens to 'give us what we want in the end'?
(2) The free H would then have to join the alkane. I get why this happens.
(3) A double bond is made. Why?

When we heat hydrocarbons to high temperatures (somewhat lower temperatures if we use a catalyst) they crack.  That means that carbon-carbon bonds are broken and smaller fragments are formed.  In the top part of the following drawing, I've picked n-heptane as an example.  You can see that there are 3

different bonds we can break and a number of different fragments will result.  Each of the fragments that result from breaking a C-C bond has an unpaired electron (there were two electrons in the bond we broke, each fragment retains one electron).  These fragments with an unpaired electron are called radicals.  The carbon with the unpaired electron does not have a stable octet configuration, therefor radicals are extremely reactive species and will react rapidly in a variety of ways to regain their stable octet configuration.
I've picked one of the possible radicals generated in the top drawing and show how it may react further in the bottom half of the figure.  Sometimes it will break a C-H bond and lose a hydrogen atom and form a stable compound with a double bond (an olefin), sometimes it will remove (abstract) a hydrogen from a hydrocarbon (heptane in our example) to generate a new alkane (pentane) and a new radical (perhaps a radical not formed in the initial cracking step).  Many of the radicals formed can rearrange to more stable radicals, this adds further complexity to the overall process.
So starting with just heptane we can get a large variety of saturated hydrocarbons and unsaturated hydrocarbons.  The situation becomes exponentially more complex if, as usual, we do have many more hydrocarbons present in addition to heptane.
The key here is that radicals are very reactive species - they want to stabilize by regaining the octet of electrons around the carbon atom that has the radical.  They can get that octet and become stable by adding a hydrogen atom (alkane formation) or breaking a C-H bond and eliminating a hydrogen atom to form an alkene.

I just want to point out that the cracking of hydrocarbons is done in an enclosed container to prevent oxygen from reacting with the hydrocarbons.

@LDC3 If you are agreeing with the answer, why aren't you casting a(n) (up)vote?

@Martin I was waiting to see if you would post an answer that was as informative.

@LDC3 I could not add anything more informative to this :D

@ron +1 Right, I've just learned about those electron dot diagrams (I'm kind of a noob), so I can now understand what you're saying about free radicals. So, I understand why C would want to react, and why it can form a double CC bond. But, are you saying that the picture I posted is just one possible outcome of many? It's 'by chance' that we get an alkene? I'm interested in the production of plastics, so I'm just curious (when they want to make alkenes on a large scale) whether these are by chance.

@User17670 Yes, your picture shows just one possible outcome.  Many different alkenes and alkanes can be formed.  However, the product distribution is not determined by chance, but rather by thermodynamics.

If you look at a single molecule and one reaction step, it is pure chance. But in the end, the product range is rather clear: For every alkane molecule  $C_xH_ {2x+2}$, you get at lease one shorter alkane plus one or more molecules with the shorthand formula $C_yH_{2y}$, which can only be cycloalkanes or olefins.

@Ron - I don't think that such reactions are "pure" thermodynamics. I'd expect that different catalysts could be used and that different catalysts would produce different mixtures. So in a continuous flow operation the reaction could be controlled.

@MaxW The simplistic view is that catalysts only change the activation energy.  If true, that would suggest that different catalysts might change the rate at which different products are formed, but once equilibrium is attained, then the same thermodynamically controlled product distribution would result.  In a flow operation all bets are off, since it is likely a non-equilibrium process.


text:
What are the most important resonance structures of 5-aminocycloheaxa-2,4-dienone?

Show, by means of curly arrows and electron shifts, the resonance contributors to the structure given below. Suggest which one would not be a major contributor and state why.

This is a question from my exam. I do not understand where I went wrong. My effort is placed at the end (along with a request).

Also, my TA removed 1 point out of 7 for a reason. Can anyone decipher for what? There's the writing in red but it is illegible to me. I don't understand to what he refers by 8 and 6 t or +, or whatever he wrote.

Wouldn't this isomerize to its tautomer m-aminophenol?

The handwriting of your TA probably stems from ancient egyptian script, I had a hard time reading it. In the end I agree with jerepierre, interpreting it as "6E" or "8E" short for 6/8 electrons. The last statement then translates to "but all have 8E."
Before continuing to answer your question, I want to make a statement about resonance structures in general. Resonance structures are used to express a complicated bonding situation in terms of simpler, Lewis type structures. None of these structures exist individually, only a mixture of all possible resonance structures at the same time exists. I am not a fan of declaring single resonance structure as having more or less impact on the overall structure. Of course one can go ahead and calculate the total wave function in terms of resonance structures, generally known as valence bond theory, but this is overkill for such simple molecules. Then you get a mathematical representation of the electronic structure which you can decompose into contributions of the individual resonance structures.
A simpler approach is to analyse the molecule with Natural Bond Orbital (NBO) Analysis. I have done this for you, but only as an educational effort. Maybe after understanding this concept it is easier to grasp why certain structures are make a smaller contribution.
Let me go ahead and explain a little NBO to you.  The electronic structure of a molecule is in general highly delocalised. All orbitals are stretched over the whole molecule. Without changing the electron density, it is possible to localise (transform) these orbitals to represent two centre interactions, that can be interpreted as bonds. This gives an approximate picture in terms of a Lewis structure. There will be electron density which cannot be assigned to a bond. The smaller this density is, the better is the fit as a Lewis structure.
For this educational approach I picked the 5 resonance structures you have posted plus an additional ionic form. (I would have detracted a point for the last structure.) There are many more ionic forms, but these will not have big effects, e.g. cleavage of a carbon carbon bond.

The following is based on NBO analyses from DF-BP86/def2-SVP.
\begin{array}{crr}\hline
\text{Structure} & \text{% Lewis} & \text{% non-Lewis}\\\hline
\mathbf{1}   & 96.4  &  3.6  \\
\mathbf{2}   & 96.8  &  3.2  \\
\mathbf{3}   & 96.2  &  3.8  \\
\mathbf{4}   & 95.8  &  4.2  \\
\mathbf{5}   & 96.3  &  3.7  \\
\mathbf{6}   & 95.2  &  4.7  \\\hline
\end{array}
The first statement that should be made is that this molecule is quite well described by Lewis resonance structures. However, delocalised bonding would describe the molecule much better. All structures are reasonably close to each other, as one would expect for a very delocalised system.
The ionic structure has the smallest impact, which is feasible because nitrogen-hydrogen bonds are better described as covalent. However, it proves that there can (in the right medium) be a proton shift from nitrogen to oxygen, and is therefore worth consideration.
The structure with the second largest error (% non-Lewis) is $\mathbf{4}$. A positively charged carbon would be by far more electronegative than nitrogen. And many imine-type structures are known, which makes $\mathbf{5}$ more important. (In this sense I would disagree with your explanation, which could also justify the detraction of a point.)
In general: Carbocation structures that are not in close proximity to (much) more electronegative elements should be considered as having a lesser influence on the electronic structure of a molecule.

Would it not be simpler or perhaps clearer to tabulate the relative energies of these structures and on the that basis determine which are expected to be greater resonance contributors?

@Buck These are all just interpretations of a single electronic structure, they all have the same energy just different localisation schemes of the election density. I guess one could do a full Valence bond calculation, but I can't do that. Probably it would be a good idea to look at this with natural resonance theory, but I don't have access to that either.

Ah, so they are degenerate. You did a nice job of arranging the results. It's so tidy it's hard to see the steps involved, even with your explanation that NBO was used. I haven't worked with NBO so it's not easy to see how you "pick the structures." I thought you fed geometries + constraints into Gaussian and it spit out a density, but in this case ... well, I guess I'll have to read up on NBO.

i think your teacher said "but all have it" refering to the split charge (?) I would personally say that the ones with less contribution would be the ones with carbon with positive charges because while N+ and O- are relatively stable, C+ is certainly not. Being a bit more specific, I would say (not 100% sure) that the 4th one because is the one in which the positive charge is in an unstable carbocation and close to a Nitrogen atom which would destabalise it even more due to its electronegativity.

I think those are "6E" and "8E". "6E" indicates that there is a carbon atom with only 6 total electrons in that resonance structure. "8E" indicates that all atoms bear octets of electrons.
When determining importance of resonance structures, maximizing the number of atoms that follow the octet rule is the priority. It may be counterintuitive, but the structure with a positive charge on nitrogen contributes more to the electronic structure of the compound than the structures with positive charge on carbon.


text:
What should be the oxidation state of oxygen in HOF (hypofluorous acid)?

What should be the oxidation state of oxygen in $\ce{HOF}$ (hypofluorous acid)?

Sources on the internet have confused me. Most state its oxidation state to be 0, while the others state that it is -2 (considering the oxidation state of fluorine to be +1, which is a bit surprising for me as it's the most electronegative element in the periodic table)

Source stating the oxidation state to be -2:

J. Chem. Educ. 1972, 49 (4), 299:
  The two possibilities brought forth in the Note are (1) assign hydrogen +1, oxygen -2, and fluorine +1 or (2) assign H +1, oxy- gen 0, and fluorine -1. On the grounds that $\ce{HOF}$ is a powerful oxidizing agent, the conclusion reached in the Note is that (1) should be favored.

Other sources stating the oxidation state to be 0:

Yahoo Answers!
Answers.com

Check here:  http://chemistry.stackexchange.com/questions/15299/how-to-find-out-oxidation-state-when-compound-has-multiple-electronegativities/15306#15306

Related: Does fluorine in FNO3 have +1 oxidation number?

Look up the electronegativities of carbon, hydrogen and chlorine.  Count electrons accordingly in the carbon-hydrogen and carbon-chlorine bonds of dichloromethane, a dichloromethane, a familiar organic solvent.  The oxidation state of carbon is ... .  Do not be afraid of zero!

Until the (recent) redefinition of the IUPAC, the concept of oxidation states was not as well defined as one would expect. I have discussed the issues of the old version and outlined the new definition in more detail in an answer to Electronegativity Considerations in Assigning Oxidation States.
When you apply the official pre-2016 definition (via the Internet Archive) from the IUPAC gold book, then you have to assign $\text{OS}(\ce{H}) = +1$ as it is not a compound with a metal, $\text{OS}(\ce{O}) = -2$ as it is no peroxide compound, and leaving the disturbing $\text{OS}(\ce{F})= \color\red{+1}$.
Going with the on electronegativity based alternative description, you will still assign $\text{OS}(\ce{H}) = +1$ as it has the lowest electronegativity. Then you assign $\text{OS}(\ce{F})= -1$, because of the highest electronegativity. This leaves a oxidation state of $\text{OS}(\ce{O}) = 0$.
This assignment is matched by the 2016 definition, the summary version of which is: The oxidation state of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds. The definition refers to the use of Allen electronegativities (see Pure Appl. Chem. 2016, 88 (8), 831–839 for more detail). Consequently the heteronuclear approximation yields:

The true charges are of course something completely different. Based on a DF-BP86/def2-SVP calculation I ran some population analyses with Multiwfn 3.4.1 (a newer version is available). The NBO charges are taken from a prior version of this answer, but I since have lost access to the program and the files used.
\begin{array}{lrrr}\hline
\text{Method}    & \ce{H} & \ce{O} & \ce{F} \\\hline
\text{Hirshfeld} & +0.18  & -0.10  & -0.09  \\
\text{ADCH}      & +0.38  & -0.29  & -0.09  \\
\text{VDD}       & +0.18  & -0.10  & -0.09  \\
\text{Mulliken}  & +0.20  & -0.06  & -0.15  \\
\text{Löwdin}    & +0.09  & +0.01  & -0.10  \\
\text{Becke}     & +0.38  & -0.21  & -0.17  \\
\text{ADC Becke} & +0.38  & -0.29  & -0.09  \\
\text{CM5}       & +0.35  & -0.26  & -0.08  \\\hline
\text{QTAIM}     & +0.61  & -0.44  & -0.17  \\\hline
\text{NBO}       & +0.45  & -0.32  & -0.12  \\\hline
\end{array}

So what conclusion can be made from this? Do they show fractional OS?

@Alpha The lesson to be learned here is, that oxidation states are hardly telling the truth, but they can be a helpful concept in understanding reactions. Unfortunately there is a very thin line between using and overusing a concept. Personally I am not a big fan of fractional oxidation states and would always go for the electronegativity argument.

So can we deduce anything about oxidation states based on the NBO partial charges?

@Jan No. I find it hard to deduce an oxidation state from any charge, but then again I do not put much stock into oxidation states in the first place.

@Martin Your answer is helpful. However, are those partial charges a representation of the actual distribution of charge within the molecule? If so, how come oxygen has more negative charge than fluorine?

@ghosts_in_the_code The charges I used are computed from the electron density distribution with the natural bond orbital theory. This is one way to compute them, but there are others. The higher negative charge of oxygen is probably because it accumulates much of the electron density of hydrogen. The difference between fluorine and oxygen is not very big, hence there is less polarisation than for the other bond. Without the hydrogen, oxygen would probably have a smaller negative charge than fluorine, but not by much.

So if we were to assign oxidation states to most closely match the real charges, the best fit to eight out of ten models is to render the oxygen-fluorine bond nonpolar so $\ce{H}=+1,\ce{O}=-1,\ce{F}=0$.  !!!

@OscarLanzi since partial charges are no observable there will always be from for interpretation. Oxidation numbers are also very crude. The methodology I used is to crude for a statement like yours, even though the language I used may suggest otherwise. I think is better viewed a little more fuzzy.

The oxidation state in my opinion will be zero.Because as flourine is the most electronegative element it will never show positive ooxidation state.It's maximum oxidation state is zero.

Could you elaborate on this a bit more? At this point, there is no source for your answer and it doesn't give much more information than is already presented in Martin's answer. @MohammedSafwanA

$\ce{F}$ is the most electronegative element, so it is always -1 oxidation state, except in $\ce{F2}$, where it is zero.
So, in $\ce{HOF}$, $\ce{H}$ has +1, $\ce{O}$ has zero and $\ce{F}$ has -1 oxidation states.
There is no way $\ce{F}$ will ever have oxidation state +1. In $\ce{OF2}$ (oxygen fluoride, and not fluorine oxide) $\ce{O}$ has +2 oxidation state!

Hi, welcome to our site! "There is no way F will ever have oxidation state +1. " I understand your sentiment, but did you have a look at the accepted answer? It clearly shows how F can have a OS of +1. Can you please explain by what reasoning have you contradicted it?

Please see my answer for a more detailed analysis. While I agree with your process, it is unfortunately not the one that is guided by the common definitions. I find this contradiction quite spurious and I am dissatisfied with the current status quo. It is true that fluorine should never have a positive oxidation state, and for $\ce{OF2}$ this is even more ridiculous. Unfortunately any attempts to change the definition have been unfruitful so far.

IUPAC is right for normal compounds. It simply fail for the case when O and F have a direct bond.

$\ce{H}$ is less electronegative than $\ce{O}$, so you pretend the electrons in the $\ce{H-O}$ bond "belong" to $\pu{O}$; $\ce{H}$ is $+1$. By similar reasoning, $\ce{F}$ is $-1$.
You can work out the oxidation number of $\ce{O}$, either by counting electrons, or by using the fact that adding up all the oxidation numbers of any chemical species has to give you the charge of that species, which in this case of course is $0$.

Please see my answer for a more detailed analysis. While I agree with your process, it is unfortunately not the one that is guided by the common definitions.