text:
Why is Aluminium Chloride a Lewis Acid if it's an ionic compound?
In my textbook it says in the reaction $\ce{AlCl3 + Cl- -> AlCl4-}$, Aluminium Chloride ($\ce{AlCl3}$) has six valence electrons and accepts a lone pair from $\ce{Cl-}$, but wouldn't that mean the aluminium atom covalently bonded with the three chlorine atoms to get six electrons (which is wrong as it's ionic)?

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if it's an ionic compound

Actually, it is debatable. $\ce{Al^{3+}}$ is a very small cation and its formation is strongly unfavored. In fact, this cation, just like many others, may exist only in solvates, like $\ce{[Al(OH2)6]^{3+}}$, or unchemical environment.
In particular case of anhydrous $\ce{AlCl3}$ the molecule is a strong Lewis acid, strong enough to coordinate with lone pair of another similar molecule, so it forms either dimer $\ce{Al2Cl6}$ with two chlorine bridges, donating one pair to both Al ions, or polymer, with Al atoms exhibiting octahedral coordination. Unlike many ionic compounds, melted $\ce{AlCl3}$ has poor conductivity.
This, however, does no apply to $\ce{AlCl3.6H2O}$ which is, again, a ionic compound $\ce{[Al(OH2)6]^{3+} (Cl^{-})3}$ and does not exhibit any Lewis acidity (though it does exhibit strong Brønsted acidity, i.e. tendency for hydrolysis)
This said, however, I have to clarify: ionicity of compound does not prevent particular ion from being Lewis acid/base. In fact, cations often are excellent Lewis acids. The only thing needed for a particle to be a Lewis acid is presence of low  vacant orbital, and sometimes presence of such orbital is not derivable from octet rule. For example, $\ce{SnCl4}$ is Lewis acid, though in it $\ce{Sn}$ has full octet. The orbital in question is describable in terms of MO theory which I strongly encourage to study.
The most often considered Lewis acid is hydrogen cation. In fact, this Lewis acid is so strong, that it never occurs free, it is always bound to Lewis base (again, except unchemical environment). Another common Lewis acid is $\ce{Na^+}$. This cation is quite weak Lewis acid, but it is sterically accessible, and usually is paired with anion, which is usually a Lewis base. This said, despite mostly ionic nature of $\ce{NaCl}$, there is considerable covalent bonding in it. (of course, on the other hand there is such anion as $\ce{BPh4^-}$ which is not a lewis base of considerable strength, so its salts are almost purely ionic in nature)

What do you mean the hydrogen cation exists in in chemical environments?

@Dissenter It exists in environments that are not domain of chemistry of condensed phases, but physics: plasma, particle beams and similar.

@permeakra thank you very much, my doubts have been cleared!


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What makes mercury vapors so harmful?
I just read an experiment in which mercury has to be heated and in the same experiment there was a note warning not to inhale mercury vapors as they are very harmful.
I just can't get the point that why are mercury vapors harmful.

Did you read wikipedia article on mercury poisoning?

Mercury poisoning (also known as hydrargyria or mercurialism) is type of metal poisoning and a medical condition caused by exposure to mercury or its compounds. Mercury (chemical symbol $\ce{Hg}$) is a heavy metal occurring in several forms, all of which can produce toxic effects in high enough doses. Its zero oxidation state $\ce{Hg^0}$ exists as vapour or as liquid metal, its mercurous state $\ce{Hg2^{2+}}$ exists as inorganic salts, and its mercuric state $\ce{Hg^{2+}}$ may form either inorganic salts or organo-mercury compounds; the three groups vary in effects. Toxic effects include damage to the brain, kidneys and lungs.
Mercury poisoning can result in several diseases, including acrodynia (pink disease) and Minamata disease.
source wikipedia

@agharehanabbas Please do not copy a source verbatim without quoting the text and without a proper citation.  Plagiarism will not be tolerated.

ok i wil look at it on my further posts

You can edit your answer to include a reference.

@jonsca i have edited my answer and added my source to it

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Difference between backbonding and hyperconjugation and conjugation
What is the difference between backbonding and hyperconjugation and conjugation?
I mean the basic differences as all are concerned with donation of electrons. What is the most striking effect of each that differentiate them from others.

This topic is extremely broad and can be explained with widely varying levels of complexity and rigor. The way your question is currently formatted, you might be asking for an encyclopedic response. If you've done any research on this topic already, I think it'd be preferable if you narrowed the scope of your question.

@GregE. How can you vote to open a question?

OK, let's step through these terms and let's add "cross-conjugation" and "homoconjugation" for good measure to round out the series.
Backbonding:
This term is most often used to describe a type of bonding that occurs in inorganic compounds.  Metals bonding with carbon monoxide provide good examples of backbonding. The carbon atom in carbon monoxide has a lone pair of electrons that can be used to form a sigma bond with a metal.  Because carbon monoxide has low-lying $\pi^{\ast}$ orbitals, it can accept electrons back from the metal and further strengthen the bonding between the metal and the carbon monoxide ligand.  This process of "accepting electrons back from the metal" is termed backbonding. Here is a good picture of the $\ce{\sigma}$ and $\mathrm{d}-\pi^{\ast}$ (backbonding) bonding interactions.

It happens most often when 1) the metal is electron rich, 2) the ligand has low-lying $\pi^{\ast}$ orbitals and 3) the metal has available a $\mathrm{d}$ orbital with the proper symmetry to overlap constructively with the carbon monoxide $\pi^{\ast}$ orbital.  Here is a link to a great, one page article on backbonding if you'd like to read more.
Hyperconjugation:
This term is most commonly used in discussions about resonance.  Hyperconjugation is a stabilizing interaction that results from the interaction of the electrons in a $\sigma$-bond (usually $\ce{C-H}$ or $\ce{C-C}$) with an adjacent (empty or filled) $\mathrm{p}$-orbital. This interaction creates an extended molecular orbital that increases the stability of the system.  Most typically it involves the breaking of a carbon-hydrogen bond and then using what were the $\ce{C-H}$ bonding electrons to stabilize an adjacent charge. In resonance structures, we can't move nuclei, but we can move electrons. For example, below is a diagram showing resonance structures that involve hyperconjugation that serve to stabilize the n-propyl carbocation. Hyperconjugation has been used as part of the explanation for the carbocation order of stability being tertiary > secondary > primary.

Recently it has been suggested that hyperconjugation, rather than steric effects, is what stabilizes the staggered conformation of ethane (very interesting!).    Here is a link to a fairly extensive, but very well written article on hyperconjugation, it covers the effects of hyperconjugation on resonance structures and conformational stability.
Conjugation:
This term is used to describe a system of continuous, overlapping p-orbitals where electron delocalization can occur.  Examples of conjugated systems include the 1,3-butadiene, acrolein, benzene, beta-carotene, etc.

Cross-conjugation:
Sometimes you'll hear the term "cross-conjugation" used.  In this case, the continuity of the overlap between the p-orbitals does not follow a simple linear pattern as seen in the examples of conjugation pictured above.  Benzophenone and 3-methylene-1,4-pentadiene are examples of molecules with cross-conjugation.

Homoconjugation:
Sometimes $\mathrm{p}$-orbitals that are not directly conjugated (e.g., that are not directly connected, there is a "spacer" in between them) can still interact or overlap in a "through space" fashion.  The solvolysis of the exo- and endo-norbornenyl chlorides shown below provides a striking example. The exo-isomer generates the corresponding carbocation thousands of times faster than its endo-isomer. This difference is attributed to homoconjugation. The $\mathrm{p}$-orbitals on the remote norbornenyl double bond can interact through-space with the developing $\mathrm{p}$-orbital that is being formed on the carbon where the exo-chloride is leaving from, dramatically stabilizing the formation of the resultant carbocation. This interaction is geometrically impossible for the endo-isomer.  Again note, there is no direct conjugation between the remote double bond and the carbon atom where the carbocation will form, it is entirely a through space interaction that generates a pseudo-allyl system.

The most basic difference lies in the kind of orbitals involved in all. In hyperconjugation, there occurs an interaction between sigma orbitals on one side and completely or partially filled orbitals on the other.
In conjugation, alternating single and double bonds are present so as to create a region of overlapping $\mathrm{p}$ orbitals having in between them bridges of sigma bonds. Electrons are said to be delocalised.
Backbonding can be best described as $\pi$-coordinate bond, wherein the excess negative charge is donated to the central atom having vacant orbitals (e.g., $\ce{BF3}$).

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Units for dissociation constant and relationship to Gibbs Free Energy
How does one "convert" between an equilibrium constant calculated from the Gibbs free energy for a dissociation reaction and experimentally determined dissociation constants?
$$
\Delta G^\circ=-RT\log(K_{eq})
$$
I think I understand the derivation of this equation and why the equilibrium constant should be unitless if it's defined in terms of activities.
But I have a binding reaction that has an estimated free energy change of $-32~\mathrm{kcal/mol}$, which yields a dissociation constant on the order of $10^{-24}$ at $298~\mathrm{K}$.
Is it correct to just multiply the equilibrium constant by the standard concentration (1 molar) to get a more typical dissociation constant with units? A yoctomolar dissociation constant seems kind of strange to me, but I'm not sure where I've made a mistake.

Is it $\ce{AB<=>A + B}$, with $\Delta G^\circ_{\mathrm{diss}}(\ce{AB})=-32~\mathrm{kcal/mol}$? Multiplying with the standard concentration just adds a unit to the same value. Also your definition of $K$ is slightly incorrect: http://goldbook.iupac.org/S05915.html

free energy change of -32 kcal/mol, which yields something on the order of 10^-24 at 298K.

Seems the main mistake is a sign error.  Also "log" must mean "natural logarthim" the way the equation in the question is written to be correct.
If the free energy change is negative, the equillbrium constant is $>1$
Should be $10^{\frac{32\frac{kcal}{mol}}{2.303\times0.00199\frac{kcal}{Kmol}\times{298K}}} = 10^{23}$

Sorry, that was my mistake, I was referring to the free energy change of the binding reaction as being negative, and so the dissociation constant should in fact be < 1. In any case, my concern is with the order of magnitude of the constant, which doesn't seem biologically or even physically reasonable. If my calculation looks correct otherwise, does that mean that my estimated free energy change is incorrect?
Thanks for taking the time to answer my question!

It is certainly physically possible.  For say dinitrogen to dissociate into two nitrogen atoms, the equilibrium constant would be much smaller than 10^-24.  Without knowing more about your system, I don't see any way to rule out it being correct.


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Can I use the hypotenuse of the pythagorean theorem to compute the work?
A friend of mine told me that she used the hypotenuse of the pythagorean theorem to compute the work in graph below and the to compute the coefficient of efficiecy in this thermodynamics triangle cycle. What do you tnink about?

Please explain yourself further, currently I have no clue where you are talking about?

Me and a friend of mine we were discussing about the work done in this graphic. In my opinion the total work is given by the area of the triangle. For she used the hypotenuse of the pythagorean theorem to compute the work.

Is this regarding a thermodynamic cycle?

Yes. it´s regarding to thermodynamic cycle.

@ViniciusL.Beserra yes you can do it and more over it might be the only way in this situation to calculate the total work done

The work would be area of triangle, which is no way related to the hypotenuse (other than $\small H^2=P_0^2+V_0^2$)$$W=\frac12P_0V_0$$
Heat is :
$$Q_{in:A-C}=(C_V+R/2)\left(3\frac{P_0V_0}{R}\right)+C_V\left(2\frac{P_0V_0}{R}\right)$$
Efficiency is :
$$\eta=\frac{W}{Q_1}=\frac{\frac12P_0V_0}{(C_V+R/2)\left(3\frac{P_0V_0}{nR}\right)+C_V\left(2\frac{P_0V_0}{R}\right)}=\frac12.\frac{R}{3C_V+3R/2+2C_V}$$
$$\eta=\frac{R}{10C_V+3R}=\frac{R}{5fR+3R}=\frac1{5f+3}$$
which for  a monoatomic,diatomic and polyatomic(linear/non-linear) are:
$$\begin{array}{|c|c|}\text{Monoatomic}&1/18\\
\text{Diatomic}&1/28\\
\text{Polyatomic-Linear}&1/28\\
\text{Polyatomic-Non-Linear}&1/33\\
\end{array}$$


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What are the oxidation states in Borohydride (BH4-)
The CRC Elecrochemical Series and the solution to our exercise suggest that the oxidation states of borohydride are -V for Boron and +I for Hydrogen.
Since H (2.2) is more electronegativ than B (2.0), I would have expected the oxidation states to be +III for B and -I for H.
What am I missing here?
The solution to our exercise lists this (WARNING: it turned out that the solution was wrong):

And the CRC Elecrochemical Series lists this (also suggesting -V for B):

I would have expected the oxidation states to be +III for B and -I for
  H.

You're right, boron is in the (+3) oxidation state in these equations.
In $\ce{B(OH)_3}$ each $\ce{OH}$ is (-1) and the boron is (+3), overall the molecule is neutral
In $\ce{BH_{4}^{-}}$ each $\ce{H}$ is (-1) and the boron is (+3), overall the ion is (-1)

Thanks for answering, I am still skeptical, however. Usually our assignments are prepared quite carefully, and how would you explain that redox potential? Boron(+III) taking up 8 electrons would also result in B(-V).

The boron does not take up any electrons.  All of the electrons are used by the hydrogens.  On the left side of the equation we have 10 $\ce{H^{+}}$, on the right side of the equation we have 4 $\ce{H^{-}}$ and 6 $\ce{H^{+}}$.  The 8 electrons reduced 4 $\ce{H^{+}}$ to 4 $\ce{H^{-}}$.

Sorry for having been skeptical ;-). It turns out that the solution was wrong after all (info from the professor): Hydrogen is being reduced, not boron. I'm glad... (voted and accepted).

@Zalumon  Glad we got it figured out!


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Who is Hræðraford, the "learned clerk" "writing in modern chemical Latin"?

This question has been re-asked on History of Science and Mathematics:
Who is Hræðraford, the “learned clerk” “writing in modern chemical Latin”?

Truesdell, C. Rational Thermodynamics. New York, NY: Springer New York, 1984. http://dx.doi.org/10.1007/978-1-4612-5206-1, p. 41 says:

… in the numerous papers and books whose titles and topics include "irreversible"—the learned clerk HRÆðRAFORD, writing in modern chemical Latin, has recently called their field "thermodynamice inreparabilis"—…

Who is this "learned clerk" Hræðraford?


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What kind of metal alloys can resist corrosion from molten salt?
Molten salt is pretty hot, 801 C. With such heat and the corrosive nature of salt, I am wondering what type of metal alloy or other material hybrid would be appropriate to handle and work with molten salt?

George Lai's High-Temperature Corrosion and Materials Applications (excerpts available on Google Books) is a good source. Very often for molten salts, Hastelloys are used; they're special-purpose alloys developed by Haynes International. I can't find a reference, but I believe Oak Ridge National Laboratories used Hastelloys for their molten salt reactors (which had fluoride salts, very reactive).
If you're working with sodium chloride (little potassium), according to Lai, Hastelloy-188 seems to be the best bet, although many others would work as well. One reality of chemical engineering is that there will always be corrosion, it's just a matter of keeping it at an acceptable rate for the application.

Oh wow, they did a heat exchange and everything, which is very relevant to my interests. :)

I think you should use tungsten as it has the highest and best melting point among all the stable metallic elements its melting point is nearly $3700~\mathrm{K}$
More over its price is also considerably lower than other metals with higher melting points and if you think of corrosive nature, then I would recommend you to use silver but it becomes very costly.
Feel free to comment for further clarification. I will definitely help you.


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Removing prenylated phenolics from clothing, shoes, and upholstery
I live in Southern California and sometimes hike in areas that can be overgrown with a plant called purple poodle-dog bush. As I found out the first time I went crashing through a stand of the stuff, it's a horrific skin irritant, much worse than poison oak. It has microscopic hairs that get stuck in your skin and act like syringes. The hairs have chemicals on them that are skin irritants. Wikipedia says:

This contact dermatitis is due to prenylated phenolics exuded by hairs (glandular trichomes) of the plant. The principal irritants are derivatives of farnesyl hydroquinone and 3-farnesyl-P-hydroxybenzoic acid.

I'm not a chemist, and all the nomenclature is Greek to me. My question is this. When I hike in an area where it will be hard to avoid the bush, I try to wear long sleeves and change my clothes immediately afterwards. Will throwing my clothes in the washing machine remove these chemicals and/or break them down into harmless ones? What if my shoes aren't machine washable? Will the chemicals break down naturally within a certain period of time? Would other methods such as exposure to sunlight help? Are these chemicals soluble in water, so that simple rinsing would help? How about making sure they aren't lurking in the upholstery of my car?

I don't have time to answer fully, but I imagine the compounds have a non-zero but rather low solubility in water. The actual amount of irritant substances present may be low enough that it could dissolve even in a regular wash, however. Another thing is that relativity caustic media will cause some degree of deprotonation, forming a salt which is likely far more soluble. The hydroxybenzoic acid derivative is particularly susceptible to this, though even the hydroquinone is slightly acidic.

The ultimate problem is likely the hairs which contain the irritants; they might not be easily removed mechanically or chemically, and may impede the access of solvent to the irritants if they are stored in cavities on the hairs.

If the solubility is low but the amount of the substance is small, would it make sense to leave the clothes soaking in water for a long time so that the substance would gradually leach out and go into solution? What would be an example of a caustic medium that I could use to soak my clothes?

Soaking could possibly slowly dissolve everything, but if you were to do that then I would recommend adding a basic compound to the water to possibly increase the speed and efficiency of the process. A safe, cheap and easily washed off base that you could use would be sodium carbonate, $\ce{Na2CO3}$. A solution containing 2 g/100 mL of water will reach somewhere around pH 11 at ambient conditions. Sodium bicarbonate $\ce{NaHCO3}$ can be used too, though it's less basic.

I cannot completely guarantee that your clothes will resist the alkaline solution entirely unharmed, though it should be quite safe. At least one commercial non-chlorinated bleaching agent for clothes actually contains sodium carbonate as part of its formulation (though the bleaching compound is actually the hydrogen peroxide). Furthermore, as I mentioned earlier, the hairs themselves may act as a physical barrier and hamper dissolution. In that case, it may be best to strip the hairs off directly using some adhesive (sticky tape?)

@NicolauSakerNeto: How about making that an answer?

As a starting point, I shall assume that the irritant compounds are freely exposed to the environment (which is not necessarily the case, as I mention further on). As such, one of the conceivably simplest ways to remove them is to wash them off, i.e. disperse them in a solvent and discard the liquid mixture. For several reasons (abundance, extensive solvation ability, non-toxicity, etc.) water is a good choice.
First of all, how well would pure water work at dispersing the prenylated phenolics? Probably not too great. The representative parent compounds of the irritants, hydroquinone and p-hydroxybenzoic acid have modest solubilities in water (59 g and 5 g per 1 L, respectively), and the addition of farnesyl chains in the compounds would quite likely cause a severe reduction in water solubility, unless the farnesyl chains are highly substituted with hydrophilic groups.
Of course, there is no substance that is entirely insoluble in any solvent, though it is very common for certain solutes to require impractical amounts of solvent to disperse completely. It could well be that a washing machine's load of water (of order 100 L) is actually capable of dissolving the farnesylated toxins, especially because natural products are generally present in rather small quantities. However, there are some techniques which can possibly help the process. I list a few in the following:

Use hot water - The majority of substances become more soluble in hot solvents because dissolution very often increases the entropy of a system ($\Delta S>0$), and hence from the Gibbs' free energy relation $\Delta G=\Delta H - T \Delta S$, an increase in temperature leads to a larger decrease in free energy, making the process more favourable. Of course, this only really works well if the water is hot throughout the soaking/washing process; adding hot water and letting it cool off will just cause part of the solute to precipitate out of solution again.
Use a surfactant (detergent) - Since the offending compounds potentially have large hydrophobic chains, they can conceivably removed by the same process we use to wash away oils and fats. An amphipatic molecule contains a hydrophobic part (such as the hydrocarbon chain of sodium dodecylbenzenesulphonate) responsible for interacting with the oils and a hydrophilic part (in SDBS, the sulphonate group), which is well solvated by water. Thus, the surfactant bridges the interaction between oil and water, allowing effective dispersion.
Use a basic solution - In this specific case, the undesirable compounds have some acidic character. In particular, the hydrogen atom in the -COOH (carboxyl) group of the p-hydroxybenzoic acid derivatives are about as acidic as acetic acid. The -OH (hydroxyl) groups in both p-hydroxybenzoic acid and hydroquinone are also slightly acidic, though weaker by some 5-6 orders of magnitude. Thus, mixing in a basic solution may allow an acid-base reaction to take place, forming salts of the organic compounds with significantly increased solubility in water. Many washing powders/liquids are already slightly alkaline, but it is also easy to prepare basic solutions separately. A cheap and very safe base would be sodium carbonate, $\ce{Na2CO3}$, though a few tens of grams may be required to prepare a decent volume of basic medium for soaking. Sodium bicarbonate, $\ce{NaHCO3}$, also works though it is a weaker base and would require a higher amount to reach a given pH value. Caustic soda, $\ce{NaOH}$, is yet another cheap option, but care must be taken to not add too much, as it is a strong base (in contrast to the previous two). I don't know how resistant different fabrics are but, very approximately, any more than a couple of grams of lye per litre of water risks partially hydrolysing the fabric and weakening its structural integrity. Use at your own risk!
Use a bleaching agent (chlorinated or non-chlorinated) - Bleaches work by partially oxidizing organic matter, breaking organic molecules into smaller chunks and making each piece more hydrophilic (due to added oxygenated functionalities). Even if the compounds aren't removed entirely, the reaction products may be much less harmful.
Miscellaneous - You suggest exposure to sunlight. Light is certainly capable of decomposing organic substances, especially UV light, but there's no guarantee that the exposure required to eliminate the irritants won't also break down fabric and bleach dyes. Waiting is also probably not very efficient, as there is no guarantee the irritant compounds are unstable enough to undergo spontaneous decomposition in ambient conditions at a significant rate.

Unfortunately, all this discussion is of limited use if the initial assumption (freely accessible irritants) is incorrect. It could be that the compounds are stored in cavities in the plant hair's surface or interior, and the hairs themselves are insoluble and impermeable to water, effectively locking the toxins away. In this case, the compounds will likely be removed only slowly, as the hairs are progressively worn out by abrasion and other processes. Thus a physical removal method may be preferable. I have heard that adhesives (sticky tape, waxing strips, etc.) can be successfully used to peel out painful hairs. This is also a useful option for surfaces which cannot be washed easily (such as car upholstery).
In practice, I am not sure how this will all work out. It's been a few days since I wrote my comments, so perhaps you've had time to test something out? If so, I would be curious to hear some feedback!

Sorry about not answering before. I felt an answer needed more polish, but I was having trouble making time for it. Even now I don't think what I wrote is as good as it should be. Also, I suspect a fair amount of the above is not particularly useful to you, but I like my answers to be as thorough as I can make them.

Thanks for the very helpful and complete answer. I soaked the clothes overnight in a sodium carbonate solution, then washed them again in the washing machine. Unfortunately it's difficult to tell whether it worked or not. My immune system is currently lit up like a Christmas tree, so I've put the clothes aside and will not try wearing them until my skin is back to normal. If they cause the reaction to flare back up at that point, then I guess we'll know the method didn't work!

Yikes, I didn't know it was that bad. Do be extra careful then! I wouldn't want to unwittingly inconvenience you a second time.

I realize it's been a year, but I thought I'd make a small addition: it may be that dry cleaning could be effective at removing the offending compounds in this case. Dry cleaning often uses non-polar chlorinated solvents like tetrachloroethylene as a washing liquid, which would excel at dissolving the fatty prenylated phenolics, again under the assumption that the solvent can reach the compounds.


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How to test for fluoride levels in drinking water?
I would like to test for fluoride levels in the water.
Are there any methods or test kits available to test the levels of fluoride in home water supply?
See also: Is there any at home test for fluoride levels in your water, and what is the best water distillation unit?

A good reliable set-up is pricey.  Here's a link to a reliable, quality probe for fluoride ion detection in water, it costs \$740  (US).  The meter that it plugs into sells for \$790 (US) if you get the less expensive hand-held device.  There also manuals and instructions on how to actually use the gear, at the website as well.


text:
What are the methods of fluoride removal from drinking water?
I would like to remove fluoride from my water supply.
Are there any methods, any test kits or water distillation units to remove fluoride from drinking water?
See also:

Is there any at home test for fluoride levels in your water, and what is the best water distillation unit?
A simple and easy way to neutralize fluorides in drinking water
Scientists Discover New Technique to Remove Fluoride from Drinking Water

Read More: http://www.trueactivist.com/scientists-discover-new-technique-to-remove-fluoride-from-drinking-water/

Freezing removes a lot of flouride. No idea where it goes. Can anyone help?

If water is partially frozen the ice is usually much purer than the water because the soluble materials remain in the remaining liquor.

Fluoride is very rarely a problem in water supply. It is added in small quantities for health reasons in many areas. There is no need to remove it (unless you hold the Pure Bodily Fluids philosophy of General Jack D Ripper).

There are many filters which do not remove fluoride from potable water. Boiling or freezing does not help either. Reverse osmosis filtration may be a solution. However they are expensive ($100+).
Activated alumina defluoridation filters are used in areas where contamination from fluoride is common. These can reduce fluoride levels from 0.5 ppm to 0.1 ppm, but these require frequent replacement and are not a solution for domestic treatment. Distillation filters can be purchased to remove fluoride.

The easiest method is probably partial freezing and collecting the ice. Most soluble material will remain in the liquid. Distillation from a basic solution should retain fluoride Add some washing soda, Na2CO3 or TSP. Reverse osmosis [check with the manufacturer] followed by mixed bed ion exchange removes almost all ionics.
Amazon sells test kits ~$30 that allegedly test for almost everything.
Another thought! Winter is approaching collect ice from a clean frozen stream or lake and boil it before drinking.

Another idea collect rain, allow the entrained dust to settle, boil before dinking. Cistern water thru a RO gives superb drinking water, use the reject for washing and flushing.


text:
Rust forming and being washed off by water
I have noticed that, when exposed to water for a few days, the steel in my wagon turns into rust. After several more weeks, the rust that formed is dissolved in the water, and once I pour the water out there isn't any rust where there previously was rust.
This is unusual because iron oxides and iron hydroxides (the components of rust) don't normally dissolve in water, and so the water just keeps being reduced and the iron keeps being oxidized.
How can it apparently happen such that steel turns to rust, then back to steel with no rust, in the presence of water if rust does not normally dissolve in water?
This happens in a cycle like this:
steel + water = rust -> rust + water = steel + rusty water

Rust appears because iron is oxidized in presence of oxygen (it is a oxydo-reduction reaction, meaning an electron transfer occurs between species), and water helps this phenomenon occur. Iron loses 3 electrons to the favour of the oxygen.
In presence of water, oxygen can oxidize iron by the mean of hydroxyl group. It leads to:
$$\ce{2Fe(OH)_3 \leftrightharpoons Fe_2O_3\cdot 3H_2O}$$
where water is not the oxidizer, rather it is a mediator for oxidation.
Hydroxides have low solubility, so water can only dissolve a small fraction of it.
On the other hand, poured water can abrade rust (via mechanical stress) and remove it of from the bulk metal, leaving the surface clean for producing more rust. This is how huge rods of metal can vanish in a few months in corrosive environmental conditions (such as wet air or water).
Preventing rust requires the use of specific alloys and surface finishing. You can also add a galvanic electrode (generally made of $\ce{Zn}$, a so-called sacrificial anode) to protect submerged metallic structures such as the hull of a boat.
Therefore, in your example water is both a mediator of oxidation and mechanical abrasion agent.

Steel isn't pure iron by a long shot! The dissolved carbon in steels creates a myriad of shorted electro-chemical cells. Any naturally occurring water is not pure and contains ionic impurities which make the water electrically conductive, and when in contact with these inherent shorted cells accelerates the consumption of the iron. The high Chromium content of Stainless steels mitigates this by isolating the carbon-iron cells from the aqueous solution. If you capture the off-gasses from water corroding steel, you will find it is $\ce{H2}$.

0.25% carbon and less in  low carbon steel- which is most carbon steels- His wagon is probably made of medium carbon steel at up to  0.6%. Neutral water rarely produces hydrogen or any gas in open atmospheric corrosion more like absorbs oxygen to produce hydroxyl ions.


text:
Solar still and how it works
I use a solar still that holds 182 cubic feet of water. Here is how it works:
There is an insulator on both sides of the metal bowl. On the outside it covers the whole thing. on the inside it covers the sides but not the bottom. This collects the rain and also heats up the rainwater during sunny days so that the evaporation rate is faster. I have the pipe nearest the metal bowl the largest it can be while the metal bowl still collects rain and gets heated up by the sun. The water and any gases pass through there and into the condenser. I have this cold enough to condense water into a liquid but not so cold that the water turns solid and the $\ce{CO2}$, $\ce{O2}$, $\ce{N2}$ etc. turn liquid. Last but not least there is a spigot at the very end and a bucket below it. This separates atmospheric gases from liquid water with $\ce{CO2}$ being the main concern.
There is also dust and pathogens from the air in rainwater.
The dust most likely will stay behind and just make the bowl dusty. The pathogens get killed by the heat that there still is when all the water has been evaporated.
I would say the maximum loss with this system is from 1 pint or so to 1 gallon or maybe more depending on the water level in the bowl.
However I think there is a way to calculate maximum and minimum loss(neither of these is 0).
How could I calculate maximum and minimum loss based on the depth of the water(because obviously at different depths there is a different amount of loss) and the size of everything(which by the way the pipe slowly tapers off to a certain size and so you have to take pressure into account as well as volume)?
Also how could I calculate the pressure in the pipe based on the fact that it tapers off and starts at the maximum size possible with the bowl still collecting rain and being heated by the sun?

I think we need more information. Does the water boil in the collector, or just evaporate? What are the possible loss points for water (is the container open, etc.?) A schematic would be very helpful.

well I think the majority just evaporates but some of the water might boil after a while. The possible loss points for water are the bowl and the pipes that come before the condenser

I really do not understand how the water gets into the bowl - is it open towards the sky? I might not understand your picture, but if your still is not closed over the evaporation container, then water just diffuses into thin air. A photograph would certainly help much more.

The water gets into the bowl by rainfall and gets heated up by the sun. Also my pipe starts off as large as it can be without blocking sunlight and I am afraid that if the bowl is closed that 3 things would happen: 1) There would be an insulator up top making the water even hotter because of raised conduction and 2) the sunlight would be blocked by the insulator and lastly 3) Rain will not be able to fall into the bowl.

I still do not understand the setup - Is the rainwater collected from a roof and then passed into the bowl?  I also think this question would fit much better in [physics.se].

the rain just falls from the sky and the bowl collects quite a bit of rainwater because it is wide and shallow and can hold 182 cubic feet of water before it overflows.

There is no way to answer your question accurately for this setup. It will depend on the temperature of the water in the bowl, the exposed surface area, the convective flow rate of air near the surface, and the relative humidity of the air - none of which are constant or even very predictable.
Your best bet is to set it up and measure the amount before and after distillation. The difference would be your loss.


text:
What dangerous gases could electrolysis with water and sodium carbonate produce?
My friend was very nervous when he noticed that I'm performing electrolysis in a room that we've just been residing. (it was under open window to let $\ce{H2}$ out) After a sharp argument, he made me shut it down, though I still thought that it was safe.
I'm using $\ce{Na2CO3}$ as electrolyte and as far as I know, it doesn't get electrolysed. I use normal pipe water. Such water surely contains some minerals however in really small amount. My friend pointed out that it also contains chlorine and that's what he's afraid of. He would disregard my notice that he's drinking this water often.
I'm using copper electrodes, producing copper hydroxide and copper carbonate. I failed to find out whether these compounds are hazardous. They're in solid state anyway.
From what I've said you can see that I'm very sure that the reaction is not hazardous. But I'm not chemist. There could be something I didn't think of in which case I've been poisoning myself for quite a while.
Could you make me sure or warn me?

My friend pointed out that it also contains chlorine and that's what he's afraid of.

The amount is laughable at. However, using distilled (deionized) water is recommended for other reasons.

I failed to find out whether these compounds are hazardous. They're in solid state anyway.

They are reasonably harmless. Of course, eating them is not recommended, but copper compounds have low toxity. Move to graphite or platinum electrodes is still recommended.

For what other reasons is distilled water recomended?

@TomášZato Because it is unknown what other compounds and in what amounts are present in tap water. Since deionized water is reasonably free available at reasonable price (<$2/liter in my city), using it is recommended.

If you're using hard water you might still end up creating a reasonable amount of chlorine, it's still worth distilling the water first.

Copper salts are toxic, which means the water you use would not be healthy after a while and be careful on disposing of it.
The minerals you have are probably in small amounts, and you can check this measuring the conductivity of the water.
You can make the gases pass through a filter of CaCO3 that will absorb any chlorine.
There is also the answers to this question for details on what is happening.

Any chlorine coming from where? Does electrolysis cause dissolved chlorine to evaporate from water faster than when you're, for example, washing yourself?

What do you mean by that second question? If there's say, NaCl, in the water, you could produce chlorine gas. But if you're doing it right next to an open window and in small amounts (which it appears you are), there should really be no worries.

You said that I could filter the chlorine. So I asked if the electrolysis causes chlorine (which is dissolved in freshly poured water) to quickly leave the water. As I've said, I don't use salt because of the toxic products it creates.

There could be some chlorine produced by the electrolysis but from what you say it isn't much and would stay in the water. The use of deionized water is so you don't have ions to consider. The idea of the filter is to tranquilize your friend. The oxygen isn't much either.


text:
What is the isomer distribution in monosubstituted fluorobullvalene?
Bullvalene (tricyclo[3.3.2.02,8]deca-3,6,9-triene) is a fluxional molecule able to interconvert any two carbon atoms through a series of degenerate Cope rearrangements (for more information, see the Wikipedia article on bullvalene).  At slightly above room temperature bullvalene produces a single line in its 1H-NMR spectrum.  Considering fluorobullvalene, there are 4 possible isomers:

1-fluorotricyclo[3.3.2.02,8]deca-3,6,9-triene, i.e. the fluorine could be attached to any of the 3 equivalent cyclopropyl carbons
3-fluorotricyclo[3.3.2.02,8]deca-3,6,9-triene, i.e. the fluorine could be attached to any of the 3 equivalent olefinic carbons alpha to the cyclopropane ring
4-fluorotricyclo[3.3.2.02,8]deca-3,6,9-triene, i.e. the fluorine could be attached to any of the 3 equivalent olefinic carbons beta to the cyclopropane ring
5-fluorotricyclo[3.3.2.02,8]deca-3,6,9-triene, i.e. the fluorine could be attached to the unique tris-allylic carbon

Again, at room temperature or thereabouts, only a single line is observed in the 1H, 13C and 19F-NMR spectra.  I remember once reading that about 80% of the fluorine is in the tris-allylic position (4), most of the remainder is attached to the two olefinic carbons (2 and 3) and a very small amount is attached at the cyclopropyl position (1). This was often cited as an example to support the argument that fluorine prefers to form bonds to carbon when the involved carbon orbital has high p-content (low s-content).
Does anyone know what the isomer distribution is?  Those 4 numbers and a reference would be very helpful.

Extensive NMR studies of substituted bullvalenes were done in the 1960's and 70's, by Oth et al. Much of their original work was published in German. Thankfully, some of these topics were revisited (in English!) during the 1990's by Luz et al., and these included some rather nice low temperature 19F and 13C NMR studies in both solution and solid state (and make it much quicker and easier for me to read).
Let us assign the following isomers 1–4, as is done in the question:

To answer your question:
Oth et al.[1] studied fluorobullvalene at −25 °C, and found no evidence by 19F of isomer 1. The equilibrium populations of isomers 4:3:2 was 78%:7%:15%. However, they could not distinguish between isomers 3 and 2, so it could be 78:15:7. In any case, predominantly isomer 4. They do state, however, that isomer 1 must exist ["Das C-Isomer muss also durchlaufen werden."], as the Cope rearrangement mechanism for isomer 4 to go to isomer 3 must go through isomer 1.
Luz et al.[2] studied the solution state Cope rearrangement of a number of monosubstituted bullvalenes.
At −30 °C, they identify the ratio of isomers 4:3:2 to be 87%:3.4%:9.6%. Again no evidence of isomer 1 by 19F at this temperature. However, at −55 °C, they observe a peak in the 19F spectrum that they ascribe to isomer 1, with an intensity of 0.2% relative to isomer 4. This peak broadens beyond detection at −40 °C. The ratios of isomers at −55 °C, then, was found to be 1.0:0.018:0.065:0.002 for isomers 4:3:2:1.
And just out of interest:
Of course, the other monsubstituted bullvalenes all have different preferred isomer distributions in solution. In the solid state, all monosubstituted bullvalenes studied by Luz crystallize as well-ordered single isomer crystals. Fluorobullvalene[3] crystallizes exclusively as isomer 4. Cyanobullvalene, bullvalenecarboxylic acid[4], and (ethylthio)bullvalene[5] crystallize exclusively as isomer 3. Bromo- and iodobullvalene[6] crystallize entirely as isomer 2.
Useful references for you:

Oth, J. F. M.; Merényi, R.; Röttele, H.; Schröder, G. Fluor-, chlor- und jodbullvalen. Tetrahedron Lett. 1968, 9 (36), 3941–3946. DOI: 10.1016/S0040-4039(00)72372-0.
Poupko, R.; Zimmermann, H.; Müller, K.; Luz, Z. Dynamic NMR Investigation of the Cope Rearrangement in Solutions of Monosubstituted Bullvalenes. J. Am. Chem. Soc. 1996, 118 (34), 7995–8005. DOI: 10.1021/ja954004t.
Müller, K.; Zimmermann, H.; Krieger, C.; Poupko, R.; Luz, Z. Reaction Pathways in Solid-State Processes. 1. Carbon-13 NMR and X-ray Crystallography of Fluorobullvalene. J. Am. Chem. Soc. 1996, 118 (34), 8006–8014. DOI: 10.1021/ja954005l.
Müller, K.; Zimmermann, H.; Krieger, C.; Poupko, R.; Luz, Z. Reaction Pathways in Solid-State Processes. 2. Carbon-13 NMR and X-ray Crystallography of Cyanobullvalene and Bullvalenecarboxylic acid. J. Am. Chem. Soc. 1996, 118 (34), 8015–8023. DOI: 10.1021/ja954006d.
Luger, P.; Roth, K. X-Ray, n.m.r., and theoretical studies of the structures of (ethylthio)bullvalene. J. Chem. Soc., Perkin Trans. 2 (1972-1999) 1989, 649–655. DOI: 10.1039/P29890000649.
Luz, Z.; Olivier, L.; Poupko, R.; Müller, K.; Krieger, C.; Zimmermann, H. Bond Shift Rearrangement of Chloro-, Bromo-, and Iodobullvalene in the Solid State and in Solution. A Carbon-13 and Proton NMR Study. J. Am. Chem. Soc. 1998, 120 (22), 5526–5538. DOI: 10.1021/ja9728029.

Let me begin by thanking Martin and long for supplying the key reference (J. Am. Chem. Soc. 1996, 118 (34), 7995–8005).  Importantly, long also provided the isomer ratio I was asking for; according to him, the ratio was estimated using the numbers at bottom of page 7997 (with the 0.002 added).
The free energy differences between the isomers of fluorobullvalene are reported in this paper as
$$\begin{align}
\Delta G(1,4) &= \pu{+3.41 kcal/mol} \\
\Delta G(3,1) &= \pu{-2.21 kcal/mol} \\
\Delta G(2,3) &= \pu{-0.46 kcal/mol} \\
\end{align}$$
Here is a diagram showing the 4 possible isomers of fluorobullvalene in equilibrium, as well as a simple example of the symmetry-allowed, 6-pi electron Cope rearrangement that occurs in bullvalene and interconverts the isomers:

In this answer, the equilibrium concentrations of the four isomers were determined to be as follows. These numbers differ somewhat from long's numbers - which were an approximation long made based on nmr peak intensities - but appear to match those shown in Fig. 4 in the paper at 300 K:
\begin{align}
[1]~ &=~~~ 0.23\% \\
[2]~ &=~ 20.27\% \\
[3]~ &=~~~ 9.37\% \\
[4]~ &=~ 70.13\% \\
\end{align}
The question left to be answered: why does the isomer distribution fall out like this, and, in particular, why such an avoidance of the cyclopropyl position by fluorine?
As we work towards an answer to this question we will discuss bonds that are $\ce{sp^{i}}$ hybridized.  “i” is referred to as the hybridization index.  The hybridization index is such that the ratio $1:i$ is the ratio of the s:p atomic orbitals mixed to form the hybrid orbital.  For example in an $\mathrm{sp^3}$ bond, $i=3$, and the ratio of s:p atomic orbital content in the bond is $1:3$.  Further, we can see from this that the percentage s character in a bond is given as
$$\text{s-character} = \frac{1}{1+i} \tag{1}$$
which is $1/(1+3) = 25\%$ in the $\mathrm{sp^3}$ case.
There are a number of ways to estimate the percent s-character in a bond.  One way is through use of the formula
$$^1\!J_\ce{C-H} = \pu{500 Hz} \cdot \text{s-character in the bond} \tag{2}$$
where $^1\!J_\ce{C-H}$ is the one-bond C–H coupling constant.
Note: $\ce{^12C}$ does not have a magnetic moment, therefore it cannot interact with a proton spin and split it. $\ce{^13C}$ is a spin-½ nucleus, it can interact with a proton spin and cause signal splitting in the carbon and proton nmr spectra.  $\ce{^13C}$ only occurs at 1.1% natural abundance, but one can still measure the resultant C–H splitting (e.g. the coupling constant).
Next, consider the polarization in a $\ce{C-F}$ bond. It will look something like this:

Because of the relative electronegativities of carbon and fluorine, electron density will shift away from carbon towards fluorine. To understand the effect of this electron shift, let's compare the hybridization in methane ($\ce{CH4}$) and methyl fluoride ($\ce{CH3-F}$).  In methane, the carbon orbital used to form the C–H bond is exactly $\mathrm{sp^3}$ hybridized – this is demanded by the symmetry of the molecule.  Using equation (1) we know that the percent s-character in the atomic orbital used by carbon to interact with and form a molecular orbital with hydrogen is 25% in this case.  From equation (2) we would predict that $^1\!J_\ce{C-H}$ in methane would be (500*0.25) Hz, or 125 Hz.  Indeed, the experimentally determined coupling constant for methane is measured as 125 Hz.
In methyl fluoride the experimentally observed $^1\!J_\ce{C-H}$ coupling constant is 149 Hz.  Using equation (2) we find that there is 149/500 = 0.30 or 30% s-character in the carbon orbital used to form the C–H bond.  Using equation 1, this is an $\mathrm{sp^{2.36}}$ bond.  There are 3 of these C–H bonds, so a total of 90% of the available carbon 2s orbital is used in forming them.  This leaves 10% of the carbon 2s orbital remaining and available for use in the C–F bond. Using equation 1 we find that i=9 and this carbon orbital is $\mathrm{sp^9}$ hybridized – very nearly pure p!
What a difference in bond hybridization between methane and methyl fluoride and it teaches an important lesson. Since the carbon 2s orbital is lower in energy than the carbon 2p orbital, the more s-character there is in a bond, the lower the energy of the electrons in that bond. If the electron density in an orbital is low (like in the carbon portion of the C–F bond in methyl fluoride), then nature doesn’t mix much s-character into that orbital and, instead, uses that s-character to stabilize orbitals that do have significant electron density.  Hence, the carbon orbitals used to form the C–F bonds in methyl fluoride only have 10% s-character because the electron density remaining around carbon in that bond is so low. Instead, the s-character is used in the 3 C–H bonds which do contain significant electron density around carbon.
Before we apply this type of analysis to fluorobullvalene, let’s make some predictions based on what we’ve learned so far.  Clearly, fluorine will preferentially bond with carbon atoms that have low s-character in the bonding orbital.  The $^1\!J_\ce{C-H}$ coupling constant in cyclopropane is 161 Hz.  Using equations (2) and (1), we find that the C–H bond in cyclopropane has 32% s-character and is $\mathrm{sp^{2.1}}$ hybridized. In bullvalene itself, carbon (4) in the tris-allylic position looks like a $\ce{sp^{3}}$ hybridized carbon whereas the olefinic carbons are probably $\ce{sp^{2}}$ (more or less).  So right off the bat, we would expect more fluorine at position 4, than at either of the olefinic positions or the cyclopropyl position.
The relevant $^1\!J_\ce{C-H}$ coupling constants in the parent molecule bullvalene are presented in the table below, along with the percent s-character derived from equation (2), the hybridization index $i$ from equation (1), the value of $\mathrm{sp}^i$, and the percentage of the isomer.
$$\begin{array}{|c|c|c|c|c|c|}\hline
\text{Carbon} & ^1\!J_\ce{C-H}\text{ / Hz}  & \text{% s-character} & i & \mathrm{sp}^i & \text{% isomer} \\ \hline
1 & 169.0 & 33.8 & 1.96 & \mathrm{sp}^{1.96} & 0.23 \\ \hline
2 & 160.8 & 32.2 & 2.11 & \mathrm{sp}^{2.11} & 20.27 \\ \hline
3 & 166.1 & 33.2 & 2.01 & \mathrm{sp}^{2.01} & 9.37 \\ \hline
4 & 133.9 & 26.8 & 2.73 & \mathrm{sp}^{2.73} & 70.13 \\ \hline
\end{array}$$
We immediately see that the amount of fluorine at a given position changes with the percentage s-character in the bond. Just as we expected, the higher the s-character, the lower the amount of that fluorinated isomer. In fact, if we measure the correlation coefficient between the percentage s-character in the bond and the percentage of the isomer with fluorine attached at that position, a correlation coefficient of –0.9982 is obtained (if we use long's suggested isomer distribution [0.18%, 5.99%, 1.66%, 92.17%] the correlation coefficient is –0.9875).  A value of 1 would be a perfect correlation, a value of –1 would be a perfect inverse correlation. In the case at hand, the percentage s-character in the bond and the percentage of the isomer with fluorine attached at that position demonstrate a near-perfect inverse correlation. An inverse correlation means that as the percentage s-character increases, the isomer with fluorine attached at this position decreases and the high value for the coefficient indicates that the movement of the two values is highly correlated.
Bullvalene provided an interesting test case where we could measure fluorine’s affinity to bond to different types of carbon. From this example we can see that fluorine strongly prefers to form bonds to carbon that utilize carbon atomic orbitals with low s-character.

I believe you have a slight mistake in your ratio calculation. I was now able to reproduce your values. I believe you forgot that the conformers 1, 2, 3 are triply degenerated with respect to 4.

@Martin Wouldn't ΔG already include the isomer degeneracy in its entropic term?  Also, my numbers seem to match the isomer percentages, near 300K, reported in the top graph in Fig 4. of Luz's paper much better than the numbers you report in your "Exptl." row (even after I normalize them to 100%).

I do not really understand anything about the procedures in the paper, certainly not how they arrived at their free energy values. It is a pity that they do not provide the actual distribution/ population numbers. From the figure I would arrive at 4:3:2:1 = [92:1:7:0]%, which would not fit your distribution either - How did you do that?

@Martin Not sure where on the graph you're reading those numbers, maybe on the far right.  In any case we want to read the isomer distribution around T=300K (27C).  The easiest point to find around that temperature is the upper (Centigrade) x-axis tic between 4.6C and 39.3C.  That tic identifies 21.1C.  Using that I read a 4:3:2:1 ratio of 71:8:18:0.22, not far off from the values I listed in my post.  In fact, judging by the slopes of the lines, if we moved a tad further to the left so we'd be closer to 27C, the isomer numbers would be even closer.

I will have to look at my notes again, but I remember finding completely different numbers - maybe if I am back at my office I can figure this out - maybe I just misunderstood the graph.

I had some free time and resources at hand and did some highly accurate quantum chemical calculations for this system. I took it as a chance to compute some benchmarks and play a little with density functional theory.
Originally I also wanted to compute transition barriers, but based on the previous findings I had to omit that, since I currently cannot afford ab initio calculations. However, I'd like to share these results.
For your reading convenience, here is the chart from longs post, in order to see the numeration ($\ce{X=F}$).

Electronic Energies $\Delta{}E^\circ$ are as usual in $\mathrm{kcal/mol}$. Thermodynamic data $\Delta{}E^\circ$ was estimated at the DF-BP86/def2-TZVP and DF-BP86/def2-SVP level of theory with $T=298.15~\mathrm{K}$ and $p=1~\mathrm{atm}$. Experimental data was taken from the cited paper. Product ratios were calculated from the electronic energies (except for BP86 where data was available) assuming that the contribution of thermodynamic corrections is neglectable, i.e. $\Delta{}E^\circ = \Delta{}G^\circ$. The level of degeneracy is for each conformer $g(\mathbf{1})=g(\mathbf{2})=g(\mathbf{3})=3$ and $g(\mathbf{4})=1$.
(I was somehow unable to reproduce the product ratio that ron calculated with his approach. I used How to obtain product ratio from energy differences via Boltzmann statistics)
\begin{array}{lrrrr}\hline
\text{method}                & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} \\\hline
\text{Exptl.}~\Delta G^\circ &  0.0   & -2.67  & -2.21  &  -3.41  \\
                       \%    &  0.4   & 38.0   & 17.5   &  44.2   \\\hline
\text{BP86/SVP}~\Delta E^\circ
                             &  0.0   & -3.3   & -3.5  &  -2.6   \\
\phantom{\text{BP86/SVP}}~\Delta G^\circ
                             &  0.0   & -3.0   & -3.0  &  -2.6   \\
                       \%    &  0.3   & 43.5   & 48.0  &  8.2    \\\hline
\text{BP86/TZVP}~\Delta E^\circ
                             &  0.0   & -3.2   & -3.0  &  -3.1   \\
\phantom{\text{BP86/TZVP}}~\Delta G^\circ
                             &  0.0   & -2.9   & -2.6  &  -3.0   \\
                       \%    &  0.4   & 48.6   & 30.6  &  20.5   \\\hline
\text{PBE0/TZVP}~\Delta E^\circ
                             &  0.0   & -3.0   & -2.8  &  -3.3   \\
                       \%    &  0.3   & 44.7   & 30.9  &  24.1   \\\hline
\text{M06//BP86/TZVP}~\Delta E^\circ
                             &  0.0   & -3.1   & -2.9  &  -3.7   \\
                       \%    &  0.2   & 38.5   & 27.2  &  34.1   \\\hline
\text{MP2//BP86/TZVP}~\Delta E^\circ
                             &  0.0   & -1.3   & -1.0  &  -3.9   \\
                       \%    &  0.4   &  3.7   &  2.1  &  93.8   \\\hline
\text{CCSD(T)//BP86/TZVP}~\Delta E^\circ
                             &  0.0   & -1.3   & -1.0  &  -3.9   \\
                       \%    &  0.4   &  3.7   &  2.1  &  93.8   \\\hline
\end{array}
It is astonishing that MO6 almost exactly reproduces the experimental values, while the ab initio calculations CCSD(T) and MP2 conclude that there is almost only the $\mathbf{4}$ conformer. This was quite unexpected, maybe frequency calculations have to be done to obtain more fitting results.

Your "Exptl." percentages don't seem to match the isomer distribution reported in Luz's paper (see the top part of Fig. 4 and look around 300K) even if I normalize them to 100%.

I fixed the numbers, there was a small typo, at least they now add up to 100% (with small rounding errors). I am not so sure about the degeneracy anymore - maybe my approach is wrong (this is why I asked the corresponding question in the first place). I can't really tell anything from that figure, I tried to obtain some numbers from this plot, but since it is logarithmic and very small, it is more guessing than anything else.

I have a pdf of Luz's paper and was able to blow the figure up to quite a large size.  I was able to interpolate the numbers with reasonable accuracy.

I did that too. From that I arrive at a distribution of 4:3:2:1 = 90:1:9:0. I now found numbers in the paper, that suggest 4:3:2:1 = 1.0:0.018:0.065:0.002 = 92:2:6:0. And this is nowhere near what either of us calculated. What I really do not understand is how they obtained the free energy values from their distribution. If I look at the CC and MP2 values I would obtain that distribution with good accuracy. I am really very puzzled here.


text:
Why are solids and liquids not included in the equilibrium constant? What about in a reaction rate calculation?
Take for instance the reaction
$$\ce{H2(g) + I2(s) <=> 2HI(g)}$$
The equilibrium constant would not include the solid $\ce{I2}$, but why is this? I have read that its concentration is a constant, but what does that mean? If you increase the amount of $\ce{I2}$ the reaction should shift towards the right so logically it does appear to play a part in the equilibrium.
For liquids, I can understand why you don't include it when you have something like water which (usually) does not play a part in the reaction, but why do you still not include when the liquid takes part in the reaction?
Also, do you include solids/liquids for reaction rates? Increasing the amount of a solid should increase the surface area and thus make for a faster reaction, so it seems like you should, but I am not too sure.

It very much depends on what definition of the equilibrium constant you are looking. The most common usage of the same has quite a variety of possible setups, see goldbook:

Equilibrium Constant
  Quantity characterizing the equilibrium of a chemical reaction and defined by an expression of the type $$K_x = \Pi_B x_B^{\nu_B},$$ where $\nu_B$ is the stoichiometric number of a reactant (negative) or product (positive) for the reaction and $x$ stands for a quantity which can be the equilibrium value either of pressure, fugacity, amount concentration, amount fraction, molality, relative activity or reciprocal absolute activity defining the pressure based, fugacity based, concentration based, amount fraction based, molality based, relative activity based or standard equilibrium constant (then denoted $K^\circ$ ), respectively.

The standard equilibrium constant is always unitless, as it is defined differently (goldbook)

Standard Equilibrium Constant $K$, $K^\circ$ (Synonym: thermodynamic equilibrium constant)
  Quantity defined by $$K^\circ = \exp\left\{-\frac{\Delta_rG^\circ}{RT}\right\}$$  where $\Delta_rG^\circ$  is the standard reaction Gibbs energy, $R$  the gas constant and $T$  the thermodynamic temperature. Some chemists prefer the name thermodynamic equilibrium constant and the symbol $K$.

It is worthwhile to note, that the first definition is always an approximation to the standard definition and in the standard definition all compounds regardless of their state are included in the equilibrium.
If you are looking at a reaction that takes place in gas phase, solid materials will play a constant role, since their partial pressure will solely depend on their vapour pressure and can therefore be regarded as constant. Therefore it can be seen as a part of the equilibrium constant. (The same applies to liquids while deriving for gas phase.) If you would increase the amount of solid in the system you would still not change it's concentration in gas phase.
For the reaction $$\ce{H2 (g) + I2 (s) <=> 2HI (g)}$$ you can form the standard equilibrium constant with activities/ fugacities
$$K^\circ = \frac{a^2(\ce{HI})}{a(\ce{H2})\cdot{}a(\ce{I2})},$$
with $$a=\frac{f}{p^\circ}.$$
The activity for a pure solid is normally defined as one ($a(\ce{I2})=1$) and therefore
$$K^\circ\approx K = \frac{a^2(\ce{HI})}{a(\ce{H2})}$$

For concentration dependent equilibrium constants, the following assumption is to be considered: $c(\ce{H2O})\approx55.6~\mathrm{mol/L}$ and is usually always much larger than that of any other component of the system in the range where the equilibrium approximation can be used. Also in most solution based reactions, the solute itself does not directly influence the reaction. Its concentration will therefore not change (significantly) and can be included in the equilibrium constant.

You are correct in assuming that the kinetics of a surface reaction will only depend of the actual area of the surface (and of course the reactants forming the products). But here again this area is most likely to be considered as constant and will result in a scalar for the reaction rate (in the area where an accurate description of the reaction is possible).
As of Greg's comment, He is of course right. In an equilibrium process the surface area will affect both directions of the reaction, hence it will not show up in the reaction rate at all. (The only effect it might have is, that the equilibrium state will be reached faster.)

So from what I understand, solids and liquids are considered to be constants because their "concentration" depends on their partial pressure which is independent of the amount present, right? And for the rate expression it is also ignored for the same reason and any effects due to surface area are included in the rate constant?

Yes you are correct. As long as the surface does not change, everything related to it is constant and can be included in the rate constant.

However, for the equilibrium expression if the solids and liquids are considered constants because their "concentration" is dependent on their partial pressures, wouldn't changing the temperature change the partial pressure and thus their "concentration"? So then why isn't their different equilibrium constants for different temperatures?

There is! Tabulated are mostly standard values at standard temperature and pressure. Have a look at the second definition and you will see the temperature dependency of $K$.

It is not true. The reaction is both direction will scale with the surface, therefore the it will not appear in the equilibrium constant.

@Greg I hope my edit fixes this.

Then what will  be the equilibrium constant of a reaction in which all the reactants and products are  say liquid will such a reaction have any equilibrium constant ?

@Freelancer As long as it is in equilibrium: Yes.

But the liquids are not included in equilibrium constant ... So as all reactants and products are liquids then  none of them will be included in equilibrium constant  so whose concentration will  we use while writing the equilibrium constant ?

@Freelancer If you have another question, please ask it by clicking the Ask Question button. As I stated: "in the standard definition all compounds regardless of their state are included in the equilibrium."

[OP] The equilibrium constant would not include the solid $\ce{I2}$, but why is this?

Let me explain this with a different example. If you have a saturated solution (e.g. lemonade with too much sugar in it) it is at equilibrium. If you add more sugar, the lemonade does not get sweeter. That tells you that the amount of solid does not matter (as long as there is some), so it also tells you it should not be a variable in the equilibrium constant.

[OP] I have read that its concentration is a constant, but what does that mean? If you increase the amount of $\ce{I2}$ the reaction should shift towards the right so logically it does appear to play a part in the equilibrium.

No, increasing the amount of a solid does not disturb an equilibrium.

[OP] Also, do you include solids/liquids for reaction rates? Increasing the amount of a solid should increase the surface area and thus make for a faster reaction, so it seems like you should, but I am not too sure.

Yes, more solid means a faster forward reaction. At the same time, it means a faster reverse reaction (more surface area for iodine to deposit on). For that reason, the equilibrium constant does not change. In the example above with the sugar, finely granulated sugar dissolves faster than coarsly granulated sugar, but the solubility (and the the equilibrium constant) is the same.

[OP] Why are solids and liquids not included in the equilibrium constant?

They are, in a way. Concentrations enter the equilibrium constant as relative to a standard concentration. For solutes, the standard concentration is $\pu{1 mol L-1}$. Speaking loosely, you take the concentration in $\pu{1 mol L-1}$ and drop the units. For pure substances in the liquid and solid state and for solvents, the standard concentration is $c^\bullet $, the concentration of the pure substance. For pure substances, this always gives a value of 1, so you might as well leave it out. Even for solvents, it will be very close to 1 for dilute solutions. For gases, on the other hand, the concentration depends on the pressure, and you have to agree on whether to use partial pressure (with a standard of 1 atm) or concentration (with a standard of $\pu{1 mol L-1}$) when defining the equilibrium constant.
Using concentrations is an approximation in any case. For more sophisticated treatments, you would use the activity, which is also defined as 1 for pure solids and liquids.

This is by far the best answer here and is wildly under appreciated.

Solids and liquids, as long as there is some available, are fully available to the reaction in a macroscopic amount that does not limit the reaction.
This is different from a solution of a certain molarity or a gas of a certain pressure. The sparseness of solutes or gases means that the availability is limited by the concentration or the pressure. For solids and liquids, the availability is considered constant as long as some macroscopic amount is present.

It's plain and simple. We use equilibrium expression to see whose concentration has what effect on the rates of reactions. If you have a pure solid, and you increase its amount you do change the rates of reactions. But a pure solid has no concentration because it is not dissolved in any solvent (that's why we call it pure solid). So, basically you just increase the amount of your pure solid, and that will have an effect on the concentration of some other reactant or product. So, the change in concentration of that reactant or product will be seen in equilibrium constant(because that reactant or product does have a concentration). And in the equilibrium constant we see the effect of change of concentration of any product or reactant, not amount. So, the change in amount of our pure solid will have an effect of change in concentration of some other reactant or product. And we will take into account the effect of increasing the amount of solid, by seeing the change in concentration of any other resulting reactant or product in our equilibrium or rate equation. Hopefully, this intuition helps!

"We use the equilibrium constant to see whose concentration has what effect on the rates of reactions." I'd be careful about tying the equilibrium constant to kinetics, it may confuse less educated readers.

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text:
How to identify an unknown compound with spectroscopic data?
I need some help identifying the following unknown compound. Any suggestions or hints regarding the possible compound would be highly appreciated.
So far I am thinking the broad peak on the IR is an alcohol that somehow mixed with possible $\ce{sp^3-CH}$ bonds. And the peak around $\pu{1600 cm-1}$ is I think a $\ce{sp-C-C}$ bond. The compound should have four distinct groups of protons. Beyond this basic analysis, I couldn't really come up with a compound.

I have updated (07/09/14) the question with a newer NMR (below) and an IR.

Some other additional info I worked out in the lab:

look: white powder like solid.
melting point: $\pu{116.3-119.0 ^\circ C}$
Solubility:

Water - insoluble
NaOH - insoluble
Sulfuric Acid - soluble

Classification Tests:

Water insoluble phenol test - positive; a dark red precipitate formed.
Chromic acid test - positive; green precipitate formed.
Bromine test for unsaturation - positive; bromine discoloration happened until about $\pu{2 ml}$ of bromine was added. White precipitate formed.

This is my first time doing these tests so please don't consider their results too strictly. My Instructor said the IR is a much safer way to determine functional groups.

The signal at 7.15 is chloroform, so you only have three signals belonging to your substance

Have you got a full display of the NMR spectrum? Specifically, I would like to see a water impurity in your solvent at 1.6ppm, and possibly any -OH peaks which would be likely to broadened significantly, but still visible in CDCl3. Is the baseline of the NMR spectrum flat, or does it contain any underlying exchange-broadened signals (as your IR would indicate)? Can you describe your process for collecting your IR data? I will collate my thoughts on the structure based on NMR, but the IR is somewhat contradictory. Are you sure your solvents were dry?

Have a look at typical IR bands (other source) and identify them.

 - 3424 phenolic $\ce{OH}$ stretch
 - 1640 aromatic $\ce{C=C}$ bending or $\ce{C=C}$ alkenyl stretch
 - 1264 aromatic $\ce{C-O}$ stretch
 -  747 aromatic $\ce{C-H}$ bending

Have a look at you spectrum, identify your residual solvent signal.

  7.15 is chloroform $\ce{CDCl3}$

Have a look at the typical $\ce{{}^{1}H-NMR}$ shifts table and identify the protons.

 These are aromatic protons

What is the summary of your first conclusions?

 It's an aromatic compound with an hydroxyl group.

How many protons?

 multiple of 5

Calculate the coupling constants via $$J=\Delta\delta_{\mathrm{ppm}}\cdot\nu_0$$ and identify the splitting pattern. (I can only assume it is a 500MHz spectrometer, hence $\nu_0=500~\mathrm{MHz}$.)
Example: Doublet at 7.55 ppm:
$$J=(7.5615-7.5461)\cdot500~\mathrm{MHz}=7.7~\mathrm{Hz}$$
Solutions:

 7.55 doublet, $2\ce{H}$, $J=7.7~\mathrm{Hz}$
 7.40 doublet of doublet (fake triplet), $2\ce{H}$, $J=7.6~\mathrm{Hz}$, $J=7.5~\mathrm{Hz}$
 The last one is tricky because there is one peak not picked
 It should be at 7.30 a doublet of doublet, $1\ce{H}$,  ${}^3J=7.7~\mathrm{Hz}$, $J=7.6~\mathrm{Hz}$

Disclaimer:
Please read long's answer and follow his thoughts. They make a lot of sense and hint that the compound we are looking at is not the one I am proposing. But due to my lack of current knowledge, I cannot make another suggestion.
Combine all you see and you will get a clue which substance it is.

 7.55 ortho
 7.40 meta
 7.30 para
 Phenol.

The IR spectrum looks rather strange, very few signals compared to a reference spectrum from the SDBS. I also thought it was phenol after a quick look, but the IR confused me.

@MadScientist I cannot open the link you posted, nor have I any reference elsewhere. But what other compound could it be? What signals are you missing?

This is a version of the image that should be accessible. I still think it is phenol, there isn't much else it could be. But the IR is weird, which confused me.

@MadScientist I totally agree, that looks rather weird compared. Maybe the spectrum given by the OP is just very thin or a different medium - it does not state how it was measured.

@MadScientist With regards to the IR, the unknown was a solid which I dissolved in dichloromethane and use NaCl plates. I washed the plates with methanol but made sure it was fully dry before using the IR machine.

Isn't chloroform 7.26 ppm?

@jerepierre As you do not see any internal standard in the spectrum, it could be off by a few ppm. Also the value for chloroform is not fixed. And since there is no signal at 7.26 it can only mean it is shifted - the display of a spectrum depends on the calibration.

By NMR, this compound is not phenol. The chemical shift order is incorrect. The ortho (H2) protons of phenol have the largest downfield shift, which is inconsistent with phenol. For phenol, the expected shifts from benzene would be approx. H2: -0.5 (6.7ppm), H3: -0.1 (7.2ppm) H4: -0.4 (6.9ppm). Admittedly, it is a confusing set of data, given the IR spectrum.

@could you make another suggestion then please. How did you derive these downshifts? I did not know any better. Did I make any mistake in my reasoning? (It is also always pretty complicated deriving a structure without at least a 13C.)

@long Hmm is the IR still contradictory ? I got a pure sample from my instructor and ran the IR. The sample was solid so I dissolved it in dicholoromethane and used NaCl plates...

@Martin - I have added some thoughts in an answer below, but am time limited this next week, and cannot provide fuller references. The main problem with the assignment of the NMR spectrum relates to the relative positioning of the H2 and H3 protons, a feature well established in mono-substituted benzene rings. I hope this is useful for you.

@long I really like your answer and your thoughts you presented. I do now understand that (and why) the order of the signals do not match with what I proposed. The available data is maybe not enough to propose an actual structure. However, for a pure educational problem, this is a bad setup. I very much hope, the OP will post the solution to this problem eventually.

@Martin I agree that this setup is not good. My main concerns about the IR data is that it is very easy to introduce unwanted -OH signals into the sample through the use of wet solvents and poor cell, and it is very hard to hide true exchangeable signals in NMR, especially in CDCl3. And NMR data is very reliable. So we are not getting all the data, and have to make guesses along the way. In any case, your answer provides a nice flowchart to follow as a general rule for solving these types of questions.  Well done.

@Martin I updated the question with a brand new NMR(below the old one-I think they are very similar) and new IR reading. Please consider the new spectroscopy. Thank You !

The data given is this question has changed sufficiently from the original post that it warrants a new answer to the latest (and much more reliable) data, although should be taken in conjunction with Martin's response, and Long's original response. Let's first look at the IR data:
IR Spectrum:
A much better spectrum, and, compared to the original spectrum, serves as a very good example of how easy it is to introduce unwanted water into a sample. Peaks around 3020, 763 and 692 indicate an aromatic. The 763 and 692 are indicative of a mono-substituted benzene ring. There are also some clear aromatic overtones between 1950-1750.
NMR Spectrum:
Still only a partial NMR spectrum, and one that we can only assume is correctly calibrated. Linewidths are better, and significantly, we see a small coupling from the peak at 7.15ppm.
The splitting pattern and peak ratio observed is indicative of a monosubstituted benzene ring (see above); 7.30 (1H, dd, H4), 7.41(2H, , H3) and 7.56(2H, d, H2). Long-range (small couplings) are ignored. We also have a peak at 7.15 (1H, d, H5). Typical coupling in these systems is 6.5-8.5Hz for ortho coupling, 1-3 for meta, and <1 for para. Shifted from benzene (7.34ppm), the substituents come at H2 (+0.22), H3 (+0.07), H4 (-0.04).
The peak at 7.15 could be one of few things. A bis-halide is a possibility, but more likely is an alkene. No real evidence in the IR supports an alkene (although my IR knowledge is limited). However, a symmetrical alkene may be a possibility, as the main -C=C- absorbance would be IR inactive, although Raman active. This gives us two possibilities that come to immediate mind:
cis-stilbene (cis-1,2-Diphenylethylene) m.p. 5-6oC

Additive prediction for the aromatics and alkene proton can be done using many basic NMR texts. I'm using Pretsch, Buhlmann and Badertscher for this case (as it's the only one I have with me at the moment):

H2 : 7.34+0    = 7.33
H3 : 7.34-0.07 = 7.27
H4 : 7.34-0.15 = 7.19
H5 : 5.25+1.38(Argem)-0.07(Artrans) = 6.56

trans-stilbene (trans-1,2-Diphenylethylene) m.p. 123-125oC

H2 : 7.34+0.16 = 7.50
H3 : 7.34+0    = 7.34
H4 : 7.34-0.15 = 7.19
H5 : 5.25+1.38(Argem)+0.36(Arcis) = 6.99

As this is clearly a problem session for you to make a final determination of structure, I will leave it for you to make that judgement. however, as a final point, this whole exercise, I hope, serves to demonstrate the importance of preparing good quality samples in order to collect high quality and reliable data. I hope you can see the difference between your original data and final data. This is a message that I continually preach to all those researchers I work with, both at an undergraduate level and postgraduate.

Thanks a lot. I think I am going to go with trans-1,2-Diphenylethylene whose melting point is fairly close to what I got in lab. I will be attempting to a Bromine derivative test tomorrow after which I should be pretty certain if this is it-but it makes sense. Also I asked a very similar question(http://chemistry.stackexchange.com/questions/14061/help-with-identifying-an-unknown-substance?lq=1). I am fairly certain that the compound is benzoic acid but I would be very appreciative if you can confirm my suspicions. Thanks a lot for your help ! –

This answer aims to build on the general approach that Martin has provided, which overall makes a reasonable summation based on the data provided. IR is not really my specialty, but there is some more information that we can get out of the NMR data which should be helpful, and more reliable (in my opinion) than the IR data. Unfortunately, I am away away from my office for the next week, so cannot provide immediate references to support some statements here, so you'll have to take some things on face value. Let's begin with an overall summary of what data we have:

A partial 1H NMR spectrum, with only some of the peaks integrated. This is an expanded region of what we can assume to be a 500MHz (based on the export path). The linewidths are broad, and there is no clear source to allow confirmation of correct calibration. We therefore need to make two assessments:

The calibration is incorrect, and the peak at 7.15, which has no integration, is in fact the residual CHCl3, and all chemical shifts need to adjust downfield (0.09-0.11 depending on what value for CHCl3 in CDCl3 you use; I use 7.26. For simplicity, let's adjust the chemical shifts downfield by +0.1ppm
The calibration is correct, in which case the peak at 7.15 cannot be discounted, and should therefore have its integral determined.
A full display NMR spectrum would be very useful here to look for underlying exchange broadened proton signals. This might occur anywhere from about 2-15ppm, and may be very broad such that they appear as a hump in the baseline, but even in CDCl3, we should see them, and

An IR spectrum which looks to have been run at pretty low concentration. I did not see your original IR spectrum, and wonder why you needed to redo it. What would be nice to know is whether the ratio of intensities for your absorbance peaks are the same for both IR data sets; particularly did the ratio of the broad stretch at 3422 change with respect to absorbances at 3019, 763 and 692?
Your sample is a solid, as you mention in one of your comments. This would be a useful peice of information to have from the start. It is soluble in dichloromethane.

So let's now start with collating information from the data provided.
IR Spectrum.
Clearly, the significant signal is the broad peak at 3422, and this is textbook-indicative of an O-H stretch. It's probably a little too high to consider a N-H group of any sort. So, it could be an alcohol or an acid, but we have no C=O peak, so it leaves us with an -OH group.
We have absorbances at 3019, 763 and 692; all indicative of an aromatic. More specifically, 763 and 692 are indicative of a mono-substituted benzene ring.
As I say though, IR is not really my thing, and that's about all I can get from this spectrum. My biggest concern is the reliability of the OH peak.
NMR Spectrum

The splitting pattern and peak ratio observed is indicative of a monosubstituted benzene ring (see above); 7.30 (1H, dd, H4), 7.39(2H, dd, H3) and 7.55(2H, d, H2). Typical coupling in these systems is 6.5-8.5Hz for ortho coupling, 1-3 for meta, and <1 for para. Looking at the H2 signal at 7.55, we can use our knowledge of coupling constants to determine the frequency of the spectrometer:
7.5615-7.5461 = 0.0154 x Spect.Freq ~ 7.5Hz => 487MHz, so close enough to 500MHz, and confirms our suspicions that it is a 500MHz, as the export path suggests.
So, we can calculate an accurate ortho coupling for H2-H3 to be:
7.5615-7.5461 = 0.0154 x 500 = 7.7Hz.
Now, mono-substituted benzene rings have been extensively studied and are very well understood; chemical shift data has been widely tabulated, and forms the basis for many chemical shift prediction algorithms. Starting with the benzene chemical shift (7.34ppm) as a basis, it is possible to use the shifts of each group to infer some information about the type of substituent.  The first thing to look for with this type of system is the order of H2 versus H3 (versus naked benzene). An electron-donating group increases shielding, and the ortho proton (H2) is typically found upfield of the meta proton (H3). Electron withdrawing groups decrease shielding, and H2 typically experiences a downfield shift from benzene, and usually resonates downfield from the meta (H3) proton. For the system you have, H2 is downfield of H3, and this is indicative of an electron-withdrawing group.  This is a very strong argument against this system being phenol. Phenol has its H2 protons upfield of H3. This is also what is so confusing about the IR spectrum you have.
So, let's now consider the possible structure for this unknown compound you have. Let's make the assumption that, as a homework/tutorial problem, this is going to be a fairly simple molecule, with a pretty common substituent. Remember we have two scenarios to consider for our NMR.
Scenario 1 (corrected for CHCl3 at 7.26ppm): the substituents come at H2 (+0.21), H3 (+0.05), H4 (-0.04). Looking at Pretsch, Buhlmann and Badertscher, this matches incredibly well for the substituent being a phenyl group [H2 (+0.22), H3 (+0.06), H4 (-0.04)]. This would give the structure biphenyl, a white solid, which has a reported H2-H3 coupling of 7.71Hz. Very strong evidence by NMR, but is not supported by -OH stretch in IR data, although all other IR data is in agreement.
Scenario 2 (spectrum already correctly calibrated): If we assume that the spectrum is correctly calibrated, then the CHCl3 residual peak comes under the H4 signal - probably could be the sharp peak which is the second peak from the right in this group. This means that the peak at 7.15 needs to be considered. By eye, its integral is roughly 1. A singlet of chemical shift of 7.15 is typical of a bis-halide, and so we could consider α,α-dichlorotoluene or α,α-dibromotoluene. Both are sufficiently electron withdrawing to give H2 downfield of H3, and  However, the former is definitely a liquid at room temp, and I suspect the latter is also. Therefore, not strong candidates.
I would like to have seen the original IR spectrum, and the full NMR spectrum to have confidence in any prediction. However, if I were just shown the NMR data, I would have confidence in predicting the structure as biphenyl. I hope you can provide the real solution to this eventually.

What I miss most in the NMR data is a signal for the OH proton that should be there considering the IR data. I usually like solving puzzles like that, but here again it gives me the creeps. I have the strong feeling, that there is missing something important. Anyway, thank you very much for having a look and sharing your thoughts - very helpful indeed.

I have submitted for new NMR and will obtain another IR for the same compound again. Hopefully, I can update the question tomorrow with the new information.

@long I updated the question with a brand new NMR(below the old one-I think they are very similar) and new IR reading which doesn't have the OH. Please consider the new spectroscopy. Thank You so much for all the help you have given already !

@PeprikaDesilva - Aaah, much better data sets. Remiss of me when I looked at the NMR before, but most likely trans-stilbene. I will put together another answer when I have a closer look shortly.


text:
How do I find out what benzene containing compound this is with IR and NMR?
I have the following IR and NMR of an unknown substance. Can someone tell me what is a possible structure of the compound or point me in the correct direction. So far I am thinking of a benzene ring based compound with an OH group. Something like 2-phenylethanol but not sure.

Some properties:

melting point: $\pu{116 ^\circ C}$.
Composition: white powdery solid.
Soluble in $\ce{NaOH}$ and $\ce{NaHCO3}$.

Based on the IR, the broad peak at $\pu{3424 cm-1}$ represents an alcohol while the 4 humps in the region $\pu{2100-2600 cm-1}$ is indicative of an aromatic ring (most likely a benzene ring). The medium peak around $\pu{1660 cm-1}$ is most likely a carbon-carbon double bond. NMR is where I get stuck; I know there should be 6 distinct types of protons and since most of the peaks are around $\pu{7-8 ppm}$, it is most likely the benzene ring protons, but I am not sure how to calculate J values and splitting as it was done in

Subtract the signals of the individual peaks of your (doub|trip|...|multi)plets and multiply them by the MHz of your spectrometer. You will then have the coupling constants, in other words $J=\Delta\delta_{\mathrm{ppm}}\cdot\nu_0$.

@Martin Unfortunately I am not sure of the frequency used at my university-my instructor got me this graph for my unknown sample. Is it possible to use this NMR without the JValues ? Thanks for any help you can offer !

Since it says bruker500, I think it is fair to assume that it operates at 500 MHz. The coupling constants are necessary to get which hydrogens are next to each other.

This is another example of how important it is to work with good quality data, which can only come from preparing good quality samples. Sample preparation comes from good technique and lots of practice. Without it, you are in a very difficult starting position. Enough of my soapbox rant.
Let's start with IR (not my strong point)
The first and second IR spectra are so different that it is hard to be confident about which one is correct. Are they collected in different media (one a solution and one a mull perhaps)? The first looks very low concentration

~3420 is clearly an -OH stretch. Probably too high to consider a N-H group of any sort. So, it could be an alcohol or an acid.
1682 a C=O group. Probably in the acid region, likely aromatic acid, conjugated or intramolecular H-bonded. Possibly amide, but not convincing based on other peaks.
There are lots of other peaks, most of which could mean many things, but I do see 749 and 708 which support the NMR notion of a monosubstituted benzene. I usually think that IR is only useful to corroborate peaks of a known structure, but then I'm an NMR advocate.

NMR
Again, the first and second spectra are very different; the obvious concern is that they are of different batches of sample - were the first and second analyses taken from the same sample, or from a different batch. In any case, let us look at the second, which is only a partial spectrum, and omits some possibly very significant information, especially in the region above 10ppm, where I would like to see a broad acid peak.

The aromatic protons shown exhibit a classic mono-substituted benzene ring pattern, labelled above. There is a strong downfield shift of the ortho (H2) protons, consistent with an electron withdrawing group substitution (such as -(C=O)R). Ortho coupling for the H2 proton is difficult to determine accurately as there are 3 peaks picked. This may be a digitizer point average, but considering the average of each peak, and the coupling from the first spectrum, we get a ortho coupling of 7.2-7.9. Not overly useful, but less than 8, so we can exclude ethers and halides. Couplings are calculated by:
(8.1651-8.1507) x 500 = 7.2
Using additive analysis for monosubstituted benzene rings, we can determine the following shifts from benzene:

H2 (8.162) 7.34 + 0.82
H3 (7.514) 7.34 + 0.17
H4 (7.651) 7.34 + 0.31

Looking at the published tabulated data for substituent effects on monosubstituted benzene rings (I'm using Pretsch, Buhlmann and Badertscher, but lots of other sources are available), we find one possible candidate that matches closely:

Benzoic Acid H2(+0.79), H3(+0.14), H4(0.28)

The caveat that I would add to this is that the data presented is not the best quality, and the solution presented is consistent with the data only (not definitive). The key concern is the missing -OH peak in the NMR spectrum. Unless this sample was run in D2O, or washed with D2O, I would expect to see an acid peak around about 12 or so. Depending on the amount of water in the sample, this may be shifted somewhat and broadened, but should still be visible, even if a broadened hump in the baseline. Additional NMR experiments could be useful to confirm the structure.

I usually do not clutter the comments section with unnecessary affirmation, but when I do, I say things like: "This is an awesome post, you're awesome, dude!"

@Martin Ha! Thanks, but don't take anything away from your own efforts, which provide a very strong stepwise approach to solving these types of problems. Unfortunately, the data sets provided are not great, and in real life we would use a range of other useful techniques to determine structure including 2D NMR (which I think is far more useful than IR for determining unknowns)

Agreed on every term. When I was doing my organic chemistry studies we were provided several spectra, including at least a 13C, sometimes COSY, NOESY and HSQC or HMQC. They provide much more insight into the chemical bonding and I believe they are absolutely necessary for modern chemistry. Also CHN and mass spectroscopy should be done for most compound. Doubt free characterisation of a compound is an art! Thumbs up for everyone who can do it right.

@long Thanks a lot ! I tried to obtain a NMR with the more down field view but unfortunately I couldn't access the NMR machine. However, I measured the melting point to be in the range 118.6-122.4 Celsius which is close enough to the Benzoic acid mp which is 122. So this is most like it. Thanks a lot !

I tried to solve it, but I horribly failed. But as I put some effort in it, I still share this with you. Maybe you can work it out eventually.
I have to admit that this spectrum is bad to work with, it does not contain an internal standard and the solvent signal is not marked. It is possible, that it is integrated. If this is an educational exercise, your supervisor clearly did a very bad, a horrible job.
The signals are way too small to see a splitting pattern, it is very hard to even put a number to the signals. The aromatic area should at least have been enlarged. If you have another possibility, you do never ever scale the integrals to an aromatic proton, especially not the one that mixes with other signals.
The following you did already: Have a look at typical IR bands (other source) and identify them.

3424 phenolic $\ce{OH}$ stretch
3054 aromatic $\ce{C-H}$ stretch
around 1600 aromatic $\ce{C=C}$ bending or $\ce{C=C}$ alkenyl stretch
1265 aromatic $\ce{C-O}$ stretch  could be ether alkoxy stretch
737 and 704 aromatic $\ce{C-H}$ bending, hints towards meta substituted phenyl

Now you could go ahead and identify the residual solvent signal.

7.14 should be chloroform $\ce{CDCl3}$
7.28 is also a good candidate

It can also be none of them.
If you have a look at the typical $\ce{{}^{1}H-NMR}$ shifts table you can identify some of the protons.

7-8.5ppm These are aromatic protons
2.20ppm allylic/benzylic proton, could also still be $\ce{sp^3-C}$, it is very low field. As it is a singlet it has no neighbour carbons with a hydrogen, but it integrates to only one, or maybe two - it is hard to tell.

From the IR you identified a phenolic $\ce{OH}$ and = there should be a signal around 5ppm, which is not there. So I am very puzzled about this.
Now the splitting pattern is completely random to guess.
You can calculate the coupling constants via $$J=\Delta\delta_{\mathrm{ppm}}\cdot\nu_0$$ and identify the splitting pattern. (I can only assume it is a 500MHz spectrometer, as it says bruker500, hence $\nu_0=500~\mathrm{MHz}$.)
Example: Doublet at 8.15 ppm:
$$J=(8.1560-8.1402)\cdot500~\mathrm{MHz}=7.9~\mathrm{Hz}$$

8.15 doublet, $2\ce{H}$, $J=7.9~\mathrm{Hz}$
7.65 multiplet, $1\ce{H}$, $J=1.4~\mathrm{Hz}$, $J=5.8~\mathrm{Hz}$, $J=6.5~\mathrm{Hz}$
7.52 multiplet, $3\ce{H}$, $J=8.3~\mathrm{Hz}$, $J=7.0~\mathrm{Hz}$, $J=7.4~\mathrm{Hz}$, $J=7.8~\mathrm{Hz}$
7.39 doublet of doublet or triplet, $1\ce{H}$, $J=7.5~\mathrm{Hz}$, $J=7.4~\mathrm{Hz}$
7.28 singlet, $1\ce{H}$ - or solvent
7.14 singlet, $1\ce{H}$ - or solvent
2.20 singlet, $1\ce{H}$

I am very sorry I could not be of any more help. A $\ce{{}^{13}C-NMR}$ would certainly have been helpful at least.

I found this summary and it might be helpful for future tasks: http://www.csupomona.edu/~psbeauchamp/pdf/424_spectra_tables.pdf


text:
Human Body and RNA/DNA
After research I found that enzymes are responsible for metabolism, specifically catabolism and anabolism. This in turn creates the necessary chemicals in your body. The only question left is: what are enzymes? I asked in Biology and the answer led me to enzymes being created by ribosomes. Ribosomes by RNA. RNA by the four chemicals, adenine, guanine, cytosine, and uracil. Okay so, my question is: what chemical elements are those four chemicals comprised of?

That is very broad, could you narrow it down please or break it up into multiple questions. Or try asking at biology.se first and then come back here for the chemistry bits...

If this is still confusing/broad please let me know. Thank you.

@Sciiiiience It is indeed too broad, an answer could cover an entire book.  See something like http://en.wikipedia.org/wiki/Metabolism to get started and try to narrow it down, please.  It's a fascinating area, but it's going to require some more specific inquiries.

@jonsca Okay thank you. Feel free to close this question or whatever it is happens.

@Sciiiiience It's a great topic area, so I'm looking forward to your edits, then it can certainly be reopened thereafter.

There are hundreds of enzymes; each one was created to do a specific task (some have a lot of leeway, others are quite narrow). Most proteins are encoded in the DNA. mRNA is created from the DNA (by an enzyme) and moves to the ribosomes where the encoded enzymes are made. [Here][1] is a chart of metabolic pathways in cells.    [1]: http://www.sigmaaldrich.com/catalog/product/sigma/m3782?lang=en&region=US

A quick search would tell you what the 4 nucleobases for RNA are made of:
Uracil: 4 Carbon 4 Hydrogen 2 Nitrogen 2 Oxygen

Adenine: 5 Carbon 5 Hydrogen 5 Nitrogen

Guanine: 5 Carbon 5 Hydrogen 5 Nitrogen 1 Oxygen

Cytosine: 4 Carbon 5 Hydrogen 3 Nitrogen 1 Oxygen