text:
How can one calculate the pH of a solution?
So, I have 1 litre of acetate buffer, and 0.1 mol of oxonium ions/$\ce{H3O+}$ is added. The task is to calculate the pH of this solution.
What I was thinking:
$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{c(\ce{Ac-})}{c(\ce{HAc})} = \mathrm{p}K_\mathrm{a} + \log\frac{n(\ce{Ac-})}{n(\ce{HAc})}$$
$$\mathrm{pH} = 4.74 + \log\frac{n(\ce{Ac-}) - 0.1\ \mathrm{mol}}{n(\ce{HAc}) + 0.1\ \mathrm{mol}}$$
But neither $c(\ce{Ac-})$ nor $c(\ce{HAc})$ (which should, in theorey, be equal) is given. The only thing I know (or I think I know) is this: $$\frac{c(\ce{Ac-})}{c(\ce{HAc})} = \frac{1}{1} = 1$$
So, is it possible to calculate the pH from that? Am I missing something?

A buffer is most effective when conjugate acid and base are in a 1:1 ratio, but that is not always the case (particularly after we start adding acid to it.)  Do you know the concentration of buffering compounds prior to the addition of acid?  If not, the question can't be answered - the pH after adding the acid could vary widely depending on the buffering capacity of the solution.

I'm tring to write the methodology
If anything is wrong please point it out to me i'll be happy to correct!
To find $\small\ce{pH} $ of a solution you can:

Find the conjugated base of your acid  (for example $\small\ce{HCOOH}$ becomes $\small\ce{HCOO-}$ (correct me if i'm wrong here)
Write the equation of the solution
Now you want to prepare an aqueous solution.
Find the concentration of $\small\ce{[CH_3COOH]}$
you use the formula to determine the concentration in solution of acid
$$
   \ce{[CH_3COOH]\ =\ C\ -\ [CH3COO- ]}
   $$
($\ce{C}$ is the concentration of brought acid in $\ce{mol/l}$)
Then you write that:
$$\qquad\ce{[H+ ]\ =\ [CH3COO- ]}$$
as both dissociate from the same acid as
$$\ce{CH3COOH\ ->\ CH3COO- \ +\ H+}$$
the $\small\ce{pH}$ concentration is given by:
$$\ce{pH\ =\ -\log(\small[H_3O+ ])}$$

$pH$ here you are! (without any unit)

There are two things you need to know that are missing
1. The strength of the buffer (i.e. how many moles of Ac are in a litre of it. This includes undissociated and dissociated forms of accetate.
2. What is the pH of the buffer i.e. what does it measure before any NaOH is added. If the pH of the buffer is equal to the $pK_a$ of the acid then indeed
$[\ce{HAc}] = [\ce{Ac^-}]$ ([·] symbolizes the molar concentration of ·). In order to solve the general problem we need to be able to calculate the fraction of the total Ac that is dissociated and the fraction that isn't.
The fraction that is dissociated comes right out of the Henderson - Hasselbalch equation and is $$f_1 = 1/(1 + r_1)$$ where $$r_1 = 10^{(pH - pK_1)}= [\ce{Ac^-}]/[\ce{HAc}]$$ as is clear from inspection of the Henderson - Hasselbalch equation. The subscript 1 indicates that $r_1$ is the ratio of the number of acid ions that have lost 1 proton to the number that have lost $ 1 - 1 = 0$. In $f_1$ the subscript is indicative that $f_1$ is the fraction of the total Ac molecules that has become singly charged by loss of a single proton. When dealing with monoprotic acetic acid the subscripts aren't that important as there is only one proton to loose. But if the acid is polyprotic we have unionized, once ionized, twice ionized etc. ions to consider. The fractions of those ions are $f_0$, $f_1$, $f_2$... and we have other ratios as well
$$r_j = 10^{(pH - pK_j)} = [H_{n-j-1}Ac^{−j}]/[H_{n-j}Ac^{-(j-1}]$$ Here "Ac" stands for the acid anion and n is the number of protons it can yield when fully dissociated. For acetic acid. $\ce{CH_3COOH}$, and Ac is $\ce{CH_3COO}$, n = 1 and j only has values of 0 or 1. With phosphoric acid, $\ce{H_3(PO_4)}$, Ac is $\ce{(PO_4)}$, n = 3 and j = 0,1,2 or 3.
So suppose now we have a polyprotic acid with values for $r_1$, $r_2$, $r_3$... and that there are x moles of the acid in a solution. Then there would be $xr_1$ moles of the singly deprotonated species, $xr_1r_2$ moles of the doubly deprotonated, $xr_1r_2r_3$ moles of the triply deprotonated and so on.Then the total number of moles of Ac would be the sum of the number of moles of each$$C_{Ac} = x + xr_1 +xr_1r_2 +xr_1r_2r_3...$$ The fraction of the total that is undissociated is $$f_0 = x/(x + xr_1 +xr_1r_2 +xr_1r_2r_3...) = 1/(1 + r_1 +r_1r_2 + r_1r_2r_3...)$$ The fraction that is singly dissociated is $r_1$ times this $$f_1 = r_1f_0$$ and the fraction that is doubly dissociated is $r_2$ times that $$f_2 = r_2f_1$$ and, in general $$f_j = r_jf_{(j-1)}$$
At this point let's remember that $f_j$ is a function of the solution pH and all the pK's of the acid in question and that it is the fraction of the anions of that acid that carry charge -j. Thus we can write an expression for the total charge on all species of Ac at a given pH. This is $$Q_{Ac} = -C_{Ac}(0f_0 + 1f_1 + 2f_2 + ...)$$
Before going on to show you how Q solves buffering problems let me stop to suggest that the simplest way to work with it is to make an Excel (or other) spread sheet. Designate a column for pKs and a cell into which the pH goes. For example, put pH into cell A1 and start the pKs list in A2, Then in B2 put =10^($A$1 - A2).  Using $A$1 lets you copy and paste B2 into as many cells as you have pKs. The B column now contains the r corresponding to the pK in the cell to the left of it. Now enter the formula for $f_0$ in a cell and make another column with the $f_j = r_jf_{(j-1)}$ formula in it. So how many pK's. I say make the spread sheet for 5 or 6. Why? Well lets go back to $\ce{CH_3COOH}$ for a minute. It doesn't have 1 proton to give, as we have been assuming. It actually has 4. Are the other three ever coming off? Not with any base in my lab but we can model those other protons simply by assigning pKs that are so high (say 50) that the f values for anything other than $f_0$ or $f_1$ are 0. The point being that if you are going to go to the trouble to make the spreadsheet you might as well make it big enough to handle any acid you may ever encounter as it easily handles anything up to its maximum size using this trick.
Now how to use $Q(pH)$. If there are a total of $C_{Ac}$ moles of Ac in a solution the negative charge on them at $pH_0$ is $C_{Ac}Q(pH_0)$. At $pH_1$ it is $C_{Ac}Q(pH_1)$. Thus to move the solution from $pH_0$ to $pH_1$ you must supply or remove charge of $$\Delta Q_{Ac}(pH_0\ce{->}pH_1) = C_{Ac}Q(pH_1) - C_{Ac}Q(pH_0)$$ by adding or absorbing protons. If $ pH_0  > pH_1$ then $Q(pH_1) > Q(pH_0)$ (less negative) and so the difference will be positive indicating that protons (acid) will need to be added to effect this pH shift.
The acid species in the solution are not the only thing that emits or absorbs protons when pH changes. The solvent does too.
$$\Delta Q_{W}(pH_0\ce{->}pH_1) = 10^{-pH_1} - 10^{-pH_0} + (10^{(pH_0 - pK_w)} - 10^{(pH_1 - pK_w)})$$ represents the number of protons that must be supplied (or absorbed) to change the pH of water from $pH_0$ to $pH_1$.
Now let's use this to solve the original question. We'll assume that someone added 0.2 M acetic acid to a flask containing some water, added NaOH solution until the pH was 4.76, added water to near 1 L, trimmed with NaOH and HCl and then made up to the 1L mark. At this point we have a liter of 4.76 buffer with $C_{Ac}$ = 0.2 mol ( 0.1 mol of Ac and 0.1 mol of $Ac^{-1}$. This implies that $f_0 = f_1 = 0.5$ Also
$$\Delta Q_{Ac}(pH_0\ce{->}pH_1) = 0.02(Q(pH_1) - Q(4.76))$$
$$\Delta Q_{W}(pH_0\ce{->}pH_1) = 10^{-pH_1} - 10^{-4.76} + (10^{(4.76 - pK_w)} - 10^{(pH_1 - pK_w)})$$
We know that if by adding 0.1 M protons to the liter we cause it to settle at $pH_1$ then $$\Delta Q_{Ac}(4.76\ce{->}pH_1) + \Delta Q_{W}(4.76\ce{->}pH_1) = 0.1$$ which simply says that the protons we added have to be equal to the protons absorbed. Or, put another way, the net protons transferred has to be 0
$$\Delta Q_{Ac}(4.76\ce{->}pH_1) + \Delta Q_{W}(4.76\ce{->}pH_1) - 0.1 = 0$$ The job now becomes one of finding $pH_1$ which satisfies this equation. That is actually quite easy to do. If you enter the formulae given here into a spreadsheet and set it up to compute the left hand side of this last equation you can then try various values of $pH_1$ until you find one that gives a result close to 0. Or you can let Excel's solver do it for you. It didn't take me long to figure out that adding 0.1 mol protons to 1 L of an 0.2 M acetate buffer at 4.76 would lower its pH to 2.74.
Now this post has gone on long enough so I won't have space to tell you about extension of this technique to much more complicated systems. You can use it to find the pH of complicated mixes of weak and strong acids and bases or any materials which have buffering capacity. I ginned it up to predict the pH of brewer's mash which is a mixture of weak acid (bicarbonate) and several malts (each of which has its own buffering properties).


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Is formaldehyde a carbohydrate?
Formaldehyde has the formula $\ce{CH2O}$, and the ratio of atoms in a simple carb is $\ce{1C:2H:1O}$. This fits the formula of carbohydrates. When I researched this, I found some sources saying that formaldehyde is the simplest carb, but other sources saying it is glycolaldehyde ($\ce{C2H4O2}$). Is formaldehyde a carbohydrate?

According to IUPAC definition:

‘carbohydrate’ includes monosaccharides, oligosaccharides and polysaccharides as well as substances derived from monosaccharides by reduction of the carbonyl group (alditols), by oxidation of one or more terminal groups to carboxylic acids, or by replacement of one or more hydroxy group(s) by a hydrogen atom, an amino group, a thiol group or similar heteroatomic groups. It also includes derivatives of these compounds [...]

But according to Wikibooks

the term is generally understood in the biochemistry sense, which excludes compounds with only one or two carbons. Natural saccharides are generally built of simple carbohydrates called monosaccharides with general formula $\ce{(CH2O)_n}$ where $n$ is three or more [...]

Because formaldehyde and glycolaldehyde (not a true sugar) has one and two carbon respectively in its structures we can exclude both as a carbohydrate.

@According of (CH2O)n formula, Di and higher saccharides do not fit, as e.g. sucrose C12H22O11=C12(H2O)11.

The term carbohydrate literally means a "hydrate of carbon". With respect to this definition, anything with the general formula of $\ce{C_x(H2O)_x}$ could be considered a "hydrate of carbon." Examples would include molecules like formaldehyde, acetic acid and, yes, glucose.
However, when most everyone uses the term "carbohydrate", they are talking about the biomolecules (like glucose, fructose, etc). In this much more common definition, a carbohydrate should be defined as a polyhydroxy aldehyde or ketone. Or in other words, the molecule must have either the ketone or aldehyde functional group AND multiple (more than 1) hydroxyl groups. So now molecules like formaldehyde and acetic acid would NOT be considered carbohydrates, but molecules like glucose, fructose, allose, mannose, glyceraldyde, dihydroxyacetone, etc would be considered carbohydrate.

source: Stack Exchange

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text:
Are pesticides used on fruit easily washed off?
I was hoping there'd be someone who studies pesticides here.

Are they water soluble?
Do they get absorbed in to the fruit?
How well would you have to wash the fruit to fully remove the residue?

Pesticides never were my main field of research, so you might want to take the following with a grain of salt:
Most pesticides i'm aware of are insoluble in water, but you might want to cross-check that for particular substances in the huge Pesticide database of the EC.
This insolubility seems reasonable, otherwise the substances would get washed off after the next rain.
Concerning your second question, the answer unfortunately is yes. To my knowledge, different pesticides are typically applied at the various state of plant growth. In order to suppress weeds, one might spray fields in an early, pre-emergent state. Later, fungicides or other substances might be applied during blossom. In order to protect the bees, it is however not recommended to apply insecticides during that time. Reasonable and minimized use of pesticides will keep the load in the fruit close to the limits of detection or beyond. Depending on the origin of the fruit, this might however not be the norm. In these cases, high concentrations of pesticides, even those banned in your home country, are detected both in the fruit and its products.
This leaves us with the question what we can do as consumers. If the fruit is highly contaminated, we can't do anything. Therefore, it might be reasonable to avoid products from countries infamous for breaking environment protection rules and go "organic" instead. Honestly, i never peel apples, but i wipe the surface with a piece of cloth. But i avoid industrial sorts, buy from local farmers, or pick them from the ground of meadow orchards.
I hope that this contribution, although more personal and less scientific, was helpful.

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text:
Differences between H-H-O and H-O-H
I'm not an expert.
In simple words how do they differ:
$\ce{H-H-O}$ and $\ce{H-O-H}$ molecules?

$\ce{H_{2}O}$ is the water molecule, two hydrogen atoms attached to a central oxygen atom, $\mathrm{C}_{2v}$ symmetry, thermodynamically minimal structure of these atoms, Adam's ale, etc.
HHO is a poorly defined term often bandied around by 'water powered car'... enthusiasts. I'm not sure that it's supposed to represent a molecule so much as a state of hydrogen and oxygen that conventional chemistry has no concept of.

Here's a post on skeptics.SE about it
And here's an acerbic article on rationalwiki that mentions 'HHO'

I think we're supposed to believe that it's a gas of hydrogen and hydroxyl radicals, or of atomic hydrogen and oxygen, or protons and hydroxide ions, or something, that is metastable at room temperature and that also happens to violate the first law of thermodynamics when produced and combusted. Why this glaringly obvious anomaly in quantum electrodynamics and thermodynamics has never been observed anywhere in nature before is anyone's guess (probably part of the conspiracy).
These guys should write up a few papers, get published in Nature, collect the Nobel chemistry, physics and peace prizes...

When I researched the mechanism for the combustion of hydrogen I did not even come about a molecule that would resemble $\ce{H-H-O}$.

There is no molecule in existence with the structure H-H-O, for the simple reason that hydrogen possesses only one orbital and is therefore chemically incapable of forming more than one bond or maintaining more than two electrons in its orbit. Therefore, the formula $\ce{HHO}$ is either a very idiosyncratic way of denoting a molecule of water (normally written $\ce{H2O}$ and occasionally $\ce{HOH}$, in order to emphasize its structure, i.e., H-O-H), or it refers to oxyhydrogen, which isn't actually a molecule at all, but rather a mixture of hydrogen and oxygen gases (the molecules $\ce{H2}$ and $\ce{O2}$, respectively) used as a fuel.

To add to this, there are exceptions to the 'hydrogen only makes one bond' rule of thumb, in the form of unusual electron-deficient bonds as found for instance in boranes, however this does not apply to 'H-H-O'. These bonds cannot really be reconciled with valence bond theory and require molecular orbital theory to make sense.

@RichardTerrett, thanks, +1. I'm at the undergrad level, so my knowledge of MO theory is limited.

I was completely unfamiliar with the notion of HHO before this question, so +1 just for bringing it to our attention.
I don't have much to add to the previous answers, but for those interested in learning more about the origins of HHO (and if you have access to these journals through your university), head on over to:

The main article by Santilli in the International Journal of Hydrogen Energy volume 31 (2006) pages 113-1128.
A discussion by J. M. Cato in the same journal, volume 32 (2007) pages 1309-1312 which points out some of the issues with interpreting the data in the main article. (Note there's some great pedagogical opportunities in this work for General Chemistry courses: the Santilli paper uses non-SI units that can be converted, and the basic thermochemistry of evaporation and oxidation/reduction reactions can be incorporated into Hess' Law type activities.)
There are two follow up discussions, in the same journal again, one by Cloonan volume 21 page 1113 and one by Kadeisvili which serve as a rebuttal to Cato's arguments.

This whole discussion makes for a great activity on the scientific method, understanding of analytical techniques, proper interpretation of data and the inherent resistance to change within the scientific community.  This reminds me of the cold fusion debacle, which is very well described in Gary Taubes book Bad Science.  Good reading for anyone interested in these types of controversial experiments.

The structure $\ce{H-H-O}$ does not technically exist in any common conditions because hydrogen does not generally form two covalent bonds at once. Such a structure would require a TON of energy to be put in because the hydrogen nucleus's lone proton would need to be able to hold electrons in the $2s$ sub level and prevent them from leaving the atom's circumference. Oxygen however readily forms two covalent bonds making $\ce{H-O-H}$ a very chemically plausible and common structure. Why can oxygen form two covalent bonds? Think of it in terms of quantum mechanics, much in the way as I explained for hydrogen.

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text:
How to extract elemental bismuth from pepto bismol?
I was trying to isolate elemental bismuth from pepto bismol (bismuth subsalicylate) using the process seen here but after I added hydrochloric acid, instead of getting a pink foamy substance, I got a white liquid with some white foam. The only reason I can think of, for why this would happen, is that the pepto bismol was expired by a month.
After I realized the reaction didn't work, I added a homemade universal indicator (red cabbage - phenolphthalein - preserved with a little isopropyl alcohol) and started adding sodium hydroxide to neutralise any remaining acid. When I started adding the sodium hydroxide, the solution started heating up and then changed color fairly rapidly from pink to a dull grey green which didn't change when I tried adding more acid.
Shortly after the color change, something started to precipitate out. I filtered off the precipitate and got a clear orange liquid. I then added a little hydrochloric acid to the orange liquid and something immediately started precipitating out. The remaining liquid then turned clear and smells slightly like mint.
My only guess for what I have made is bismuth hydroxide (the first precipitate) but I have no idea what the second precipitate is. Here are pictures of the final precipitates.

The initial filtration is purely awful lab technique. You cannot isolate a tiny bit of impure bismuth for less than it costs to buy a lot of very pure bismuth. http://www.metalshipper.com/bismuth.html
http://www.amazon.com/Bismuth-Ingot-Chunk-99-99-Pure/dp/B001QUVMC8
http://www.rotometals.com/Bismuth-s/4.htm
http://www.youtube.com/watch?v=s1CNGMElWfA http://www.youtube.com/watch?v=MMEkA2_N6vY  Oh, yeah!

Can you give more details about your lab procedures? What concentration of HCl did you start with? What about the pepto? Did you use tablets or liquid? Etc.

Look at Nile red on youtube. He's done it

I'm not sure you did do the experiment wrong. I performed this experiment and it closely matched this video which had a light yellow liquid and foam.
Here are my results

I filtered it a couple more times to remove more foam. Then added the aluminum foil.

This is what I got at the end before I filtered the liquid out at the end.
In the end I was able to extract 6.2 grams of bismuth powder from 67 tablets of pepto (a bit less than the theoretical ~10 grams I was hoping to get).

Also, the wiki page on Bismuth chloride show it is a yellow crystal in dry form. http://en.wikipedia.org/wiki/Bismuth_chloride

Your second, white precipitate is salicylic acid. You created a solution of sodium salicylate (soluble) when you added the sodium hydroxide; when you added HCl to that, the salicylic acid (relatively insoluble) precipitated, leaving NaCl in solution.
The mint smell is probably due to some esterification between the salicylic acid and the alcohol in your indicator.
As to why your initial attempt failed, I'm still a little unclear on the color of your solution and the amounts used. You describe it first as white and then later as pink. One possibility is that you did not use enough HCl. The foaming that is normally seen is due to some carbonate used as a filler or secondary antacid in the tablets, which reacts with the HCl and will consume some of it.
Would suggest doing the whole thing again, paying careful attention to the initial pulverization and the quantities specified in whatever instructions you have on hand.


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What is the chemical composition of a Rubbermaid Roughneck container and is it feasible for me to make my own molds?
I'm looking for more information on the composition of a Rubbermaid Roughneck container.  The description says "Roughneck containers are made with a durable polyethylene material for a lifetime of durability and strength." The Wikipedia entry indicates that polyethylene is the most common type of plastic with chemical composition $\ce{(C2H4)_{n}H2}$. But if this is so how are these containers able to flex instead of break like other plastic containers as well as tolerate greater temperature changes? I have a project that requires a custom shaped material of similar attributes. How feasible would it be to make my own mold at home? The shape I want to mold is very simple: a very thin rectangle about 35 width and 23 length. Are the chemicals involved dangerous to work with?

Or just buy the Rubbermaid container and use a heat gun to thermoform it into the shape you want.

This is a pretty broad question and might be better split into several, however a general comments on polymers, such as poly(ethylene), is that the properties of polymers can be tweaked quite a bit by control over the mean molecular mass and dispersity (probability distribution of molecular mass) of the polymer chains, as well as the addition of all kinds of stabilisers and plasticisers. Plasticisers especially may dramatically alter the flexibility of plastics. The exact composition of the Rubbermaid Roughneck is unlikely to be forthcoming.
Molding of thermoplastics by injection molding or casting is carried out by melting down plastic pellets and squirting the melt into or onto a mold. A mold release agent is often used to help get the cooled plastic out of the mold. The safety of the plastic depends on its specific composition (read the material safety data sheet for the pellets) and how hot it is (don't get molten plastic on yourself, wear protective clothing etc.). I don't know anything about the safety of mold release agents except for the fact that miniature model kits typically still have the agent on the plastic when purchased. This suggests they are fairly safe, though their presence may interfere with paint adhesion which is why pros scrub kits down before painting.
The geometric simplicity of your custom part, however, suggests that you could cut it from a plastic sheet, obviating molding and the potentially expensive equipment that goes with it.


text:
How can one find products when one knows the reagents?
I'm going to use an example (I could have used a different one);
On the document it is about the study of the chemical balance of an acid-basic solution.
The chemists have written the chemical reaction equation to make a table of the evolution of pH during the reaction.
If I understand with this equation I'll be able to understand others.
$\ce{CH3COOH + H2O ->CH3COO- + H3O+}$
Here is what I already understand:

This is the equation of methanoic acid with water
I understood why the general formula of methanoic acid is $\ce{CH3COOH}$

What I want to know is how do we know that

$\ce{CH3COOH}$ changes into $\ce{CH3COO-}$ ?
$\ce{H2O}$ changes into $\ce{H3O+}$?

In general, how does one find the products when one knows the reagents?
Does reagents molecules always correspond the same ions?

Correct me if I'm wrong here -- you want us to explain how chemists come up with reactions?  Currently this question is hard to understand, and possibly too broad. Please edit it and make it clearer, and I'll reopen it if it's up to the mark.

It is still rather broad, but I'm not sure how it can be made more specific. I'll leave this open for now --despite being broad, it is a nice, conceptual question. We'll see what sort of answers it gets :) (I may write one up myself later)

A note: I really doubt that understanding how your specific reaction was determined will explain how reactions are determined in general. Tere are various ways for predicting reactions, your methanoic acid reaction above is predicted from one of the simpler ones.

If I went into specifics, this answer would span a few thousand books. Instead, I'll give an overview:
There are many ways of determining the products given the reagents:
Actually reacting the reactants
One can, of course, actually make the reaction take place. Once you've done that, you need to identify the contents of the resulting "soup". A few methods for this  (You may have to separate the products first) :
NMR Spectroscopy
Here, a nifty technique is used to determine the magnetic properties of a compound. From these properties, one can make reasonable guesses at the substructures existing in a molecule, and check against a list of possible products. This is a very powerful technique, and usually gives good results.
Qualitative analysis
Here, we use various chemical reagents on the products, and analyse these products, making conclusions about the nature of the original products.
Microscopy
Here, using an STM, one figures out the structure of the molecule. This isn't used that often, but is an interesting technique.
There are many other such techniques, with varying accuracies and applicabilities.
Predicting the products in the absence of experiment
A lot of chemical research is into improving its (already vast) predictive power. The end-game of most sciences, after all, is to be able to predict stuff. So, in many cases, we can just predict how the reaction proceeds, without actually making the reaction occur in a lab.
Figuring out a reaction mechanism
This generally occurs in organic reactions. Here, you break the reaction into elementary steps. Each step can be explained by looking at electronic attraction/repulsion, as well as the increase in stability (which can generally be estimated by taking into account the inductive, mesomeric, hyperconjugation, and steric effects, though other things may effect the overall stability as well). From these steps, one gets a proposed mechanism, which may be verified.
Knowing solubilities and other numerical values
In ionic/inorganic reactions, the path a reaction takes generally hinges on the solubility of its final products. This is sometimes the case for other parameters as well. Knowing various numerical values lets one compare two possible reaction paths
Computational chemistry
Here, one uses the equations of quantum mechanics (Schrodinger's equation and a few of its cronies), and solves them for the given problem using numerical methods.

The particular reaction you have given
This one can be easily explained by resonance. In $\ce{CH3COOH}$, the $\ce{H}$ is loosely bonded, as the $\ce{C=O}$ bond resonates with the $\ce{C-O}$ bond. This makes it easy for the $\ce{H}$ to float away as $\ce{H+}$, leaving behind a pretty stable $\ce{CH3COO-}$ (made stable by resonance of the negatively charged lone pair over the oxygen atoms).
So far, we know that $$\ce{CH3COOH -> CH3COO- + H+}$$
Now we know that it is behaving like an acid. It is extremely well known that water forms a Hydronium ($\ce{H3O+}$) ion by self-dissociation, and this is accelerated by making the medium acidic (All free $\ce{H+}$s find the nearest $\ce{H2O}$ molecule and form $\ce{H3O+}$). $\ce{H3O+}$ is stable due to resonance, again.
So we now know that $$\ce{H+ + H2O -> H3O+}$$
Put the two reactions together and you get $$\ce{CH3COOH + H2O -> H3O+ + CH3COO-}$$
Though, in this case, the historical method of finding out the reaction may have been different -- acetic acid has been well known and studied for a very long time.


text:
Carbon with 5 bonds?
I've heard that, even though according to Molecular Orbital Theory there is no chance of having nobel gases bonded to each other, it is not totally impossible. For example, under extreme conditions, Ar2 can be synthesised.
So I am wondering whether a carbon can make 5 bonds if the required conditions are provided.

Here it is a nice example with TMS of a carbon covalently bound to 5 other atoms: http://onlinelibrary.wiley.com/doi/10.1002/anie.200601164/abstract

There are informations on this topic in question about hyperlithiation.

Carbon cannot have more then 4 double-electron bonds in reasonable conditions. However, in can form a bond with 5 or 6 atoms, like $\ce{Fe6C}$ fragment, where iron atoms form octahedron around the carbon atom. However, the sum of orders of 6 $\ce{C-Fe}$ bonds will be still 4.
The situation is different if we consider exited states. Indeed, it is possible for hight excited state of carbon to be able to form 5 or 6 bonds. The resulting structure, however, will die quickly.
Noble gases are able to form molecules in excited states, or, to be precise, their molecules are stable only in excited state. This kind of molecules is known as excimer (excited dimer).

The hypervalent carbon do exist. Look at this article:
Kin-ya Akiba, et. al., J. Am. Chem. Soc., 2005, 127 (16), 5893–5901.

X-ray measurements confirmed the 10-C-5 structure.

I believe the term "hypervalent carbon" is especially unfortunate for this compound (or class of compounds). They even argue with MO theory and just call it "pentacoordinate carbon" later. Nevertheless, this publication is of course a solid (pun intended) example, +1.

Carbon can make 5 bonds on methanium, $\ce{CH5+}$. It is a kind of carbonium ion and superacid. Methanium can be produced in the laboratory as a dilute and low-temperature gas.
Reaction:
$$\ce{CH4 + H+ -> CH5+}$$

I'm afraid that there are three "normal" and one 3-center-2-electron bond in methanium - 4 total

@Mithoron   I would consider the 3-center bond to involve 2 bonds to the central carbon atom.  Do you see it differently?

In 3-center 2-electron bond three atoms are bound to each other. It could be drawn with three lines between them, like it would be delocalized bond, but Lewis structures fail here IMO

There are stable crystal structures where one could notionally see carbon atoms with more than 4 'bonds'. One example that comes to mind is a carborane unit; e.g. see
http://en.wikipedia.org/wiki/Carborane
For example. But of course, such 'bonds' are a notional thing really and don't reflect the true nature of the underlying electronic structure.

Your comment on notional bonds can be said of any bond unless you use a theory such as Natural Bond Order analysis, which does not necessarily contradict the penta-bonded carborane.

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


text:
Could you estimate the rate of reaction from the chemical structure of the reactants? If so, how?
Say you have two compounds with their own unique chemical structures and you know the environment (e.g. temperature, pressure, presence of enzymes) could you then, using this information, approximate the rate of reaction? And if so, how would you? Please provide a mathematical answer, not just use some software.

Are you looking for an academic/theoretical answer, or are you interested in concrete practical methods (i.e., modeling software, crude back-of-the-envelope calculations, etc.)? I'm aware of software suites that can simulate certain classes of reactions, but if you're looking for an exact rate equation to model something other than an elementary mechanistic step, then I suspect you're out of luck (even in those instances, the rate constant of the equation has to be determined experimentally [or approximated computationally] and potentially varies with multiple parameters).

A mathematical model, i.e. an academic/theoretical answer. I knew this would be a tricky question, especially since from what I can tell most chemists know hardly any math beyond the basics, but still I'm a mathematician and if someone knows the answer that'll be much appreciated.

theoretically it is possible, but will require  scan of potential energy surface near reaction path. After this, using molecular dynamics it is possible to scan over reaction path and say, how many collisions of which energy will result in reaction. With this, you will be able to estimate speed of reaction at any reasonable temperature. This task, however, is extremely calculation intensive and unlikely to be pragmatic, so it is a comment, not an answer.

Permeakra's answer is good, as are the ones by Home and Greg E.  But the sad truth is that in general the answer to the question is NO, you cannot estimate the rate of the reaction.

As Paul J. Gans correctly pointed out in his comment, in general it is impossible.
In practice you can calculate reaction rates only for quite restrictive class of reactions - the unimolecular gas-phase reactions - using RRKM theory.
The whole potential energy surface is actually not needed, as far as I know, you just need to identify reactant, product, and transition state.
RRKM is based on few assumptions, with the first one being quite obvious: since we are talking about potential energy surfaces, the electronic motion and the nuclear motion is separated by the Born-Oppenheimer approximation.
The nuclear motion is also assumed to be well separated into rotational, vibrational, and translational one. The internal degrees of freedom are treated classically with quantum mechanical corrections.
RRKM is quite useful for predicting gas phase dissociation and isomerization rates  under low pressure (low enough so that reaction can be safely assumed to be unimolecular). For that reason, this theory is ofthen used in mass-spectrometry, since the above mentioned conditions are quite often met here, but it is of no use for reactions in the condensed phase.


text:
Can flexible plastics have reflective coating (vacuum metalized)?
Is it possible to achieve an optical mirror coating to a flexible plastic part?
If vacuum metalization is possible, would the reflective coating layer be durable enough to withstand the bending of a flexible plastic such as polyurethane or a thermoplastic elastomer, without cracking over time?

I think this is fine for here, honestly.  Interesting question.

read this: http://en.wikipedia.org/wiki/Metallised_film  || I expect, that for some polymeres cooling of the plastic part may be an issue, but still.

There are many techniques for metal-coated polymers, as linked by permeakra Generally, vacuum deposition is an issue because the metal is hot. Cold deposition techniques (e.g., sputtering, ink coating) often work better.
Common, commercially-available materials include:

Pyralux - copper-coated polyimide from DuPont.
Mylar - DuPont's name for biaxially-oriented polyethylene terephthalate (BoPET)

As mentioned in other answers, adhesion is an issue, and to my knowledge, the processes to bind Pyralux and metabolized Mylar are proprietary.
Chemistry.About.com does mention vapor deposition of metals into Mylar, though.

It might be possible to deposit a metallic film on a sheet of plastic. It is however, a problem with respect to the bonding of that layer. Depending on the plastic used, it might not have a strong bond with the metal, which would then over time develop defects.
On the other hand, it might be possible, when carefully manufacturing this, that the plastic has 'functional groups' (parts of the molecule with a specific function) that would bind a metal. Note however, that in this case, the functional groups might also react with other chemicals. Almost no functional group will only under go one reaction, especially not when metals are involved. Therefore, this would not be the most durable of options. Furthermore, because of the interface between, most likely, an amorphous solid (wikipedia, amorphous solids) and a metal structure is bound to create some defects.
A last option I can think of, is the use of conductive polymers. These polymers, because in many ways they mimic metals, also have a similar 'shininess' to them. However, I'm not sure about their really being a mirror.
Hope this helped.

I would expect that gilding - classic manual gold plating using gold leaf - works.
(But as the frequency, amplitude and radius of flexing is not known, it's hard to tell)
It would be starting with the polished polymer surface.
That is coated with a special glue,
the gold leaf is just layed down on it,
and the gold surface is polished/burnished to create the mirror surface.
The difficult part may be finding the right glue substance, which depends on environment conditions. But a standard product should work.
So - why do I propose this ancient method?
I see the advantage in using an actual (microscopic) compact layer of sheet metal.
Gold leaf is just a thin gold foil.
But there is a special property in gold: it's very ductile. It's so ductile, that gold leaf can actually be produced roughly by taking a piece of gold and hammering it flat.
Now, I would think that the gold layer will not have much problems with the flexing polymer under it.
Another important property of gold here is the innertness of the metal, because we have the mirror layer directly exposed to air - not behind glass like in "consumer mirrors". With silver, that would be a problem
Gold makes a very good mirror for many wavelengths - you did not say which you are interested in. But if the yellow color tint of reflected visible light is a problem, one could try the same with other metals. No other metal has the level of ductility of gold, so they are possibly less suitable.
What makes me think that creating a sheet of metal first, and putting it on the surface in a second step is that any kind of metal plating where the metal is deposited on the surface will follow any roughness of the surface, which will cause irregular thickness and mechanical strength. On the other hand, the layer following the contours has a larger area, when looking closely, which shoud help with mechanical stretching.
The typical thikness of deposited metas ist thinner than a foil. It can be made as thick for sure, but this could make coating expensive in terms of processing time.

Yes, OCLI (optical coating labs inc) coated plastic film in 5' wide rolls that were 1000's feet long.
This was done in the early 80's and I'm sure there is a lot of coating companies that do it now. One of my favorites was a nickle alloy film that we made into a recording tape. But they can coat both dielectric materials and metals.
OCLI was bought up by JDS uniphase and later by somebody else.


text:
Which Chemical Database Software Are Available?
By chemical database software I mean software with similar functionalities to Instant JChem in that they have a large database of different compounds and have the basic capabilities you, or at least I expect from a chemistry software. Namely:

Can approximate the chemical and physical characteristics of the compounds in question.
Can approximate products of reactants and their corresponding proportions.
Can show the 2d and 3d structure of compounds in the database.
Can connect to online chemical databases like PubChem, ChemSpider, NCBI, etc.
Has a molecule editor.

I would also prefer it if it is free of charge.

Your question is rather vague, what properties and level or approximation you are looking for 2) You are asking for rather unrelated functionalities 3) It is not clear what you don't like with JChem. 4) You may want to check out keywords like "chemoinformatics", and look what you are specifically looking for

A widely used database in academia and industry is ChemACX.The only one of your requirements that I don't believe it does is the product prediction.
http://www.cambridgesoft.com/Ensemble_for_Chemistry/ChemACX/Default.aspx

Not a database, more a toolset, but there is KNIME in wich you can make workflows that do all kinds of things.
With the right plugins you can also work with molecules in various ways, such as reading and writing, descriptors, manipulation, reactions, etc.


text:
Why do big thermoplastics (like Polytetrafluoroethylene) behave like thermosets?
Why do big thermoplastics (like Polytetrafluoroethylene) behave like thermosets? I know for sure it has to do with the van der Waals force and the fact that the molecules are huge, but why? Why does a big molecule break down before the van der Waals forces break?

You may want to register your account (go here and click "add logins")

Thermosets are liquids or soft solids which irreversibly 'set' or 'cure' when heated. This 'setting' produces covalently  bonded cross-linkages between adjacent molecular chains, resulting in in their mechanical rigidity.
Thermoplastics too are formed from linear polymer chains, but their molecules typically have very long chains with high molecular weight, with very few cross-linkages. The weaker van der Waals bonds which link adjacent molecules in thermoplastics are easily broken when heated, allowing the chains to 'slide' past each other, thereby giving them the property of 'softening' when heated.

Why does a big molecule break down before the van der Waals forces
  break?

In the case of polytetrafluoroethylene (PTFE), we have a very long, linear polymer chain of carbons covalently bonded with fluorine. PTFE is quite unique because it doesn't have any hydrogen atoms. Instead, it has these large highly electronegative fluorine atoms.

The bonds within the PTFE molecule are very strong (C-C bonds are 607kJ/mole and C-F bonds 552kJ/mole, much stronger than the typical C-H bonds found in other plastics such as polyethylene).  In effect, the PTFE molecule is like a 'rod' of carbons bonded to a surrounding 'blanket' of fluorine atoms (arranged in a slight helical twist).
Although the van der Waals bonds which form between adjacent the adjacent linear 'rod' molecules of PTFE are much weaker than the C-C and C-F bonds within the molecules, these van der Waals forces are able to combine in an additive manner, so the long straight PTFE 'rods' are able to be 'packed' together in a hexagonal arrangement, a bit like pencils in a box. This results in PTFE having a relatively high density and melting point of $327^o C$.
But when PTFE melts, its extremely long molecular chain have a high probability of entanglement, resulting in a extremely high melt viscosity. Although it's melt viscosity varies with the shear stress applied and temperature, in the temperature range between $360^o C$ and $380^o C$ and shear stresses of $0.1MPa$, commercial samples of the granular polymer display viscosities of the order of $10^{10}Pa.s$. This is exceptionally high compared with most other engineering thermoplastic melts. In fact, at atmospheric pressure, it hardly seems to 'melt' in the conventional sense of 'becoming a flowing liquid', although it becomes slightly softer, it forms a near-clear gel with very high elasticity.
Even below it's melting point, PTFE will start to decompose at $260^o C$, releasing toxic compounds through sublimation. When heated above $360^o C$ thermal decomposition is more rapid and the PTFE degrades noticeably. As a consequence, PTFE can not be processed by many 'conventional' thermoplastic techniques, such as injection molding. Instead, it is processed in a finely powdered or granular form, using modified powder metalurgical techniques such as cold-molding, sintering, compression and extrusion, to fabricate rods, tapes, wire, tubing, sheeting and other profiles which are typically suitable for further machining.
Above $480^o C$ PTFE thermally decomposes in air to release a mixture of vapours and particulates, including carbonyl fluoride, hydrogen fluoride, tetrafluoroethylene and other fluorinated hydrocarbon products, including a small amount of some highly toxic species such as perfluoroisobutylene.
The precise nature of the reactions responsible for this decomposition is still subject to research, but it appears likely that it is initiated by random-chain incisions in the PTFE chain, followed by depolymerization.


text:
What is the reaction between Magnesium chloride and Sodium bicarbonate?
For some reason it has been very difficult to find the right answer for this question as there's so much conflicting information to be found online. Some mention a reaction where carbonic acid is made, others mention carbonates and hydroxide end product but nothing seems to be precise.
But can the reaction be as simple as this (a double displacement reaction) where magnesium bicarbonate and sodium chloride are the end products:
$$
\ce{2NaHCO3 + MgCl2 ->  2NaCl + Mg(HCO3)2}
$$

Is the reaction to occur in water (aqueous)?

Having actually perform this reaction, I can say with certainty that when you mix aqueous sodium bicarbonate with aqueous magnesium chloride, there is no apparent reaction. This is consistent with an equilibrium reaction (actually, just ions):
$$\ce{2 NaHCO3 + MgCl2 <=> 2 NaCl + Mg(HCO3)2}$$
However, upon brief heating in a microwave (for example), a white precipitate of magnesium carbonate develops indicating the reaction has moved to the right.
$$\ce{2 NaHCO3 + MgCl2 ->[Heat] 2 NaCl + MgCO3(s) + CO2 (g) + H2O}$$

The chemicals must be in water, since magnesium bicarbonate decomposes when dried. So I will assume that the mixing is done in water.  To start with, both salts are strong electrolytes and so they immediately dissociate to give $\ce{Na+}$ ions,
$\ce{HCO3-}$ ions, $\ce{Mg^{2+}}$ ions, and $\ce{Cl-}$ ions.  As Brenton Home has pointed out, none of the possible compounds formed from these are insoluble and there is no gas formation due to the decomposition of the bicarbonate because the resulting solution is not acidic.
So the answer is that there is no reaction.

I thought I would start out in saying that I think this is a random, but great question. This is because as you said there's so little information out there regarding this sort of reaction so it was an interesting little puzzle for me to figure out for you.
My guess would be that the reaction you detailed would not occur. This double replacement reaction would occur in aqueous solution as otherwise the product of magnesium bicarbonate probably wouldn't have a chance of forming since at least one other alkali earth metal bicarbonate salt, calcium bicarbonate, does not exist outside of aqueous solution. I tell you this piece of information instead of whether magnesium bicarbonate exists outside of aqueous solution because I just don't know for certain whether it does. Evidence for this theory that it doesn't exist out of aqueous solution includes (in addition to the point raised above about calcium bicarbonate):

Magnesium bicarbonate's predicted partition coefficient is -0.809 according to chemspider which is identical to the predicted partition coefficient for calcium bicarbonate which makes it very likely that it is highly water soluble.

Since it occurs in aqueous the only driving force to the reaction would be the production of an insoluble salt, in your reaction, however, there is no insoluble products that results.
Therefore I suspect an alternate reaction would occur. Namely:
$$\ce{2NaHCO_3 + MgCl_2 -> 2NaCl + MgO + H_2 O + 2 CO_2}$$
As this reaction would lead to three insoluble products and thus would invoke a greater drive for the reaction.

I think that there is an oxygen missing to equilibrate your reaction (and that it is likely to be as MgO, since magnesium is unlikely to be reduced in the process). And I agree that it would be necessary that the initial poster provide some background on the conditions.

Not magnesium oxide, which is too sttongly basic to form along with carbon dioxide.  If you get a reaction at all it's magnesium carbonate.

There can be a reaction, maybe.  A simple double displacement gives nothing, but perhaps magnesium bicarbonate can "disproportionate" in an acid base sense:
$\ce{Mg^2+ + 2HCO3- -> MgCO3(s) + CO2 + H2O}$
where the reaction would be driven by the magnesium carbonate precipitating.  I would need to look up the relevant equilibrium constants (acid-base for carbonate amd bicarbonate ions, precipitation of the magnesium carbonate) and be given solution concentrations to see whether we really get this reaction.


text:
Prediction of solubility (from chemical structure) at different pHs
I would like to know What is the best method for predicting the water solubility (and in other solvents) of a compound given its molecular structure at different pHs. What I know now is based on the seminal paper from Eisenberg et al.;
Eisenberg, D. & McLachlan, A. D. Solvation energy in protein folding and binding. Nature 319, 199–203 (1986).

where partition coefficient can be estimated depending on each atom type and its exposed surface to the solvent (SASA). This seems to work well for many substances, but I have the problem that I need to know the solubility dependence of some substances against pH, and I the only thing I found now as Chemaxon tools, but their results are not very accurate. It seems that the best option might be quantum mechanical calculations and tools such as COSMO-RS. At the moment we can not pay its license.
So in summary, I wonder what approaches do you suggest for calculating solubility of small chemical compounds depending on pH. Any method/software/server would be fine for us.

Welcome to Chemistry.SE. Please provide some examples, to show what effort has been put in the query by you.

http://pubs.acs.org/doi/abs/10.1021/ci800058v Would be a good start for reading, but its paywalled.

I'll reopen this once you add some examples and shown us some of your own thought process on the matter (see the [faq] and [ask])

I'm reopening this for now. It's borderline "not constructive", however, since it asks for "the best" method (and is effectively a software recommendation). Please add your additional improvements from your other questions here as well.

sorry again, you can close or delete this and let open this one;  http://chemistry.stackexchange.com/questions/4251/prediction-of-solubility-from-chemical-structure-at-different-phs#comment6417_4251


text:
Why does electrolysis take so much time?
What makes electrolysis or electroplating take so long ? What stage takes longer when ions are attracted to electrodes or when they gain or lose electrons at electrode.
And if we are talking about an electrolyte composed of positive ions and electrons , which is faster, electrons entering the circuit at positive electrode or gaining electrons by positive ions at negative electrode?

'A watched electrode never plates', as the saying goes. Seriously though, whilst I can't answer the question I would venture that due to dielectric screening, ions in solution only 'feel' the electrodes at very short range, which would imply that electroplating is for the most part diffusion-limited.

The limiting part is ion diffusion. There is an upper limit to voltage applied, or wrong ion will be reduced, and with that the speed of diffusion is limiting stage.

It depends: The rate of electrolysis is limited either by the electric source or by the ion concentration. I am inexperienced with electrolysis but I imagine that quality may increase as the the process slows, much like crystals grow with more imperfections when the rate of crystallization is fast.
There is no such thing as an electrolyte solution composed of positive ions and electrons.

"Activity coefficients of single ions cannot be measured
  experimentally because an electrolyte solution must contain both
  positively charged ions and negatively charged ions." -Debye–Hückel
  theory

Electroplating systems tend to use an electrode source that replenishes the solvent ions plated onto the target electrode.


text:
How to determine which gas is more soluble
I have a chemistry homework problem that asks whether $Br_2$ or $O_2$ is more soluble in water in room temperature. However, I can't seem to find a way to figure it out, short of looking it up.
Is there a guiding rule for such figures, such as relating to molar mass?

The answer the question is seeking might be simpler than you think. The most obvious difference between oxygen and bromine is that bromine is a liquid not a gas a room temperature. Might this be significant?

You could answer this question by looking at the possible interactions between a gas molecule and water molecules. The only I can see are van der Waals interactions (because these gas molecules cannot participate in hydrogen bonding).
$\ce{O_2}$ and $\ce{Br_2}$ have no permanent dipole so the only interactions are Debye forces (interaction between water permanent dipole and a gas molecule induced dipole) and London forces (between instantaneous dipoles).
The induced/instantaneous dipoles strength increases with the size of the atom because the bigger the electron cloud the easier it is to modify its shape.It happens because the electrons far away from the nucleus are less attracted by it and can move around a bit more easily than electrons very close to the nucleus.
This is the reason why van der waals forces are stronger for bigger atoms.
This would explain why $\ce{Br_2}$ has a better solubility than $\ce{O_2}$ in water.
As a side note:
London interactions are very important if you look at liquefaction temperature of gases. For instance, it explains why $\ce{Ar}$ becomes a liquid below 87K while $\ce{He}$ become a liquid only below 4K.


text:
Why does water favour nucleophilic substitution over elimination?
The two broad ways a halogen can leave a haloalkane are nucleophilic substitution and elimination reactions.
I was told that elimination reactions (that form a double bond as a $H^+$ is also removed) do not occur when the haloalkane is dissolved in water; here nucleophilic substitution (anion replaces halogen only) is favoured.
Why is this? I guess it's not just water - what are the general conditions (including type of solvent) that prefer nucleophilic substitution over elimination, and vice-versa?
I understand there are different variants of both elimination and substitution ($S_N1, S_N2, E1,E2$); are any of them more prone to occur than others of the same type in water?

The nature of the substrate (in particular the degree of substitution of the α-carbon) is of principal importance in determining which mechanism predominates. Generally, $\mathrm{S_N1}$ and $\mathrm{E1}$ are both less common than $\mathrm{S_N2}$ and $\mathrm{E2}$, partially as a consequence of their first-order kinetics and the stability constraints imposed by the charged intermediate species the mechanisms generate. $\mathrm{S_N1}$ and $\mathrm{E1}$ both require tertiary, allylic, or benzylic α-carbons in order to be viable (although secondary carbons can occasionally react by these pathways as well, but slowly, and only if conditions disfavor competing reactions), mainly because the carbocation intermediate that forms during the rate-determining step can be stabilized by the neighboring electron-donating alkyl groups via hyperconjugation (or via resonance in the case of carbons in allylic or benzylic positions). Tertiary carbons also disfavor $\mathrm{S_N2}$ because the approach of the nucleophile is sterically hindered by the β-alkyl groups. Under those conditions, $\mathrm{E1}$ and $\mathrm{S_N1}$ reactions would be in competition. $\mathrm{E1}$ would be favored if there were no strong nucleophiles present, as well as by the application of heat. Obviously, $\mathrm{S_N2}$ reactions are favored by the presence of strong nucleophiles (ideally, ones that don't also act as strong bases) and a primary α-carbon substrate. $\mathrm{E2}$ reactions are favored over $\mathrm{S_N2}$ if strong bases, especially non-nucleophilic, are present. $\mathrm{E2}$ also prevails when either the substrate and/or the base is sterically hindered, since abstraction of a proton is vastly less impacted by steric factors than is the approach of a nucleophile attacking a carbon with simultaneous loss of the leaving group. Finally, $\mathrm{E2}$ reactions impose an additional restriction in that the β-hydrogen and the leaving group must be anti-periplanar to one another, in order for simultaneous loss of the leaving group and formation of the π-bond to occur in a coordinated fashion. If the molecule can't adopt such a conformation (for example, it may be rotationally locked due to the presence of π-bonds, or lack the kinetic energy necessary for a ring flip, etc.), then $\mathrm{E2}$ is categorically ruled out.
(Parenthetically, I should add that elimination reactions, in general, require higher temperatures than substitution reactions. The naive explanation for this is based on the equation for the free energy change of a reaction: $\Delta G = \Delta H\ -\ T\Delta S$. Elimination reactions, at the most simplified level, involve two reactant molecules [namely, a base and the substrate], and yield three products [the protonated base, the substrate molecule that has undergone elimination, and the substrate's leaving group]. Consequently, elimination reactions are entropically favorable [i.e., the $T\Delta S$ term is positive]. However, elimination reactions are also endothermic [i.e., $\Delta H$ is positive], because heat energy is required to break the two σ-bonds of the abstracted proton and the leaving group, while the resulting products form only one σ- and one π-bond, the latter of which is less stable. The implication of this is that, in order for the overall change in free energy, $\Delta G$, to be negative [a requisite condition for spontaneity of the reaction], $T\Delta S$ must be greater than $\Delta H$, so the temperature needs to be sufficiently high if the reaction is to take place.)
The solvent is of only slightly less importance. Polar protic solvents (water, alcohols, weak carboxylic acids, etc.) have partial-positive hydrogens, which serve to stabilize negatively charged species. Consequently, $\mathrm{E2}$ and $\mathrm{S_N2}$ mechanisms, which are aided by having negatively charged bases or nucleophiles in high concentration, are somewhat dampened by protic solvents, allowing otherwise disfavored $\mathrm{E1}$ and $\mathrm{S_N1}$ pathways to compete. Polar aprotic solvents (THF, DMF, DMSO, chloroform, ethers, ketones, etc.) cannot stabilize negative charges as effectively, making them the preferred choice when $\mathrm{E2}$ or $\mathrm{S_N2}$ mechanisms are desired.
Finally, to your specific question about haloalkanes. I've explained above why various combinations of conditions might favor one class of reaction over another. In water, specifically, a few distinct factors, beyond the general principles outlined above, might be relevant:

Due to the leveling effect, no base stronger than hydroxide can exist in water at any appreciable concentration (that is, any base much stronger than hydroxide will simply deprotonate the water molecules, forming hydroxide and the base's conjugate acid). However, strong, non-basic nucleophiles are not subject to this effect, and though they may be solvated and weakened by hydrogen bonding, this effect would be less drastic than the protonation that strong bases would undergo.
Hydroxide is actually a much stronger base in many other solvents than it is in aqueous solution, which is chiefly the result of it being very thoroughly solvated and stabilized by hydrogen bonding. Hence, competing nucleophiles become stronger by comparison, while stronger bases simply end up being consumed, producing more hydroxide.
Because elimination reactions require heat, substitution reactions will be heavily favored and immediately take place unless the reaction vessel is heated prior to mixing of the reagents. The problem is compounded by the fact that, with hydroxide as the nucleophile, an alcohol would form, which would be stabilized by hydrogen bonding with water molecules. Further, because haloalkanes are poorly soluble in water, and immiscible liquids in the same container will boil at lower temperatures than they would unmixed, it could become very difficult in practice to achieve the temperatures necessary to favor elimination reactions while also maintaining the agitation necessary to keep the haloalkane and the aqueous phase from completely separating.

Thanks for the brilliant answer, but why in E2 must the hydrogen and leaving group be anti-periplanar?

Also, do I understand correctly that the haloalkanes separate from water when heated because the enthalpy change in boiling would be higher than the entropy change (multiplied by temperature) of separating?

@Alyosha, Re: E2, there's a simple steric explanation: the transition state of an E2 mechanism is more stable when the β-H and LG are anti-periplanar because they are staggered, while they would be eclipsed in a syn-periplanar mechanism. There's a more rigorous explanation from molecular orbital theory, which, if I remember correctly, argues that the antibonding σ* orbital of the LG can only be effectively accessed by the electrons of the bonding σ orbital when the geometry of the relevant groups is anti-periplanar. As for your other question, I don't have an answer, sorry.


text:
Saturated vs unsaturated fats - Structure in relation to room temperature state?
I'm sure most of us have heard that saturated fats are solid at room temperature, and unsaturated fats are liquid at room temperature. I'm wondering how this relates to their chemical structure -- saturated fats contain only single bonds between carbons, yet to qualify as an unsaturated fat a C=C double bond must exist.
Since a double bond is stronger than a single bond, and the length of the C=C double bond is shorter than that of the single bond, why is it that the fat containing a double bond is a liquid and saturated fats are solids at room temperature? Seems like the double bond would inhibit movement and the resulting substance would be less like olive oil and more like butter.

In the solid state, the individual triacylglycerol molecules are interacting with each other primarily through Van der Waals interaction. These weak bonds between molecules are broken at the solid-liquid transition. The amount of energy needed to disrupt these interactions (which determines the melting point of the fat or oil) is determined by the energy associated with all of these bonds added together. In a saturated fat, the acyl chains are able to align perfectly right along their length, maximizing intermolecular interactions. This effect is reflected in the fact that the melting temperature of a pure triacylglycerol increases as the chain length increases.
You can see this effect clearly in the melting temperatures of individual fatty acids. (C18:0 means an 18 carbon molecule with zero double bonds in the acyl chain):

C18:0 (stearic acid) 70°C
C16:0 (palmitic acid) 63°C
C14:0 (myristic acid) 58°C

So the addition of a single -CH2- group in the acyl chain increases melting temperature by a few degrees.
When a cis double bond is introduced into the acyl chain this creates a kink in the structure. Because of this, the acyl chains cannot align completely along their length - they don't pack together as well. Because of this, the sum of the energy associated with intermolecular Van der Waals interactions is reduced. Again this is seen clearly in the melting temperatures of fatty acids:

stearic acid C18:0 70°C
oleic acid C18:1 16°C

As you can see from these numbers, the effect of introducing a double bond is large compared to the chain length effect.
A typical fat or oil will, of course, be a mixture of different triacylglycerols, but the underlying principle is the same.

This has to do with the chemical structure of the fatty acids. Single bonds create a straight molecule, which can be easily packed together to form a solid, while the double bond introduces a kink in the structure.

Fats, which are mostly from animal sources, have all single bonds between the carbons in their fatty acid tails, thus all the carbons are also bonded to the maximum number of hydrogens possible. Since the fatty acids in these triglycerides contain the maximum possible amount of hydrogens, these would be called saturated fats. The hydrocarbon chains in these fatty acids are, thus, fairly straight and can pack closely together, making these fats solid at room temperature. Oils, mostly from plant sources, have some double bonds between some of the carbons in the hydrocarbon tail, causing bends or “kinks” in the shape of the molecules. Because some of the carbons share double bonds, they’re not bonded to as many hydrogens as they could if they weren’t double bonded to each other. Therefore these oils are called unsaturated fats. Because of the kinks in the hydrocarbon tails, unsaturated fats can’t pack as closely together, making them liquid at room temperature.

The source website is no longer accessible, but biochemistry textbooks will state the same thing.

image doesn't work

I updated the image - the link is unfortunately dead - but the information is correct so it's fine


text:
Can ammonia be detected by a 1D-1H-NOESY experiment?
I'm not detecting ammonia in my samples while using a NOESY pulse program for a 1H-NMR experiment.
Is it possible for ammonia to be detected by a 1D-1H-NOESY experiment?

Are you trying to irradiate at the ammonia 1H frequency? What are you expecting to see?

No. I want to check it's concentration in a sample.

Then why are you measuring a NOESY, and not a regular 1D?

Because I use a software to handle the spectra which is based on NOESYs. Their spectra database was built using NOESYs, and they don't guarantee reliable quantification with another pulse seq.

I've updated my answer, you're likely using a different NOESY than the one I explained previously.

In the 1H,1H-NOESY experiment you are using the nuclear Overhauser effect to transfer magnetization from one proton to another proton in close proximity (< 5 Å).  This only works within a single molecule, unless you have a reasonably stable complex you can't observe signals in the NOESY between protons of different molecules in your solution.
Ammonia has three protons which are at least chemically equivalent, their signals will appear at the same chemical shift. In a 1D NOESY you irradiate at the frequency of a specific proton and then observe the transferred magnetization at other frequencies. You can't observe magnetization transfer between protons at the same frequency that way, as your irradiation will destroy any signal at this specific frequency.

You can ignore the explanation above, as the NOESY experiment you are using is different. In metabolomics the NOESY 1D is used for water suppression, this is a different experiment than the 1D NOESY I explained above.
The common variant of this I found also uses presaturation to suppress water. This presaturation will also affect all signals that exchange quickly with the solvent, and the protons of ammonia are exchanging with water. So I would expect the ammonia signal to be far weaker or even not visible in such an experiment.
If you want to observe ammonia, you have to use a water suppression method that doesn't affect exchanging protons.

So, additional info: I'm talking about a complex mixture with lots of compounds to be quantified. Ammonia is just one of them. And it's 1(one) dimensional proton NMR.

Multiple compounds don't change anything about my answer, but I'm really wondering why you are measuring a NOESY in this case. It doesn't seem to make any sense, are you sure you are not mistaken about the type of experiment you want to run?

First off, NOESY is a tool that gives information more useful for structure determination and confirmation (as Mad Scientist has explained). A simple and quick 1H NMR experiment should be enough to quantify your compounds assuming you have sufficient peak resolution (often a problem with multiple compound quantification).
How do you know that you cannot detect the ammonia? Where do you expect to find its peak (I believe it is a triplet found around 0 ppm, depending on solvent/conditions)?. Remember that the chemical shift may be affected by temperature, concentration, or hydrogen bonding, so it may not be where you expect to find it.
I suspect your problem could also be with the deuterated solvent you use to prepare samples. For example, if you are using D2O, you will not see any OH, NH (ammonia), or SH protons as they will rapidly exchange with deuterium from the solvent, effectively making your ammonia "invisible". You may try to dissolve your compounds in deuterated solvents that do not exchange as readily (CDCl3, DMSO-d6).
Another possibility which I already mentioned is your peak resolution. If you have multiple compounds in this sample at various concentrations then you will likely have many peaks where the ammonia signal could "hide" under, so be careful there as well.
I have found by experience that when trying to quantify mixtures in 1H NMR (especially as part of a routine) it is extremely valuable to first collect a reference spectrum of each individual compound (when possible) using identical preparation techniques and experimental conditions in order to make peak identification and quantification of mixtures a breeze.
Hope this helps you sort out your problem!


text:
Why isn't water acidic?
The definition of an acid is a compound with a hydrogen cation and a non-metal anion. Water is a hydroxide bonded to a hydrogen atom. This hydroxide has a net negative charge, too, since the negative oxygen's charge is stronger than the positive hydrogen's. Water has a pH of 7, though, meaning that it's neutral. Why isn't water acidic

It's is our baseline. It is defined to be neutral at room temperature. Anything more acidic than water is acid and vice versa.

If you look at what happens in the equilibrium dissociation of water, both a hydronium ion (Arrhenius acid) and a hydroxide ion (Arrhenius base) are formed. Since these two ions are  formed in equal quantities ($\ce{10^{-7} mol/L}$ each in pure water), the water will be neither acidic nor basic.
$$\ce{H2O_{(l)} + H2O_{(l)} <=> H3O+_{(aq)} + OH^{-}_{(aq)}}$$  (from wikipedia)
Wikipedia can explain more about acids and their definitions if you are interested: http://en.wikipedia.org/wiki/Acid.

In addition to this, pure water is defined as neutral, as it is a most common solvent in nature.

Just keep in mind also that water's pH is often a bit less than 7 because of the carbone dioxide in the air dissolving in the water forming carbonic acid.

Also the pH is not related to neutrality. Neutrality is when the number of hydroxide and hydrogen ions are equal. pH is just a measure of hydrogen ion concentration, and this is temperature dependent. At higher temperatures, water's pH decreases because it dissociates more but it remains neutral because each dissociation produces one hydrogen ion and one hydroxide.

pH is a measure of hydrogen ions, just as pOH is a measure of hydroxide ions. a pH or an pOH of 7 DOES mean that that the number of hydrogen and hydroxide ions are in balance - or "neutral" in that the solution is not acidic or basic.

Water has “the neutral pH” because pH = 7 (a meme only useful in aqueous chemistry) is used to estimate acidity/basicity of a substance with respect to water namely.
In Brønsted–Lowry approach, water is an acid and a base simultaneously: H2O sometimes dissociates to OH− (the conjugate base of H2O) and H+, i. e. we see deprotonation. The latter immediately reacts with surrounding H2O to form aqueous varieties of hydrogen cation (conjugate acid of H2O, simplistically called “hydronium”), i. e. we see protonation. But there is an equilibrium, namely OH− and (H2O)nH+ react back to form the covalent water. This results in certain stable (under constant conditions) concentration of the hydrogen cation, that is used to define pH.
When we dissolve in water some substance that (preferentially) protonates it, an excess (H2O)nH+ forms and pH became lower than of pure H2O. Thus a Brønsted–Lowry acid (wrt H2O) becomes an Arrhenius acid. When we dissolve in water some substance that (preferentially) deprotonates it, (H2O)nH+ is consumed, an excess OH− forms, and pH became higher than of pure H2O. Thus a Brønsted–Lowry base (wrt H2O) becomes an Arrhenius base (or alkali).
But the paragraph above make sense only in the aqueous framework. When we mix H2O with NH3, a huge amount of ions forms:

OH−, the conjugate base of H2O;
NH4+, the conjugate acid of NH3 (in other words, an ammonia-based form of H+).

In Arrhenius picture, we see an alkali: there is almost no (H2O)nH+ and a lot of OH−, whereas NH4+ isn’t “hydrogen cation”. But in NH3-centric picture we see indeed a strong acid! There is a lot of NH3’s conjugate acid, whereas NH3’s conjugate base (NH2−) is absent. In “symmetric” Brønsted–Lowry picture we see that H2O acted as acid (it deprotonated) and NH3 acted as base (it was protonated).
Hence, acidic and basic solutions are all relative. Water is acidic, with respect to substances that are bases wrt H2O.

In your example, the hydrogen (+1) bonded to the hydroxide (-1) will balance out to a net charge of 0. Because there is no overall charge, the pH will be neutral.
However, pure, neutral water can behave as a Lewis Base (lone pair donor) due to the lone pairs present on oxygen. For example, water behaves as a Lewis Base and Al3+ as a Lewis Acid (electron acceptor) to form a hexaaquaaluminum(III) complex. That is: Al3+ with 6 water molecules surrounding it in a cubic fashion.

There seems to be a lot of confusion here - both in the question and in some of the answers. Let's try and clear a few things up.

The definition of an acid is a compound with a negatively charged hydroxide group

This is definitely wrong. There are many different definitions of acids and bases. The three most commonly used ones are:

Arrhenius theory: This is one of the oldest theories of acids and bases. It says that an acid is a species which dissociates in aqueous solution to give hydrogen ions in solution. A base is a species which gives hydroxide ions.
Brønsted-Lowry theory: This is really a generalization of the Arrhenius theory beyond aqueous solutions. It says that an acid is a proton donor (whether in aqueous solution or not) and a base is a proton acceptor.
Lewis theory: This is a much more general theory of acids and bases which has strong links to molecular orbital theory, although it was actually developed prior to this. It states that an acid is an electron pair acceptor and a base is an electron pair donor.

The most relevant of these theories to our discussion are the first two. Both of them state that an acid must have labile protons which it can easily lose. In both theories a hydroxide ion would usually be considered a base, not an acid.

Water has a pH of 7, though, meaning that it's neutral

There are two concepts here; neutrality and pH. These are often confused as meaning the same thing but they are actually completely independent. An aqueous solution is neutral when it contains equal numbers of hydrogen and hydroxide ions. pH is a measure of the hydrogen ion concentration in aqueous solution and is defined as: $$\mathrm{pH} = -\log_{10}(a_{\ce{H+}})$$
where $a_{\ce{H+}}$ is the hydrogen ion activity. For dilute solutions this is approximately numerically equal to the hydrogen ion concentration.
If we consider just pure water then there will always be a small amount of hydrogen and hydroxide ions in solution, due to the self-ionization of water. $$\ce{H2O + H2O <=> H3O+ + OH-}$$
As you can see from the equation, the hydrogen (or actually $\ce{H3O+}$) ion concentration will always be the same as the hydroxide ion concentration and so pure water will always be neutral. However, as you heat water, the dissociation equilibrium shifts towards more dissociation so the concentration of hydrogen ions goes up and the pH goes down.

Why isn't water acidic

Using the Brønsted-Lowry definition, we can describe any reaction involving proton transfer in terms of an acid and a base. Therefore, calling something an acid or base is only strictly relevant in the context of a specific reaction. $$\ce{H2O + HCl -> H3O+ + Cl-}$$ $$\ce{H2O + NaNH2 -> NaOH + NH3}$$
In the first reaction, water accepts a proton from hydrogen chloride, so it is acting as a base. In the second reactions, water donates a proton to the amide ion, so it is acting as an acid. So you can see that acidity is all relative.
A way of quantifying this relationship is to measure the $\mathrm{p}K_\mathrm{a}$ of a species. This is a measurement of the equilibrium position of the following reaction. $$\ce{HA + Solvent <=> A- + SolventH+}$$
It is defined in a similar way to pH: $$\mathrm{p}K_\mathrm{a} = -\log_{10}K_\mathrm{a}$$ $$K_\mathrm{a} = \frac{a_{\ce{A-}}a_{\ce{SolventH+}}}{a_{\ce{HA}}}$$
Tables of $p\mathrm{K_a}$ values can be found in places such as here.

Water shows its acidic nature when strong bases are used. Group I elements reacts with water (like any other acid) to liberate hydrogen gas.
However, you have to remember that water is a polar compound. Oxygen has enough electronegativity to pull the bonded pair of electrons toward itself but hydrogen being a partial member of group 17, has the power (I mean electron withholding property) to resist the oxygen's effort to pull all electron's toward itself. Hence water is unable to show its acidic nature.
Moreover, water has low ionisation potential to liberate hydrogen ions. Even if it dissociate, equal number of $\ce{OH-}$ and $\ce{H+}$ are produced to cancel each other's effect and remain neutral.


text:
How is the time for a test tube in a centrifuge determined?
I came across some hairy math that's used for analytical centrifugation, mostly for things like determining the sedimentation rate for liquids.
However, what I don't understand is how someone determines how long a test tube should be inside a centrifuge? Is it based off of others experimental data or is there a fixed way of doing it?

It is likely determined by experimental trial and error. There may be some standard ways for estimating the speed and time for a given solvent, particle type, and particle size.

"Samples were centrifuged at 5000rpm for 10 min; if we do it longer a funny smell comes out the back of the centrifuge #OverlyHonestMethods" ( https://twitter.com/GavinHub/status/288601830257209346 )

The short answer: check out equation 5 below
I will explain the whole story in terms of sedimenting particles, but you can easily read this as sedimenting droplets as well.
Necessary condition for sedimentation
If you want particles to sediment the first thing to check is whether their density is higher than the density of the liquid, if not: don't even bother. This can be easily shown on the basis of a force balance with buoyancy and gravity acting on the particles:
$$\tag{1} F_{net}=(\rho_{p}-\rho_{l})V_{p}g $$
where subscripts $p$ and $l$ stand for particle and liquid respectively.
What I should also mention here is that I neglect Brownian motion which can, for very small $V_p$ become a significant effect that avoids sedimentation even when the density of the particles is above that of the liquid.
For now, let's assume the particles are big enough to ignore Brownian motion. If we have a centrifuge it is typical to specify the centrifugal acceleration in terms of $a=Ng$ where $N$ is a multiplication factor dependent on the rotation speed of your centrifuge and the distance of the test tube to the center of rotation ($a=\frac{v^2}{r}$, see the wiki for details). The net force acting on the particles in a centrifuge is thus:
$$\tag{2} F_{net}=(\rho_{p}-\rho_{l})V_{p}Ng $$
Settling velocity
After the initial acceleration of the particles there will quickly be a force balance between the net force introduced by the centrifuge and the drag force on the particle. For a perfectly spherical particle that doesn't settle to fast this is simply the Stokes drag:
$$\tag{3}F_{drag}=6\pi\mu_l R_p v_s$$
where $\mu_l$ is the liquid viscosity, $R_p$ is the particle radius and $v_s$ is the settling velocity of the particle.
Balancing the drag with the centripetal force we obtain:
$$\tag{4} v_s=\frac{2(\rho_p-\rho_l)R_p^2Ng}{9\mu} $$.
Finally, to determine the settling time we need the liquid height in the test tube $h$ so we find $$\tag{5} t_s=\frac{h}{v_s} =  \frac{9\mu h}{2(\rho_p-\rho_l)R_p^2Ng } $$
Note that with respect to the settling height we take the worst case scenario here because we want all particles (also those close to the surface of the liquid) to settle.
Equation 5 nicely shows the influence of various parameters on the time a test tube has to stay in the centrifuge. Smaller particles means longer times, lower density difference also and obviously higher centrifugal acceleration means shorter times. Plugging in some typical values for $1 \mu$m particles in water $(\Delta\rho=100$kg/m$^3)$ and a strong centrifuge ($N=10^4$) we find that we need $O(10 s)$.
Discussion of the simplifications
Obviously there are quite some assumptions in the derivation above.

Brownian motion might not be negligible therefore either increasing the settling time, or avoiding settling altogether.
Particles need to accelerate to their settling velocity first, this will also increase the settling time a bit.
In general particles will not be spherical. If, for example, the particles are more rod-shaped then the orientation will determine whether the settling time increases or decreases with respect to the spherical particle prediction.

For many types of particles a Svedberg coefficient is tabulated in literature which will allow you to calculate the settling time without having to assume sphericity of the particles. In this document they indicate the use of Svedberg coefficients.
Clearly, these simplifications can mean that the prediction of equation 5 is of by a factor 10 (or maybe even 100 if you're unlucky) therefore in most practical situations you would use this equation as a starting point for some trial-and-error experiments. You can also use it to determine what will happen to the settling time when you e.g. run a new experiment with a different liquid or particle type.


text:
How can oxygen atoms suddenly appear on the product side of an equation?
The question states the following  equation:
 $$ \ce{C(s) + 2H2(g) -> CO2(g) + 2H2O}$$
 And then asks you to calculate the enthalpy change for it, given enthalpies of combustion.
The stated values are $C(s)$ -393,  $\ce{H2}(g)$ -286,  and $\ce{CH4}(g)$ -890
I'm curious as to why oxygen atoms have suddenly appeared on the products side. The only thing I have been able to think is that it refers to the enthalpy of formation of methane, and that the answer would be -393 + (2*-286) + 890. But it still doesn't make sense to me considering the equation provided. I understand the products are the products expected from combustion, but no oxygen has been added on the reactants side.

There's a serious problem with the equation.  You've already noted that.  If the problem came from a book, check the book.  If it came from a homework assignment, ask about that.

They probably want you to calculate the enthalpy change required to convert the reactants into the products. Another way of looking at it is the enthaply difference between the products and reactants.
Remember, $\ce{O2(g)}$ has 0 enthalpy of formation in standard state. (so do $\ce{C(s)}$ and $\ce{H2(g)}$, but we don't need that)
So, adding $\ce{2O2}$ to the reactants side won't change $\Delta H^o=H^o_{products}-H^o_{reactants}$. So calculating $\Delta H^o$ for the given "reaction" is the same as calculating it for $$\ce{C(s) + 2H2(g) +2O2 -> CO2(g) + 2H2O}$$. And $\Delta S^o$ for this reaction can be calculated very easily using enthalpies of combustion. You don't even need to use the given value for methane.
So, while the question is wrong in calling it a reaction, it is still asking us to do something reasonable -- that is, calculate $$\Delta H^o=H^o_{\ce{CO2(g) + 2H2O}} - H^o_{\ce{C(s) + 2H2(g)}}$$. Since we can subtract $H^o_{\ce{2O2(g)}}=0$ easily, the problem can be solved by taking a combustion path.


text:
Is it possible to form bubbles of ionized hydrogen and ionized oxygen under the right pressure/temperature conditions during electrolysis of water?
When we run a current through water (pure or not), $\ce{O_2}$ forms at the anode while $\ce{H_2}$ forms at the cathode. If I understand correctly, as water is already slightly ionized, $\ce{H}$'s snap together (and so do $\ce{O}$'s) during the electron exchange. Is it possible to apply enough heat or enough pressure to water to prevent $\ce{H}$'s and $\ce{O}$'s from snapping together and so we could thus form bubbles of ionized hydrogen and ionized oxygen?

The electrolysis will not form any ionized oxygen or ionized hydrogen, but instead proceeds through radical species. Some experiments have attempted to show the presence of $H^{\bullet}$ and $OH^{\bullet}$ radicals at the anode / cathode respectively (see P.H. Kasai, D. McLeod J. Am. Chem. Soc (1978)). Provided that the surface concentration is sufficient, these radical species will very quickly recombine to lead to $H_{2}$ and $O_{2}$. The surface of the electrode is very likely to influence the process, as the radicals species are likely to interact with it and not be released in high concentration in the solution.
Other factors are also influential, and the radical species might also react with other constituents of the solution (see for instance T. Mokudai,  K. Nakamura,  T. Kanno,  Y. Niwano, PLOS One (2012) on the formation of hydroxyl radical vs formation of hypochlorous acid through electrolysis).
Heating the solution is likely to increase radical concentration, but the temperature should be extremely high to prevent radical recombination within the gas. As for pressure, my guess would be that a reduced pressure is likely to improve radical isolation by displacement of the solvation equilibrium.


text:
Questions about intramolecular interactions in AMBER forcefield
I am trying to run a molecular dynamics simulation with AMBER forcefield. However, it's the first time I use is, and while I know the form of the various intra- and intermolecular energy terms, some details are not clear to me. So, for the AMBER forcefield:

What pairs are excluded in the summation of intramolecular Lennard-Jones interactions? Typical choices are “excluding bonded atoms”, “excluding 1-2 and 1-3 pairs”, “excluding 1-2 and 1-3, applying a factor 1/2 to 1-4 pairs”, and “excluding 1-2, 1-3 and 1-4 pairs”.
For torsion potentials, is the potential scaled by the number of such torsions? (as is done in the DREIDING forcefield, to ensure a constant torsional barrier for the central bond)

As for your first question, this site states that 1-4-interactions are scaled by a factor of 0.5, which I suppose implies 1-2 and 1-3 interactions being excluded.
The answer to your second question is yes, as can be seen here (section TORSIONS). The potential is divided (and thus scaled) by Ntors.


text:
Why do single elements in a molecule have an oxidation state of zero?
For example, take $\ce{O2}$.
The oxidation state of oxygen is -2, yet once its in a molecule its oxidation state becomes zero?
How is this so?

Very strictly speaking, I don't think this must necessarily be true. It is possible to imagine allotropes of a pure element which would consist of ions and thus would require non-zero oxidation numbers. For example, the hypothetical octanitrogen $\ce{N8}$ allotrope could be composed of the azide anion $\ce{N3^{-}}$ and the pentazenium cation $\ce{N5^{+}}$. Granted, I do not believe this to be the case for any element, at least not in ambient conditions.

By definition oxidation state or oxidation number is related to the number of electrons that an atom, molecule or any substance either gains or loses while forming a new substance in a chemical reaction. It is positive when electron leaves and negative when electron enters an atom. In some cases it can be zero because of sharing the electron pair to form a coordinate covalent bonds between identical atoms having same electro negativity values.

Oxidation states are something of an "accounting fiction", used to describe where electrons are spending most of their time in a molecule.  This is turn is dictated by electronegativity: the more electronegative, the more the element attracts electrons.
So when you say oxygen's oxidation state is -2, this is only with respect to other elements of lower electronegativity. Now in the case of oxygen, the only other element with higher electonegativity is fluorine. For example, the molecule Oxygen Difluoride ($\ce{OF2}$) has oxygen in the +2 oxidation state because fluorine is more electronegative!  For any other element in combination with oxygen, however, we assume the electrons spend their time with the oxygen filling its outer valence shell, thus the -2 oxidation state.
But two oxygens in a molecule of gas have the same electronegativity, and there is no way to say that the electrons want to spend more time with one or the other oxygen.  Therefore, the "average" electron state for an oxygen molecule is as for the individual atoms, and so the oxidation state is 0.

We do have at least one single-element species where the atoms have nonzero oxidation states: ozone.  Unlike most single-element substances, ozone has oxygen atoms in nonequivalent positions (from https://commons.wikimedia.org/wiki/File:Ozone-resonance-Lewis-2D.png#mw-jump-to-license; created by Ben Mills; license information in the link):

The central oxygen atom shares more of its electrons and retains fewer unshared pairs than its neighbors and thus, according to the oxidation state formalism of nonpolar oxygen-oxygen bonds, takes on a positive charge.  This atom thus has an oxidation state of +1.  The other atoms, which take turns sharing fewer than normal electrons, are each considered to have an oxidation state of -1/2.
The charge separation implied above shows up experimentally as a nonzero dipole moment.  Even with nonpolar bonds, the unequal distribution of nonbonding electrons would create a dipole.
Sulfur forms a similar triatomic molecule, $\ce{S3}$, with nonzero oxidation states of the individual atoms, which has been identified on Jupiter's moon Io.

Also while being 0 to the outside we know some compounds that are so called inner alloys. Take manganese for example. In its elemental form you have a very complicated structure consisting of various Frank-Kasper polyhedra. And there are some effects, which can be explained more easily if we assume different oxidation states for the Mn atoms in the Mn metal. To the outside it still is 0 but the amount of electrons each Mn gives into the conduction band is not the same for each Mn atom. At least this is how my Prof once explained the strange structure of Mn to me. You can even manipulate the amount of some of these oxidation states and make alloys with increased Mn(II)-high spin character for example, which show stronger magnetic effects.

Sounds interesting, but did your professor find it interesting enough to publish citeable results? Having such references is a good way to attract reads, upvotes, and even citations on answers elsewhere in SE.

authors: Vladimir F Героям слава


text:
Are all the phenyl protons in benzyl alcohol equivalent in the ¹H-NMR spectrum?
Below is a $\ce{^1H}$ Nuclear Magnetic Resonance (NMR) spectrum for benzyl alcohol. The hydrogen in the alcohol group gives rise to a unique signal. The two hydrogens in $\ce{CH2}$ give rise to a unique signal with around twice the signal area. What I don't understand is why all hydrogens in the phenyl group are equivalent. My guess would have been there would be 3 signals from the phenyl ring. One pair of $\ce{H}$ that are equally distant from the methanol, another pair that is also equally distant, and last, a single hydrogen that is the furthest away from the methanol. Clearly this is not the case, and the 5 hydrogen give rise to a single signal only.
Why? What is a good way to think about equivalent and non-equivalent atoms in NMR spectrometry? I would rather avoid rules like the one that says "if you can substitute the hydrogen with another atom and it gives the same compound, the hydrogens are equivalent". I would rather like an intuitive way to think about it, if it's possible!

You're right, those five hydrogens are not magnetically equivalent. You would expect to see three signals for the aromatic protons.
If you look closer you'll see that the aromatic signal is not a perfect singlet, but is rather more complicated with an additional small peak and a broader base, it is a multiplet. You have three distinct signals, but they are so close in chemical shift that they overlap and result in this multiplet.

The three signals are very close in chemical shift, and have a strong coupling. If you have heard about $AB$ spin systems, where the CS difference is similar to the coupling constant, which make the outer two lines of the two doublets become weaker ("roof effect"), and the inner two stronger and move closer to one another: This is an $AB_2C_2$ system. The inner lines coalesce completely, and the outer lines become an unshapely foot around the central line.
If you had a stronger magnet, where the CS difference (in Hz) is proportionally larger, you would get multiple peaks and might even be able to integrate and identify them.


text:
Primary amine versus phosphoramidate reactivity with succinate esters
I have posted two reactions below. The top reaction is well known and occurs readily and near quantitatively (e.g. in protein modification), while the bottom reaction to my knowledge does not exist in the literature, but would be valuable to me. I would like to ask the experienced organic chemists out there if you think 2 and 4 will have similar reactivities with 1, or whether the phosphonate moeity will cause trouble or  reduce reactiveness (resonance stabilization or side reactions?).
Forgot to add it to the image, but I normally run the top reaction in dichloromethane at a little above 0 C, with N-hydroxysuccinimide produced as a by-product.

Note: I currently don't have the time/resources to test this, but plan to in the future if I think it will work.

amine group in the second reaction is more like amine group in amides. If the first reaction does work with amides, the second reaction has a chance.

this will probably be a no go then, since amides are not strong enough nucleophiles to react here... :( back to the drawing board (thank you for your comment though)

You can probably deprotonate the phorphoramidate and get a decent reaction.


text:
What is the difference between the titration of a strong acid with a strong base and that of the titration of a weak acid with a strong base?

Which of the following statements best characterizes the difference between the titration of a strong acid with a strong base and that of the titration of a weak acid with a strong base?
A) Both the strong acid and the strong base are completely dissociated in solution whereas the weak acid dissociation equilibrium constant is small enough that there is a measurable amount of the un-dissociated acid present in aqueous solution.
B) The equivalence point of a titration of a strong acid with a strong base may be observed with an indicator whereas the equivalence point of a titration of a weak acid with a strong base can only be observed with a pH meter.
C) The titration of a strong acid with a strong base is a fast reaction whereas the titration of a weak acid with a strong base is generally a very slow reaction.

This website seems to indicate that choice C is correct.
For B, all I know is that the pH at the equivalence point does not equal 7.00 (pH > 7.00) for the weak acid titration.
For A, I know the first part of that sentence is true. Strong acids and bases are characterized by completely dissociating whereas weak acids do not dissociate to the same extent.  But is it true that there's a "measurable amount" of the un-dissociated acid present in aqueous solutions?

The answer is A.
Let's look at a weak acid: acetic acid. It has a $K_a$ of $1.8\cdot  10^{-5}$. From this you can calculate how much it will dissociate in water. Let's say we have a concentration of $1~\mathrm{M}$:
\begin{align}
K_a &= \frac{[\ce{CH3COO^{-}}][\ce{H+}]}{[\ce{CH3COOH}]}\\
1.8 \cdot 10^{-5} &= \frac{x \cdot x}{1~\mathrm{M} - x}\\
x &= 4.233 \cdot 10^{-3}\\
\end{align}
This means, at equilibrium, $[\ce{CH3COOH}] = 1~\mathrm{M}- 4.233 \cdot 10^{-3} = 0.995~\mathrm{M}$. As you can see, by far the largest part is not dissociated!
B is not true, you can still observe the equivalence point. Have a look at the following picture, that shows the titration curve for weak / strong acid titrated with strong base. When choosing an indicator for colorimetric titration select one so that the pH jump at the equivalence point contains the interval $\mathrm{p}K_a\pm1$. Phenolphthalein has a $\mathrm{p}K_a \approx 9$, so to decide if it is a suitable indicator you would check if the pH jumps from 8 to 10 at the equivalence point. It does:

C is not true either. If it was a very slow reaction, you could not titrate accurately, as you would have to wait a long time between every drop of $\ce{NaOH}$.