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Electrolysis of Water
What is the mechanism for the electrolysis of water? I understand that hydrogen and oxygen gas are made, but how exactly does this happen when electrons are passed through water? A diagram would be very useful.
Would this have anything to do with electronegativities?

Related: If electrolysis splits water, why does only either hydrogen or oxygen (but not both) bubble up on one pole?

Nobody has given an answer yet, so I'll give it a crack.
In electrolysis, as you probably know, there are two electrodes – the cathode (negative) and the anode (positive). We can deal with what happens at each separately, then tie them together.
At the cathode:
$$\ce{2H2O + 2e- -> H2 + 2OH-}$$
You're interested in the mechanism of these reactions. We know that water exists in solution in an equilibrium:
$$\ce{H2O + H2O <=> H3O+ + OH-}$$
At the cathode, electrons are supplied to the $\ce{H3O+}$ ions, reducing them:
$$\ce{H3O+ + e- -> H2O + H* }$$
With many reactions happening at once, the $\ce{H*}$ radicals react together, forming $\ce{H2}$. By removing the $\ce{H3O+}$ ions, it shifts the equilibrium of water to the right, to oppose the change – removing $\ce{H3O+}$ ions and increasing the concentration of $\ce{H2O}$:
$$\ce{H2O + H2O <=>> H3O+ + OH-}$$
Because the concentration of $\ce{OH-}$ ions doesn't change, the local solution around the cathode becomes basic – this can be done in a school lab. Add indicator. Electrolyse. Observe the color change near the cathode.
At the anode:
$$\ce{2H2O -> O2 + 4H+ + 4e-}$$
Here I suspect the mechanism is a bit similar:
$$\ce{H2O + H2O <=> H3O+ + OH-}$$
We take the $\ce{OH-}$ and oxidise it:
$$\ce{OH- -> OH* + e-}$$
This results in a radical – the hydroyxl radical – being formed. The steps from here, to get to water being formed are probably quite complicated as a mechanism.

I would very much doubt the formation of (stable) radicals in aqueous solution. Especially not the ones we have here.

Hmm. Yes. It has to be said that I would also doubt that, after a bit more research. However, surely radicals could form in aqueous solution, but would immediately react with other surrounding molecules - for example, in the body, hydroxyl radicals formed by radiation, are in aqueous solution - just they react very quickly, causing a free-radical chain reaction.

I'm going to update my answer with a more likely reaction mechanism - it seems to involve the molecules adsorbing onto the metal, and hydrogen bonding.

Maybe have a look at the Grotthuss Mechanism, I think that could explain the conductivity and electrolysis without radicals. (But in principle I thought your explanations made a lot sense.)

Swedish Architect's answers already proposed quite thoughtful insight into this question. The only thing I was missing there were some solid research results on the matter to backup these reasonable claims. In course of my own research I came across a website, that provided me with a quite recent density functional study of the subject: "Electrolysis of water on (oxidized) metal surfaces" by J. Rossmeisl, , A. Logadottir and J.K. Nørskov. Chemical Physics 2005, 319 (1–3), 178–184.
I will here try to outline their reaction mechanism, which they used for their quantum chemical treatment. In their work they considered platinum and gold surfaces as electrode material. I will not go into detail in analysing their results, since this exceeds the scope of this question.
Water splitting follows the general net reaction process of $$\ce{H2O -> 1/2O2 + H2}.$$
At the cathode the following mechanism is proposed in acidic media:
\begin{align}\ce{
2H+ + 2e- &-> H^{$*$} + H+ + e- \\
          &-> H2
}\end{align}
The asterisk superscript denotes, that the formed Hydrogen $\ce{H}^*$ is chemically adsorbed to the active site at the metal surface. They continue that hydrogen is only weakly bound to the metal surface and therefore the hydrogen evolution is much faster than the rate of oxygen production. Hence they focus on the reaction on the anode.
The oxygen evolution takes place at the anode and follows a more complicated mechanism. Again the asterisk denotes that the molecule is chemically adsorbed the the metal surface.
\begin{align}\ce{
2H2O &-> HO^{$*$} + H2O + H+ + e- \\
     &-> O^{$*$} + H2O + 2H+ + 2e- \\
     &-> HOO^{$*$} + 3H+ + 3e- \\
     &-> O2 + 4H+ + 4e-
}\end{align}
The key step in this reaction series is the formation of the active peroxide intermediate as they found out. It is therefore also the rate determining step. The doubly bonded  $\ce{O}^*$ and singly bonded $\ce{HO}^*$ species are relatively strongly bound to the surface. They also found out, that a coverage of up to $2/3$ of the surface with a monolayer of oxygen is necessary in order for the peroxyl species to become more easier than $\ce{O}^*$ and $\ce{HO}^*$.
In an alkaline electrolyte they propose the following mechanism, which only slightly deviates from the above. The intermediate species are the same, that's why they did not need to treat them separately. On the cathode they just summarised as $$\ce{2H2O + 2e- -> 2{}^{-}OH + H2}$$ as the mechanism is bound to be the same. On the anode on the other hand the following steps should be observed:
\begin{align}\ce{
4{}^{-}OH &-> {}^{$*$}OH + 3{}^{-}OH + e- \\
          &-> {}^{$*$}O + H2O + 2 {}^{-}OH + 2e- \\
          &-> {}^{$*$}OOH + H2O + {}^{-}OH + 3e- \\
          &-> O2 + 2H2O + 4e- \\
}\end{align}
Additional note and criticism
Water is only liquid at room temperature due to a strong network of hydrogen bonds. As the included monolayers of water molecules it is sufficient to say that the system is treated extensively. However, what they failed to describe in their model is the conduction of electricity through the electrolyte. The reason why electric current passes quickly through an aqueous medium can be benefitted to the Grotthuss mechanism. This being said, I would think that the treatment of more than one monolayer should be considered, however I understand that this might not be possible due to technical limitations.
Another detailed description of the mechanism based on cyclovoltametrie can be found in Catalysis Today 2013 202, 105–113: "Water dissociation on well-defined platinum surfaces: The electrochemical perspective" by Maria J.T.C. van der Niet, Nuria Garcia-Araez, Javier Hernández, Juan M. Feliu and Marc T.M. Koper.

Is there a way to access thi "Electrolysis of water on (oxidized) metal surfaces" by J. Rossmeisl, , A. Logadottir and J.K. Nørskov. Chemical Physics 2005, 319 (1–3), 178–184. without buying or being a member?

@SwedishArchitect Maybe you find a mirror of a preprint somewhere, or you find a university with access near to you. But apart from this I guess there is no way.

It cannot be stressed enough that O-O formation is necessary for electrolysis to occur (electrolysis is kind of my specialty). This is quite clear from oxygen formation alone. Another factor that must be strongly reinforced is that the role of the electrode surface is the most important part of these reactions (including peroxide formation) and is the primary reason why the mechanism is highly elusive.

Sorry for reviving this old post, but for the splitting of water, wouldn't it occur forever? The hydrogen and oxygen gas produced exits the water, and it is no longer in equilibrium.

@Mathelogician Electrolysis of water is not an equilibrium process to start with. If you supply current, you'll split water, if you don't you don't. It does not matter whether hydrogen and oxygen exit the water, as long as there is enough to be electrolysed.

@Martin-マーチン Yes, but if I just throw a 9-volt battery in there, the electrolysis would happne forever, right? because the gases are produced, and instantly leaves the water, ad infinitum.

@badatchemistry Only if you had an infinite supply of water and an infinite battery the reaction would carry on infinitely in an open system. In all other cases it won't. And that has little to nothing to do with the gasses escaping.

@Martin-マーチン Then what stops this reaction? (and also can we move it into a chatroom because it's getting a bit long?)

If you don't have starting materials, you cannot proceed with the reaction. You can find and ping me in [chat], sure. @badatchemistry

I'm adding another answer, representative of another mechanism, which I think is probably more likely than the one I proposed. It follows a similar line to the previous mechanism, but involves the adsorption of water molecules onto the cathode and anode. I will consider the reactions at the anode and the cathode separately, like in my previous mechanism.
Water is a polar molecule - the oxygen atom is more electronegative (the AOs on oxygen are lower in energy than the 1s orbital on the hydrogens, which contribute to the bonding MOs) than the hydrogen atoms, and thus the molecular orbitals will have a higher electron density on the oxygen than on the hydrogen. This gives the oxygen a small negative charge, and each hydrogen a small positive charge - not a full, ionic charge.
Because of this charge liquid water has a network of hydrogen bonding, as the positively charged hydrogen attracts a lone pair from another oxygen, thus producing a hydrogen bond - an attraction between the water molecules.
At the cathode (-ve electrode) the water molecules align themselves with this negative charge on the metal cathode. The hydronium ions are reduced by the negative metal electrode. Thus we have $$\ce{H3O+--M + e- -> H2O + H-M}$$
This explains the production of hydroxide ions - turning the solution basic at the cathode. The hydrogen is bound to the metal , using a d-orbital. (If you're interested look up hydrogen adsorption onto metal surfaces). These hydrogen atoms are able to diffuse across the metal surface, in two dimensions, kind of like skating across an ice rink - you can't jump, but you can skate easily across it (unless you're really good!). These hydrogen atoms can then react together, forming hydrogen gas. Consider the equilibrium of water:
$$\ce{H2O + H2O <=> H3O+ + OH-}$$
Decreasing the concentration of $\ce{H3O+}$ means there is proportionally more $\ce{OH-}$ than $\ce{H3O+}$, which results in the solution near the cathode becoming basic. The hydrogen bound to the metal is not particularly stable - therefore a pair of hydrogen atoms can react, becoming $\ce{H2}$, as the hydrogens are able to diffuse across the surface of the metal.
At the anode (+ve), we know that $\ce{O2}$ is produced. Again the water molecules align themselves to the charge - the negative oxygen is attracted to the positive metallic anode. The positive metal will interact with the electrons involved in bonding in water. We can envisage this as the positive metal "pulling" electron density from the oxygen. This in turn pulls electron density from the two bonds, joining the oxygen and the two hydrogen atoms. This is the inductive effect - where a positive charge is transmitted through atoms - the electron withdrawing effect. This weakens the $\ce{H-O}$ bonds. The hydrogen atoms are then solvated, to form the hydronium ion, $\ce{H3O+}$. Reaction:
$$\ce{2H2O + H2O -> 2H3O+ + O + 4e-}$$
The 4 electrons are then used to reduce the hydronium ion at the cathode - remember this is a circuit. The oxygen remains bound to the metal, and then reacts with another oxygen to form $\ce{O2}$.
I think this mechanism is more likely than the other I proposed. I am not sure about how the oxygen is bonded to the metal - in a normal coordination complex the metal donates electrons, forming a coordinate/dative bond, instead of oxygen donating a lone pair to form a dative bond. However, in this case the metal is positively charged, with an active potential difference - energy is being put into the system. I would be interested for more experienced/knowledgable people to improve this mechanism and point out reading on this.

I would add to the first answer that the formation of the hydroxyl radical can also form a small amount $\ce{H2O2}$ via
$\ce{OH⋅ + OH⋅ <=> H2O2}$.
The unmatched hydrogen radicals could react forming $\ce{H2}$ and leave the solution. The newly form $\ce{H2O2}$ could attack, for example, a copper electrode contributing to anodic (or cathodic) corrosion.
Galvanic cells frequently have many non-electrochemical based side reactions occurring as well.

The consensus in the research community is that peroxides form on the surface of the electrode, and not in the bulk. It has been posited for quite a while that O-O formation is necessary for electrolysis to occur. See any general review paper for water oxidation, such as: http://onlinelibrary.wiley.com/doi/10.1002/cctc.201000126/abstract

In pure water at the negatively charged cathode, a reduction reaction takes place, with electrons ($\ce{e^{−}}$) from the cathode being given to hydrogen cations to form hydrogen gas (the half reaction balanced with acid):
Reduction at cathode: $\ce{2 H+(aq) + 2e^{−} -> H2(g)}$
At the positively charged anode, an oxidation reaction occurs, generating oxygen gas and giving electrons to the anode to complete the circuit:
Oxidation at anode: $\ce{2 H2O(l) -> O2(g) + 4 H+(aq) + 4e^{−}}$
The same half reactions can also be balanced with base as listed below. Not all half reactions must be balanced with acid or base. Many do, like the oxidation or reduction of water listed here. To add half reactions they must both be balanced with either acid or base.
Cathode (reduction): $\ce{2 H2O(l) + 2e^{−} -> H2(g) + 2 {}^{-}OH(aq)}$
Anode (oxidation): $\ce{4 {}^{-}OH (aq) -> O2(g) + 2 H2O(l) + 4 e^{-}}$
Combining either half reaction pair yields the same overall decomposition of water into oxygen and hydrogen:
Overall reaction: $\ce{2 H2O(l) -> 2 H2(g) + O2(g)}$
See also: http://en.wikipedia.org/wiki/Electrolysis_of_water

This is straight out of Wikipedia

Yes, I know it's a very late reply, but this post is still useful in that learning that sci-hub has many freely-available papers that are usually locked behind a paywall is useful.
We'll not get into why those papers are behind a paywall, nor why the researchers creating those papers generally do not benefit from that paywall, only the publisher does.
In reply to Swedish Architect's query as to where to find that paper:
Catalysis Today 2013 202, 105–113: "Water dissociation on well-defined platinum surfaces: The electrochemical perspective" by Maria J.T.C. van der Niet, Nuria Garcia-Araez, Javier Hernández, Juan M. Feliu and Marc T.M. Koper.
https://sci-hub.st/10.1016/j.cattod.2012.04.059

AFAIK using SciHub is controversal, regarding legality of free access. It may be illegal in many countries. //  CH SE prefers avoiding direct links to its sites. Better providing just article and DOI URL reference. Who want to obtain it from Sci-hub, he can, but CH SE site prefers not promoting it.

One question not yet addressed is what happens when the electric potential applied is less than the dissociation potential of water. We know that water swaps protons (deprotonation / reprotonation) naturally all the time. Would a low electric potential cause the OH- and H3O+ to 'gravitate' in different directions, thus not allowing them to recombine back into H2O? Given that H3O+ has a higher molecular mass than H2O (and OH- has a lower molecular mass), could buoyancy be used to augment the electrical potential to keep the OH- and H3O+ apart?

I ask because OH- is an integral part of the initiation of the combustion process... and creating a faster flame front in-cylinder of an internal combustion engine (by injecting OH- along with the fuel) should allow spark timing to be retarded (so the piston isn't fighting the combusting, expanding gases for as much of the rotational cycle), thus increasing engine torque while ensuring more complete combustion.


text:
Does the arrangement of atoms around a chiral carbon matter?
In other words, are these two structures equivalent?

Yes, they are both (S)-2-chlorobutane.

The difference between these structures is a rotation of the molecule by 60 degrees. You could build a model to convince yourself that they are the same. The arrangement of the atoms around the chiral center does matter.
The utility of the R/S convention is that you can unambiguously decide whether the two compounds are the same or different, which can be difficult depending on the orientation of the compound.

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How to rotate a chiral carbon

I know that to correctly assign R,S priorities I have to rotate the above figure until the H is projecting into the page (away from me). However, I don't know how to rotate the H and the rest of the molecule when the H is in the plane of the page.
When I imagine tilting the molecule back I get a see-saw shaped molecule ... obviously incorrect.

Envision that you are rotating the molecule on the C-Ethyl bond with the hydrogen going into the page. Remember, it won't lie flat.

Ah, I think I need to buy a model, or convert this to a Fisher projection.

You can download ACD ChemSketch, and draw the molecules. Then you can rotate them in 3D.

The "1" and "3" should be solid lines in the plane of the paper; the "4" going back as a dashed wedge as you've shown it, and the "2" being a solid wedge projecting towards the observer.

No matter whether someone is a beginner or advanced, models are extremely helpful.

One advantage of ACD ChemSketch is that you can generate the name for small molecules such as the one above: (2S)-butan-2-ol.

Thank you all for the help. I'll have to check out the mentioned software. I'll also try rotating this with a model.

A recommendation three years after the fact. If you are redrawing the structure to orient the lowest priority group in the rear, don't do it. There's a good chance you might mess it up. Leave it where it is. Use your HANDS! Point the thumb of your left hand from the chiral carbon to 4 (lowest priority). Now your fingers will point in the direction 1-->2-->3. Your left hand is sinistral. S-configuration.

Imagine viewing it from right side:

H is upwards and CH3CH2 is downwards.
OH is leftwards and CH3 rightwards.
H and CH3CH2 are backwards(from right) so they must be on vertical line of Fischer projection and similarly for OH and CH3.

Alternatively:

Put the dotted wedge CH3 in any vertical position let it be bottom.
Then imagine all other groups to be planar, we see in clockwise order:

CH3CH2 , H , OH

Draw them similarly on the Fischer projection in clockwise sense.!

Note that both are (S)

Swap number 4 for the number of the bond/molecule/atom that is pointing away from you. Decipher whether it is R/S by drawing from 1 to 3. The answer is opposite to this.
https://www.youtube.com/watch?v=x28IcxUhu3A (From 9 minutes 25 seconds on)

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How does amine inversion affect basicity?
What is amine inversion, and how does it affect basicity?
For instance, why is $\ce{Et3N}$ less basic than (2.2.2.)cyclohexyl-amine despite the +I from ethyl groups? The answer to the question is option 3.

  Compare the basic strengths of the compounds given:

(i) > (ii) > (iii)
(ii) > (i) > (iii)
(ii) > (iii) > (i)
(iii) > (ii) > (i)

Amine inversion is also known as nitrogen inversion. You can find more here: http://en.wikipedia.org/wiki/Nitrogen_inversion
It has nothing to do with relative basicity of amines, that is defined by 1) energy of particular N-H  bonding (increases with +I effect) and 2) energy of ammonium cation hydration. (decreases with alkyl group number and their size). The relative balance of said factors is fragile and cannot be easily predicted without reference to experimental data except cases when amine has lone pair involved into bonding interactions (aniline, pyrrole, amides) or connected with extremely strong electron acceptors.

https://i.stack.imgur.com/CJPxj.jpg

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Field Effect and it's effect on basicity on amines
What is the order of basicity between $\ce{Me2NH}$, $\ce{Me3N}$, $\ce{MeNH2}$ in a protic solvent.
It has something to do with field effect which I could not understand and could not find anywhere.
Can someone please explain?

Try searching for the inductive effect.

Here's the IUPAC definition on the field effect: http://goldbook.iupac.org/F02358.html

In case of Ethyl in protic solvent the basicity is:
$$\ce{Et2NH \gt Et3N \gt EtNH2}$$
This is a result of two opposing factors, inductive Effect and Solvation. In case of other alkyl groups than Ethyl:
$$\ce{R2NH \gt RNH2 \gt R3N}$$
And it is as the following when inductive effect becomes dominant in gaseous state:
$$\ce{R3N \gt R2NH \gt RNH2}$$
Field Effect (or I may say inductive effect) as stated on Wikipedia
The 'Inductive Effect' is an experimentally observable effect of the transmission of charge through a chain of atoms in a molecule.
The permanent dipole induced in one bond by another is called inductive effect.
The electron cloud in a $\sigma$-bond between two unlike atoms is not uniform and is slightly displaced towards the more electronegative of the two atoms. This causes a permanent state of bond polarization, where the more electronegative atom has a slight negative charge ($\delta^{–}$) and the other atom has a slight positive charge ($\delta^{+}$).
If the electronegative atom is then joined to a chain of atoms, usually carbon, the positive charge is relayed to the other atoms in the chain. This is the electron-withdrawing inductive effect, also known as the $-I$ effect.
Some groups, such as the alkyl group, are less electron-withdrawing than hydrogen and are therefore considered as electron-releasing. This is electron releasing character and is indicated by the $+I$ effect.
In short, alkyl groups tend to give electrons, leading to induction effect.

The $\ce{R}$ groups are $+I$ that give $\ce{N}$ a more electron density and hence $\ce{N}$ becomes more basic, and hence, we can say that $\ce{R3N}$ would be most basic as it has the most alkyl groups, and thus the basicity in gaseous state can be predicted. But in solution, the water molecule tend to solvate amines and decrease its basicity, the solvation depends, here, mainly on size as difference between $\ce{-R}$ and $\ce{-H}$ groups and thus the two orders of basicity arise for alkyl amines.

also theoretically, it is a permanent effect and not a temporary one and thus can be seen experimenally with electrostatic potential models

Wikipedia is not a solid source. The goldbook however provides a similar definition and can be referred to as a standard. Please provide sources in your answers whenever possible.

The basicity of amines has to do not only with the degree of substitution with electron-releasing alkyl groups but also factors such as steric hindrance and degree of solvation.
The electronegative nitrogen is able to isolate electron density from the alkyl groups. So having more electron-releasing alkyl groups to stabilize the positive formal charge created by the abstraction of a proton from the protic solvent might seem like a good thing.
On the other hand steric hindrance due to bulky alkyl groups reduces basicity.
Also, having more alkyl groups reduces solvation due to hydrogen "bonding." Consider ammonia, $\ce{NH_3}$. This molecule can form four hydrogen bonds. On the other hand, a tertiary amine that has been protonated can form only "good" hydrogen bond; one might argue that the hydrogens on the alkyl groups can form hydrogen "bonds", but these are obviously going to be weaker "bonds" compared to the ones among ammonia molecules.

So according to you what will be the order of basicity?

I don't know for sure.


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What is the complete mechanism of reaction for sodium benzoate and ascorbic acid?
What are the complete steps of reaction from sodium benzoate and ascorbic acid to benzene? I can't seem to visualize it, and, much less, find anything online regarding the actual steps of the reaction.
Additionally: how does it make use of, say Cu(II) or Fe(III) to complete the reaction? It seems that the reaction can only proceed with these present.

From Benzene production from decarboxylation of benzoic acid in the presence of ascorbic acid and a transition-metal catalyst:
$\ce {Cu^{2+} + H_2Asc \rightarrow Cu^+ + HAsc^.}$
$\ce {Cu^+ + O_2 \rightarrow Cu^{2+} + O_2^-}$
$\ce {2O_2^- + 2H^+ \rightarrow H_2O_2 + O_2}$
$\ce {Cu^+ + H_2O_2 \rightarrow Cu^{2+} + OH^- + OH^.}$
Then the hydroxyl radical reacts with benzonic acid as opposed to benzoate.

The hydroxl radical abstracts hydrogen from the acidic proton of the benzoic acid, producing water and a benzoate radical.  This benzoate releases $\ce{CO2}$, producing a $\ce{C6H5}$ phenyl radical, which can abstract a hydrogen from water to regenerate the hydroxyl radical and produce the benzene. https://en.wikipedia.org/wiki/Benzene_in_soft_drinks

Which hydroxyl radical? Answers should be self-contained. Please [edit] it to include how to generate the radical or at least reference another answer if it is explained there.


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Electride salt preparation
In preparation for the Birch reduction, I am trying to prepare an electride salt, $\ce{[Na(NH3)6]+, e-}.$ This salt should be brilliant blue, as far as I know, but I am getting a black sludge when adding sodium to the liquid anhydrous ammonia at −60 °C or so.
Is this because the salt is missing the electron?
How do I remedy this?

electride salts are said to be blueish black at bulk and I beleive its solutions appear black if concentration is high enough.

As long as your experimental setup did not contain any leakages or impurities through which air or moisture could have gotten into the mixture, the reaction should have worked fine under these conditions. The color of the hexaamminsodium electride solution is actually deep blue and intense, so at higher concentrations it can appear black at first sight (see picture below).

Note the blue coloration in a thin section along the rim.


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Major form of methyl 3-hydroxy pent-2-enoate?
I came across this question in a test:

The enol form is the ________ form for 3-hydroxy pent-2-enoate?  (major/minor)

My argument is that the enol form must be more stable because of hydrogen bonding in the enol form and subsequent six membered ring formation (chelation). However the keto form has no such stabilization factor. Hence according to me the enol form must be more stable. But sadly, that is the wrong answer. What's going wrong?
Structure for reference:

quick googling shows that for a very similar compound ethyl acetoacetate enol and ketoforms are in equilibrium with 10% of enol form. However, this equilibrium is influenced by solvent significantly. Actually, the percentage shows that the difference in stability between these two forms is veeery little.
The difference may be attributed to the difference in stability of the bonds, i.e.   C=O + C-H is better then C=C + C-O + O-H. However, internal hydrogen bonding DOES exist and participates in stability of enol form. For example, for hexan-1,3-one, the percentage of enol form is much lower.

Is my thinking about the hydrogen bond stability incorrect? Also, the solvent wasn't mentioned in the question.

@Apurv see edit. It is correct, it exists, but it is not strong enough in this particular case

why isn't it strong in this case?

@Apurv Unfortunately, no idea.

actually the question was asked in a national level exam. I need to send the doubtful questions for review to the exam committee. So I decided to clarify this out first on SE. Do you think that this question deserves a review by the committee? (considering my thinking and that you are unsure about the answer)

I believe, question "is enol form is more stable then keto-form for this particular compound?" cannot be answered without specific knowledge about this one or similar compound and cannot be reasoned from general principles taught in common organic chemistry course. However, asking to determine realtive stability of enol form (i.e. for wich compound percentage of enol form is more)  for several clearly differend compounds, say, cyclohexa-1,3-one and acethylacetone, would be a valid question that can be answered from general principles. Is this answer OK with you?


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Water purification procedure that does not involve a flocculant of any kind?
I have read about "electroflocculation" that uses less flocculant/alum, but is there a method – no matter how expensive or impractical – that uses no flocculant at all?

This of course depends on the scale and intent of the purification/treatment process, but flocculation is but one step in the water treatment process. So, I'm not sure whether you are asking if you can purify water without flocculation, or if you can get flocculation/aggregation to occur without the addition of floccing agents.
The simplest answer to water purification, on a small scale, is filtration and distillation. Flocculation serves to remove solid particles from the water which contribute to turbidity, but which may also include pathogens that are of real concern to human safety. On larger scales (household to community), incredibly effective methods for removing these types of contaminants include various forms of filtration. Some of the first successful water treatment plants on a large scale used sand filtration, and fast and slow sand filters are still used widely today. There are many commercially available drinking water filters for the removal of sediment, microbes and chemicals which are highly effective. Of course, on an industrial scale, filters of this nature would require frequent changing, and this is where sediment removal via flocculation is most effective in reducing costs. You could Google 'water purification in developing worlds' for a range of water purification methods that do not involve flocculation.
On the other hand, inducing flocculation without the addition of flocculant again depends on nature of the contaminant you wish to remove. The biofuel industries are pursuing this goal, developing methods of harvesting valuable microalgae stocks. Autoflocculation (or bioflocculation) of microalgae is known to occur as a function of pH, which can be controlled by restricting CO2 levels. Again a Google search on 'microalgae autoflocculation' will give you many hits. Another method is the light induced autoflocculation of microalgae. There is an interesting post on ResearchGate that may be of some interest to you if you are focused on flocculation of microalgae (if not, then ignore!)

Sure.  We could imagine, for instance, that a strong electrical field (anode/cathode pair) and a special stirring, would cause the migration of all non-H2O molecules to nucleate together and/or adhere to walls and be removed from solution.
Flocculants just start this process going, and are used in context to other processes, as @long mentions.


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Is there always a need for a salt bridge to let a galvanic cell work continuously?
I read in a book that "Sometimes, both the electrodes dip in the same electrolyte solution & in such cases we do not require a salt bridge".
How is this possible?

If both electrodes are in the same vessel of the same solution, ions are free to flow between the electrodes. The purpose of a salt bridge is to allow for a complete circuit by enabling the ions to move freely, so would be extraneous in this case because they can already move freely.

It depends where and how the electrodes are positioned. The UC Davis ChemWiki page Electrochemistry 2: Galvanic cells and electrodes describes the possible arrangements.
For example, if the electrodes are placed in the same vessel, a salt bridge is not required (as shown below). Ions can pass through the porous barrier.

If, however, the electrodes are are placed in separate vessels (such as in the diagram below), then a salt bridge is required to facilitate ion movement.

You need a salt bridge / barrier to maintain electrical contact (without metal) but to prevent copper from plating onto the $\ce{Zn}$.
A super-saturation of metallic copper would produce a small voltage in a homogenous electrolyte with but $\ce{Zn}$ and $\ce{Cu}$ ions. For any $\ce{Cu}$-positive voltage it is be energetically favorable for $\ce{Cu}$ to plate onto the $\ce{Zn}$. However, there is activation energy in nucleating a new phase ($\ce{Cu}$ metal). Thus we will get super-saturated: Conceptually, the "concentration" of $\ce{Cu}$ metal at the Zn electrode exceeds the concentration of $\ce{Cu}$ metal at the copper electrode before the $\ce{Cu}$ phase appears! However, once $\ce{Cu}$ plating begins, the voltage would drop to zero since the cell is now "shorted". Copper will continue to build up and zinc will erode away but no voltage will appear. If this nucleation voltage never is reached (preventing voltage-buildup requires an electrical connection between the two strips), you can get a small amount of energy from a cell in a homogenous solution, but it probably will be much less than 1.1V, and maintaining a low enough voltage would require current to keep flowing.


text:
Nomenclature with Complex Substituents

For the above molecule, I know that the longest carbon chain has 7 carbons, so the backbone is heptane.
The two substituents are at the 4 position on the heptane.
One of the substituents is a methyl group. That much is obvious.
The other appears to be a complex substituent and is a 1-chloroethyl group.
Thus, would the name be:
4,4-(1-chloroethyl)methylheptane?
I think the issue is what happens when we have different substituents with the same loctant. I've looked in one book and at all its practice problems, and this scenario never occurs.
I know that 4-methylheptane is a real compound and that its name is correct. So I am leaning toward 4,4-(1-chloroethyl)methylheptane as being a correct name; the complex substituent is correctly named; both locatants are present; everything is alphabetized; my only issue is whether I am missing some convention (e.g. should there be a dash somewhere, a comma somewhere, or what?)

Close. But each (different type of) group added to the heptane backbone is numbered separately: 4-(1-chloroethyl)-4-methylheptane. Also, you have an asymmetric carbon, so the chirality needs to be specified (I will leave that to you).

Actually we are both wrong. The longest, most substituted chain in this compound is the 6-carbon chain (hexane).

This 6 carbon chain as a 2-chloro substituent, a 3 methyl substituent, and a 3 propyl substituent; therefore the name is (2S)-chloro-3-methyl-3-propylhexane.

Actually I'm not sure anymore about the above name (the one my professor gave). I just read in Wade that we use the longest chain; if there are two chains of equal length, use the one that is more substituted. The longest chain is unequivocally the 7-carbon chain.

@Dissenter I believe this answer is as correct as it gets. And your last comment also substantiates this claim. Maybe it is time to accept this answer ;D (And you could clear some of your comments)


text:
R,S in Cyclic Compounds - Path of Highest Priority?
Consider (R)-3-bromo-1,1-dimethylcyclohexane.
I wish to clarify how I assign R,S priorities in a cyclic compound. My book says to compare the atoms along the "path of higher priority." The problem is in a cyclic compound, if you go far along this path of higher priority without finding a point of difference, you may collide (picture below):
I'm assuming that because the path taken initially results in no difference, we can still assign the top path a higher priority than the lower path because the top path has two methyl groups. My book says that if the paths are identical then choose the one with more of the same atoms.

If you "collide" without finding some difference, then the molecule is not chiral.  In the example at hand, it appears that you have analyzed it correctly.  First, you look at the 2 carbons adjacent to the chiral center.  They are both $\ce{CH_{2}{'}s}$; no difference, therefore move out to the next carbon on each side of the ring.  Here we find a $\ce{C-CH_{2}-C}$ on one side and a $\ce{C-CMe_{2}-C}$ on the other side.  The latter has higher priority than the former.  Stop at this point, no need to go further, and complete the "1-2-3-4" prioritization.

Good idea to keep my wits about me; if we really do collide and find that the molecule is symmetrical then by definition there cannot be chirality! Thanks for breaking it down for me. It is very much appreciated, ron.

So to conclude in the R configuration of the above specified molecule, the bromine must be equatorial?

No, if you flip the ring you'll still have (R)-3-bromo-1,1-dimethylcyclohexane, but the bromine will now be axial.  In other words, (R)-3-bromo-1,1-dimethylcyclohexane has 2 conformers, one with the bromine equatorial and one with the bromine axial.  The mirror image compound (S)-3-bromo-1,1-dimethylcyclohexane also has 2 conformers, one with the bromine equatorial and one with the bromine axial.  The 2 bromine-axial compounds are mirror images of each other as are the two conformers with the bromines equatorial.

Good point. I'll draw the other conformer too. Shouldn't have forgot about conformational isomers.


text:
Hybridization in Light of Resonance
According to this video:
http://www.youtube.com/watch?v=ns4Py96Bkn8&NR=1

1) Is this true? (You might recall I asked a similar question about $\ce{sp^2}$ hybridization implying $\ce{sp^2}$ hybridization for all directly linked atoms a while back, but I couldn't remember my prof's exact wording). I think this YouTube video is saying the same thing my prof said.
2) Does this rule imply that both the oxygens in the below diagram are $\ce{sp^2}$ hybridized? Each is connected to an $\ce{sp^2}$ carbon atom. Each has lone pair(s).

3) Even if we didn't know that the rule existed, we should still know that both oxygens are $\ce{sp^2}$ hybridized because of the following resonance possibility, correct?

ETA:
If the above is true then this doesn't make sense:

Are you referring to the OCS on the blackboard?  You might want to post this as a separate question, making it concise and clear.

I mean the chlorate ion on the blackboard.

The molecule is pyramidal, so the chlorine cannot be described as being $\ce{sp^2}$ hybridized.  Nor can the oxygen as it is presumably (based on the resonance structures) some mix of $\ce{sp^2}$ and $\ce{sp^3}$ hybridization.

I agree. I'll have to see my prof to get his exact wording on this issue. And if necessary I'll counter with the chlorate ion example.

1) Is this true?

It's more or less true in the cases were hybridization methods \ techniques work.  For example, when the hybridization of water is discussed it is often said that the central oxygen is more or less $\ce{sp^3}$ hybridized and therefor the two lone pairs on oxygen are in (roughly) $\ce{sp^3}$ orbitals.  Current thinking based on photoelectron spectral data suggests that, instead of this traditional view, one lone pair appears to be in an orbital that is predominantly $\ce{P}$ and the other in an orbital that is predominantly $\ce{S}$.  So hybridization models don't always give the correct answer.  I'm not sure what the hybridization of an atom with a lone pair that's attached to a double bond really is.  But it is probably OK to say what the teacher said in an introductory class in order to keep things simple for a while.

2) Does this rule imply that both the oxygens in the below diagram are
  sp2 hybridized?

Assuming that the traditional hybridization approach is applicable here (e.g. is the ester oxygen behaving like water or is it "traditionally" $\ce{sp^3}$ hybridized?) then yes, both oxygens are roughly $\ce{sp^2}$ hybridized.  Remember, hybridization is not a step function.  There are hybridization indices between 2 and 3.

3) Even if we didn't know that the rule existed, we should still know
  that both oxygens are sp2 hybridized because of the following
  resonance possibility, correct?

Some data on esters suggests that the $\ce{C-O-C}$ angle is around 115 degrees.  Somewhat between 109 degrees ($\ce{sp^3}$) and 120 degrees ($\ce{sp^2}$).  And maybe an intermediate hybridization is what we should expect based on the resonance structures for an ester.

I love that first statement! I just want to add, that an $\ce{sp^2}$ center cann $\pi$ interact with any $\ce{sp^{x<2}}$ center.


text:
Specific Heat of Water Solution
If we add an ionic solute to water, will the specific heat of the water be increased or decreased?
I understand that the specific heat is a measure of how much energy we must add to water for the average kinetic energy of the water molecules to go up by a certain unit.
I also understand that hydrogen bonding is what limits the kinetic energy of the water molecules. This accounts for water's high specific heat.
However, when adding an ionic solute to water, such as $\ce{KI}$, we disrupt the hydrogen bonding. So we may expect the specific heat of a water solution to be lower than that of pure water.
On the other hand, there are ion-dipole interactions created, correct? These will act to at least partially counteract the specific heat depression, correct?

possible duplicate of http://physics.stackexchange.com/questions/2066/why-does-adding-solutes-to-pure-water-lower-the-the-specific-heat

Actually that link does not bring up ion-dipole interactions; it is simply a discussion about one person's empirical findings. Could anyone shed some light on this?

From this website:

Every substance has its own specific heat and each phase has its own distinct value. In fact, the specific heat value of a substance changes from degree to degree, ...

It would mean that a solute in water would change the specific heat from plain water.
My guess is that if the density is greater than water, then the specific heat would be greater (because you are heating the water and the solute). To be certain, you would need to determine it for yourself.


text:
What's a natural substitute for baking powder or baking soda?
What's a natural substitute for baking power or baking soda? I am doing a project where I need a raising agent in liquid form. Naturally, I can't use anything carbonated so baking powder/soda are out of question.

Yeast? It causes bread to rise.

Could you elaborate a little more, what you are trying to accomplish? For example: What ingredients do you use, what is the purpose, should it be eatable, are you applying heat?

The three sentences in your question seem completely unrelated. Why would does requiring that your ingredient be 'natural' (whatever that means) imply that you can't use baking soda, or that it be liquid, or that it can't be carbonated?

I never heard about chemical rising agent in liquid form. You can use monocalcium phosphate that was an ancient leavening agent.

But maybe you should follow the advice of Dissenter and use yogurt, Sourdough Starter or Kefir all these produce carbon dioxide through fermentation, and are more "natural".
You can look at this wikibook for a list of leavening agent.

Carbonate is very natural.  Wood ash has a significant amount of potassium and sodium carbonate.  You can see this by collecting white powdery wood ash and adding vinegar.
Rock formations like limestone naturally contain carbonate, although limestone is calcium carbonate.

While this is all correct and true, the OP is asking for a substitute.


text:
Cis/Trans Isomerism in Cycloalkenes
How do I tell which one is cis/trans? I know the definitions: cis is same side; trans is across; but
1) What am I comparing here.
2) How do I compare/visualize?

In simple molecules like what you've drawn above, see if the hydrogens are on the same side of the double bond or not.  If they are on the same side, then the compound is cis; if the hydrogens are on opposite sides, then the compound is trans.  In more complex cases, prioritize the four substituents on the double just as you've been doing with "R\S" nomenclature.  If the two highest priority groups are on the same side then the compound is cis, if they are on opposite sides, then the compound is trans.  The terms "cis" and "trans" are still used, but especially in the more complex cases they have been replaced with the first letters of the German words "entgegen" (E, opposite) and "zusammen" (Z, together).  So cis-2-butene could also be called Z-2-butene.

Cool; would the added picture in the OP be an accurate depiction? The first one is a cis; the second one is a trans isomer?

Correct, BTW the same terminology can be used for double bonds involving other atoms, like imines, azines, etc.  Lone pairs are given a lower priority than hydrogen.


text:
Calculating the pH of a dilluted aqueous KOH solution

Calculate $\mathrm{p}\ce{H}$ of an aqueous solution of $0.01\:\mathrm{nM}~\ce{KOH}$ in water.

I think that water also has $\ce{OH-}$ ions in concentration of $10^{-7}\:\mathrm{M}$ (since $\mathrm{p}\ce{H}=7$)
So the total concentration of $\ce{OH-}$ after mixing the solution with water would be:
$0.01\:\mathrm{nM} + 10^{-7}\:\mathrm{M} = 10^{-11}\:\mathrm{M} + 10^{-7}\:\mathrm{M} \implies 1.0001 \times 10^{-7}\:\mathrm{M}$
and given that $\mathrm{pH} = 14 - \mathrm{p}\ce{OH}$
$\implies 14 - \log(1.0001\times10^{-7})$
$\implies \mathrm{pH} \approx 7$
Does this make sense?

Thanks, I am glad to hear! Has been some time since I had chemistry in high school and couldn't find any example on the internet. Glad that I didn't provide any false information :)

Since this is my first post here on stack chemistry, I am wondering how you guys handle questions like this. Shall I delete it, or can someone just post a "yes" answer that I accept - so that it isn't kept marked as "unanswered". (maybe someone might find this question useful as future reference, maybe it is so trivial that it isn't necessary?)

ok, I added actual answer

Yes, that's right, the concentration of $\ce{{}^{-}OH}$ from dissociation of water will be much greater than the amount added from $0.01\:\mathrm{nM}~\ce{KOH}$.
$0.01\:\mathrm{nM}~\ce{KOH}$ will not significantly the affect $\mathrm{p}\ce{H}$ of water.

Your result is correct but I don't agree with your method. Your method would fail under certain circumstances. Which circumstances you might ask? When the pH of the solution is actually between 6.8 and 7.2. In these cases, your assumption that water auto-ionizes to create exactly $\ce{[HO^-]} = 1.0\times10^{-7} M $ will have significant figure implications).
About your assumption:
$\ce{[HO^-]} = 1.0\times10^{-7}$
This is not true when you add any amount of acid or base to water because the addition of either (or both) will disrupt the equilibrium that exists in water (referred to as $\ce{K_{w}}$, the auto-protolysis of water, or the auto-ionization of water):
$\ce{2H_2O <=> HO^- + H_3^+O}$
$\ce{K_{w}=[HO^-][H_3^+O]}=1.0\times10^{-14}$
All you know about your system is that $\ce{[HO^-]_{water} = [H_3^+O]_{water}}$. This is in fact true for any water solution; as we can see from the auto-ionization reaction above, there exists a 1:1 ratio between $\ce{[HO^-]}$ and $\ce{[H_3^+O]}$.
You also know that because $\ce{KOH}$ dissociates completely in water solution, $\ce{[HO^-]_{KOH}} = 1.0 \times 10^{-11} M$
Therefore, the correct way to solve this system would be using $\ce{K_{w}}$. Work below:
$\ce{K_{w}=\sum{[HO^-]\sum[H_3^+O]}}=1.0\times10^{-14}$
$\ce{=([HO^-]_{water}}+\ce{[HO^-]_{KOH})([H_3^+O]_{water})}$
Because $\ce{[HO^-]_{water} = [H_3^+O]_{water}}$, we can make this substitution:
Let $x=\ce{[HO^-]_{water} = [H_3^+O]_{water}}$
Therefore,
$\ce{=(x}+\ce{[HO^-]_{KOH})(x)} = 1.0\times10^{-14}$
And you can solve for x from here. Keep in mind what x equals!

Sorry, I am unfamiliar with the Latex markup here and had to keep refreshing to see the effects of my syntax.

You have a preview under your edit window, there you can see how your post will look like.

You should also use \times ($\times$) when you want to multiply terms.

I never noticed that it changed! You're right; I'll just scroll down next time to see my changes.

@LDC3 or even better \cdot($\cdot$) as $\times$ refers to a vector product

So you want me to edit it again ;)?

@Martin When there are 2 numbers, I'm sure nobody will mistake them for vectors (i.e. $2.4 \times 10^{-5}$).

@LDC3 I agree, that there will be no confusion, bit this should be no excuse to use the correct notation. The main principle behind the notation $\times$ is that a vector product does not necessarily commute, while a product of scalars does.

Glad it helped!


text:
Double Newman Projections
How do I draw a double Newman projection for cyclohexane? Which bond(s) do I sight down?

From what I understand, if we have 1-methylcyclohexane, we sight C1-C2 and C5-C4, as shown in my attempt below:

Would this be correct? I wish to achieve level 4!

Note that -Cl and -CH(CH3)2 are on equatorial positions(The slant lines suggest that)
And the -H is on axial position(The perfectly vertical line suggest that)
Note that it is not just a coincidence, rather it is intentional. I think it would suffice further.


text:
Isopropyl and Butyl Groups - Relative Priorities
This does not compute for me:

Why would the four-carbon butyl group have a lower relative priority than the 3 carbon isopropyl group?
If we compare the carbons one by one between the isopropyl and the butyl group, we'd run out of carbons in the isopropyl group first! How then can isopropyl possibly have a higher priority than the butyl group?

Comparing IsoPropyl and n-Butyl

On first position both have C.
On second position IsoPropyl has two C's whereas n-Butyl has only one C.
Thus IsoPropyl has higher priority and we do not see the further chain.

Thank you. I referred again to a text (a different one) and it became more clear to me that we first compare the ones attached to the carbon one away from the chiarlity center, rather than simply span outward.

If two groups have the same first atom, then compare the second atom from the chiral center.  If there are multiple second atoms, then compare them in order of atomic number.
from http://www.chem.sc.edu/faculty/shimizu/333/Chem_333/5a.vii.html, step 1, part 2


text:
Valence Shell for Transition Metals
In transition metals, is the shell with the highest energy considered the valence shell? For example, in copper the electronic configuration is ${[Ar]\text{ } 3d^{10}\text{ } 4s^{1}}$. However, in accordance with the afbau principle the $3d$ subshell actually has higher energy than the $4s$ one. Does this mean the $3d$ subshell is considered the valence?
Second, why is gold not considered to be a reactive substance. It's electronic configuration is not stable: $[Xe]\text{ } {4f}^{14} {5d}^{10} 6s^1$, as it has an unfilled $6s$ orbital. So why is it not highly reactive like some other transition metals?

Nope, as wiki tells us, only outermost electron shell is considered
valence shell, though d-electrons for transition metals are
considered as valence electrons. No logic here, only tradition.
as you can see, there are three electron subshells above obviously inert Xe core. 5d-electons, that most of the time are far from the nuclei because of the form of their orbitals, feel dramatic increase in effective core charge through the third d-row. For example, for Ti effective charge, keeping its d-electrons is 4. For gold it is 11, so the electrons are well-held.
For s-electrons the form of the orbital puts very significant part of electronic density near the nuclei. This means, that for heavy elements (staring from as far as Ga) a so-known 'inert electron pair effect' is observed: in comparison with p- and d-electrons, due the form of the s-orbital, s-electrons are much less shielded from the charge of the nuclei by underlying, leading to much more strong bonding of said electrons. As effect, Bi is highly unstable in valence state +5, perbromates are much stronger oxidizing agent than perchlorates, but same is not true for bromates and chlorates, and Ag, Au and Hg have very tightly bound s-electons, leading to relative inertness of said metals.
For Au and Ag, however, strongly bound s-electrons participate in extremely powerful (due to relative small size of s-orbitals) s-s interactions, while for Hg only relatively weak s-p and d-p interactions are possible because of filled d- and s- subshells. Thus, metallic Ag and especially Au in metallic state have s-electrons involved into effective covalent interactions, leading to relatively inert nature. Beware, however, that they are far from unreactive: Ag have high affinity to sulfur, and both Ag and Au dissolve in cyanide solutions when oxygen is present, not saying about such agressive reagents as chlorine and bromine.
Still, Au, Ag and Cu share one trait with Ia (I main subgroup) elements. Due to only spherical s-orbitals involvement in atom bonding in metal state, all three metals have high plasticity. Other d-elements have significant d-d bonding in metal states, and d-orbitals are not spherically symmetrical, leading to much less plastic behavior.

For the answer to #1, why do metals use both $d$ and $s$ subshells for bonding, in comparison to other elements? And as for #2, I don't quite understand what is meant by -electrons are much less shielded from the charge of the nuclei by underlying, leading to much more strong bonding of said electrons. Also, if the s-s interactions in Ag and Au make both effectively inert, why does silver tarnish whereas gold does not?

@user2612743 #1 because d-orbitals for them arelarge enough to be comparable in size with s/p orbitals of next shells. #2 shielding is about inner shells reducing nuclei effective charge for outer shells. For Li, effective charge felt by the outer electron should be 3(nuclei) - 2 (celectrons on the inner shell). Electrons on s-orbitals are shielded less because most e-density on it is near nuclei, but for p- and d-orbitals it is far from nuclei. || Both Ag and Au has ~0 affinity to sulfur because of comparable size of orbitals, bit >0 for Ag, a bit <0 for Au. H2S is always present in air.

How does the fact that the s orbital is less shielded and therefore more strongly bound to the nucleus result in s-s interactions? And also, if I understand correctly it is this s-s covalent-like interactions that essentially fill the s suborbital and result in the inertness?

@user2612743 The fact of little shielding does not make s-s interactions possible, but it does make s-orbitals smaller and as result their bonding becomes more effective. This does not prevent Cu/Au/Ag from forming effective covalent bonds with other elements, but common elements can't form effective covalent bonds with Au because of vastly different orbitals size and strong bonding of Au with it's valence electrons makes forming of ionic bonds problematic. As result, Ag/Au oxides and halogenides are relatively unstable.


text:
Balanced equation for synthesis of p-nitrophenol from p-nitrodiaznoium chloride salt and sulfuric acid/water?
I am trying to write a balanced equation for the above mentioned reaction but having some trouble figuring it out. So far, I have:
$$\ce{([C6H4N3O2]+ Cl^{-}) + H2SO4 + H2O -> C6H5NO3 + N2 + ?}$$
How do I "incorporate" sulfuric acid and water into the reaction? Thanks for any help.

The sulfuric acid is not consumed in the reaction pathway to product, so there's no reason for it to appear in the equation for the reaction.  Therefor, the overall equation for the reaction is $$\ce{([C6H4N3O2]+ Cl^{-}) + H2O -> C6H5NO3 + N2 + HCl}$$

I have no problem with your answer, but I'm just curious, how is sulfuric acid catalytic here? My understanding has always been that its only role is to acidify the solution from the start, so as to avoid azo-coupling side reactions.

That's what I recalled, but I couldn't find a reference to support sulfuric acid having a large effect on the rate of diazonium salt decomposition.  I'll modify my answer accordingly.  Thanks!

The water is the source of the hydroxyl ($\ce{-OH}$) group in the phenol, which accounts for one oxygen and one hydrogen. The other hydrogen is ultimately abstracted, probably by a second molecule of water to form $\ce{H3O+}$. The simplest way of representing this, however, is to simply write $\ce{HCl}$ on the product side, though it's tacitly understood that it's unlikely to actually exist undissociated in any significant quantity.
As for the $\ce{H2SO4}$, I'm fairly certain its only purpose is to maintain the acidity of the solution, with the aim of avoiding azo-coupling side reactions. (Although, with a para-substituted deactivating nitro group, this is probably not much of a concern.)


text:
Reactivity of thiosemicarbazide
How to form Schiff's Base at both the ends of the thiosemicarbazide?
$\ce{NH2}$ of hydrazine part reacts easily to form Schiff's base i.e. imine. But $\ce{NH2}$ attached to $\ce{C=S}$ does not. How to do that?
One approach is by using NCS but that forms imine at $\ce{NH2}$ attached to $\ce{C=S}$ only and not at the other end.

(thio)amides have extremely low nucleofility, I think it is impractical to try for the shiff base you want directly, it better to try inderect approach

I have read papers which say thioamide is more reactive than amides.  Could you suggest any indirect methods?


text:
Find the uncertainty in position
The question says:

A proton is accelerated to one tenth of the velocity of light. If it's velocity can be measured with a precision $\pm 1\%$. What must be its uncertainty in position?

Therefore,
$v=0.1 \times c =3\cdot 10^7\ \pu{m/s}\\
 \Delta(v)=\frac{1}{100}\\
 m= 1.6\cdot10^{-27}\ \pu{kg}$
Then I directly substituted these values into the formula:
$$\Delta(v)\cdot\Delta(\text{position})\cdot \text{mass}=\frac h{4\pi}$$
To get the uncertainty in position, however, the answer I got was approx. $3.5 \cdot 10^{-6}\ \pu{m}$ which is way too different from the correct answer:

$0.5\cdot10^{-13}\ \pu{m}.$

Can anyone please explain me how to solve this question?

The question tells you
\begin{aligned}
&&v&=3\cdot10^7\:\mathrm{m/s}\pm1\%\\
&&\Delta v &= |(v+1\%v) - (v-1\%v)|\\
\implies&&\Delta v &= \frac{2}{100}v\\
\therefore&&\Delta v&= 2\cdot3\cdot10^5\:\mathrm{m/s}
\end{aligned}
Hence
\begin{aligned}
&&\Delta x\cdot\Delta p &\geq \frac{\hbar}{2}\\
\implies&&\Delta x\cdot\Delta (v\cdot m) &\geq \frac{\hbar}{2}\\
\implies&&\Delta x&\geq \frac{\hbar}{2\cdot\Delta v\cdot m }\\
\therefore&&\Delta x &\gtrapprox\frac{1.05\cdot10^{−34}\:\mathrm{J\cdot s}}{2\cdot6\cdot10^5\:\mathrm{m/s}\cdot1.6\cdot10^{−27}\:\mathrm{kg}}\\
&&\Delta x&\gtrapprox5.5\cdot10^{-14}\:\mathrm{m}\\
\end{aligned}


text:
Since aqueous acids release Hydrogen cations into the solution, will an electric current create Hydrogen gas?
I was wondering that since acids dissociate in water to make a lot of  $\ce{H+}$ cations. Is it possible that introducing a supply of electrons via an electric current, would create $\ce{H2}$ gas?

Yes.  In fact the first electrolysis was done with acidified water.  Pure water is only slightly ionized and is a poor conductor.  Adding ions to water increases its conductivity and speeds up the electrolysis process.  So by adding a source of protons to pure water the following reaction will occur at the cathode $$\ce{2H+ (aq) + 2e^{-}->H2(g)}$$  Of course a parallel reaction will occur at the anode $$\ce{4OH- ->2H2O + O2(g) + 4e-}$$ otherwise a current wouldn't flow and the electrolysis would not continue.


text:
Why are radicals unstable?
Why are unpaired electrons especially reactive? Does pairing electrons decrease the reactivity of the electrons? But then doesn't forcing two electrons into the same orbital cost energy?

Free radicals are usually very reactive.  If we bring two free radicals or a free radical and a molecule with electrons available for bonding together a bond will usually form.  Yes, it does cost energy to bring those electrons close together.  But look at the payoff, a new chemical bond is formed and this is usually quite exothermic (stabilizing), usually more than enough energy to offset the energy cost of bringing those electrons close enough to form a new bond.  Because of the exothermicity of bond formation the reaction profile usually has a small activation energy.  Translation: a small activation energy means a  fast reaction.  A caveat, at the outset I said that "free radicals are usually reactive".  There are many known free radicals that are quite stable.  This is usually due to steric factors (the reactive free radical center is surrounded by bulky groups and just not sterically accessible for reaction) or electronic factors (some free radicals exist in very large delocalized systems, hence the spin density at any one atom in the system is so small that reaction is unlikely).  One further point, most of what I've said relates to free radicals, which was the title of your question, however you did mention "unpaired electrons" in the beginning of your post.  Molecules with unpaired electrons (note the plural, electrons) are different than free radicals with (usually) just one unpaired electron.  Some molecules like $\ce{O2}$ are quite stable even though they have two unpaired electrons.

How does diatomic oxygen have 2 unpaired electrons? Does this have to do with molecular orbital theory?

There are 2 degenerate molecular orbitals available for the last 2 electrons to go into, so no need to spend the extra energy and pair them up, just leave them unpaired in separate orbitals.  That's why oxygen is paramagnetic (see http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Magnetic_Properties).

Ah okay; how do we reconcile the MO bonding diagram with the usual Lewis structure?

Well, you can draw a Lewis dot structure for $\ce{O2}$ that has two unpaired electrons, it just doesn't have an octet around each oxygen.  Sometimes MO diagrams are a poor approximation of reality, but they are handy and quick.  Same can be said for Lewis dot structures, and they're an even more primitive approximation methodology.  It is important to know where these various approximations break down.  So examples of failures ($\ce{O2}$ with Lewis dot structures) can be very helpful in this regard.

Thanks :). I remember that being mentioned - the failure of Lewis structures to show that oxygen is paramagnetic. Any other exceptions to keep in mind?

I'd need to think for a while about it.  That might be an interesting question on its own.

@Dissenter I think Lewis theory would fail any time a molecule has a non-singlet ground state.

@Dissenter $C_2$ molecule is also paramagnetic though hard to observe. $Be_2$ molecule is unstable according to MOs, but Lewis structure is drawable. $(CH)_4$ square (not tetrahedron) is paramagnetic.

What is the square molecule called?

Also for the paramagnetic representation of oxygen, is that a resonance form?

@Dissenter asked,"for the paramagnetic representation of oxygen, is that a resonance form" Yes, you can draw more than one resonance form for $\ce{O2}$.  The one with a single bond has unpaired electrons.

for @ron 's answer -- exothermic bonds can be formed even for non-radicals ... what's  so special about radicals that makes them more reactive ..?


text:
What solvent would have the fewest ions?
We know that water auto-ionizes (to a rather small extent).
My question is which solvent would have the fewest ions (or at least fewer ions than water).
I'm thinking that perhaps some kind of hydrocarbon, as these are both weaker acids and weaker bases than water and thus even less reactive. I've seen computers submerged in oils (as well as distilled water) due to their low conductivities.

Does Mercury count as a solvent? I believe that it is a good solvent for most metals? how ionic would pure Mercury be?

Well if it's ionic then it isn't pure ;)

Ron's answer is a good option, however autoionization is not necessarily equivalent to self-protonation. There may be some solvents where autoionization is even more suppressed than any self-protonation reaction. Perhaps the liquid with the lowest tendency to ionize may be something composed of molecules with extremely strong bonds, for example liquid nitrogen or liquid carbon monoxide. Of course, in Ron's acid table there already are several compounds which would be expected to ionize less than once in even reasonably macroscopic volumes of liquid.
Edit: As Greg points out in the comments, why not go all out and consider noble gasses. The only plausible ionization reaction for the pure liquid elements is:
$$\ce{Ng(l) <=> Ng+_{(Ng)} + e-_{(Ng)}}\ ,$$
where $\ce{Ng}$ indicates a noble gas and the $\ce{(Ng)}$ subscript indicates solvation by the noble gas atoms. The electron doesn't attach to anything, as $\ce{Ng-}$ would actually be unbound (no barrier to dissociation, or at least this is true for non-ultraheavy noble gasses such as ununoctium and beyond).
Given that noble gas atoms likely make for an extremely poor solvent, we can get a very rough measure of the equilibrium constant by assuming the solvation is nil. Thus, the reaction would be equivalent to the gas-phase ionization of the atoms. Assuming the entropy change is a small factor relative to the very large ionization enthalpies, then the reaction $\Delta G_\mathrm{ion}^\circ$ is simply equal to the ionization energy.
For helium, this equals a massive $\pu{2372.3 kJ mol-1}$. Applying $\Delta G^\circ = -RT\ln K_\mathrm{ion}$, then for liquid helium at its boiling point under atmospheric pressure ($\pu{4.2 K}$), the ionization constant is as low as $10^{-29500}$ (though most of the reason the exponent is tiny is because of the really low temperature; for comparative purposes the value at room temperature would be around $10^{-400}$). A more accurate calculation would drive this value up quite a bit, but it would still be many, many orders of magnitude more difficult to ionize than anything else.

Good point; I forgot about non-liquid solvents.

Yes, I was thinking about this as well. My first thoughts for substances particularly unlikely to auto-ionize were noble gases, and possibly perfluorinated alkanes.

@GregE. Ah yes, I had considered liquid noble gasses for a second but somehow forgot to put them in the answer. Liquid neon or helium are probably the two top candidates, I would say far above PFAs.

@NicolauSakerNeto, yes, I definitely agree with that. I mention PFAs only because, as a practical matter, they're more readily available and can be liquids under standard conditions.

Would a PFA be Teflon?

I was thinking about nitrogen and polysulfur, but how do you measure their dissociation constant in a way that allows  comparison to the hydrocarbon cases? I

@Dissenter, yes, essentially, although it's a solid polymer. I had in mind short-chain PFAs that are liquids under standard conditions.

2-Methylpropane has a $\mathrm{p}K_\mathrm{a}$ of 53 ($\mathrm{p}K_\mathrm{a}$ table).  That sounds pretty non-ionic.  Can't think of any common all-carbon (no substituents of any kind) liquid compounds.


text:
Hydrolysis of salt
The hydrolysis of salt tells you whether a salt is acidic or basic. But it does not tell you how acidic or basic a salt is. Is there a way to determine this?
My guess is that taking of the weak base part of a salt( presuppose that it is acidic) and see how many hydronium ion it could produce. So the more hydronium ions, the more acidic this salt is?

Why not just use pKa or Ka values for the ion of the salt with acid-base properties?

Why for pKa you take the negative log of the ionization constant for acid? What the log represents? Why do this. I know the definition of log, just asking for the reason to take the negative log.

The negative log returns a positive pH value (generally), since concentrations of hydronium and hydroxide ion are typically very low, so they are expressed using negative exponents - i.e. $1.0 \times 10^-6$. The negative is mainly there for convention - it's a bit easier dealing with positive values in general.

Alright, then is my guess correct or not?

Why would you take the base part of the salt and see how much hydronium ion it creates? Why not hydroxide ion?

Only strongly acidic salts have anions that are part of weak acids

Steps for determining acidity or basicity of a salt by hydrolysis:

Write the neutralization reaction in reverse (salt and water makes base and acid).
Break up the salt into its ions.
To write the base, add an $\ce{OH-}$ to the cation (the positive ion) for each + charge.
To write the acid, Add $\ce{H+}$ to the anion (the negative ion) for each negative charge.
If you for a strong base, the salt is basic.  $\ce{(NaOH, KOH, LiOH, Mg(OH)_2,  ...)}$
If you form a strong acid it is acidic.  $\ce{(HCl, HNO_3, HClO_4, HBr,...)}$
If you form both a strong acid and a strong base it is neutral.

EXAMPLE:
$\ce{NH_4Cl}$
$\ce{NH4Cl + H_2O <=> NH4OH + HCl}$
Since $\ce{HCl}$ is a strong acid it breaks up and yields $\ce{H+}$, the salt is acidic.

The MathJax module implemented here makes use of the mhchem latex package. You can learn more about formatting in the [help]. I have formated your post with these style commands, if you want to have a look. Thank you.

Isnt NH4OH not really a real thing?


text:
Microbial Fuel Cell
Does anyone know of a step by step guide to building a microbial fuel cell? Could someone here on Chem SE possibly provide one? I am looking for something like THIS but for the whole fuel cell and not just the cathode.
Descriptive papers on the topic would be helpful also.

You can watch YouTube video.

Materials:

2 1l Plastic Containers
Cotton rope
Aluminum mesh
Paper clips
Copper wire
Alligator clips
Electrical tape
Glue gun
Glue sticks
Aquarium air pump
Duct tape
Sludge
Hand spade
Bucket
Water
Salt
Small Pot
Spoon
Voltmeter
Drill gun

Procedure

Collect sludge from the bottom of a still creek or pond into a bucket. Such a source will most likely have plenty of anaerobic bacteria.
Drill one hole for copper wire on lids of containers. On one of the two lids, drill one hole for the air pump tube and one small hole for ventilation (this will not be sealed). Drill one hole on one side to both containers for salt bridge.
Prepare the electrodes. Fold aluminum mesh a few times over and bind with large paper clips. Strip ends off of copper wire and attach to both electrodes.
Insert copper wire and air pump tube into drilled holes on lids. Seal with hot glue or caulk.
Prepare the salt bridge. Heat water over stovetop and dissolve in as much salt as possible. Twist a long rope around itself to create a thicker rope. If necessary, cut the   rope to approx. 15 cm. Soak the rope in the salt water. Once damp, wrap the rope in one layer of electrical tape and one of duct tape, but keep the ends exposed.
Insert each end of the salt bridge into the drilled holes on the sides of the containers. Seal with hot glue or caulk and extra tape (as needed).
Fill one container almost to the rim with sludge and the other with water.
Submerge the electrodes into the sludge and water. Close the lids of the containers, and make sure the one over the sludge is airtight.
Anaerobic bacteria should be exposed to as little oxygen as possible.
Attach alligator clips to each loose end of copper wire. Clamp the alligator clips onto the voltmeter probes. Make sure the red probe is attached electrically to the electrode in the water. The black probe should connect to the electrode submerged in the sludge.
Turn on the aquarium air pump.
Turn the dial on the voltmeter to 20 VDC (can also be marked by the symbol,      ) to measure the force of electricity moving through the circuit in terms of volts. Turn the dial to 2000m to measure it in terms of millivolts, a thousandth of a volt, to observe a more accurate reading.
Turn the dial to 10A. This number is the flow of current the microbial fuel cell is generating, measured in amps.

source:Instructables

Here this might help
http://www.sciencebuddies.org/science-fair-projects/project_ideas/Energy_p026.shtml#materials
But you might want to use different electrodes.

Welcome to Chemistry.SE! For formatting help visit the [help] and for more information about this page, take the [tour]. Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference.


text:
In which of the following chemical compounds are the building blocks molecules?

In which of the following chemical compounds are the building blocks molecules?
(A) $\ce{AgNO3}$
(B) $\ce{HNO3}$
(C) $\ce{KNO3}$
(D) $\ce{NaNO3}$

(B) is the correct answer, but I have no idea why. Could you briefly explain why this is the correct answer?

Strange question, is this complete? as written in the book?

@GM: I agree. Yes, that's all information that's given in the book.

Can you give some more context, or tell us which book? The question as written does not make sense.

I highly doubt this is correctly copied from the textbook. It isn't even good English AFAIK.

@Jori: It's because I translated it in English. Again, that's all information that's given, nothing else. The question looks exactly the same as the one given in the book, the only difference is that this one is translated in English.

Would you tell us what the original text was (in the native language)? Maybe there was a translation error.

I decided to call my chemistry teacher and he said that possible explanation could be that $HNO_{3}$ is a (nitric) acid, whereas e.g. $KNO_{3}$ is an ionic compound, therefore, it's buildnig blocks are not molecules, but ions.

Answer B is correct because nitric acid is the only compound whose building blocks satisfy the IUPAC Gold Book definition of a molecule, a neutral (uncharged) entity consisting of at least 2 or more atoms. The building blocks of all other given compounds are atomic ($\ce{K+}$, $\ce{Na+}$ and $\ce{Ag+}$) or molecular ions ($\ce{NO3-}$), which by their nature do not fulfill the requirement of being electrically neutral. Per definition, they are molecular entities, i.e. distinct particles which can be identified as separate entities. This term is more general and also includes charged species like ions.


text:
What do '(6S)' and '5' mean in (6S)-5-formyltetrahydrofolate?
Tetrahydrofolate:

I guess (6S) means some sixth S-center and 5 some other kind of numbering.

I've redrawn a truncated version of the title molecule below with the "5-formyl" group in place.  I've also inserted small numbers so you can see how the base system is numbered.  So "5-formyl" means that a formyl group ($\ce{-CHO}$) is attached to the "5" position.  The 6-position has 4 different substituents attached to it so it is a chiral carbon atom.  The R-S nomenclature system is used to describe the 2 mirror image possibilities at chiral centers.  The enantiomer you've drawn is the "6S" stereoisomer.


text:
Any suggestions how to solve this?
$m(\ce{FeSO4. $x$~H2O})=20.0\:\mathrm{g}$
$N(\ce{H2O})=3.03 \cdot10^{23}$

$x=?$
What I've done:
(1.)

\begin{aligned}
   m (\ce{H2O})&= \frac{N(\ce{H2O}) \cdot M(\ce{H2O})}{N_{A}}\\
   &= \frac{3.03 \cdot10^{23} \cdot 18.02$\:\mathrm{g/mol}}{6.02 \cdot 10^{23}/\:\mathrm{mol}}\\
   &= 9.06987\:\mathrm{g}
  \end{aligned}

(2.)

\begin{aligned}
   m(\ce{FeSO4})&= m(\ce{FeSO4} \cdot x \ce{H2O}) - m(\ce{H2O})\\
   &= 20.00\:\mathrm{g} - 9.06987\:\mathrm{g}\\
   &= 10.9301\:\mathrm{g}
  \end{aligned}

Any suggestions what to do next?

Calculate $n(\ce{FeSO4})$ with $M(\ce{FeSO4})\approx 136\:\mathrm{g/mol}$

 $n(\ce{FeSO4})\approx0.08\:\mathrm{mol}$

Calculate $n(\ce{H2O})$

 $n(\ce{H2O})\approx0.5\:\mathrm{mol}$

Set them into relation $n(\ce{FeSO4}):n(\ce{H2O})= 1:x$

 $x\approx 6$


text:
Why don't alkaline earth metals lose only one electron when they are ionized?
Why doesn't alkaline earth metals lose only one electron when they are ionized ? I know that magnesium atoms like to have the electronic configuration of neon but I don't understand the reason that we can't find $\ce{Mg+}$ atoms .

Magnesium can definitely exist as $\ce{Mg^{+}}$. Hence the availability of first ionization energy data for magnesium.
The same goes for other metals; there are only uncommon valences, such as $\ce{Na^{3+}}$.
If you had enough energy, you could make $\ce{Na^{4+}}$ if you wanted to. It would just cost, according to the table below, 9543 + 6912 + 4562.4 + 495.8 kJ/mol.

I think your question is why isn't $\ce{Mg^{+}}$ more commonly encountered. That has to do with stability. Losing two electrons and becoming isoelectronic with a noble gas is more stable than having an unpaired electron. An unpaired electron is available for bond formation, and bond formation tends to be exothermic.
If you want to read up on the +1 valence of magnesium, such as in $\ce{MgH}$, you can read more here:
http://bernath.uwaterloo.ca/media/24.pdf

Actually you have to add up all the IE to generate the highly charged species, so for $\ce{Na4+}$ that would take about 21500 kJ/mol.

Good point! Will fix in text.

How is the magnesium being ionized?  If it is reacting with oxygen or another oxidizer, look at the Born-Haber cycle.
The magnesium losing electrons is endothermic.  Oxygen gaining one electron is exothermic but the 2nd electron affinity of oxygen is endothermic.
$\ce{Mg^{2+}}$ and oxide making magnesium oxide is exothermic.  Removing the second electron from magnesium and adding the second electron to oxygen is "paid for" by the increase in lattice energy due to having 2 doubly charged ions instead of 2 singly charged ions.

$\ce{Mg+}$ is not commonly encountered because compounds with $\ce{Mg^{2+}}$ are more stable. This does not mean that $\ce{Mg^{2+}}$ is more stable on its own, in gas phase even $\ce{Mg+}$ is less energetically favorable than $\ce{Mg^0}$. However, when interacting with other atoms, stabilization due electrostatic interaction is much higher for $\ce{Mg^{2+}}$ than for $\ce{Mg+}$ because of much smaller radius and double charge of the former.
Still, at least some compounds of $\ce{Mg(+1)}$ are known, even if in exotic conditions, like $\ce{MgCl}$ is known in gas phase.


text:
Is Bond Formation "Strictly" Exothermic?
Is bond formation "strictly" exothermic? The IUPAC definition of exothermic doesn't make any reference to bond formation. However, I have seen the aforementioned statement before - that bond formation is "strictly" exothermic.
I suspect that the answer is "it depends" and that "it depends" at least on how accurately bonds are characterized. For example, the first reaction below is endothermic, while the second one is exothermic:
$\ce{2NO -> ONNO}$.
$\ce{2NO_2 -> O_2NNO_2}$.
Lewis structure analysis for the first reaction at best misleads because one might suspect the nitric oxide molecule as having a bond order of 2.

However, the lone electron is actually delocalized. So the actual $\ce{N-O}$ bond order should be 2.5.

In the $\ce{ONNO}$ molecule however there is no delocalization of electrons and the bond order of the $\ce{N-O}$ bond is 2. So the $\ce{N-O}$ bond is weakened, and even the formation of the $\ce{N-N}$ bond cannot compensate for this weakening.
Also, the above example reminds me of certain unstable homonuclear species. Such as $\ce{Ne_2}$, which according to MO theory has a bond order of 0 - i.e. there is no bond. So I suppose that bond formation is not always exothermic. On the other hand, if the bond order is 0, is a bond really "formed"?

Bond formation is alway strictly exothermic in the sense of the change of enthalpy.

exothermic reaction A reaction for which the overall standard enthalpy change $\Delta H^\circ$ is negative.

A bond can only exist, if it needs energy to break it, i.e. the bond dissociation energy is always positive.

bond-dissociation energy, $D$ The enthalpy (per mole) required to break a given bond of some specific molecular entity by homolysis, e.g. for $\ce{CH4 -> .CH3 + H.}$, symbolized as $D(\ce{CH3−H})$ (cf. heterolytic bond dissociation energy).

This has absolutely nothing to do with a reaction being exothermic/endothermic or exergonic/endergonic, because this is defined by the rearrangements of bonds.

Regarding noble gas diatomics, it is quite clear from MO-Theory, that there is no bond. However, even these non bonded elements have a non-zero dissociation energy. Please refer to "Why are noble gases stable" and to answers and comments within.

+1 for bringing up the need to consider bond breaking as well as bond breaking when determining whether or not a reaction is exothermic.

What do you mean noble gases have a non-zero dissociation energy? How can non-bonded elements have "dissociation" energies?

Admittedly they are very, very small, but not negligible. There are weaker forces at play, i.e. London Dispersion and van der Waals. These are also part of the reason, why noble gases can be liquid close to zero Kelvin.

Martin, it does have to do with a reaction being exothermic or endothermic, because you need to define another state that you are comparing to.  Here you are comparing methane to methyl radical and hydrogen radical, which is a specific reaction.  Weak bonds do form between noble gas atoms: http://scitation.aip.org/content/aip/journal/jcp/98/4/10.1063/1.464079 http://www.sciencedirect.com/science/article/pii/S0009261401010880#  What about the bonds of trigonal bipyramidal transition states? http://en.wikipedia.org/wiki/Transition_state Do they require energy to break?

@Dave I am not comparing anything with anything else, I am just quoting the IUPAC Goldbook. This definition is complete in a concise and theoretical way. And with this given, the BDE is always positive. Experimentally you do not need to actually break the bond, you can derive it from spectroscopy. Thank you for the literature. As for the 'bonds' in the transition states, they do not fit within the definition of bonds or chemical bonds.

Well, this is getting tricky if one gets into details. Hmm, maybe you did well not going the path of Dave. Yes, bond creation proper, even in He2 dication is exothermic. There's the thing that closing two He+ requires a lot of energy, much more then bond releases, that means that molecule is endothermic, but not bond itself. That means your answer is OK, but there's a good reason for "controversies".

@Mithoron Yes, tricky it is indeed. And obviously there is room for debate, or at least more elaboration. If I had time I'd go over that again, but I simply cannot do it.

At very high pressure, say $>\pu{100 GPa}$, diatomic molecules such as $\ce{H2}$ and $\ce{O2}$ are no longer favorable relative to monoatomics, even at low temperature where Gibbs free energy $G$ is dominated by enthalpy $H$.
See Phase Diagram of Hydrogen.
For example the cores of Jupiter and Saturn are monoatomic hydrogen.
Diatomic bond formation would be endothermic under such conditions.

Dave do you know why that is?  I would have thought that a diatomic occupies less space than two monoatomics and so at high pressure the diatomic side of the equilibrium would be favored.

The diatomic will have a certain optimal separation between the nuclei (bond length).  My thinking is that at high enough pressure/density it is no long possible to have the nuclei so separated.  Also, the monoatomic state is metallic so it is bonded in that sense, just not covalently bonded.  This is related to my question here: http://chemistry.stackexchange.com/questions/10226/maximum-pressure-of-covalently-bonded-material

Why is the monoatomic state metallic?

Even though metallic hydrogen may be called monoatomic, it is bound to many other atoms at once. It's not that bonding becomes endothermic, but the localization of bonding (breaking the distributed metallic bonds to create localized covalent bonds) is. [I took too long to write this and it seems you've made the comment yourself already]

So still, at the end of the day, bond formation is an exothermic process.  If the reaction coordinate for bond formation is endothermic, then bond formation simply does not occur.

ron, I wouldn't say bond formation simply does not occur, but instead equillibrium concentration is less than 50%.

Looking back, I don't like my old answer so I'm adding a totally different answer.
A specific (theoretical) example of endothermic bond formation is described in Prediction of a Metastable Helium Compound: HHeF J. Am. Chem. Soc. 2000, 122, 6289-6290.
As reported in table 2:
The energy to dissociate HHeF to H + He + F is negative.
The energy to dissociate HNeF to H + Ne + F is negative.
The energy to dissociate HHeF to HF + He is negative.
The energy to dissociate HNeF to HF + Ne is negative.
However, HHeF is stabilized by being in a potential energy well, the activation energy to dissociate being relative high.
The authors conclude: "Remarkably, HHeF is also predicted to be a metastable species, which represents the first neutral compound containing a helium chemical bond."
The bond formation is endothermic.

Do the author's actually say that bond formation is endothermic?  I can understand that formation of HHeF could be endothermic (just like the formation of dewar benzene from benzene is endothermic) because you break an HF bond and make weak H-He and He-F bonds.  But do the author's claim that formation of the actual H-He or He-F bond is endothermic (the formation of the actual 1-4 bond in Dewar benzene is exothermic even though the reaction is endothermic).  Do the author's distinguish between reaction endothermicity and bond formation endo\exothermicity?

@ron the best I can do is quote: "The H +
Ng + F dissociation limit is strongly dependent on the nature of the noble gas: exothermic for He and Ne, near thermoneutral for Ar and endothermic for Kr (Table 2). On the other hand, fragmentation to HF + Ng is predicted to be a strongly exothermic process for all species." (where Ng is generic for noble gas).

It seems to me they are discussing bond breaking (fragmentation, dissociation) not bond formation.  Do you read it differently?

yes it is talking about dissociation, and agreeing with Martin's criteria for whether or not bond formation is exothermic (that the dissociation energy to neutral species being positive means bond formation is exothermic), that the dissociation to H + He + F has negative dissociation energy necessitates an endothermic bond formation.

Besides these detailed explanations a picture sometimes helps to understand bond energy.
The picture below shows what happens to the potential energy when two atoms approach one another and a bond forms. The separated atoms are at an energy of zero. The bond is formed at the minimum (negative) energy (ignoring zero point energy of the bond vibrations). The dissociation energy D$_e$ is always positive. In a complex molecule this picture should apply to each individual bond.
Note that there is no scale, the bond could be normal covalent or due to dispersion forces, it makes no difference in principle to the general shape of the potential energy. Of course, if the bond energy is small, say, comparable to $k_BT$ at room temperature, the bond would only be stable at low temperatures.


text:
Existence of Hyponitrite Molecule
Does the hyponitrite molecule exist? I know the hyponitrite ion exists (http://en.wikipedia.org/wiki/Hyponitrite) but what about the molecule?
I have found no reference to the molecular form on Wikipedia. But it does seem that such a molecule could at least come together through the dimerization of nitric oxide.

Apparently, $\ce{NO}$ does indeed dimerize at low temperature. Plenty of references come up in a search, and this is mentioned on the Wikipedia page for nitric oxide.

Yes, see Structural and Stablity of Nitric Oxide Dimer.
Just don't call it "hyponitrite molecule", it might get offended.


text:
Atmospheric Humidity Effects on Non-water Evaporation
Does evaporation of non-water liquids into the atmosphere vary with atmospheric humidity?

Yes, because though the liquid phase initially contains no water, water molecules may enter the liquid phase from the gas phase.  The more hygroscopic the liquid is, the more the evaporation rate will be affected by humidity.  If the liquid is very hygroscopic and has a low vapor pressure, the initial evaporation rate may even be negative in humid conditions, as water enters the liquid phase at a rate greater than liquid enters the gas phase.


text:
Explanation for the reactions in a saltwater battery with zinc and copper electrodes
I am a physicist, not a chemist. I'm trying to get a basic understanding of the reactions taking place in a battery using a saltwater electrolyte with copper and zinc terminals. I'm writing a general science level article about it. I have found that the reactions taking place at the terminals are:
\begin{aligned}
 \text{Zinc terminal:}&&\ce{Zn(s) &-> Zn^{2+}(aq) + 2e^-} &(\Delta G = -0.76\:\mathrm{V})\\
 \text{Copper Terminal:}&&\ce{Cu^{2+}(aq) + 2e^- &-> Cu(s)} &(\Delta G = -0.34\:\mathrm{V})
\end{aligned}
I think I understand the first reaction in terms zinc leaving the terminal and going into solution kind of as if it were dissolved $\ce{ZnCl2}$ (does that make sense?). But I need help understanding the second reaction. I've inserved a copper terminal into the solution. But not copper ions. Am I to understand it like this: when I put the copper terminal in the solution, copper ions immediately drift off, perhaps for the same reason the zinc ions do, but then deposit back onto the terminal in the presence of the electrons released when the zinc goes into solution?
I'd like an explanation somewhere between that it just happens and full on redox theory, if that is possible.

This answer deals with the situation, when but electrodes are in the same salt water solution.

Assuming your salt solution contains sodium chloride
$$\ce{NaCl (s) ->[\ce{H2O}] Na+ (aq) + Cl- (aq)},$$
then your reaction at the zinc terminal will certainly be
$$\ce{Zn (s) + 2Cl- -> ZnCl2 (aq) + 2e-}.$$
On the copper terminal water will be electrolysed
$$\ce{H2O + 2e- ->[\ce{Cu}] 1/2 H2 ^ + {}^{-}OH},$$
or using autoprotolysis
\begin{aligned}\ce{
2H2O &<=> H3+O + {}^{-}OH\\
H3+O + e- &->[\ce{Cu}] 1/2 H2 ^ + H2O
}.\end{aligned}
This will happen since sodium has a much lower electron affinity than water/ hydronium
\begin{aligned}
\ce{Na+ + e- &-> Na} &E^\circ&=−2.71\:\mathrm{V}\\
\ce{2H2O + 2e- &-> H2 + 2 {}^{-}OH}&E^\circ&=−0.83\:\mathrm{V}\\
\ce{2H+ + 2e- &-> H2}&E^\circ&=0\:\mathrm{V}.
\end{aligned}
Depending on what salt solution you are using, different cations might be reduced. You can find a collection of values at wikipedia.

In a battery (Galvanic cell) there are usually coupled half cells, and in this case this would probably be $\ce{Zn|Zn^2+||Cu^2+|Cu}$.
This means, that the Zinc terminal is in a zinc containing solution and the copper terminal is in a copper containing solution. In this case your assumption is correct
\begin{aligned}
 \text{Zinc terminal:}&&\ce{Zn(s) + ZnCl2 (aq)&<=> 2Zn^{2+}(aq) + 2Cl- + 2e^-} \\
 \text{Copper Terminal:}&&\ce{CuCl2(aq) + Cu (s) + 2e^- &<=> 2Cu(s) + 2Cl-}
\end{aligned}
There has to be a salt bridge to neutralise charge differences (ion flow).
Note that in that case zinc ions will go into solution, while copper ions will be removed from the solution. If the concentration of copper drops close to zero, then no current can be measured. The galvanic cell can be charged by "reversing" the process, apply opposite current. Then zinc ions will be removed from solution and copper will go into solution.
In reality, this is a equilibrium process.

thanks!  That really clears up a lot. In my article I simply said that one electrode sheds electrons and that the other picks them up and that the electrolyte fascilitates this process. The formula you give show that I got this right. Without the Cl-, the zinc wouldn't be oxidized and the Cu++ not reduced, at least not as much.  You did go over my head with the equations with the Cu over the arrow and the 1/2 H2 and the up arrow. Remember, I'm not a chemist ;)

The Copper over the arrow simply means, that this is catalysed by copper/ takes place at the copper terminal. The arrow pointing upwards simply means, that this compound evaporates/ is a gas and leaves the system. (An arrow pointing downwards would mean that a solid precipitates.)

Very helpful answer. Thanks! Could you clarify why autoprotolysis only happens at the copper terminal and doesn't happen at the zinc terminal (for sodium chloride solution)? Why does hydronium ion not steal an electron from the zinc as well?

@akhmed Autoprotolysis will happen everywhere, but hydrogen will only evolve at the copper terminal since there the electron flow is right. Also interesting in regards to water and conductivity is the Grotthus mechanism. In total the concentration of hydronium is too low for it to react with zinc directly.

Awesome. Thanks a lot! I was also pointed to this paper. As I understood it, the hydronium's electron stealing process goes like this: [step 1] a proton must detach from hydronium and form M-H bond first (so M-H bond must be strong enough), then [step 2] two M-H meet each other on the metal surface and $\ce{H_2}$ gas forms (so M-H bond must not be too strong either or hydrogen gas will never form). It appears $\ce{Zn-H}$ is weak so step 1 doesn't occur often.

The key point you are missing is that the thermodynamic values you report are for standard conditions, meaning that there would also be 1 M $\ce{Zn^2+}$ and 1 M $\ce{Cu^2+}$ in the solutions.  The Nernst equation can be used to determine the non-standard condition potential (in your case, sticking a copper wire into a solution with effectively no copper ions.  In practice, this would be very hard to do, as adventitious ions (and other reactions not considered) will contribute to the cell potential when the intended ion concentrations are so low.
This concept can be used to determine the concentration of copper ions in solution, and may be of interest in explaining the phenomenon to a general audience.  Two $\ce{Cu}$/$\ce{Cu^2+}$ half-cells with the same concentration of copper ions will result in a cell potential of zero; however, if one of the cell concentrations is known (for argument's sake, say 1 M) then the "unknown" ion concentration could be found using the Nernst equation.

A quick answer to get you started - I'll be happy to elaborate if necessary.

I really appreciate you taking a look! Unfortunately, you went over my head and well over that of my audience. It sounds like my understanding is basically correct that once in solution, both $\text{Zn}^{2+}$ and $\text{Cu}^{2+}$ ions end up in solution, but that the copper has a higher affinity to rejoin its terminal than the zinc ions do. I still don't understand how this gets free electrons onto the copper terminal. Nor do I fully understand the role of the electrolyte. Everyone says that the ions carry the electrons from the cathode to the anode, but how?

Go back to the salt bridge question and note the figure I made. Each beaker is a half-cell which, in order to function properly as an energy source must contain the oxidized and reduced forms of an electron-transfer reaction.  What you refer to as terminals (and Chemists refer to as electrodes) are the reduced forms and the ions are the oxidized forms.  The reason electrons flow through the wire is because the thermodynamically favored reaction is for Zn to be oxidixed and Cu(2+) to be reduced.

(part 2) If you were to place a piece of Zn metal into a beaker filled with Cu(2+), you would "see" Cu metal plate onto the Zn and the blue color of the copper solution would fade. The electrolyte is there solely to decrease the resistance of the solution.  Do you mind elaborating on your audience and purpose of the presentation, which will help me improve this answer.

Thanks @bobthechemist. That clears up a lot. I had never before thought of the fact that there would be both zinc and copper would go into the solution as ions, but that the copper has the higher propensity to go back on the electrode. I like how you explained that both the oxidized and reduced species exist, but there is a thermodynamic difference in which is favored for each species. That really clears up a lot.

The article is for eHow, and I didn't really end up using much of this discussion, except to know that I can have more confidence that I know what is really going on.


text:
How to determine the molecular formula based on molecular weight, amount of substance, and partial masses of some, but not all contained elements?
I am working on a chemistry assignment and can’t figure out part of the problem.
$\pu{0.2 mol}$ of a compound with a molecular weight of $82$ contain $\pu{9.6 g}$ of carbon, $\pu{1.2 g}$ of hydrogen and the rest is nitrogen.
The question is asking for the empirical and molecular formula which I can solve, but how do you determine the amount of nitrogen?
First I assumed a $\pu{100 g}$ sample but that didn't get me anywhere. The empirical formula answer is supposed to be $\ce{C2H3N}$ but I couldn’t achieve that. Unless the book is wrong.

If you have $0.2\ \mathrm{mol}$ of a compound with $M_\mathrm{r}=82$, then you have $16.4\ \mathrm{g}$ of material. If there is only carbon, hydrogen and nitrogen present, and the carbon and hydrogen total up to $10.8\ \mathrm{g}$, then the amount of nitrogen must be $16.4\ \mathrm{g}-10.8\ \mathrm{g}=5.6\ \mathrm{g}$.

ron's answer starts out with 0.2 mole substance.
If instead you start out with 1 mole of substance, you can get the molecular formula directly. One mole would contain 48 g of carbon and 6 g hydrogen (five times more than 0.2 mole would). That would be 4 moles of carbon and 6 moles of hydrogen, i.e. $\ce{C4H6N_x}$. Because the molar mass is $\pu{82 g/mol}$ and the carbon and hydrogen only accounts for $\pu{54 g/mol}$, we are still missing $\pu{28 g/mol}$ worth of nitrogen, so $x$ is equal to 2. This gives a molecular formula of $\ce{C4H6N2}$. You get the empirical formula by dividing all subscripts by two, i.e. by the largest common divisor.
To understand this answer, you have to know that the molar masses of C, H, and N are $\pu{12 g/mol}$, $\pu{1 g/mol}$ and $\pu{14 g/mol}$, respectively.