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How exactly is energy created by oxidative phosphorylation
I understand how the protons and electrons move around through various carriers, shuttles and proteins, but there's this talk of electrons being in "lower in energy" at the end. What does that mean? How is an electron lower in energy?
Also why do they need an electron acceptor? What would happen if there was no $\ce{O2}$ to accept the electrons? Wouldn't they just bind to something else?

It seems to me me it could be called catalytic analog of combustion - very effective and complicated.

How exactly is energy created by oxidative phosphorylation?
Energy is "created" by the syntesis of ATP. The ATP molecule is considered an energy carrier, because it is used in other reactions as a source of energy. The released energy comes from the loss of one or two phosphate gropus to form ADP or AMP.
Remember that energy is not created, but simply transferred. The oxidative phosphorylation is a mechanism to recreate the ATP using the energy from other sources (O2, nutrients, etc).
Also why do they need an electron acceptor? What would happen if there was no O2 to accept the electrons?
I'm not sure I can give you a precise answer, but I'll try. In chemistry reactions, the system is always working towards equilibria. In general, if you take away one reagent of a reaction, there won't be a reaction at all. If there's no oxygen available, ATP won't be formed, and as a consequence the body resorts to other sources of energy, or cells will die.
Wouldn't they just bind to something else?
I don't think so. The electrons aren't free particles moving around the cell, the transfer only happens if all the reagents for the reaction are complete. Enzymes are, generally speaking, very specific, and what could happen is a "saturation" of NADH's which wouldn't be converted back to NAD and hence the cycle would stop.
But there's this talk of electrons being in "lower in energy" at the end. What does that mean? How is an electron lower in energy?
I don't remember hearing or reading that. Can you post some context or source?


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Predicting order of nucleophilic substitution reactivity

What order of reactivity do you predict will be observed when each alkyl halide is mixed with sodium iodide in acetone?

1-Chlorobutane
1-Bromobutane
2-Chloro-2-methylpropane
Bromobenzene
2-Chlorobutane
1-Chloro-2-butene

I know that $\ce{I-}$ is the best leaving group, followed by $\ce{Br-}$ and $\ce{Cl-}$.
For the $\ce{NaI}$ reaction, tertiary halides should react fastest and primary halides should react slowest.

What order of reactivity do you predict will b observed when each alkyl halide is mixed with silver nitrate in ethanol?

1-Chlorobutane
1-Bromobutane
2-Chloro-2-methylpropane
Bromobenzene
2-Chlorobutane
1-Chloro-2-butene

For the $\ce{AgNO3}$ reaction, primary halides should react fastest and tertiary halides should react slowest.
Which takes preference, leaving group or rank? Also, how do I rank 1-chlorobutane vs. 1-chloro-2-butene? They are both primary and the halide is $\ce{Cl-}$ for both.

possible duplicate - http://chemistry.stackexchange.com/questions/27734/nucleophilic-substitution-reactions

Question one is basically which one does Sn1 fastest (aprotic solvent and strong nucleophile). So in question 1 you're looking for the least substituted species with the best leaving group. Question 2 is basically which one does Sn2 fastest (protic solvent and weak nucleophile). More subbed and better LG.

For the Chlorobutane/ene situation think about resonance stabilization in the intermediate/transition state.

Acetone a is aprotic polar solvent which is often used as solvent for $\ce{S_{N}2}$ reactions (notably in conjunction with $\ce{NaI}$ as in the Finkelstein reaction). For $\ce{S_{N}2}$ reactions steric hindrance and leaving group capability is of great importance, so primary halides will be usually easier replaced by iodine then tertiary.
$\ce{AgNO3}$ will probably react as a base forming nitric acid. Even though nitric acid is a rather strong acid this is more likely than $\ce{NO_3-}$ acting as nucleophile. Besides that, $\ce{Ag+}$ is known to have a unique selectivity for halogens and will attack them readily, forming $\ce{AgX}$ and a carbocation, hence promoting $\ce{E 1}$/$\ce{S_{N}1}$.

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Testing for the presence of Hexamethylenetetramine / Formaldehyde
I read in Russian popular articles that it is allowed to preserve herring/fish with Hexamethylenetetramine in Europe. Since Soviet times, I like to buy salty herring and marinade it with Vinegar. The food-safety article says that Hexamethylenetetramine + Vinegar = Formaldehyde, which is a serious poison.

How can I easily test presence of Hexamethylenetetramine or Formaldehyde in food?

The salty herring that I buy is sold loosely, without packaging, so that I cannot simply read the ingredients.

There is two procedures in this link, (google hit #2) if you are into chemistry. Though it sounds like the chemicals required aren't very friendly or easily obtainable either. It does mention that Formaldehyde can get largely cooked off over extended time.

I don't believe the premise that E239 (the E number designated to hexamethylene tetramine) is used to preserve fish within the EU is correct - in January, 2014, the European Food Safety Authority published an article in their journal about this preservative where it states quite clearly that the only permitted use for this additive is in Provolone cheese. There are concerns that over-consumers of this particular cheese may accrete more levels of E239 within the body than are healthy.
E239 does appear to be used in the UK for preserving fish, or at least, it is listed as one of its intended uses. Whether its actually used or not I haven't been able to determine, but given we're in the EU, technically, it shouldn't be. If I source any other research regarding this, I'll add it later, but currently, it seems it depends where you're buying your fish from in the first place, that is, the country of origin.

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Why does Silver form bonds with covalent character?
Compounds of silver form particularly strong bonds which is accounted for by the significant covalent character of the bonding. Furthermore, is the tendency to form covalent bonds linked to the fact that $\ce{Ag+}$ is a soft acid?

related: http://chemistry.stackexchange.com/questions/16261/metal-compounds-that-bond-covalently?rq=1, http://chemistry.stackexchange.com/questions/429/why-do-magnesium-and-lithium-form-covalent-organometallic-compounds?rq=1

Silver compounds, if anything, are usually more ionic than those of its closest neighbors.


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Obtain precipitated iron?
Is it possible to obtain pure, precipitated iron with no oxidation by some chemical process?
Alternatively is it possible to electrically purify iron using an anode and cathode?

Is it possible to obtain pure, precipitated iron with no oxidation by some chemical process?

Very pure iron can be obtained by pyrolysis of iron pentacarbonyl.  Iron metal obtained in this manner is in fact called carbonyl iron and is commercially available.
$$\ce{Fe(CO)5 -> Fe(s) + 5 CO}$$
Doing this yourself is very dangerous, however, because iron pentacarbonyl is volatile and toxic.  The decomposition reaction to produce iron requires high temperatures, and the byproduct is carbon monoxide gas, which is also notoriously toxic.

Alternatively is it possible to electrically purify iron using an anode and cathode?

This is also possible.  I don't know about other salts of iron, but to form iron from iron oxides, very high temperatures are required.  The other answer mentioned Dr. Sadoway's work.  Let me highlight a recent article of his unveiling new anode materials for electrolytic decomposition of iron oxide.  At temperatures as "low" as 1538°C, molten iron oxide can by electrolyzed to pure oxygen gas and to molten iron metal using electrochemistry.
Water is not an appropriate solvent for electrochemical formation of iron metal.  At potentials low enough to favor $\ce{Fe^0}$ formation from $\ce{Fe^{+2}}$, water is unstable and would be reduced to hydrogen gas.

Why is iron different from zinc for electroplating?  Zinc requires an even lower potential to deposit than iron and yet we electrogalvanize steel all the time.  I'm guessing it has to do with electrochemical reaction kinetics (overvoltage), but I'm looking for documentation that this factor really does reverse the thermodynsmic expectation.

Sure... here's a an article on it and a patent for the process. In brief,
"[Donald Sadoway, the John F. Elliott Professor of Materials Chemistry at MIT] found that a process called molten oxide electrolysis could... [produce] steel as a byproduct... Sadoway's method used an iridium anode... But after more research... the MIT team identified an inexpensive metal alloy that can replace the iridium anode in molten oxide electrolysis."

Can you give a (brief) summary of what's at the link, please?  Otherwise the answer isn't really self-contained.  Thanks!

Blockquote
using iron sulphate you can use the same technique used for etching to produce iron powder. You attach the positive side of the current to your piece or iron you want to produce the iron from, and the negative side to a piece of iron.

Did you quote this from somewhere, since I can't find anything? If so, please include a link to your source.

that was from personnel experimenting. People that do etching of metals in this case iron remove the metal by this process http://mordent.com/etch-howto/ the iron quickly rusts though so I'm not sure how quickly you would have to remove the deposit. Having said that when I reread your question properly you mean precipitate from a molten state not a solution I think?


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Does the magnesium 2+ ion prefer tetrahedral or octahedral coordination?
In organic chemistry I was told that the magnesium ion which is part of Grignard reagents is coordinated to several ether molecules. The overall structure around the magnesium is tetrahedral, which implies that two ether molecules have coordinated to the magnesium center. The professor said something about fulfilling magnesium ion's octet with this configuration.
In inorganic chemistry and biochemistry however I learned about hexacoordinated magnesium - i.e. $\ce{Mg(H2O)6}$. Magnesium ion tends to form octahedral complexes; this implies a coordination number of 6.
So -

Does magnesium ion adopt both these configurations? Can it be found as part of tetrahedral and octahedral complexes?

Which is preferred? I suspect this answer depends on a variety of factors.

Have a look at "ligand-field theory"...

You have to make a clear distinction between octet (electron configuration) and octahedral (coordination geometry).

Magnesium bonded to most purely nonmetallic elements is predominantly ionic or ion/dipole, even hydrogen with its low electronegativity (but poor covalent overlap with the rather diffuse magnesium valence orbitals). So the "octet rule" or "valence electron count" applies less than the balance between magnesium being attracted to negative charges against steric effects/repulsions between like charges. That balance, at least with magnesium, is usually struck with octahedral coordination.
Two compounds where magnesium shows different coordination geometry:
Magnesium diboride, $\ce{MgB2}$. In this moderately high temperature superconductor ($T_c = 39$ K) the boron is organized into graphene-like sheets, which effectively fixes the coordination geometry as hexagonal-prismatic. Each magnesium ion is coordinated to a hexagonal space between boron atoms on each surrounding graphene-like sheet. The bonding orbitals of the graphene-like sheets are at lower energy than isolated boron atomic orbitals, so the boron is effectively more electronegative and the magnesium-boron bonding here is predominantly ionic.
Magnesium silicide, $\ce{Mg2Si}$. This narrow-gap semiconductor, along with its counterparts having heavier G14 elements in place of silicon, has drawn interest for thermoelectric applications. Here the magnesium is tetrahedrally coordinated, in part because this already corresponds to a high coordination number (eight, in an antifluorite structure) for the silicon. The high coordination number on silicon again implies enhanced ionic character (silicon would form only four bonds in a purely covalent mode), so steric factors related to the silicon coordination apparently play more role in limiting the magnesium coordination than covalent bonding "rules".

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What are the resonance structures for chlorite anion?
What are the possible resonance structures for $\ce{ClO_2^-}$? Assigning one double bond to the structure makes for formal charges of $\ce{O(-1)-Cl(0)=O(0) <-> O(-1)=Cl(0)-O(-1)}$. It appears that the same set of formal charges can be achieved with two double bonds (expanding the octet on $\ce{Cl}$ to 12 electron capacity, using the d orbital). That is, $\ce{O(0)=Cl(-1)=O(0)}$.
Is this feasible, and if not, why not?

Disclaimer
The here used methods are also only approximations and I do not claim that the provided numbers are accurate. However, they are used here to make the point, that bonding situations involving resonance are not easily explained.

Oxygen is slightly more electronegative than chlorine, therefore the bonds will always be slightly polarised towards the oxygen.
The following resonance scheme is based on an NBO analysis on the DF-BP86/def2-SVP level of theory. The percentage value gives an estimate how close the given conformation is to an idealised Lewis structure. From this you could also derive how much such a structure would contribute to a resonance situation (in terms of Valence Bond theory).
What we do see here is, that the electronic structure is better described with more formal charges. The contribution of a π bond is best described by an ionic donor-acceptor structure, rather than breaking the octet rule at chlorine.
Noteworthy is also that the doubly ionic donor-acceptor configuration is as equally good a lewis structure than any that breaks the octet rule.
The worst representation is the one with a four electron three centre bond (which is also not part of the Lewis bonding picture anyway).

Videos like the one mentioned in the Wikipedia article are easy to follow, but often they are too simplified to be correct.

This is quite a curious answer, I had no idea calculations such as these were possible. It took me a minute to understand the bottom three structures though, because the arrows look almost like dative covalent bonds rather than electrostatic attraction between the ions.

@Nicolau In principle they are dative covalent bonds. These are just representations of the same electron density. So in general all the presented structures are different viewpoints of the same situation. NBO localises and then partitions the electron density - in principle you could go about and construct valence bond configuration from that and actually calculate the mixing, but that is tedious and takes a lot of time and I cannot do it.

Then I'm not quite sure how chlorine has a positive formal charge even though it's being given electron pairs from the oxygen atoms. The bottom-right representation looks particularly strange to me. The charges seem to be calculated as if there were no covalent bond present at all, dative or otherwise.

@Nicolau In the original Lewis picture there is no differentiating between bonds and in this sense only the very first structure is a true Lewis structure. The next two structures break the octet rule. The last in the first row has delocalised bonds. The dative bond does not exist in the Lewis picture, but extending this formalism, it only counts as electrons from where it originates. In the Lewis picture it does not matter if there is a bond in that case or not.

Ahh, now I see. I thought that might be it, but counting electrons from where they originate is rather unusual, at least to me. Thanks for taking the time to explain.

@Nicolau I always try to take the time. The Lewis picture is flawed, I guess we agree on that, but it is a nice way to represent molecular structures...


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Which is most reliable excitation energy: LUMO-HOMO energy or TD-DFT excitation energies?
After running a TD-DFT , we can calculate the HOMO-LUMO difference and then calculate the the excitation energy.
Another approach is just simply pick the energies calculated in the TD-DFT and use them.
For me the latter makes more sense as that's the reason we do TD-DFT , otherwise we could access the HOMO-LUMO difference by any other methods but I have seen papers using the first concept. I also have noticed that the energies in second concept are closer to the experimental values.
Also when I run TD-DFT, I can see Oscillator Strength. Should I pick the energy that has the largest Oscillator Strength for calculation of the maximum wavelength ?
Here is an example. Should I pick number 8 for maximum wavelength calculation?

I'm curious about the example - the oscillator strengths are very low!

Why do you need to run TD-DFT to calculate the energy difference of HOMO and LUMO levels?

Geoff Hutchinson  Sure, they can be forbidden transitions. It depends what OP wants to calculate.

When you can, use TDDFT, in part because of the oscillator strengths
I like to point out to my group that optical excitations consist of two things consistently (e.g., in a UV/Vis spectra). There are the excitation energies (the x-axis) and the intensity or extinction coefficient (the y-axis). To some degree, the peak widths give a bit more information (e.g., into vibrational structure).
The example you give above is a great example. Most of the excitations are forbidden and have no oscillator strength. I'd pick the same excitation, although I'm curious what the molecule is - usually there are much higher oscillator strengths.
You're right that there are papers and groups that use the HOMO and LUMO eigenvalues. While some DFT methods (e.g. optimally tuned range-separated hybrids) do provide HOMO-LUMO energies that correspond nicely with experiments, there's nothing physical about Kohn-Sham orbital eigenvalues. Moreover, in general, unfilled orbitals are not handled well by DFT methods.
It's been a while since I looked for reviews, but general TDDFT excitation energies are ~0.2-0.3 eV from experimental absorption energies after correcting for solvent effects, etc.
In my group, we also do what I call a "consensus" model. That is, when you run a TDDFT calculation, you get the HOMO and LUMO energies anyway. Since there are slightly different random and systematic errors from each approach, you can calibrate both against experiment and use a multivariate linear regression. The result decreases the average error significantly.

Thanks a lot Geoff, as always your answers are perfect. Actually this molecule was just a scratch I was doing ( I think I was adding multiple explicit solvent molecules to my system) .

I got really interested in your "consensus model" and I think that is exactly what I need.  Have you ever published a paper on it that I can use that model as a citation in my paper ?

We mention the idea first in this Macromolecules article. We haven't published a paper on the method itself (yet).


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How to find the mass of dissolved AgCl when placed in a solution of NaCl?
I think the following problem in my textbook is malformed:

Solid $\ce{AgCl}$ is placed in $1\,\mathrm{L}$ of $0.55\,\mathrm{M}$ $\ce{NaCl}$. Find the mass of $\ce{AgCl}$ which dissolves.

I know the method to solve the problem. I think the problem is malformed: When we get an answer, say $x$ grams $\ce{AgCl}$ dissolves, but if we placed less than $x$ grams solid $\ce{AgCl}$ in the solution initially, then it would be impossible for $x$ grams to dissolve.
So we obviously have to make certain assumptions, but I am looking at the solution in my textbook and I don't see in which part of the solution is the assumption used (implicitly). And by the way, how do we solve the problem if we have place less than $x$ grams solid $\ce{AgCl}$?

Is that all the data that was provided, or were any equilibrium constants given?

When you have solved the problem, you will notice that there is less than 1 g of AgCl dissolved (probably even less than 1 mg). I think the question should be reworded to ask What is the maximum amount of AgCl that can dissolve in 1 L of 0.55 M NaCl (at 25°C)?

The question is most likely badly worded and should have been

What is the maximum mass of $\ce{AgCl}$ that can be dissolved?

As we all know, $\ce{AgCl}$ is damn insouble in aquaeous solutions — and due to the equilibrium constant it doesn’t get better at all if you add chloride ions (reducing the silver ion concentration). So without having done the maths, I’m going to assume that a small spec of nothingness is all that will dissolve. That spec is so tiny that you will have a hard time weighing less so you can actually perform the experiment you suggested.
Because it’s just a flick of dust, and because the authors likely

knew that, too
assumend you knew that, too

they may have gone with a less obvious wording. I mean it’s clear: You can’t dissolve what you didn’t put in, and to-be-dissolved silver ions aren’t going to magically appear with Alakazam.

Since the already present sodium chloride highly ionic, it's dissociation can be considered 100% and thus the contribution of chloride form silver chloride would be thus negligible (in comparison to rock salt's) or suppressed. Thus we can treat the final chloride concentration as $0.55\rm M$. Now the solubility product is given as:
$$K_{\rm sp,AgCl}=\ce{[Ag+][Cl- ]}$$
Now we can find out $\ce {[Ag+]=\frac{K_{\rm sp,AgCl}}{[Cl- ]}}$. This would actually be all from silver chloride and hence the solubility of it.

What about the second part of my question (the last sentence)?


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Chalcogens' hydrides as acids?
I've noticed that chalcogens never form binary acids, and instead end up like water, with very little acidity. Why is this? Or am I wrong and there's a counterexample, if so please give it.

You have a point here but even water has pH=7 and all others are stronger, so it would be better to ask why they are weak.

I don't understand what you mean by "little to no pH value". Furthermore, the hydrogen chalcogenides get more acidic as you go down the group; hydrogen telluride is acidic as phosphoric acid.

related: http://chemistry.stackexchange.com/questions/27784/why-does-h%E2%82%82se-have-a-bigger-ka2-than-h%E2%82%82te

@NicolauSakerNeto well I do ask if I'm wrong...

To start off, all chalcogens are known to form binary hydrides of the form $\ce{H2X}$ which can then behave as binary acids, however weak or strong, by the equation $\ce{H2X <=> HX^- + H+}$.
Now to compare the strength of the acids. The most straightforward way of doing this is comparing gas phase acidities, which removes the complication of solvation effects. For this, we look at the proton affinities of the conjugate bases $\ce{HX^-}$. The definition of a substance's proton affinity is the $\Delta G$ when it is protonated in the gas phase (always an exergonic reaction for neutral species and anions), so gas phase acidity is simply the inverse reaction, with the opposite sign for the $\Delta G$. According to these two sources, the gas phase acidities of the hydrogen chalcogenides are:
$$\ce{H2O_{(g)} <=> HO^{-}_{(g)} + H^{+}_{(g)}}\ \ \ \ \ \ \ \ \ \ \ \ \Delta  G = +1635\  kJ\  mol^{-1},\ K^{25ºC} = 2.5\times 10^{-287}$$
$$\ce{H2S_{(g)} <=> HS^{-}_{(g)} + H^{+}_{(g)}}\ \ \ \ \ \ \ \ \ \ \ \ \Delta G = +1477\  kJ\  mol^{-1},\ K^{25ºC} = 1.2\times 10^{-259} $$
$$\ce{H2Se_{(g)} <=> HSe^{-}_{(g)} + H^{+}_{(g)}}\ \ \ \ \ \ \ \ \ \ \ \ \Delta G = +1417\  kJ\  mol^{-1},\ K^{25ºC} = 4.1\times 10^{-249} $$
$$\ce{H2Te_{(g)} <=> HTe^{-}_{(g)} + H^{+}_{(g)}}\ \ \ \ \ \ \ \ \ \ \ \ \Delta G \simeq +1386\  kJ\  mol^{-1},\  K^{25ºC} \simeq 1.1\times 10^{-243} $$
As you can see, when going down the group, deprotonation becomes less endergonic, so the binary hydrides definitely become stronger acids, even if against what one might expect from electronegativity considerations. Calculating the equilibrium constants allows a more direct comparison. Don't worry that the constants are all extremely small, what matters to us is the ratio of the equilibrium constants. Thus, in the gas phase, $\ce{H2S}$ is a whopping 28 orders of magnitude more acidic than $\ce{H2O}$, while $\ce{H2Se}$ and $\ce{H2Te}$ are 38 and 44 orders of magnitude more acidic than $\ce{H2O}$, respectively. It is certainly noteworthy that $\ce{H2O}$ is a much, much weaker acid than its heavier homologues.
With the data presented, we ask, why does acidity increase down the group? There are two main factors presented in different sources. One of them is that the chalcogen-hydrogen bond becomes weaker for heavier chalcogens, and this is ascribed to the poor overlap between the tiny $\ce{H}\ 1s$ orbital and the increasingly larger chalcogen orbitals which weakens the covalent bond and presumably makes ionization easier. This can be seen from the dissociation energies of the $\ce{X-H}$ bonds ($\ce{O-H} = 428\  kJ\  mol^{-1}$ | $\ce{S-H} = 344 \  kJ\  mol^{-1}$ | $\ce{Se-H} = 305\  kJ\  mol^{-1}$ | $\ce{Te-H} = 268\  kJ\  mol^{-1}$). The second factor is the size of the chalcogen atoms. The heavier atoms are larger, meaning they spread the same charge in the conjugate base $\ce{HX^-}$ over a larger volume, which makes reprotonation less favourable. This latter explanation seems to be the most common.
Up to now we've been talking about acidity in the gas phase, but now let's add the solvent, specifically water. The presence of a solvent stabilizes ions immensely relative to gaseous ions (especially protons), and water is a particularly good solvent for ions, so all the dissociation equilibria will be shifted massively to the right compared to the gas phase. Let us compare the acid dissociation constants in water:
$$\ce{H2O_{(l)} <=> HO^{-}_{(aq)} + H^{+}_{(aq)}}\ \ \ \ \ \ \ \ \ \ \ \ \ k_w^{25ºC} = 1.0\times 10^{-14}$$
$$\ce{H2S_{(aq)} <=> HS^{-}_{(aq)} + H^{+}_{(aq)}}\ \ \ \ \ \ \ \ \ \ \ \ \ K_a^{25ºC} = 1.0\times 10^{-7} $$
$$\ce{H2Se_{(aq)} <=> HSe^{-}_{(aq)} + H^{+}_{(aq)}}\ \ \ \ \ \ \ \ \ \ \ \ \ K_a^{25ºC} = 1.3\times 10^{-4} $$
$$\ce{H2Te_{(aq)} <=> HTe^{-}_{(aq)} + H^{+}_{(aq)}}\ \ \ \ \ \ \ \ \ \ \ \   K_a^{25ºC} = 2.3\times 10^{-3} $$
Some things to note:

All the binary chalcogen hydrides still increase in acidity after solvation when going down the group. This is not necessarily always the case. A classic example of relative acidities changing on solvation is the dimethylammonium ($\ce{Me_2NH2^+}$) and trimethylammonium ($\ce{Me_3NH^+}$) cations; in the gas phase the former is more acidic, but in aqueous solution the latter is more acidic as the $\ce{Me_3NH^+}$ ion is not as well solvated.
As I mention in the comments, aqueous hydrogen telluride ($pK_a=2.6$) is about as acidic as aqueous phosphoric acid ($pK_{a_1}=2.15$). Compare also with a common weak acid such as acetic acid ($pK_a=4.76$), a borderline strong acid such as nitric acid ($pK_a \simeq -1$), and a recognized strong acid such as hydrochloric acid ($pK_a \simeq -6$).
The $K_a$ of water is a little tricky, since it is both the reactant and the solvent. Its value depends on the definition used. I quoted the $k_w$, but many people use a $pK_a$ of 15.74 which implies $K_a = 1.8\times 10^{-16}$. The difference is not too important here.
$\ce{H2O}$ is a much stronger acid than expected from gas phase data, relative to its heavier homologues. This is ascribed to the exceptionally strong aqueous solvation possible only for $\ce{OH^{-}}$ among the listed conjugate bases, stemming from strong hydrogen bonding interactions, which helps weigh the equilibrium to the right side.

Unfortunately the site is currently having trouble displaying mathjax, so my answer looks horrible. I'm afraid there's nothing that can be done for now.

Nicolau it renders correctly now (even more horrible ;)

I think providing reference values of a strong acid, such as $HCl$ or $H_2SO_4$ would improve the answer?


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Why does the US not manufacture sodium thiopental themselves?
A shortage of a given chemical (sodium thiopendal) has prompted Utah to allow firing squads for executions. This shortage is due to mostly European chemical factories refusing to export the drug to the US.
I am against the death penalty, but can't help wondering why USA, arguably the most powerful nation in the world, can not manufacture this chemical themselves. Is it difficult to produce ?

Missouri is considering starting a state-funded lab to produce its own lethal injection drugs http://www.washingtonpost.com/news/post-nation/wp/2014/05/29/missouri-attorney-general-wants-the-state-to-produce-its-own-lethal-injection-drugs/

I […] can't help wondering why USA […] can not manufacture this chemical themselves.

They can and they did!
Back in the nineteen thirties, Ernest H. Volwiler and Donalee L. Tabern filed a couple of patents on sedative thiobarbituric acid derivatives for the Illinois-based Abbott Laboratories. Among these was US 2,153,731 for sodium thiopental.
Abbott Laboratories produced it for a long time and when the company was restructured, the still Illinois-based split-off Hospira continued the production.
Note that it is not a "bad compound" as such and actually, it is still listed as an alternative to propofol on the WHO Model List of Essential Medicines.
According to a 2010 article in the German ÄrzteZeitung (online, 29.09.2010), the FDA was aware of the thiopental shortage and a speaker of Hospira told the NYT that his company is rather unlucky about the fact that this anaestetic found it's way into death row.
Around this time, the production of thiopental at the facilities in Illinois was seemingly stopped.
In a 2011 article in the ÄrzteZeitung (online, 24.01.2011) it is stated that Hospira had initially announced to continue the production at a facility in North Carolina and in Liscate, Italy.
However, it seems that the production in North Carolina never started. Moreover, the Italian authorities demanded a declaration that the thiopental produced in Liscate would not be used for lethal injections. Obviously, Hospira could not comply.
Personally, I think that Hospira wasn't overly unlucky about the legal situation in Europe. They are a health care company and I suppose that they prefer to make profit by improving the medical situation of their customers, rather than by killing them!


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What is the maximum molarity of aqueous HCl?
Here is a problem a made up:

We have some aqueous $\ce{HCl}$. What is the maximum possible molarity of the $\ce{HCl}$?

How do I solve this? I don't have any good ideas.

The HCl molecule is, at standard conditions, a gas. The reagent used in laboratories is HCl dissolved in water, which is why you'll find in the label that it is around 37% HCl in weight. The other 63% is water.
Considering 37% as the maximum solubility of HCl, you can calculate the molarity using the solution density (1.2 g/mL) and HCl's molar mass (36.46 g/mol).
For 1000 mL of solution, you will have 1200g of weight, of which 37% is HCl: 444g.
In 444g of HCl you have 12.18 moles (444/36.46), which means the concentration of P.A. HCl is around 12.18 moles per liter.
Note that because P.A. HCl is a solution saturated with a gas, vapors are likely to be released and escape the solution (which is why P.A. HCl shouldn't be handled outside fume hoods), which makes this concentration an approximation, and it should never be used to prepare a solution which you need to know the exact acid concentration.

According to "Hydrochloric acid" IARC Monographs volume 54:
The solubilty of HCl in water is:
82.3 grams HCl per 100 grams water at 0 degree C
67.3 grams HCl per 100 grams of water at 30 degrees C
and the density of 39.1% HCl aqueous solution is 1.20.
So for example at 0 degrees C, 1 liter of solution is 1200 grams and contains (82.3/182.3)(1200 grams) = 542 grams which corresponds to 14.9 moles HCl.
For 30 degrees C, this works out to 13.2 moles HCl.
So the maximum concentration is about 15M at 0 degrees C decreasing to 13M at 30 degrees C.
For data outside this temperature range and not limited to atmospheric pressure see A STUDY OF THE SYSTEM HYDROGEN CHILORIDE AND WATER J. Am. Chem. Soc., 1909, 31 (8), pp 851–866

Adding hydrogen chloride to water reduces the melting point of the solution, so it's possible to chill further and increase $\ce{HCl}$ solubility without solidification. This phase diagram suggests that hydrogen chloride monohydrate exists as a liquid above -15°C, containing 67% $\ce{HCl}$ by weight. I don't have the density figure so I can't calculate the molarity, but it would likely be above 20 M.

yes, there is density information for that here http://pubs.acs.org/doi/abs/10.1021/ja01938a001 would be about 1.27

Molx's answer is very good. However, the way I learned it was a little different and may make more sense to some people. Instead of using 444g of HCl, which is perfectly fine, we can assume we have a sample of 100 g, much like we do in empirical formula calculations as to have every percent be out of 100.
To start, just like Molx mentioned, the maximum solubility at room temperature of HCl is 37% by mass. From this we can assume 100 g of HCl solution. From that assumption we know that 37% (the mass in HCl) of 100g of the HCl solution is 37.0 grams. The rest has to be water, as HCl is a gas at room temperature that is dissolved in water to form Hydrochloric acid.
Ultimately, to find the maximum molarity of the HCl in water, we must use the concentration equation for molarity, which is defined as moles of the solute divided by the liters of the solution (M = moles of solute/ liters of solution).
We now know that we have 37.0 g of HCl, which is necessary for the Molarity calculation. When you multiply 37.0 g HCl by (1 mol HCl/36.46 g HCl) as a conversion factor, we get 1.014 mol of HCl.
To find our liters of solution, we can use our density. Multiply 100 g of our HCl solution by our density conversion factor of (1 mL/1.19g) to get 84 mL, and then divide by 1000 to get 0.084 L.
When you substitute 1.014 mol HCl and 0.084 L into the molarity equation, we find that the molarity is around 12.07 M HCl, which is the highest concentration of HCl that we can obtain at room temperature.
Like Molx said, it "fumes" because it is so concentrated with HCl. As such, it should be treated as an approximation like Molx said as well. I hope this version makes sense too.


text:
Does this equation - involving calcium carbonate and a lead (II) cation - balance?
$$\ce{CaCO3 + 2H2O + Pb^{2+} -> Pb(CO3) + Ca(OH)2 + H2}$$
Does this equation balance?
On the right you have lead with 2+ charge, whereas on the right its completely neutral. So my question is do you need to balance the charges in order to make the equation correct?
If that equation is wrong, what is the right equation?
I need to show that hydrogen gas is formed (or some other gas but I don't know what else it can be), that the solution will become more basic (proven with $\ce{Ca(OH)2}$), and that the lead has to replace the calcium.

Unfortunately your equation does not add up, let me briefly explain why.
Calcium carbonate is a not very soluble salt. There is ionic binding between the calcium cation and the molecular carbonate anion. If you dissolve it in water you, the following equilibrium will establish: $$\ce{CaCO3~(s) <=>[\ce{H2O}] Ca^{2+}~(aq) + CO3^{2-}~(aq)}$$
Lead carbonate is also a salt, it is even less soluble. Only a little bit will dissolve according to the following equilibrium:$$\ce{PbCO3~(s) <=>[\ce{H2O}] Pb^{2+}~(aq) + CO3^{2-}~(aq)}$$
Calcium hydroxide is also a salt. It is a little bit more soluble than calcium carbonate. The following equilibrium is established when put in water:$$\ce{Ca(OH)2~(s) <=>[\ce{H2O}] Ca^{2+}~(aq) + 2{}^{-}OH~(aq)}$$
You write the following:

$$\ce{CaCO3 + 2H2O + Pb^{2+} -> Pb(CO3) + Ca(OH)2 + H2}$$

So in principle you have $\ce{Ca^{2+}}$, $\ce{Pb^{2+}}$ and $\ce{CO3^{2-}}$ on both sides of the equation. What is left is not a balanced equation, since there are two electrons missing:
$$\ce{2H2O ~\nrightarrow 2{}^{-}OH + H2}$$
You need a source for the electrons, one way of balancing the equation is
$$\ce{CaCO3 + 2H2O + Pb -> Pb(CO3) + Ca(OH)2 + H2}$$
Given the insolubilities of all the species involved, it might be hard to see any reaction at all. And I have not checked if the redox reaction is favourable, with potentials.

The reaction would make more sense, and be balanced for charge, if the $\ce{H2}$ were replaced by $\ce{2H+}$:
$$\ce{CaCO3 + 2H2O + Pb^{+2} -> Pb(CO3) + Ca(OH)2 + 2 H+}$$
Even that is a bit weird because it suggests that the water would be acidified by this (non-redox) process.  You could also write it as:
$$\ce{CaCO3 + 2OH- + Pb^{+2} -> Pb(CO3) + Ca(OH)2}$$
Lead hydroxide is apparently a compound that has never been isolated: the hydrated lead oxides are more stable than a true hydroxide salt.

Yup, your equation falls apart like dominoes.
Nothing balances the charge in lead ion, so you need to eliminate that reaction.
Nothing is left to balance the lead in lead carbonate, so that has to go away.
Nothing is left to balance the carbon in calcium carbonate, so good-bye to that reactant.
Nothing is left to balance the calcium in calcium hydroxide, so there can't be that product either.
Nothing is left to balance the oxygen in water, so -- no water.
And nothing is left to balance the hydrogen in elemental hydrogen gas, so that goes away.
Everything that's left, meaning zero, works.
You could have avoided this disaster by rendering the lead as the element not lead ion.  But Curt's answer is more likely as an actual chemical reaction.


text:
How to find partial pressures from a given equilibrium constant Kp?
I answered wrong on my quiz the following question on finding an equilibrium constant:

For the following reaction, at $1~\mathrm{bar}$ and $300~\mathrm{K}$, partial pressures of $\ce{NO2}$ and $\ce{N2O4}$ are both $0.5~\mathrm{bar}$. If pressure is increased to $2~\mathrm{bar}$ (this is the final pressure), and temperature is increased to $400~\mathrm{K}$, calculate the final composition of the system. $\Delta H$ for this reaction is $57.2~\mathrm{kJ/mol}$; assume it is independent of temperature.
$$\ce{N2O4 (g) <=> 2NO2 (g)}$$

My attempt:
I first found $K_\textrm{p}$:
$$K_p (300~\mathrm{K}) = \frac{P^2(\ce{NO2})}{P(\ce{N2O4})} = (0.5^2)/0.5 = 0.5$$
Then I used
$$\ln\left(\frac{K_p(400~\mathrm{K})}{K_p(300~\mathrm{K})}\right) = - \frac{\Delta H}{\mathcal{R}} \left( \frac{1}{400~\mathrm{K}} - \frac{1}{300~\mathrm{K}}\right) $$
I solved for $K_\textrm{p}$ at $\pu{400 K} = 92695.44$
This is where everything went wrong. I can not figure out how to get from $K_p$ to partial pressures. All I got to is the following line:
$P_\text{total} = 2P(\ce{NO2}) + P(\ce{N2O4})$, since $n(\ce{NO2})$ to $n(\ce{N2O4})$ ratio is $2:1$. I assumed they exert pressure proportionally.
Can some one guide me in the right direction?

Multiple choice answers:

$P(\ce{N2O4(g)}) = 0.0252~\mathrm{bar}$; $P(\ce{NO2(g)}) = 1.9748~\mathrm{bar}$
$P(\ce{N2O4(g)}) = 0.039~\mathrm{bar}$; $P(\ce{NO2(g)}) = 1.961~\mathrm{bar}$
$P(\ce{N2O4(g)}) = 1.45~\mathrm{bar}$; $P(\ce{NO2(g)}) = 0.098~\mathrm{bar}$
$P(\ce{N2O4(g)}) = 1~\mathrm{bar}$; $P(\ce{NO2(g)}) = 1~\mathrm{bar}$

I just calculated the answer by doing some answer to question maping.
essentially,  I solved for ln(Kp at 400K) on following equation
ln(Kp at 400k / Kp at 300k) = - (delta H / R) ( 1/ 400K - 1/300K)
Than substitute all the values from multiple choices. current answer should be 1.
But I still would like to know how to actually solve it.

Please verify the given numbers!

At the initial conditions, the mole fractions of $\ce{N2O4}$ and $\ce{NO2}$ are both 0.5. Take as a basis 1 mole of gas to start with at the initial conditions. Let $x$ represent the number of moles of $\ce{N2O4}$ that react to form $\ce{NO2}$. In terms of $x$, what are the new numbers of moles of these species? What are the new mole fractions? What are the new partial pressures?

Please note that in reality $K_p=0.14 bar$ for the reaction in your problem, not 0.5. This of course should not matter assuming this is considered only an exercise. It does matter however whether you provided the right answers to check the validity.

$\require{cancel}$
This is a partial solution to this question which involves a bit of using the state of the answers to answer the question.
As OP has suggested, the first step in this is calculating $K_\mathrm{p}(\pu{400 K})$, for this we need $K_\mathrm p (\pu{300 K})$
$$K_\mathrm p(\pu{300 K}) = \frac{P(\ce{NO2})^2}{P\ce{(N2O4)}} = 0.5$$
$$\ln\left(\frac{K_\mathrm p(\pu{300 K})}{K_\mathrm p(\pu{400 K})} \right) = -\frac{\Delta H}{R}\left(\frac{1}{\pu{400 K}} - \frac{1}{\pu{300 K}}\right)$$
Upto this point the answer is correct. However, there is a mistake in the calculation, taking $R = \pu{8.314 J K-1 mol-1}$, and $\Delta H = \pu{57200 J mol-1}$, we get:
$$\ln\left(\frac{K_\mathrm p(\pu{300 K})}{K_\mathrm p(\pu{400 K})} \right) = \frac{572\cancel{00} \times \cancel{100}}{8.314 \times \cancelto{3}{300} \times \cancelto{4}{400}}$$
Upon further solving, we get,
$$\ln\left(\frac{K_\mathrm p(\pu{300 K})}{K_\mathrm p(\pu{400 K})} \right) = -5.7333$$
Therefore, $K_\mathrm p(\pu{400 K})$ can be found to be:
\begin{align}
K_\mathrm p(\pu{400 K}) &= K_\mathrm p(\pu{300 K})\cdot e^{5.7333} \\
& = \pu{154.49 bar}
\end{align}
Now, the intended (and lengthy solution) intends for you to find the change in moles of $\ce{NO2}$ and $\ce{N2O4}$ and continue the solution.
As Chet Miller very well pointed out,

At the initial conditions, the mole fractions of $\ce{N2O4}$ and $\ce{NO2}$ are both 0.5. Take as a basis 1 mole of gas to start with at the initial conditions. Let x represent the number of moles of $\ce{N2O4}$ that react to form $\ce{NO2}$. In terms of x, what are the new numbers of moles of these species? What are the new mole fractions? What are the new partial pressures?
(emphasis mine)

Answer those two questions, find the value of x, and find the number of moles of each compound which would give you your answer about the intermediate steps that are involved in this solution. However, it is lengthy and arduous to show those steps.
A much shorter (and possibly intended) method involves using the state of the options provided. You firstly need to notice that there are only two possible options that provide an answer to this question. They are (A) and (B). Why? This is because the value of $K_p$ is greater than one which means that $P(\ce{NO2}) > P(\ce{N2O4})$. Now substituting these values, we get:
$$K_p (\textbf{A}) = \pu{154.76 bar}$$
$$K_p (\textbf{B}) = \pu{98.603 bar}$$
Within error of round off, the correct answer would be (A).

Let:
$A$ represent $N_2O_4$
$C$ represent $NO_2$
Then the reaction becomes:
$$\ce{A <=> 2C}$$
As noted from previous answers:
$K_{p1}=0.5 bar$ (@ 300K)
$K_{p2}=154.76bar$ (@ 400K)
Let:
$P_{A2}$ = final equilibrium partial pressure of A
$P_{C2}$ = final equilibrium partial pressure of C
$P_2$ = Final total equilibrium pressure
Then:
$P_2=P_{A2}+P_{C2}=2$
We can then put $P_{A2}$ in terms of $P_{C2}$:
$P_{A2}=2-P_{C2}$
We then substitute into the expression of $K_{p2}$:
$K_{p2}=\frac{P_{C2}^2}{P_{A2}}=\frac{P_{C2}^2}{2-P_{C2}}=154.76$
Solving for $P_{C2}$, and then for $P_{A2}$:
$P_{C2}=1.9748$ bar
$P_{A2}=0.252$ bar
So the final answer would be (1)


text:
Ammonium nitrate in fuel-air bombs
Are there any fuel-air bombs / thermobaric weapons that utilise ammonium nitrate as a fuel?
I know that BLU-82 is often mistakenly referred to as fuel-air bomb, so I became curious as to whether or not there were any actual fuel-air bombs that used ammonium nitrate.

Ammonium nitrate is an oxidizer, not a fuel.

Explosives that use ammonium nitrate and a fuel are mostly referred to as ANFO explosives, you might find more detailled information by using that term.
Most explosives used in military (especially bombs) are based on TNT, mostly in the form of Tritonal, a mixture of TNT and aluminium powder. Generally, only outdated bombs use ANFO explosives (due to their low price and ease-of-use). (See BLU-82, which does indeed use ammonium nitrate)
ANFO explosives are rather used in an industrial context, especially mining.
In the common mixture of ammonium nitrate with nitromethane (ANNM), no oxygen from the air is needed to sustain the explosion:
$\ce{3NH4NO3 + 2CH3NO2 -> 4N2 + 2CO2 + 9H2O}$
Since ANFO explosives carry their oxygen with them by definition (the ammonium nitrate molecule contains 3 oxygen atoms), they do not need oxygen from the surroundings and are therefore not thermobaric or fuel.air weapons.

I'm not aware of any FAE devices using ammonium nitrate as a fuel.
Oxirane is rather common, the thermobaric hellfire (AGM-114N) uses aluminium dust and the original WWII designs of Mario Zippermayr's anti-aircraft device used coal dust.


text:
How do I write out the chemical reaction of ammonia with strong acid?

Write equations for the reactions of each of: (a) nitric acid (b) sulfuric acid *(c) hydroiodic acid, with each of: (i) sodium hydroxide (ii) zinc oxide (iii) ammonia.

So, the reaction of nitric acid with sodium hydroxide and zinc oxide is simple because I know that acid + base = salt + water. Therefore:
$$
\begin{align}
\ce{HNO3(aq) + NaOH (aq) &-> NaNO3 (aq) + H2O (l)}\\
\ce{2HNO3(aq) + ZnO (s) &-> Zn(NO3)2 (aq) + H2O (l)}
\end{align}
$$
It makes sense. In the second part, for example, $\ce{Zn}$ is 2+ and $\ce{H}$ is -1. So using the criss-cross method, you get $\ce{Zn(NO3)2}$ and water. I understand this.
However, with ammonia:
$$\ce{HNO3(aq) + NH3 -> NH4NO3}$$
How am I supposed to know this equation? It's not metal + acid = salt + water. Is there word equations for everything?

In the second part, for example, $\ce{Zn}$ is 2+ and $\ce{H}$ is -1. So using the criss-cross method, you get $\ce{Zn(NO3)2}$ and water. I understand this.

I don't agree on the -1 for hydrogen, let's look again at that example:
$$\ce{2HNO3(aq) + ZnO(s) -> Zn(NO3)2(aq) + H2O}$$
Metal oxides, such as zinc oxide, are basic anhydrides.
Imagine that

$\ce{ZnO}$ was made by heating $\ce{Zn(OH)2}$ until it loses water:

$$\ce{Zn(OH)2 ->[\Delta] ZnO(s) + H2O}$$

$\ce{ZnO}$ is a salt of $\ce{Zn^2+}$ and $\ce{O^2-}$

The reaction between nitric acid and zinc oxide and just a neutralization reaction between two protons and $\ce{O^2-}$ (which, as such, isn't present in solution).
Now for the reaction with ammonia: For this you may want to remember that there are different acid-base theories: The one that helps here is the Bronsted-Lowry theory:

Acids are proton donors

$\ce{HNO3}$ definitely is that.

Bases are proton acceptors

How can something else other than the $\ce{OH-}$ that you knew and the $\ce{O^2-}$ from $\ce{ZnO}$ be a proton acceptor = form a bond to a (proton)?
Is that possible if this acceptor doesn't even have a negative charge?
(Yes, otherwise I wouldn't buld up the tension like in a suspense story :D)
If the proton can't provide the electrons for the bond, the acceptor has to deliver them! And that can happen in the form of an electron pair, a double filled, non-bonding orbital.
Therefore a reaction, such as
$$\ce{H+ + NH3 <=> NH4+}$$
is perfectly valid. The rest of the example is just arranging the counter ions.

How am I supposed to know this equation?

You have to memorise it. In aqueous solution, $\ce{NH3}$ deprotonates a small fraction of the water to give ammonium and hydroxide
$$
    \ce{NH3 + H2O <=> NH4+ + OH−} \, ,
$$
and while it is usually impossible to isolate $\ce{NH4OH}$ itself, the above mentioned ions are there in the solution. And thus, an aqueous solution of $\ce{NH3}$ is a weak base and reacts with acids in the similar way as other weak bases.

How can I access all the word equations in chemistry so I am able to write every single equation given to me like these.

Again, learning general principles plus memorising all the special cases & exceptions is the only way.$%edit$

Why then do I not have to memorise other reactions, such as acid + base = salt + water? Why did my textbook simply say "write out this equation". What is the point of making it an exercise if it simply requires memorisation? How am I supposed to know if I should be able to deduce this or memorise this?


text:
What is the viscosity of menstruation blood?
I am writing my bachelor thesis about alternative materials for feminine hygiene products and for testing I need to know what the viscosity is of menstruation blood. Can I use water as replacement for menstruation blood because this kind of blood exists out of a big part of water? And what is the composition of menstruation blood in percentages?

H. Larsson et al., Studies on blood viscosity during the menstrual cycle and in the postmenopausal period in healthy women, Acta Obstet.Gynecol. Scand., 1989, 68, 483-486 might be interesting to get a hand on. From the abstract, it seems that for plasma viscosity increases up to day 21 of the menstruation cycle.

I wonder if menstrual blood viscosity correlates well with circulatory blood viscosity, since the former has to do with shed tissue from the uterine walls rather than blood exposed from a cut.

It is difficult to give a typical value as the composition and hence the physical properties of menstrual blood varies over the course of a woman's period. This reference has a good discussion which you may find useful; you can find on p. 159:

It is fair to say that a large proportion of menses samples are considerably more viscous than blood or water - about four times as viscous as venous blood and 35 times as viscous as water.


text:
Second ionization potential of Gadolinium
I notice that Gd has a second ionization potential which is significantly higher than would be expected from the general trend in the lanthanides (see this paper p. 945 for a graph). What is the reason for this effect ?
The transition corresponds to $[Xe] 4f^{7} 5d^{1} 6s^{2} => [Xe] 4f^{7} 5d^{1}$. I understand that the 5d is occupied to prevent inter-electron repulsion between antiparallel f electrons, and this can be used to rationalize why the correspnding Tb transition requires less energy, but I'm not sure why the potential should be significantly higer than that for Eu.

Have a look at this question.

The transition corresponds to $[Xe] 4f^{7} 5d^{1} 6s^{2} \rightarrow [Xe] 4f^{7} 5d^{1}$.

Well actually, the second ionization energy being plotted in the graph is only the energy to lose the second electron so the transition is
$[Xe] 4f^{7} 5d^{1} 6s^1 \rightarrow [Xe] 4f^{7} 5d^{1}$.
For all the other elements Pr through Yb, there is no 5d electron, and instead an additional 4f electron. In all cases, it is the 6s electron that is being lost in the second ionization.
The more the nuclear charge is screened, the easier it is for the 6s electron to be lost.
The 5d electron is less effective than a 4f electron at screening the 6s electron from the nuclear charge, because 4f electron density is concentrated much closer to the nucleus (see Fig. 2 of below reference).  Therefore it takes more energy to loss the 6s electron for gadolinium.
For more information on nuclear charge screening see Atomic Screening Constants from SCF Functions. II. Atoms with 37 to 86 Electrons J. Chem. Phys. 47, 1300 (1967).


text:
Spontaneous Reaction and Gibbs Free Energy
I know how to derive the expression for the equilibrium constant starting from $\mathrm dG=0$,
$$\ln(K)=-\frac{\Delta G ^\circ}{RT}$$
However, if we are not at equilibrium, $\mathrm dG$ is not necessarily zero. In this case:
$$\frac{\partial G}{\partial \epsilon} = \sum \mu_i \nu_i = \Delta G ^\circ + RT \ln \left( \prod a_i ^{\nu_i} \right)$$
Where $\nu_i$ are the stoichiometric coefficients, $\mu_i$ are the chemical potentials, $\Delta G^\circ$ is the Gibbs free energy of reaction at standard state and $a_i$ are the activities.
Hence,
$$\frac{\partial G}{\partial \epsilon} = RT \ln \left(\frac{Q}{K} \right)$$
Where $Q=\prod a_i ^{\nu_i}$ is the reaction quotient.
So the question is, if we want to kown if a reaction is spontaneous, do we look at the ratio $Q/K$? If this is correct, if $Q > K$ then $\frac{\partial G}{\partial \epsilon} > 0$ and the reaction would be non-spontaneous, whereas if $Q < K$ then $\frac{\partial G}{\partial \epsilon} < 0$ and the reaction is spontaneous.
If so, if we start with reactants only, $Q=0$ (since the activites of the products are zero because there are none) and the reaction would be spontaneous, and we would always get at least a little bit of products.
However, various pages in internet say that if $\Delta G ^\circ < 0$ then the reaction is spontaneous, and it is not if $\Delta G ^\circ>0$.
So how do you know, then, if you start with only reactants, if a reaction is going to happen?

I think you are confusing two related concepts here. Spontaneity of a reaction is determined by the net change in free energy, $\Delta G$ - notice the capital Delta.
Which direction you will move along a reaction coordinate is determined by $dG$ - note the little $d$.
$\Delta G$ is the overall change in free energy when comparing products to reactants. It tells you whether the reactants are "uphill" or "downhill" of the products.
$dG$ is the slope of the free energy surface at your current position on the reaction coordinate. In the equation you gave, that's given by Q/K. K basically tells you what the equilibrium ratio between reactants and products is, and Q tells you what the current ratio is. The ratio of Q/K is an indication of how far to the left or right you are of equilibrium.
This picture might help explain it a little better:

Image the free energy surface as a gravitational potential energy surface (we can get away with this because free energy is the thermodynamic potential for systems at constant T & P). The ball represents the reacting system, and the x-axis is a reaction coordinate.
$dG$ (the differential) tells us the slope at a given point - it tells us where the system is going to go right now. $\Delta G$ tells us the overall change - it tells us where the system is going to eventually end up.
When you look up standard state $\Delta G$ values for a reaction, you are determining whether an overall reaction is spontaneous or not - you are figuring out which side of the equilibrium has more stuff - reactants or products.
When you look at Q, you are figuring out which direction things need to go to get back to equilibrium, based on where you currently are.

Your analysis is correct.  It is silly to say that a reaction is or is not spontaneous based soley on $\Delta G^{\circ}$.  If you have only reactants present and no products present, then the reaction will always be spontaneous, irrespective of $\Delta G^{\circ}$.

So you are saying that the reaction $\ce{O2 -> 2O}$ is spontaneous when all you have is $\ce{O2}$? That doesn't seem right...

thomji, it is right.  But the equilibrium constant might be so small that at "room" temperature, the expected amount of O atoms would be less than 1, even in a system the size of the atmosphere.

There is a difference between saying a reaction can happen and saying it is spontaneous.

I think that @thomij is correct. The reaction  O$_2$ + N$_2$=2NO , for example, has a $\Delta G^0$  of +180 kJ/mol. In a thermodynamic sense then O$_2$ + N$_2$ are stable as there is no process that can lead to a diminution in free energy.

IMHO, I think you had the right answer: "whereas if Q < K then ∂G∂ϵ<0 and the reaction is spontaneous."
So if you have only reactants, Q = 0. You then look up K for this temperature. Since K > 0 (even if it's very close to it), ΔG<0 and so the reaction will occur spontaneously. When products builds up, you will reach a point where Q = K, and at that point  ΔG will be equal to zero. (ΔG depends on the relative concentration of reactants\products and changes as products build up from ΔG < 0 to ΔG = 0 at equilibrium).


text:
Frost diagram and acidic/basic conditions.
The Frost diagram for sulfur shows the relative stability (in terms of cell potentials) of the possible aqueous oxidation states. What is the reason for the instability of higher oxidation states in the acidic regime and their stability in the basic regime ? Something I have always noticed but never been able to apply chemical intuition to.

Negative charges aren't happy in acidic media.

Yes, but the Frost diagram shows us the difference between the S oxidation states; degree of protonation doesn't affect that.

Write out the half-reactions and you will see that oxidation of S almost always creates lots of protons.  So you can think of Le Chatelier's principle: if there are already lots of protons around, it will be harder to create more of them.  Here's the example for complete oxidation of sulfide to sulfate at pH ~7:
$$\ce{HS- + 4 H2O -> 8e- + 9H+ + SO4^{-2} }$$
Conversely, under alkaline conditions, the equilibrium favors the products even more, because of the "removal" of the protons by the alkaline conditions.


text:
Determine molecular weight of biomass based on ultimate analysis
The ultimate analysis (by weight percent) for beech wood is:
$$\ce{C}=49.05\,\%,\; \ce{H}=5.83\,\%,\; \ce{O}=45\,\%,\; \ce{N}=0.12\,\%$$
Using the ultimate analysis values, the molar mass of each element, and assuming 100 grams of wood (therefore C = 49.05 g, H = 5.83 g, etc.), a representative formula for the beech wood is:
$$
\ce{C} = 49.05/12.01 = 4.0841\; \ce{mol} \\
\ce{H} = 5.83/1.01 = 5.77\; \ce{mol} \\
\ce{O} = 45/16 = 2.81\; \ce{mol} \\
\ce{N} = 0.12/14.01 = 0.0086\; \ce{mol} \\
\ce{C}_{\,4.0841\,}\ce{H}_{\,5.77\,}\ce{O}_{\,2.81\,}\ce{N}_{\,0.0086\,}
$$
What is the molecular weight of the beech wood based on the representative formula?

Simply put: you can't.

As a general rule, it is not a good idea to treat complex materials like wood as something you can calculate a mole of. You can of course analyze the material and get elemental rates, but that does not mean they are bonded in such way that this rate forms a minimum fraction which represents the whole thing.
When you calculate the number of moles of each atom in 100g, what you have is an estimation of the expected quantities in that sample. Undoubtedly, if you took different samples of the wood you would find different values, specially if sampling from different parts of the material.
As an exaggerated example, think about the elemental constitution of the human body, which is mainly 65% Oxygen, 18% Carbon, 10% Hydrogen and 3% Nitrogen in mass (source). With that in mind you can calculate a mole ratio of these elements and create a "representative formula". Obviously, though, there's no such thing as the human molecule, and you can't say that someone's body is X or Y moles of human.
Yet, like I originally answered, if you define what the representative formula is, you can calculate it's molar mass.
$$\ce {M}=4.0841*12.0107 + 5.77*1.00794+2.81*15.9994+0.0086*14.0067 = 99.947 \ \ce{g/mol}$$
Knowing ${\ce M}$, the number of moles is ${\ce 700,000/99.947 \approx 7004}$, and the concentration of biomass is $7004\ \ce{mol/m^3}$

Does it matter what is assumed for the mass of the wood? For example, to get the formula for beech wood I assumed 100 grams. However, that formula would be different if I assumed a different mass such as 25 grams. In the end, the mole concentration would be different too.

No, the molar mass of a compound does not depend on how much of a sample you take. The formula for beech wood just gives the ratios of how many atoms are in a molecule, not their number in a sample.

@andynitrox But the formula I gave for beech wood $C_{4.084},H_{5.77},O_{2.81},N_{0.0086}$ was assumed to be from 100 g.  If you assume it's from 25 g then the formula becomes $C_{1.02},H_{1.46},O_{0.7},N_{0.0021}$ which will give a molecular weight of $24.94;g/mol$ which gives a mole concentration of $28,067;mol/m^3$.

But does the particle density of your sample change only because you select a different quantity?

Like @andynitrox said, the molecular formula is independent of quantity. Just because you divided the amount by 4, the formula shouldn't change. This is very tricky when it comes down to representative formulas without integer proportions, since we can't simply draw molecules. Honestly, what I'd recommend is to avoid usin moles for these kind of complex molecules. I'll update my answer with that.

I added some discussion on the subject, I hope clarifies things a little.

In my own experience with Humic Substances (which are in fact related to wood, their structures usually are compaired to lignin), the best approach is simply use basic descripts like you already did. If in solution, you can use g/L for concentration. If solid, you can mention the mass, deinsity and elemental composition (in percentage like you did).
For humic substances, which are frequently used dissolved, it's also common to use the carbon molar concentration, i.e. 10 mmol C/L.

One more thing.  When calculating the molecular weight, should it be the original mass divided the moles in the representative formula?  For example, if assume 100 g sample then MW = 100 g / ( 4.0841+5.77+2.81+0.0086) = 7.89 g/mol.  This would be the same no matter what you assume for the sample mass.  If you assume 25 g sample then you get MW = 25 g / (1.0208+1.443+0.7031+0.0021) = 7.89 g/mol.

@Gavin not really. You get the same result because you divided the denominator and numerator by 4. Remember what we said in our answers: the molecular formula is the same regardless of the amount of substances. This is because the molecular formula (generally speaking) is related to the structure of the substances. For example, diborane has the molecular formula ${B_2H_6}$, because the molecule has this numbers of atoms, and not ${BH_3}$. Just because you can calculate a elemental ratio in mols of atoms, it doesn't mean you should use this to represent the material.

Ok, I renamed this question (so it isn't a duplicate of my other question) for just determining the molecular weight.  So if you want to adjust your answer then I'll accept it.  Thanks again for the help.

Using your mass percentages and the molar masses of the elements in your sample, you have correctly found out how many mol of each element are in 100g of your sample.
However, you need to keep in mind that wood is more than one simple molecule but instead rather a collection of quite a number of molecules. The masses of the elements that show up in all of those molecules add up to the percentages you found by this analysis. You see now that trying to formulate this distribution as a molecular formula has its limitations. Depending on how you select your sample, you will once include maybe $\ce{1 mol}$ of molecule A and $\ce{18 mol}$ of molecule B, whereas if you take twice the amount, you will count a total of 38 molecules as one.
As @Molx stated in his answer, if you always select the same mass for comparision, this might be a reasonably suitable unit.
To rephrase this: if you only had one molecule or a specific ratio of known molecules in your sample, it would be clear how the different elements distribute themselves to the different molecules. Since you do not know what molecules beech wood is made of, this molecular formula is not very useful for such a case.

As @Molx suggested in his comments, using moles may not be the best approach.  Do you have any recommendations on other approaches for representing different types of biomass?

I am not from this field, but I would suggest the density, humidity and fractional composition like in the question, that would probably specify the material sufficiently.

Please refer to my other question instead of this one, the new question will hopefully clarify what I'm trying to do, http://chemistry.stackexchange.com/questions/27895/representing-biomass-for-kinetic-reactions

Also, delete your answer so I can delete the question. Refer to the new question if you would like to comment. Sorry for the confusion.

You could ask a moderator to move the answers, otherwise Molx's is lost too


text:
Why is only one proton transferred in the reaction between sulfuric acid and sodium chloride
$$\ce{NaCl + H2SO4 -> NaHSO4 + HCl}$$
The above reaction happens, but why doesn't the following happen? Or does it but only when one of the reactants is in excess?
$$\ce{2NaCl + H2SO4 -> Na2SO4 + 2HCl}$$
The second reaction seems plausible to me. How can I predict whether a similar type of reaction will be like the first equation or the second equation?

The first reaction takes place when sulfuric acid is concentrated.

related: http://chemistry.stackexchange.com/questions/5048/hcl-synthesis-from-sulfuric-acid-and-sodium-salt, http://chemistry.stackexchange.com/questions/4750/reactions-of-sodium-halides-with-sulphuric-acid?lq=1

Sodium bisulphate (pKa ~ 2 - https://en.wikipedia.org/wiki/Sulfuric_acid) is a much weaker acid that hydrochloric acid (pKa = ~ -7), which means that chloride is not a strong enough base to deprotonate sodium bisulphate.
However all acid pbased reactions are at equilibrium.  Roughly one part in 10^9 of the species involve sulphate anion, which for all practical purposes is negligible.
Table of pKa charts: http://www.lauher.com/che142/acidbase.pdf

https://en.wikipedia.org/wiki/Sulfuric_acid says II pKa is ~2

better data, changing answer.  Although what those pKas mean is a bit questionable.


text:
Changing solubility of metal complex by adding ethanol
After creating a solution of copper(II)-sulfate-pentahydrate, water and concentrated ammonia, the newly formed complex (tetraamminecopper(II)-sulfate-monohydrate) is completely dissolved in water. By adding ethanol, one is able to decrease the solubility of that complex in water and have crystals percipitates out of solution.
Since the complex is not soluble in ethanol, I am having trouble understanding why adding ethanol changes anything about the solubility of the complex. Would adding some organic phase like petrol ether or a non-soluble solid such as ceramic pellets do the same?

Why tetraamminecopper(II)-sulfate-monohydrate is formed and not $\ce{[Cu(NH3)4(H2O)2]^2+}$?

Would adding some organic phase like petrol ether or a non-soluble solid such as ceramic pellets do the same?

No. But adding other organic solvents which are miscible with water would.
The idea behind the "trick" is to reduce the polarity of the aqueous phase. With other words, you're making water more ethanol-like.


text:
How to synthesize Boron tri-fluoride using Boron Oxide
I am finding different equations for the synthesis. My lecture handout suggests a reaction with Sodium tetrafluroborate and Sulfuric acid. This is fine but I have to remember this for an exam and that reaction looks extremely hard to remember/balance in an exam. A second way seems to be reaction of boron oxide with Calcium fluoride and Sulfuric acid. However, I cannot find consistent equations for this...
Wikipedia simply suggests that Boron oxide reacts with $\ce{HF}$ which is made in situ by the reaction between sulfuric acid and calcium fluoride:
$$\ce{B2O3 + 6HF -> 2BF3 + 3H2O}$$
It's easy to remember but I am reluctant to accept this because it excludes two of the reactants (sulfuric acid and calcium fluoride). Would it be acceptable to merge the formation of $\ce{HF}$ into the above equation - thus including all reactants:
$$\ce{B2O3 + 3CaF2 + 6H2SO4 -> 2BF3 + 3H2O + 3Ca(HSO4)2}$$
It's a little tricky to balance but not too hard to remember. Is there anything wrong with the above equation? I haven't seen it anywhere but only variations on it.

$\ce{CaF2} + {H2SO4}$ will give you $\ce{HF}$ so why bother if you add directly the $\ce{HF}$ instead ?

I don't know but that is how it is synthesised

@Babounet Because anhydrous hydrofluoric acid is more expensive and more difficult to handle than a mixture of fluorspar and sulfuric acid.

I agree Klaus, but since it's an exam and not a lab experimentation, who cares ?

The reaction of boron trioxide with sulfuric acid and calcium fluoride to obtain boron fluoride is given here.
$$\ce{B2O3 + 3CaF2 + 3H2SO4 ->[\Delta] 2BF3 + 3CaSO4 + 3H2O}$$
Conditions

Boron trioxide react with calcium fluoride and sulfuric acid to produce boron trifluoride, calcium sulfate and water. Sulfuric acid - concentrated solution.

The reaction will be easy to understand if you break the reaction into two reaction.
Babounet says:-

$\ce{CaF2 + H2SO4}$  will give you $\ce{HF}$ so why bother if you add directly the $\ce{HF}$ instead ?

Indeed $\ce{CaF2 + H2SO4}$  will indeed yield  $\ce{HF}$ along with calcium sulfate as indicated here.
$$\ce{CaF2 + H2SO4 ->[\Delta] 2HF + CaSO4}$$
This $\ce{HF}$ will now react with boron trioxide to yield boron trifluoride as indicated here.
$$\ce{B2O3 + 6HF → 2BF3 + 3H2O}$$
Thus the  products are boron trifluoride, calcium sulfate and water and not calcium bisulfate.


text:
Why does H₂Se have a bigger Ka2 than H₂Te
Why does $\ce{H2Se}$ have a bigger $K_\text{a2}$ than both $\ce{H2S}$ and $\ce{H2Te}$?

What do you mean with "$Ka_2$"?

@Philipp $K_{a2}$ refers to the second dissociation constant i.e. the equilibrium constant of $\ce {HA^{-}<=> H+ + A^{2-}}$

Would you mind to provide the data and the source?

The following data are from the Rubber Book (CRC Handbook of Chemistry and Physics), 89th edition, 2009, p 8-40.
\begin{array}{lll}
\mathrm{acid} & \mathrm{pK_{a1}} & \mathrm{pK_{a2}} \\
\hline
\ce{H2S} & 7.05 & 19\\
\ce{H2Se} & 3.89& 11\\
\ce{H2Te} & 2.6 & 11\\
\end{array}
Note that $\mathrm{pK_{a1}}$ of $\ce{H2Te}$ was determined at 18 °C, all other data at 25 °C


text:
Conversion of methane to propane
Is it possible to chemically convert methane into propane?
If so, what is involved and is it economical?

Also, can you please describe what you mean by involved? The catalyst (if any)? The process? What? (The answer could cover all of them, though)

Yes, it can be done by mimicking the conditions in the earth's mantle.
It was shown by Alexander Goncharov and coworkers [1] by placing methane in a diamond anvil cell, applying a pressure of more than 2 GPa and heating everything to a temperature around 1000-1500 K with a laser.
Is it economical? Hardly, but very interesting anyway!

Kolesnikov, A.; Kutcherov, V. G.; Goncharov, A. F. Nature Geosci 2009, 2 (8), 566–570. DOI 10.1038/ngeo591.

Can we use WURTZ reaction to convert   methane to propane?

@NeilRoy You might want to have a look at this older question for possible problems with the Wurtz reaction.

Generally, this reaction is run in reverse.
Hydrogen reformers are used to generate hydrogen by the following reaction scheme:
$$\ce{CH4 + H2O -> CO + 3H2}~~~\mathrm{(Endothermic, 1000C)}$$
This is frequently followed by the shift reaction
$$\ce{CO + H2O -> CO2 + H2}~~~\mathrm{(Exothermic, 300C)}$$
Hydrogen reformers that run on heavier hydrocarbons than methane are frequently fitted with a pre-reformer in which the reaction starts at a lower temperature. In the pre-reformer the following reaction occurs, amongst others, including some that generate hydrogen:
$$\ce{C3H8 + 2H2 -> 3CH4}$$
Whether this reaction is useful is debatable, but it does indeed occur because it is mildly exothermic and has no change in the number of moles of gas.
Running the reaction in the way that you want would require the removal of some of the hydrogen, to push the equilibrium in favour of propane. This might be useful in the case of stranded gas: i.e natural gas which is economic to extract (or which is extracted as a byproduct of oil extraction) but is not economic to ship.
Frequently stranded gas is converted to methanol for shipping. Synthesis gas (a mixture of $\ce{H2}$ and $\ce{CO}$ is an intermediate step.) Methanol can be converted to heavier hydrocarbons such as gasoline if desired.
However, as mentioned in the same reference we can also perform a more direct conversion of natural gas to heavier hydrocarbons via the Fischer-Tropsch process (using the synthesis gas step but not the methanol step). Main article
Partial oxidation of the natural gas eliminates the excess hydrogen and ensures the correct balance of of $\ce{H2}$ and $\ce{CO}$ in the synthesis gas to ensure heavier hydrocarbons are formed. We have two reactions that can vary the ratio of $\ce{H2/CO}$ in the feed to the Fischer-Tropsch process:
$$\ce{2H2 + O2 ->  H2O}$$
Consumption of $\ce{H2}$ in preference to consumption of $\ce{CO}$, decreases $\ce{H2/CO}$ ratio.
$$\ce{CO + H2O -> CO2 + H2}$$
Shift reaction, increases $\ce{H2/CO}$ ratio.
This composition control and several other parameters enable the desired mixture of hydrocarbons to be obtained from the Fischer-Tropsch process, though a single compound such as propane is rarely if ever obtained.
In summary, $\ce{3CH4 -> C3H8 + 2H2}$ is not possible under normal conditions, but $\ce{3CH4 + O2 -> 2H2O + \mathrm{a~mixture~of~hydrocarbons~of~average~formula~C3H8}}$ is.
Adding air to the mixture for oxidation is problematic because it adds nitrogen which must be purged away later, but I am guessing it is much cheaper than running a pressure swing absorption plant to remove excess hydrogen.
Here's an example of a gas to liquid hydrocarbon plant. I'm guessing it doesn't make propane, because heavier hydrocarbons are more valuable and easier to ship, but the principle remains the same.


text:
How can I determine the pH of a solution when I add more analyte after the equivalence point?
In titrating 25 mL of a 0.2 M weak acid, HA, with 0.5 M of a strong base, say NaOH, the equivalence point is reached after 10 mL of base is added. I can calculate the pH by first setting up an ICE table to calculate the moles of A- remaining at the equivalence point, and then setting up an additional ICE table for the hydrolysis of A- to calculate the moles of OH- at the equivalence point. From that, I can determine the pOH and thus the pH.
My question is this: what would be the effect of adding more weak acid? Would I need to set up an ICE table with non-zero initial values for HA, OH-, and A-? If that's the case, then how do I determine the pH?


text:
Thermal decomposition products of polyurea?
Are the thermal decomposition products of polyurea polymers known, especially of the urea group, in the absence of water?
Looking at the formula
R-NH-C=O
     |
    HN-R

there appears to be no obvious candidate for a decomposition product - not enough oxygen for CO2, not enough hydrogen for Ammonia.
Any ideas or, even better, references?

Here's one abstract:

The primary fragmentation processes in the thermal decomposition of polymers were studied in detail on a series of structurally related polyureas by direct pyrolysis with a mass spectrometer. Our results indicate that polyureas I–III undergo a quantitative depolycondensation process analogous to that observed for N-monosubstituted polyurethanes. The thermal decomposition of polyureas IV–VI proceeds by intramolecular hydrogen transfer processes that occur at higher temperatures with respect to depolycondensation. Polycarboxypiperazine VI is decomposed by a single-stage decomposition mechanism that leads to fragments with amino end groups and carbon oxide.

I took a look at the article.  Depolycondensation is accurate: most of the polyureas examined decomposed to the "parent" isocyanates and amines.  The temperature range examined was 150°C to 350°C.
If you're interested in what happens at higher temperatures, I would guess a decomposition to mixtures of CO, ammonia, water, and possibly methane and/or hydrogen.  But I didn't find any references that would support me after a quick look

Thanks a lot, this was indeed the sort of answer I was looking for.


text:
In infrared spectroscopy, why is there no C=C reading for E alkenes?
For the (E)-2-pentene, there was no C=C reading, about 1600. For the (Z)-2-pentene, there is a C=C. Why is there a reading on the Z but not on the E?

The $\ce{C=C}$ stretch is responsible for this ir peak. For an ir absorbtion to occur, the absorption must result in a change in dipole moment.  If we examine the $\ce{C=C}$ stretch in cis- and trans-2-butene we find that

image source

cis-2-butene has a dipole moment (0.33 D) and upon stretching the double bond the dipole moment will change; therefore we would expect to see an ir peak for the $\ce{C=C}$ stretch in this molecule
trans-2-butene does not have a dipole moment  (0 D) and upon stretching the double bond the dipole moment will remain zero, there is no change; therefore we would not expect to see an ir peak for the $\ce{C=C}$ stretch in this molecule

A general rule based on these dipole moment arguments is that, for symmetrically di-substituted olefins $\ce{C=C}$ stretches in trans olefins are non-existent or very weak, the absorption is much stronger in the cis isomers.
We can extend this reasoning to molecules with dipole moments, if the dipole moment is very small.  Trans-2-pentene does have a dipole moment but it is very small and it is less than the dipole moment in the cis-isomer and the dipole moment in the trans-isomer (being smaller to begin with)  will change less upon stretching the double bond than it will for the cis isomer.  Therefore we would expect a very weak absorption in trans-2-pentene, while the cis isomer should show a stronger absorption.


text:
Hyper conjugation in acetic acid?
While comparing the relative stabilities of acetic acid and formic acid why don't we consider the 3 possible hyper conjugative structures which would stabilize the positive charge on carbon thus making the conjugate of acetic acid more stable.
My professor told me hyper conjugation does not operate here. But what is the reason behind this?

Why do you think the carbon has a positive charge on it ?

Do you mean "making the conjugate base of acetic acid more stable"?

#rajesh. draw the resonating structures and observe it

Hyperconjugation does play a role in your example.  Acetic acid ($\mathrm{p}K_\mathrm{a}=4.75$) is a weaker acid than formic acid ($\mathrm{p}K_\mathrm{a}=3.75$).  To understand this we must consider the relative stabilities of both the ionized and unionized acids.

In acetic acid there is a partial positive charge on the carbonyl carbon.  Both the inductive effect and resonance effect (e.g. hyperconjugated resonance structures like B) of the methyl group act to stabilize this charge, thereby stabilizing the unionized form of acetic acid.  In formic acid, without a methyl group, similar stabilization of its unionized form is not possible.
Now let's look at the ionized acids.  The negative charge in the anion is delocalized as shown in resonance structure C. This negative charge on the two oxygens makes them much less electron withdrawing.  Consequently, there is much less positive charge on the "carbonyl" carbon in the ionized form.  In turn, there is less need for stabilization of positive charge by the methyl group.  The methyl group does not stabilize the ionized form as much as it stabilizes the unionized form.
Since the methyl group stabilizes the unionized form of acetic acid more than a hydrogen and since the methyl doesn't stabilize the ionized form much more than a hydrogen, there is a higher energy barrier for acetic acid to ionize compared to formic acid.  This means that acetic acid will be a weaker acid than formic acid.

so in case of ionized acteic acid there will be no hyper conjugation because there is no need as the positive charge on carbon has already been minimized due to resonance.only +I effect of methyl will operate or the converse will hold, ie no +I effect only the hyper conjugation?

Everything you said up to, "or the converse will hold, ie no +I effect only the hyper conjugation" is correct.  In the unionized acid both inductive and hyperconjugation effects occur.

so in case of ionized ones only +I of methyl is taken into consideration while in case of unionized ones +I as well as hyper conjugation will take place

Yes, that's correct.

I wonder how mUch structure C contributes, it could be almost 0 in which case HC is negligible.


text:
What's happening at the beginning of a weak acid titration?
At the beginning of a Weak Acid/Strong Base titration curve there is a sharp decrease in $[\ce{H+}]$ and a sharp increase in pH before the ratio between the weak acid and it's conjugate base becomes low enough that we have a buffer solution.

What's happening at that point?
My interpretation is that, given the balanced equation:
$$\ce{
CH3COOH~(aq) + H2O(~l) <=> CH3COO- ~(aq) + H3O+~(aq)
}$$
The addition of $\ce{OH-}$ would shift equilibrium to the left by Le Chatelier's principle making the newly generated $\ce{CH3COO-}$ react with $\ce{H3O+}$ to make more $\ce{CH3COOH}$, thereby increasing pH. After that, the $[\ce{H3O+}]$ is so low that the addition of the anion is not enough to keep shifting the equilibrium position to the left, so the remaining weak acid resists the changes in pH with its dissociation.
Does this make any sense?

Unfortunately this is not correct.

$$\ce{CH3COOH~(aq) + H2O(~l) <=> CH3COO- ~(aq) + H3O+~(aq)}$$

If you have a pure acetic acid solution the equation you state governs the pH of the solution.
If you are now adding a strong base like $\ce{{}^{-}OH}$, you are removing the hydronium ion from the right hand side of the equation. According to Le Châtelier, the equilibrium adapts and shifts more to the right, i.e. more acetic acid dissociates. As you remove $\ce{H3O+}$ the pH increases.
Another way of looking at this is, that you remove acetic acid from the left hand side (similar to what you stated).
$$\ce{H3CCOOH + {}^{-}OH <=> H3CCOO- + H2O}$$
According to Le Châtelier, the equilibrium (in the first equation) has to shift to the left, to regenerate more acetic acid, removing once again $\ce{H3O+}$ and the pH increases.
In any case you choose to look at, you increase the concentration of acetate in the solution and another equilibrium comes into play.
$$\ce{H3CCOO- + H2O <=> H3CCOOH + {}^{-}OH}$$
This equilibrium is essentially responsible for raising the pH.

Isn't the strong base react both with the acid and hydronium ions?

@adosar Yes, but that won't make a difference in terms of equilibrium.

So a reaction of a strong base with the acid isn't always $100%$?

@adosar I don't understand what you are asking. It is an equilibrium. I don't understand what you mean with 100%. Please ask a new question for clarification.

@Martin-マーチン Sir from the answer you wrote are there actually four equlibrium coexisting together ? The fourth one being the H3O+ + OH- -> 2H2O ? And if yes then ig when we derive pH formulas for this titration at some stage why we just wrote only some of them to get the formula why not all of them required to get the exact formula ? And its true right we cannot deduce the whether H+ is decreasing from starting onwards as such the rate at which the H+ is getting formed from CH3COOH , we cannot deduce more than rate its getting decreased through reaction with OH- ?

These systems are more complex than they might seem at first. I more or less agree with Martin's answer above but there are some subtleties at play. There are two acids present which can react with added hydroxide ions: the CH3COOH molecules and the hydronium ions.
Initially, as OH- is added, it reacts with and removes the very small amount of hydronium ions since this is by far the faster reaction (H3O+ + OH- proton transfer is probably the fastest reaction that occurs in any aqueous solution).
After that, it starts deprotonating CH3COOH molecules and builds up the CH3COO- concentration. As this acetate ion concentration increases it drives the equilibrium to the left, thereby suppressing any further dissociation of CH3COOH to produce hydronium ions.
Similarly, as the CH3COO- concentration builds up, some of it reacts with water to produce OH- ions, and this is in general what causes the pH to rise.


text:
Why are C6 dispersion forces named C6?
I'm reading about water models and their dispersion coefficients, and going back to S. C. Wang's work according to the citation of an equation (Google books here)

As you can see one citation's to a biographical work, and the other to a German work.
Nowhere in the chapter I'm reading (chapter 5, ‘Resolution’, in Rowlinson’s 2002 Cohesion: A Scientific History of Intermolecular Forces) does it say what $C_6$ stands for, and it doesn't seem to be a series of $C_n$, so I'm a little confused.
There's the equation, apparently a Casimir-Polder integral
$$E_\mathrm{DS} = ‒\sum_{n=6,8,10,...} \left(\frac{C_n}{r_n}\right)$$
in something else I'm reading, but it's biochemistry and so they're skipping over actually defining these things (so I can see DFT would be density functional theory but can't actually figure out what either $E_\mathrm{DS}$ is, nor the meaning of a $C_6$ coefficient. Looking for $E_\mathrm{DS}$ is just bringing up a bunch of articles' editors in search results (Eds.).
Can anyone enlighten me as to the latter - the meaning of a $C_6$ coefficient with respect to [London] dispersion forces?
My own answer to this after more reading

The oscillating electrons in a molecule generate not only instantaneous dipoles but also quadrupoles and higher multipoles. It is to be expected, therefore, that the London dispersion energy is only the first term in a series expansion for the attractive energy

that starts in $$u(r) = −C_6r^{−6}$$
The term comes originally from the wave equation (as Rowlinson's text does eventually describe), the number six arising from a Cartesian coordinate system for each of the two interaction partners

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

http://en.wikipedia.org/wiki/Lennard-Jones_potential
contains terms r^-12 and r^-6 with respective coefficients, could it help?

I've looked into it in some detail now, should probably delete this... :-/ Apologies, all looked a bit too quantal but wasn't so bad

It may be useful for other guys and you can post your solution as an answer

This is good, but you should cut out the portion that is an answer and post it an an answer (this is not just better practice here but gets you more points per up vote).

From Calculation of Coefficients in the Power Series Expansion of the Long‐Range Dispersion Force between Atoms J. Chem. Phys. 56, 2801 (1972)

Highly accurate calculations
  of $C_6$, the coefficient of the $R^{-6}$ term, resulting from
  an induced dipole-induced dipole interaction, have
  been made for many atomic systems, while values of
  $C_8$ (dipole-quadrupole interaction) and $C_{10}$ (dipoleoctopole
  plus quadrupole-quadrupole interactions) are...

In other words, in London dispersion interaction is usually represented as induced dipole - induced dipole interaction, the energy of which is proportional to $R^{-6}$, but there are other components of the interaction.

Nice, I found my way to that paper just a few minutes ago! My seconday source says their estimation of C6 was only ~14% over current accepted value :-)

Actually no I was looking at their 1971 paper, improved error bounds for the long-range forces between atoms, which just calls it "the constant C(6)ab" - thanks for this

@DavePhD is there a way to calculate "c6" with gaussian or gamess? I  am trying to parameterize a molecule I finished bond-angle-dihedral-improper values. But I don't know how to calculate "well depth". All articles talking about "c6" value but I don't know how to obtain it. Could you suggest me any source or any way to calculate it?

authors: Евгений Артеменко


text:
How to lose as little heat as possible to the surroundings from an endothermic reaction?
For an experiment, I'm adding several substances (as powders) to hydrogen chloride, and measuring the temperature changes to calculate enthalpy. What is a container I could perform it in (other than a styrofoam cup) that would minimise the heat lost to the surroundings?
I looked up using a vacuum flask of some sort, but would that be appropriate for this sort of reaction? The temperature is being measured via a temperature probe.

In endothermic reaction, if you start it at room temp., you'd have heat gain from the surroundings.

Yeah, some are endothermic and some are exo. I'm trying to minimise the effect of heat losses/gains on the change temperature.

What is the scale?
What is the budget?

Reactors are available from 250 mL up.
It might be cheaper to place a flask in a cut-off bottom of a suitable PE bottle and insulate it with PU foam from a DIY market.
UPDATE
A small Dewar cylinder might be an option too.
The model 00C has a volume of 100 mL and lids are available too. Usually, these Dewars are used to store liquid nitrogen and cooling mixtures (2-propanol + dry ice, etc.) so I don't know how resistant the lid is towards $\ce{HCl}$ over time. In addition, this setup isn't as tight as a flask with hollow stoppers and/or a silicone seal if that's an issue.

The scale is pretty small, only $50cm^3$ of HCl. Would a vacuum flask, the sort shown here, reduce heat loss?

@deusy The Dewars might work too. I've updated my answer.

Thanks! One last thing that you might have a suggestion for: What is the best way to stir the reaction once I add the powder? I'm measuring temperature with a temperature probe, but stirring with that is probably not a good idea because it may affect the temperature recordings.

Does the powder dissolve immediately? Are measurements  on an inhomogenenous sample useful? Can you measure the temperature before, then add the powders to a stirred solution, stop stirring and start the real measurements? Maybe this is good enough for a separate question on [tag:experimental-chemistry].

The new question is here.


text:
How to store methanol
We will be buying methanol by a about 200 liters at a time, I was wondering what I should consider when storing it.
So far I know that it should be

in the shade
earthed with a copper stake
in a secondary plastic container
wash empty containers and drain them

But I am wondering what else should be considered? How far should it be away from a house?

It is quite toxic, so I would add good ventilation and very clear toxicity labels, as it tastes the same as ethanol. And explain to people working with it, what the concerns are. See http://en.wikipedia.org/wiki/2012_Czech_Republic_methanol_poisonings

Eh, there's something good to read here, if you felt bored anytime!

Methanol burns with a colorless flame , so if there is any  chance, situation in your storage or handling that it may ignite you might want to consider infrared flame detectors and alarms as a precaution.

One of the first places to look at for advice is the Materials Safety Data Sheet (MSDS, or just SDS). Make sure these are kept in a handy place and are up to date. These should be provided by the supplier (but are easy enough to get online).
A pertinent and thorough guide is produced by the aptly named Methanol Institute, the document is called Methanol Safe Handling Manual, the chapter Health and Safety, especially the Safety Precautions subchapter outlines the following:

Smoking must be prohibited.
Vehicle access should be strictly controlled.
Ventilation must be sufficient to cope with the maximum expected vapor levels in buildings.
Positive pressure may be required for methanol-free areas, such as control, switch, and smoking rooms.
Storage tank vents to atmosphere should be sized for fire-heated emergency vapor release.
Electrical equipment must be explosion-proof to meet national electrical code requirements.
Alcohol-resistant Aqueous Film-forming Foam (AR-AFFF) with 6% foam proportioning (with water) equipment is advised for use on methanol
  fires.
Dry chemical extinguishers should be accessible for small fires. An adequate supply of handheld and wheeled types should be available.
Hydrants should be strategically placed with adequate hoses.
Small spills should be remediated with sand, earth, or other non-combustible absorbent material, and the area then flushed with
  water. Larger spills should be diluted with water and diked for later
  disposal.
Lighting should be grounded. Tall vessels and structures should be fitted with lightning conductors that are securely grounded.

These kind of directives should be on signs that are easily read alongside hazard signage.
I would suggest having a readily accessed first aid kit and make a poster with first aid instructions, also included in the handbook - First Aid Measures
In terms of storage, the subchapter Methanol Storage, the handbook states that

In general, fire protection for gasoline tanks is sufficient for
  methanol tanks, provided extra precaution is made for leak detection, toxic hazard, and
  availability of alcohol-compatible fire suppression foam.

In terms of proximity to the house, I would consider storing these tanks as far as feasible from the house.

When I read heading I thought it is general question for which 1 answer is okay. But when I read 200 litres I thought I will dig out a bit and write few more stuff.
I found few related question on Methanol Institute

What kinds of drums are recommended for methanol storage, only steel or are some plastics acceptable?

Well there is no Global Standards, but according to Methanol Institute following care should be taken.

Mild steel is usually selected as the construction material for methanol storage containers.

If moisture and trace amounts of inorganic salts are expected to exist within the container from time to time, then one should consider upgrading from carbon steel to 316 L stainless steel, or even a titanium or molybdenum stabilized grade of 316 L stainless steel

plastic storage containers are more susceptible to physical damage than metal ones (e.g., if dropped or hit by a forklift).
Pure anhydrous methanol is mildly corrosive to lead and aluminum alloys, and more so to magnesium, and platinum. Methanol-water solutions can be corrosive to some non-ferrous alloys depending on application and environmental circumstances, including copper alloy, galvanized steel, and aluminum alloy components, and some plastics and composites.

What types of extinguishers should be used to put out small methanol fires?

Large methanol pool fires are best extinguished with Alcohol Resistant Aqueous Film Forming Foam (AR-AFFF) with 6% foam-water proportioning equipment.

accidents involving methanol spills, what is the recommended evacuation distance?

The 2008 edition of the Emergency Response Guidebook (ERG2008) recommends an immediate isolation distance of 150 feet (50 meters) in all directions.

What precautions are needed when welding or torch cutting tanks or containers that were used to store methanol?

It has the potential to catch fire when hot work such as, welding, brazing, soldering, cutting, heat treating, grinding, and using power-actuated tools, is performed near methanol sources. When mixed with air, methanol can burn in the open. Methanol spills may travel short distances (yards or meters) along the ground before reaching a point of ignition and flashing back. The hazards associated with hot work can be reduced by implementing an effective hot work program that includes prior work authorization, safe welding practices, and a fire watch.

I think you should read all the questions I have provided in the beginning of Answer

Nice answer! I think we've covered it all between our two answers.