text:
Why is carbon dioxide nonpolar?
I understand that polarity corresponds to an electronegativity difference and that the larger the electronegativity difference, the more polar the bond.
However, I have read that carbon dioxide is nonpolar. This doesn't make sense to me.
In carbonyl compounds, the carbonyl carbon is partially positive and thus carbonyls are polar. $\ce{CO2}$ is simply two such units bonded together, so wouldn't it be that you would have an even more positive carbon and two partially negative oxygens causing $\ce{CO2}$ to be overall polar?
On top of that, $\ce{CO2}$ dissolves to a much larger extent in water compared to $\ce{O2}$, despite both molecules supposedly being nonpolar.

so wouldn't it be that you
  would have an even more positive carbon and 2 partially negative
  oxygens

Yes, your analysis is correct to this point.  A chemist would say that the bonds in $\ce{CO2}$ are polar (or polarized) and therefor each $\ce{C=O}$ bond has a bond dipole moment.  However the molecule itself is linear and the two bond dipole moments are oriented 180 degrees with respect to one another and cancel each other out, so overall the molecule does not have a dipole moment and is non-polar.

EDIT:  There are a couple of reasons why $\ce{CO2}$ is more soluble in water than $\ce{O2}$.  Because the two $\ce{C=O}$ bonds in $\ce{CO2}$ are polarized (whereas in $\ce{O2}$ the bond is not polarized) it makes it easier for the polar water molecule to solvate it and to form hydrogen bonds.  Both of these factors will stabilize a $\ce{CO2}$ molecule more than an $\ce{O2}$ molecule in water; stabilization translates into greater solubility.  Another factor enhancing the solubility of $\ce{CO2}$ in water is the fact that $\ce{CO2}$ reacts with water to set up an equilibrium with carbonic acid.
$$\ce{CO2(aq) + H2O <=> H2CO3(aq)}$$
This reaction will also enhance $\ce{CO2}$'s solubility in water compared to oxygen which does not react with water.

but this does not explain why more of this dissolves in water than how much O2 dissolves in water.

@caters You should probably ask about that in another question.

I accept this answer as valid and correct. However, I really have problems with this definition of polarity solely based on on the dipole moment. I would consider toluene to be an unpolar solvent, yet the molecule itself has a dipole of 0.36D, it is small, but it is there. Does that mean toluol is a polar molecule? On the other hand $\ce{CO2}$ has no net dipole moment, but its reactivity towards nucleophiles/ electrophiles is similar to polar molecules. Therefore I'd rather consider it polar. This is in no way a critique on your answer, I'd just value your opinion on the matter.

@Martin The dipole moment or polarity of a molecule never mattered that much to me.  I was usually interested in the reactivity of a molecule, which usually translates into the reactivity of a bond.  So the polarity of a bond is what I would consider.  If the hybridization in the bond wasn't symmetrical, then the bond was polarized; the asymmetry of the hybridization told me how polarized.

In understanding molecular polarity you need to take the whole structure into account.
Your reasoning is correct as far as the parts of the molecule are concerned. The individual bonds are polar.
But, a molecule can only be polar if it has a net dipole moment (that is, the charges don't balance out in direction across the whole molecule). So CO is polar as the polarity of the carbon-oxygen bond is unbalanced and the distribution of partial charge on the bond affects the whole molecule (giving the whole molecule a dipole moment). But CO2 is a linear molecule and the partial dipoles of the two bonds are in exactly opposite directions. Hence they exactly balance out giving a molecule which is non-polar.
If CO2 were not linear (like SO2 which is bent) it would be a polar molecule.

This may be the source of the OP's confusion, yes. Otherwise what would the difference between O=C=O and, say, H-O-H be? Maybe you could explain why O=C=O is linear and H-O-H is "bent".

@MrLister The why particular molecules have certain shapes is a much bigger question. If you want to know whether a molecule is polar then you have to know the structure (which you can look up and worry about why that is the structure in another class).

@MrLister well in water molecules the angle between the 2 H's is only 105° as opposed to 180° for the 2 O's in carbon-dioxide thus the negative O in water is not balanced out. A lot of the properties of water come from this inbalance - including that its normally liquid and has a way higher melting point than carbon-dioxide despite being the lighter molecule (by all other means it should be gaseous) - because the dipole character allows forming of bridge groupings of water molecules .. which doesnt occur for carbon-dioxide

The formula for the net dipole moment $\vec{\mu_{net}}$ of an overall neutral system of $n$ charged point particles is given by:
$$
\vec{\mu_{net}} = \sum\limits_{i}^{n} q_i\vec{r_i}
$$
where $q_i$ is the charge of the $i$th point particle and $\vec{r_i}$ is the position vector for said particle; each individual dipole moment points from a negative charge to a positive charge. Carbon dioxide is a linear and symmetrical molecule, meaning that in the ground state the bonds between each respective oxygen atom and the carbon atom have the same lengths, and each oxygen atom bears an identical (partial) negative charge. All of this ultimately means that the two individual dipole moments between the carbon and oxygens one and two, respectively, perfectly oppose each other geometrically and have the same magnitude, i.e., $q\vec{r_1} = -q\vec{r_2}$, hence the vector sum $q\vec{r_1} + q\vec{r_2} = \mathbf{0}$.

It's like they are at tug-of-war with each other, and they both have the same strength; it's not going anywhere. Since the oxygen is more electronegative than carbon, it has a partial negative charge and the carbon has a partial positive charge (twice of the oxygen since there are 2 oxygen).

The lone pairs on both oxygen atoms cancel each others effect, as simple as that. And the structure is symmetrical from all aspects, which again leads to zero dipole moment. Plus there is no net negative charge or lone pair of Carbon as well, therefore the compound cannot be polar. REMEMBER polarity is a vector.

A molecule with a centre of inversion (centre of symmetry) has no dipole since exchanging one atom with its opposite leaves the molecule unchanged. Thus CO$_2$ and, for example,  benzene have no dipole, but toluene does even if small.
As CO$_2$ vibrates, depending upon the motion of the atoms a transient dipole can be formed. This transient dipole is the cause of ir absorption in CO$_2$ and hence its importance as a greenhouse gas.
A symmetrical vibration causes no transient dipole, O$\leftarrow$ C $\rightarrow $O,  an asymmetrical vibration O$\leftarrow$ C$\rightarrow$ $\leftarrow $O does produce a transient dipole. (In the asymmetrical vibration one C=O bond shortens as the other lengthens). There are also two degenerate bending motions about the C atom causing transient dipoles; $\uparrow$O = $\downarrow$C=O$\uparrow$


text:
If CO2 is nonpolar how come much more dissolves in water than O2?
Why is it that $\ce{CO2}$ is considerably more soluble in water than $\ce{O2}$ is?
$\ce{CO2}$ is nonpolar but dissolves in water which like $\ce{CO2}$ being nonpolar, doesn't make any sense.
Is it the bond polarity and that the bond polarity is more important in solubility?
If not than why is $\ce{CO2}$ more soluble in water than $\ce{O2}$.

Just because a molecule is non-polar, it does not mean that there are no dipole moments in the molecule. Carbon dioxide has 2 dipole moments in opposite directions and of equal strength (the dipole moment of the molecule is 0; therefore it is non-polar).

@LDC3 I would say it slightly differently. CO2 has no dipole moment, but that doesn't make it nonpolar. The combined opposed dipole moments give the whole molecule a "quadrupole moment" meaning that if there is a 4-pole electric field with positive at north and south and negative at east and west, the CO2 molecule will tend to turn to a north-south orientation. Larger molecules may have hexapole, octupole or higher moments (with progressively less effect.) CO2's polar bonds enable it to absorb infrared light much more easily than N2 or O2, which is what makes it an important greenhouse gas.

@steveverrill In collage I was taught that dipole refers to the polarization between 2 bonded atoms, not the molecule. From this definition, $CO_2$ has 2 dipoles (in opposite directions). The net effect results in $CO_2$ being non-polar. A non-polar molecule without any dipoles is oxygen, nitrogen, hydrogen, sulfur ($S_8$), ...

@LDC3 dipole just means a separation of charges into 2 ends, a positive end and a negative end. Bonds between dissimilar atoms have a dipole. A molecule can also have a dipole. H2O has both. A molecule's "dipole moment" is its quantitative tendency to align itself with an electric field, and tables exist of these. As CO2 has no dipole moment, many textbooks classify it as "non polar" but it does belong to the next group: molecules with quadrupole moments. When discussing solubility, equating dipole moment with "polarity" is not always helpful. For example hexamine is massively soluble in water

@steveverrill Since a dipole is actually a vector, then I guess that when you sum the vectors, you could still call it  a dipole, although I would call the molecule polarized. I never mentioned anything about solubility in my posts (I'm not familiar enough with that topic). What I was pointing out in my first comment was that not all non-polar molecules are absent of dipole moments.

let me guess. The quadrupole is from 2 equal dipoles and the hexapole is from 3 equal dipoles

Taken from my answer to your original question
There are a couple of reasons why $\ce{CO2}$ is more soluble in water than $\ce{O2}$.  Because the two $\ce{C=O}$ bonds in $\ce{CO2}$ are polarized (whereas in $\ce{O2}$ the bond is not polarized) it makes it easier for the polar water molecule to solvate it and to form hydrogen bonds.  Both of these factors will stabilize a $\ce{CO2}$ molecule more than an $\ce{O2}$ molecule in water; stabilization translates into greater solubility.  Another factor enhancing the solubility of $\ce{CO2}$ in water is the fact that $\ce{CO2}$ reacts with water to set up an equilibrium with carbonic acid.
$$\ce{CO2(aq) + H2O <=> H2CO3(aq)}$$
This reaction will also enhance $\ce{CO2}$'s solubility in water compared to oxygen which does not react with water.

This is a good answer because it covers both possibilities - the enhanced solubility due to polarized bonds as well as the reaction of $\ce{CO2}$ with water. I think it is important to note though, that the reaction with water is primarily responsible for the increased solubility. For proof, take a look at the solubility of $\ce{CO}$ vs. $\ce{CO2}$ - $\ce{CO}$ is polar, but is 60 times less soluble than $\ce{CO2}$.

Good point thomij.  Note, though, that in $\ce{CO2}$ it is the oxygen that is negatively polarized, while in $\ce{CO}$ the oxygen is positively polarized.  I suspect this has a significant effect on hydrogen bonding and, consequently, solvation.  As is so often the case, too many variables, too few experiments.

@Jori "Both of these factors"  I mean like this...Because the two $\ce{C=O}$ bonds in $\ce{CO2}$ are polarized (whereas in $\ce{O2}$ the bond is not polarized) it makes it 1)  easier for the polar water molecule to solvate it and 2) to form hydrogen bonds.  Private communication is fine, give me your e-mail and I'll contact you, or we could use SE Chem chat.

sent you an e-mail

why would the oxygen in carbon monoxide have a partially positive charge? That only seems possible for oxygen flourides.

@caters First off, note that the direction of the experimentally determined dipole moment has the positive center on oxygen (see  http://en.wikipedia.org/wiki/Carbon_monoxide#Bonding_and_dipole_moment).  Second, draw the resonance structures for CO that have octets.  You should come up with 2, one of which places positive charge on the oxygen (see structures J & K here  http://www.chem.ucla.edu/harding/tutorials/resonance/imp_res_str_key.html).  BTW, if my answer above was helpful, please mark it as accepted.

Doesn't the water induce a dipole in the O2 making O2 more soluble in water than it would be otherwise? and I don't think that where the positive and negative charge is affects H bonding if you compare 2 scenarios where 1 is according to the electronegativity and 1 is opposite. the number of polar bonds, the size of the atoms, and the electronegativity difference are more important factors to consider in H bonding.

"Doesn't the water induce a dipole in the O2 making O2 more soluble in water than it would be otherwise?"  Yes;  "I don't think that where the positive and negative charge is affects H bonding" I basically agree with that.

That is because $\ce{CO2}$ will react with water in equilibrium to form $\ce{H2CO3}$, which is an acid and thus will dissociate to form ions that can easily fit in the polar solvent (i.e. water). $\ce{O2}$ on the other hand does not react with water because it is not a very good electrophile in comparison to $\ce{CO2}$. The $\pi$-bond is weak in $\ce{CO2}$ (i.e. high in energy), partly because the $\ce{C=O}$ bond is polarized and because of delocalisation in remaining $\pi$-bond of $\ce{CO2}$. $\ce{O2}$ does not have that luxury.

Though the whole molecule has a net dipole moment of zero due to its geometry, the C=O bond has a dipole moment. Compared to the O=O bond which has no dipole moment.

Some folks throw around the formula H2CO3 as if it were a real compound.  The assumption is that it is a weak acid, and thus does not ionize to a great extent.  The truth of the matter is that there are no H2CO3 molecules in aqueous solution.  They don't exist.  What we call "carbonic acid" is CO2 dissolved in water in equilibrium with slight amounts of H+ and HCO3^-.
CO2(aq) + HOH(l) <==> H+ + HCO3^- .............. Ka is small
A small Ka means that the equilibrium lies far to the left, and that almost all of the CO2 stays CO2 and certainly doesn't form molecular H2CO3.

Welcome to the site.  Be sure to read through the tour if you have not done so already.  http://chemistry.stackexchange.com/tour  I think the asker wants a comparison between CO2 and O2.  O2 is not mentioned at all in this answer.

You are right but I think carbonic acid makes sense as chemical entity. I repeat, you are right in practical contexts. But if a CO2 molecule and a water one should met in an otherwise empty space I think that the result is carbonic acid, eventually in a similar equilibrium depending on the exact conditions.

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IBrF2 - van der Waals repulsions - and number of unique bond angles
I understand that the molecule $\ce{IBrF2}$ is T-shaped, with three ligand atoms and two non-bonding pairs.
However, how many unique bond angles can this molecule exhibit?
My professor argues for the existence of only 2 unique bond angles in this molecule, with the somewhat $< 180^\circ$ axial $\ce{F-I-F}$ bond angle and the two equivalent and somewhat $< 90^\circ$ $\ce{F-I-Br}$ bond angles.
However, can't we also arrange this molecule as to have this axial arrangement of atoms: $\ce{Br-I-F}$?
My professor argues that this arrangement is impossible due to van der Waals repulsions. I suppose this makes sense; the two lone-pairs $\ce{<->}$ large bromine atom repulsions would make this shape unfavorable.
Impossible? I'm not sure about that. Dammit, nothing in chemistry is impossible.
What do you think? I'm arguing for the existence of either two or three bond angles because we can either have a $\ce{F-I-F}$ axial arrangement or a $\ce{Br-I-F}$ axial arrangement. Plus, axial bonds are longer than equatorial bonds. So that would lower the energy of the $\ce{Br-I-F}$ axial conformer. Side note: why are axial bonds longer than equatorial bonds?
In sum:

Certainly not impossible.  I bet they're relatively close in energy.  With the bromine axial you replace a 90 degree F-lone pair interaction with a 90 degree Br(longer bond)-lone pair interaction - that might be energetically favorable.  Maybe Martin can do a calculation?

Right, I was saying that the axial bonds are generally longer than the equatorial bonds. This length also minimizes repulsions. Side question: why are axial bonds longer than equatorial bonds?

I could very well be mistaken here, and literature on ternary interhalogens is virtually nonexistent, but my suspicion would be that there's a non-trivial energy difference between the two conformers, with the equatorial bromine favorable chiefly for steric reasons.

@Dissenter Equatorial bonds have fewer destabilizing 90 degree interactions than axial bonds.  Therefor, they are at lower potential energy, e.g. more stable.  That translates into shorter bond length.

Right, equatorial bonds are affected by two axial attachments, while axial atoms are affected by three equatorial atoms - is this your point?

Yes, that's it.

Are there any exceptions to this statement that axial bond lengths are longer than equatorial bond lengths?

@Dissenter Yes. Jahn-Teller distortion in octahedral coordination compound usually, but not invariably, works in the direction of a lengthening of axial bonds (here defined as the 2-out-of-6 ligands, as opposed to the equatorial 4 remaining).  Exceptions include $\ce{KCu3(OH)2(H[AsO4]2)}$ (H. Effenberger, Z. Kristallogr., 1989, 188, 43); see Figure 5 here: http://pubs.rsc.org/en/content/articlehtml/2003/dt/b309242a

I suspect the molecule $\ce{IBrF2}$ does not exist, and is purely conjectural (theoretical) at this point.  If it ever had been synthesized or observed, it would have made the cover of Science or Nature.  "First neutral ternary interhalogen compound found!" (A handful of anionic  ternary interhalogen species are known.)
But my hunch is that, if the molecule did exist, the conformation your professor describes would be favored because of symmetry. Having the same atom in (roughly) opposite directions maximizes the amount of coupling because the opposite orbitals have same energy and can combine better than if it were two different atoms.  Hence greater stabilization.

Side note: why are axial bonds longer than equatorial bonds?

This is true in binary compounds like $\ce{PCl5}$, and apparently a sufficient explanation is that each axial chlorine wants to avoid 3 other chlorines at 90$^\circ$ angles, while the equatorial chlorines have 2 at 90$^\circ$ but the other 2 are further away anyway, at 120$^\circ$.  In other words, if you just want to maximize the average Cl-Cl distance, subject to the constraint that the P-Cl distance is reasonable, you're better off with a slightly elongated bipyramid instead of one inscribed in a sphere.
In the hypothetical $\ce{IBrF2}$ of course the axial I-F bond would be shorter than the "equatorial" I-Br bond, since Br is a bigger atom.

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text:
Entropy - "Wiggle"?
The title is not a reference to a Jason Derulo song.
In any case:
1) How is change in entropy measured, experimentally? I've Googled this for a bit and I've found all sorts of mathematical equations such as delta S = q/T as a way to get a grasp on entropy. However, I want to know how entropy is actually determined in the lab (as opposed to on paper). I understand that the units on entropy are in terms of energy/temperature. So if I had to guess I would suppose that some sort of calorimetry is used.
2) What is the best way to teach entropy (at an introductory level)? I know that textbooks like to say it's a "measure of disorder." On the other hand I have lots of websites telling me that entropy is best NOT described at a measure of disorder. So what is it really, and what do you think about the term "disorder" as a synonym? My professor refers to it as the degrees of freedom in a system; he also notes that longer carbon chains tend to have higher entropy values due to a greater ability to "wiggle" along their carbon chains. What do you think of this?

Discarding the archaic idea of "disorder" in regard to entropy is essential. It just doesn't make scientific sense in the 21st century and its apparent convenience often is flat-out misleading. As of November 2005, fifteen first-year college texts have deleted “entropy is disorder” although a few still retain references to energy “becoming disorderly”. (This latter description is meaningless

I really like Boltzmann's statistical view of entropy $$S=\mathcal{k}_\mathrm{B}\cdot\ln W$$
in which $W$ is the number of microstates, that contribute to a macrostate defined through $S$. In this way you can relate it to the degree of freedom of a system, as well as the overall uncertainty of finding a particle in a particular state. It is derived for the ideal gas, but I think the philosophical touch is quite general. But I cannot answer your question. (Esp. not 1)

I found classical thermodynamics utterly and completely incomprehensible back when learned it. Luckily, a year later I took course in statistical physics, that provided me with really working insight into meaning of thermodynamical functions, entropy included. It was much more intuitive for me, so if you have troubles with classical thermodynamics, statistical physics may be a way to go.

You can't measure entropy directly, any more than you can measure interatomic distances.  You measure other quantities -- for instance often you can measure energy gain/loss and temperature, and then you integrate $dS=dE/T$.
How to explain it?  One of the best expositions I know is The Second Law by Henry A. Bent.  It is full of insightful examples, lays the ground carefully and avoids woolly talk, unlike many other thermodynamics books.
Some elementary but valid comments about the link between information-theoretical entropy and thermodynamical entropy can be found here:
http://www.madsci.org/posts/archives/1998-10/909712896.Ph.r.html
Edit: Greg reminds us that energy and enthalpy are not directly measurable either.  That's why I was careful to write energy gain/loss, which are somewhat more accessible to measurement; but of course even when doing a calorimetric experiment you're not measuring heat per se but other quantities: how much gas you burned or how intense the current was and how long it was left on. And even some of these quantities again are indirectly measured.
I agree 100% that "entropy is not more abstract than energy" -- they are all abstractions.
Ultimately, however, I do think the measurement of thermodynamic entropy is one step more indirect than that of energy, if only because it involves the notion of temperature, which (if you think about it carefully) is subtle indeed.

Just for the record: you measure heat and such, and not energy or enthalpy or whatnot. Yes, if you do the experimental design well than heat will be equal to energy, or at least a good estimation of it. Entropy is not more abstract than energy or free enthalpy, and its measurement is not indirect than those.

Wait are you sure you can't measure absolute entropy? Don't perfect crystals have 0 entropy? So can't we actually measure absolute entropy?

Perfect crystals have 0 entropy only at 0 K, and this value is not measured (you can't get to absolute zero).

As an interesting bit of history, Boltzmann was the first one to describe entropy as a "measure of disorder" of a system. It's worth noting that he didn't know this was an oversimplification. In reality, entropy is best described as a measure of the number of ways that energy can be distributed in energy levels between or within particles.
From this, it's clear how the number of possible particle arrangements correlate with entropy: a greater number of particle arrangements means that more energy distributions are possible. I would avoid conflating particle disorder with entropy/microstates though; they are correlated, but not one and the same.
As for how entropy changes are measured experimentally, I would think the easiest way to do that would be dividing the heat lost by the system by the temperature of the surroundings. This essentially tells you how much energy has dispersed.

Here are two common ways of measuring the entropy change in a reaction:

Measure the equilibrium constant $K$ at multiple temperatures. This gives you the Gibbs energy (via $\Delta_r G^\circ = - R T \ln{K}$) and, via the van't Hoff relationship, the enthalpy. You can calculate the entropy from those two.
Use microcalorimetry in a titrating mode. You get the enthalpy directly, and the equilibrium constant by fitting the trace (which tells you the extent of reaction at different concentrations). From those two, you obtain the entropy.

The entropy change is concentration-dependent, so once you have the standard entropy of reaction, you might have to correct for concentrations that are different from those at standard state.

2) What is the best way to teach entropy (at an introductory level)?

It depends. However, my favorite analogy to use when guessing which state has higher energy is "freedom of atoms to move". This works for solid vs. gas, for bound vs. in solution, and for constrained conformations vs. rotatable bonds. It also hand-wavingly makes the connection to the number of microstates that Martin references in a comment. The other term you sometimes find is "dispersion of energy", which nicely explains how entropy increases when a rock (that fell off a cliff) hits the ground.

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Is viscosity proportional to the number of hydroxyl groups?
Is the following statement always true?

The more $\ce{-OH}$ functional groups in the molecule, the more is its viscosity?

I think it is true, because it is known that weak intermolecular forces lead to lower viscosities and strong intermolecular forces lead to higher viscosities
and because the more $\ce{OH}$ functional groups in the molecule will give rise to a molecule having stronger molecular forces. As the case in glycerol ($\ce{C3H8O3}$) and water ($\ce{H2O}$)
Therefore the statement must be true. What do you think?

Is the following statement always true?  The more −OH  functional
  groups in the molecule, the more is its viscosity?

It's hard to find anything that is "always" true in chemistry, but I'd bet that within a series of molecules where the only variable is the number of $\ce{OH}$ groups, your statement is generally true, so I basically agree with your position.  Your analysis and examples are also good. As you pointed out, it is about intermolecular forces.  Specifically, molecules with hydroxyl groups can form intermolecular hydrogen bonds (see first picture below).  These hydrogen bonds cause the molecules to "stick" together and act as if they had a higher molecular weight.  Sugars have many hydroxyl groups that give rise to many intermolecular hydrogen bonds and cause sugars to flow in a slow, syrupy manner.

Here is a comparative series of alcohols all involving a 3 carbon chain.  When the number of hydroxyl groups remains the same and we just vary their position on the chain the viscosity hardly changes.  On the other hand, note how the viscosity increases markedly each time we add another hydroxyl group to the chain.  It appears that the total number of hydroxyl groups is what really matters, so your statement does appear to be generally true as long as the number of hydroxyl groups is the only variable.

It is a reasonable rule of thumb, but certainly not always true.
Compare for example the viscosities of Dodecane ($\ce{C12H26}$, $\mu=1.374 \ce{\; mPa\cdot s}$, no $\ce{OH}$ groups) and Ethanol ($\ce{C2H5OH}$, $\mu=1.095 \ce{\; mPa\cdot s}$, 1 $\ce{OH}$ group) (source). There you can see that ethanol has a higher number of $\ce{OH}$-groups, but a lower viscosity.
That said, I do think you are on to something regarding the strenght of intermolecular interactions, it is just not purely the number of $\ce{OH}$-groups that matters for that. The length of the carbon chain for example (like in the example I mention above) also plays a role.

I think the OP was asking about viscosity vs. the number of hydroxyl groups in a molecule.

@ron Yes, that is what I answered right?! OP wants to know whether more hydroxyl groups means higher viscosity, I show that that is not necessarily the case.

basically i wanted to compare the viscosity of molecules which have different number of oh groups .. but the piece of information that you have presented is really good to know ! Thanks

@maher ok, I didn't get that you wanted to compare molecules that are identical except for the OH groups, since you mentioned glycerol and water in your example I thought you were just thinking about all molecules

I'm sure someone could come up with a molecule where the addition of an -OH group decreases viscosity (maybe by forming an internal hydrogen bond that stabilizes a less "sticky" conformation?). But yes, I'd say that, all else being equal, it's almost always true, because, as ron notes, -OH groups like to stick to each other.


text:
Relationship between rate equation and equilibrium constant
I have read mannaia's answer on the question of 'rate order and confusion', as well as Nicolau Saker Neto's answer to a related question. They have both been very helpful.
As I understand, the equilibrium is actually derived from the rate constants of a reaction. At dynamic equilibrium, the rate of the forward reaction is equal to the rate of the backward reaction, hence in a hypothetical reaction
$$
\ce{pA + qB <=> rC + sD}
$$
with the rate equations $Rate_{fwd} = k_{fwd}[A]^{a}[B]^{b}$ and $Rate_{bck} = k_{bck}[C]^c[D]^d$, with $a, b, c,$ and $d$ not related to $p, q, r$ and $s$.
The equilibrium is derived as such:
\begin{align}
  R_{fwd} &= R_{bck}\\
  k_{fwd}[A]^{a}[B]^{b}&=k_{bck}[C]^c[D]^d\\
  \frac{k_{fwd}}{k_{bck}} &= \frac{[C]^c[D]^d}{[A]^a[B]^b}\\
  \therefore k&=\frac {k_{fwd}} {k_{bck}}\\
\end{align}
Based on the derivation, and the fact that the stoichiometric coefficients of the reactants in the rate-determining step (RDS) of the reaction is equal to the order of the reactants in the rate equation, is it correct to state that the reaction equations given as the equilibrium equation is actually the rate-determining step of two reactions?

At equilibrium it does not matter if the reaction has multiple steps; your equilibrium constant only has to involve the initial reactants and final products.
For instance, in an S$_\mathrm{N}$1 reaction, obviously the RDS is the departure of the leaving group, but you don't have to (and usually are not in a position to) know the concentration of the intermediate carbocation in order to calculate the equilibrium of the overall reaction.  You can simply write an  equilibrium equation for the initial reactants and final products: $$K=\frac{[\ce{LG}][\ce{product}]}{[\ce{nucleophile}][\ce{substrate}]}.$$
Of course finding $K$ may be not be easy, even if you have access to good thermodynamic data for the reactants and products, because $\Delta G_0$ (and therefore $K$) depends on solvation, temperature and other conditions. But given identical conditions, $K$  is constant over a wide range of concentrations, so you might be able to find it in the literature or measure it with a single experiment.
All this assumes that the RDS is fast enough that you're able to reach  equilibrium before the solvent evaporates, or your reactants and products decompose via other pathways, or you die of boredom!
Note that you can also write an equilibrium equation for the RDS alone:
$$
K'=\frac{[\ce{LG}][\ce{carbocation}]}{[\ce{substrate}]}
$$
But since you seldom know the concentration of the carbocation, this equation isn't as useful in practice (it is of course extremely useful for mechanistic studies).
Edit: This page has a more complete and very nice answer!


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Electric field effect on reactions
What would be the effect of putting a solution in a strong electric field? Would the reactions in the solution be affected?
I'm thinking yes – lots (if not all) – chemistry is dependent on charge (partial/real charges) and by applying an electric field, we'd be affecting the very property of the molecules/ions that drives reactions.
Has this been researched?

Would the reactions in the solution be affected?

Yes, they can be.  Here are some nice examples.  Electric fields also exert a significant effect on electron transfer reactions.  Basically you would expect charged molecules to be more affected than polar molecules, which should be more affected than non-polar, but polarizable molecules.  It's just a question of how big a field you can produce and how polarized and long-lived are the various entities (starting materials, intermediates, etc.) that exist along the reaction coordinate.

is this same with magnetic field?


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Electronic model with highest prediction rate
Amongst many models, including the valence bond model (VB) or the molecular orbital (MO) model, which are the ones with best predictive power? (e.g. the MO is thought to predict spectroscopic properties better than the VB)

The so called gold standard of quantum chemical calculations of ground state energies and properties is usually considered to be Coupled Cluster Singles Doubles (with perturbative) Triples as a method. This method is based on ab initio molecular orbital theory and it is size consistent, but not variational. A basis set of at least a triple zeta quality with additional valence and polarisation functions should be employed, good examples for this are Dunning's (augmented) correlation consistent basis sets, e.g. (aug-)cc-pV{T,Q,5,6}Z, or Weigend's def2-basis set family, def2-TZVPP, def2-QZVP, but there are various others. (If you like to have some citations, drop a note in the comments.) This level of theory is usually denoted as CCSD(T)/TZVP, where TZVP is a placeholder for one of the various basis sets.
In principle CC theory is exact at the complete basis set limit, but in practise, this is not solvable. Another theory that is in principle exact, because they become equivalent, is Full Configuration Interaction, F-CI. The truncated variants CISD, CISD(T), CISDT, ... are variational, but not size consistent. For chemical systems it is much more important, that the calculations are size consistent, therefore these methods lost their importance. An attempt to solve this was to introduce Quadratic Configuration Interaction, QCI, but this is almost the same as coupled cluster theory, so it was not successful either.
The importance of Many Body Perturbation Theory, MBPT, of high orders, like Møller Plesset Perturbation Theory of fourth order, MP4, has very much decreased because of it inefficiency.
There are some cases, which cannot be accurately described by these models. Those are cases, where the ground state cannot be well defined by one determinante, the initial guess wave function obtained from Hartree-Fock, HF. Here you have to use a different description. A Multi Configuration, MC, or Multi Reference, MR, ansatz is necessary. The qualitative equivalent to HF is Complete Active Space Self Consistent Field, CASSCF. Based on this you can use Perturbation Theory to include more correlation effects. The best possible description so far (judging accuracy versus computational cost) is second order perturbation theory on a CASSCF wave function, for short CASPT2. It is also very reliable for spectroscopic calculations. There are also various developments on Multi Reference Configuration Interaction as another theory. (I have limited expertise on that field.)
Another way of treating chemical systems is Valence Bond theory, or with the application to modern computational aspects. Here you propose the mixing of various Lewis like structures, to obtain the ground state energy and properties. The formalism can get quite complicated, maybe this is why its performance for excited states properties is not very accurate at the moment. It is still a very active field of research, by a few scientific groups. (I am not one of them.)
However, affordable spectroscopic calculations have still huge errors and sometimes the numbers can vary quite a bit. Currently most used is a variety of methods derived from Time Dependent Density Functional Theory, TDDFT. An ab initio method of great importance in this field is Configuration Interaction Singles. Also fairly well implemented are approximate coupled cluster techniques, e.g. CC2. The struggle for finding an accurate, robust and computationally cheap method are one of the main goals of modern quantum chemistry. (I can apply these methods fairly well, but I have no deeper insight how they work.)
For extended systems like surfaces and solids there are various approaches. Since they get very demanding very soon, the most commonly used practical approaches are based on plain or hybrid density functionals. They often also employ plane wave basis sets with periodic boundary condition. (Unfortunately I am not an expert on that field either.)
As you can see, there is no definite answer to your question and the development of new and efficient methods is still a very active field. Depending on what type of description you want or need, there are many possibilities. The most advanced models are obviously for ground state energies and properties, since they have been researched for about a century now.
I hope this helps you, I apologise for the chatty answer. If something remains unclear, drop me a note in the comments.

Very good summary. I would like to add three things: * Not all of these methods works nicely when you have open-shell systems (spin). * Many, if not most, of these methods can be computationally prohibitive in practical studies, especially if you are not in a computational group with good access to computational power. *All these methods are about isolated entities in vacuum, however solvent / environmental effects can often add a whole new layer to the problem.


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Why is the Chlorine missing an electron in Chlorine Monoxide?
I was asked to draw the lewis-dot structure for ClO and I drew this:
   ..  ..
  :Cl-  O.
   ..  ..

The correct answer is apparently this:
   .   ..
  :Cl-  O:
   ..  ..

How would I know that? In the first example, all the formal charges are zero, but in the second, Oxygen has -1 and Chlorine has +1.

Lewis structures are models that help us predict things and understand things about ions and molecules.

In some problems there is no "right" answer, or be.tter to say, there are several answers, neither of which are really "right".

I like your reasoning about formal charge, by the way.
I would ask an additional question:  What about CLO can we learn, predict, or understand better with one of those structures vs the other?

intermolecular forces - would they be higher with the second due to the formal charge?

Maybe they would be weaker.  The dipole moment of CO is .1D or so since C has a -1 and O a +1 formal charge.  One might expect a C-O triple bond to be quite polar but it isn't.

In fact the unpaired electron belongs neither to Cl nor to O, but to a molecular orbital spread over both atoms [but not equally: see below].  The Lewis structure is simply not a complete representation of what is going on.  (And for what it's worth wikipedia gives the same Lewis structure you drew: http://en.wikipedia.org/wiki/Chlorine_monoxide)
Edit: Actually it seems the highest occupied molecular orbital (where the unpaired electron is) is mostly localized on the oxygen atom.
MO Diagram (taken from here):

That is to say, your structure is more correct than the one in the answer key.  However the justification involves molecular orbital arguments outside the scope of a general chemistry course.
Another edit: I too like your argument about formal charges and would accept it as a justification that the unpaired electron belongs to O rather than to Cl.

You can come to the same generalization using electronegativity. O is more EN than Cl; O preferentially takes the remaining electron.


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can sodium persulfate etch or damage a titanium heater?
I use Sodium Persulfate to etch copper on a PCB process. I have been using glass heater on the etching tank but the glass is too fragile; it breaks too easily.
I read that titanium heater is used commercially (in aquariums as well). I don't know much about chemistry, could anyone tell me if it is okay to use titanium with Sodium Persulfate?
Sodium Persulfate to water ratio is 1:10 and the maximum temperature is about 50 C.

what do you mean by 5dm2? Five liters of water (liter = dm$^3$)?

I copied pasted from the description on https://www.elfa.se/elfa3~se_en/elfa/init.do?item=49-577-26&toc=0&q=etching+powder   I keep acid/water ratio 1:10

So what it says there is that 140g of powder is enough to etch 6 dm$^2$ (about 100 square inches).  So 6dm2 is not part of the formula; can you edit your question accordingly?

If the surface of the immersed element is all titanium, the answer appears to be that it won't get corroded unless the temperature is much higher than 50$^\circ$C; see http://www.azom.com/article.aspx?ArticleID=1336
The case is changed if there are joints or parts made of something else, as in this illustration: http://www.saltwatereddies.com/content-popup_image/pID-2199/popup_image.html
I would think that would not be so safe.
Of course, if you've been breaking glass heating elements right and left, you have little to lose by trying the titanium one...


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Why is diboron (B2) paramagnetic?
Why is $\ce{B2}$ paramagnetic? It's just 2 borons with a triple bond between them. Aren't all the electrons paired?

For the same reason $\ce{O2}$ is paramagnetic: degeneracy.
The filled orbitals are $\sigma(\mathrm{1s})^2$, $\sigma^*(\mathrm{1s})^2$, $\sigma(\mathrm{2s})^2$, $\sigma^*(\mathrm{2s})^2$.  The half-filled orbitals are $\pi(\mathrm{2p}_x)^1$ and $\pi(\mathrm{2p}_y)^1$.
But you're right to think that this is unexpected:  why are the $\pi(\mathrm{2p})$ orbitals  lower than the $\sigma(\mathrm{2p})$? That's the reverse of what we are told to expect! I've never encountered a rationalization that satisfied me.
See Matt Law's lecture notes for the expected and observed ordering:
MO Diagrams for Diatomic Molecules; Chapter 5; Friday, October 9, 2015; PDF via the Internet Archive.

The rationalisation you are looking for is called sp-mixing, and it is one of the more simpler items in quantum chemistry.

I disagree with Silvio Levy that the triplet ground state of $\ce{B2}$ is completely unexpected.
The rationalisation also is not that difficult, and is known as s-p mixing. After all, many or most, if not all, explanations in organic chemistry rely on this very principle; you might know it as hybridisation.
Before we get there it is worth while knowing a generic valence molecular orbital diagram where no s-p mixing occurs. This one pretty much applies to all main group elements heavier than nitrogen.

The core orbitals, in case of lithium to neon these are the 1s orbitals, sodium to argon these are 1s, 2s, and 2p orbitals, are not included, as they rarely ever play a significant role.
If you would assume the same for boron, then according to Hund's rule you would indeed expect all electrons to be paired up.
Now s-p mixing comes into play. There are two different explanations, which lead to the same conclusion, available. The more common one involves the formation of hybrid orbitals, which then form the molecular orbitals. I'm going to skip over this and focus on a hopefully clearer explanation.
When combining atoms to molecules, the symmetry changes. An isolated atom has the highest possible symmetry: spherical. The symmetry of diatomic molecules is always D∞h. The orbitals also transform according to this symmetry. We find σ orbitals, which have no nodal plane between the bonding partners, π orbitals, which has one nodal plane along the bonding axis, δ, φ, etc. orbitals.
This is clearly marked in the above scheme. Orbitals of similar energy and symmetry may again combine. The lighter the element, the closer these energy levels are together. For the elements from lithium to nitrogen this leads to a linear combination of 1σ + 3σ that drastically drops in energy. On the other hand, the linear combination of 1σ - 3σ raises in energy above the π orbitals. (You can do the same for 2σ and 4σ, the effect is less pronounced.) In the end, all σ orbitals will have contributions from s and p orbitals. The hybridisation scheme will arrive at the same conclusion.
The resulting diagram looks (in approximation) like this:

If you now fill the three valence electrons of each boron into the diagram according to Hund's rule, you will see that each π orbital will get one electron. This results in a triplet ground state. The finished valence molecular orbital diagram is pictured below.

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


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Chemistry of Walter Mitty's negative developer/stop bath/fixer?
I asked a question here How can Walter Mitty examine the negatives of photos in the bright sunny day?
Commonly known, there are three steps using chemical baths to bring out the images of negatives:

developer to bring out the image,
"stop bath" to halt the process,
"fixer" to make the image permanent.

So in details what are the Chemical procedure/compounds/reaction for the negative's developer/stop bath/fixer?
Wish the answer is in depth.

basics here: http://science.csustan.edu/nhuy/chem1002/photoexp.htm
In depth: there are many good books about the chemistry of photography, starting with this 120-year old classic:
https://archive.org/details/chemistryphotog02meldgoog

Here is another that's fully available online and recent enough to cover every major aspect of the subject:
https://openlibrary.org/books/OL5953553M/Photographic_chemistry_in_black-and-white_and_color_photography
You can buy more recent books, especially for developments in color.

@ Silvio Levy, thanks a lot, +1. Thanks for the Refs. If you can summarize brielfy, it will be even more wonderful.

It's not that complicated of an answer.

developer to bring out the image

As I'm sure you know photographic film is coated in light-sensitive material (usually silver bromide). The thing is for black and white silver halide films, that the light does not react that intensely with the film, only a very few silver atoms get reduced. During development these reduced silver atoms catalyze and seed the reduction of surrounding silver ions to silver atoms by the developer chemicals (ussually hydroquinone). The more of these silver ions that are reduced, the darker the photo. Thus more light exposure gives more more initial silver atoms gives darker negative.

"stop bath" to halt the process,

As you can imagine, if the development keeps going after enough time all of the silver will be reduced and the negative will just be black. The stop bath prevents this and stops development by using a dilute acid to stop the reaction of any residual developer chemicals which only react in basic conditions.

"fixer" to make the image permanent.

Thus far the film has reduced silver to a desired point, but there is still light sensitive silver halide on the film. This silver halide is still light sensitive and over time with exposure to light will darken. The fixer bath uses sodium thiosulfate to make the silver halide soluble and dissolves the remaining silver halide from the film so that the image doesn't darken (it fixes the image in the sense that it can't change). At this point the film is no longer photo sensitive and can be view in daylight.


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How do I find out if ions are oxidised/ reduced based on half cell potentials?

Assuming standard conditions, answer the following questions.
  (yes/no)

Is $\ce{Fe^{2+}(aq)}$ capable of reducing $\ce{Cr^{3+}(aq)}$ to $\ce{Cr^{2+}(aq)}$?
Is $\ce{Sn^{2+}(aq)}$ capable of oxidizing $\ce{Fe^{2+}(aq)}$?
Is $\ce{Cr}$ metal capable of reducing $\ce{Fe^{2+}(aq)}$?
Is $\ce{VO^{2+}(aq)}$ capable of oxidizing $\ce{Fe^{2+}(aq)}$?
Is $\ce{H2(g)}$ capable of reducing $\ce{Ag+ (aq)}$?
Is $\ce{Cr^{3+}(aq)}$ capable of oxidizing $\ce{Fe^{2+}(aq)}$?

I have

yes
yes
no
yes
yes
no

What am I doing wrong?

Refer to the reactivity series and remember that a metal can reduce those below it.

What about oxidation, metal can oxidize those above it?

Yes, as by definition of oxidation and reduction, if A reduces B, then B is oxidized by A

Look at this table of half-reactions: http://bilbo.chm.uri.edu/CHM112/tables/redpottable.htm
I'll explain how it contradicts your answer to the first question; you can then rework your answers and update your question with justifications.

Is $\ce{Fe^{2+}(aq)}$ capable of reducing $\ce{Cr^{3+}(aq)}$ to $\ce{Cr^{2+}(aq)}$?

One entry of the table says $$ \ce{Cr^{3+}(aq) + e- -> Cr^{2+} (aq)} \quad –0.424.$$
The $-0.424$ is what's called the standard potential, expressed in volts.  It is a measure of how favorable the half-reaction is (compared with a standard half-reaction).  The higher the number, the more favored the products will be.
Now you have to find a half-reaction involving $\ce{Fe^{2+}}$.  Since you want Fe to reduce something, it will itself be oxidized.  The table shows no half-reaction where $\ce{Fe^{2+}}$ is oxidized, but that's because by convention only reduction half-reactions are shown; to get the opposite, oxidation, half-reactions, just swap the terms and negate the potential.  That is, take this line from the table:
$$
\ce{Fe^{3+}(aq) + e^{–}  -> Fe^{2+}(aq) } \quad 0.771
$$
and turn it into
$$
 \ce{Fe^{2+}(aq) -> Fe^{3+}(aq) + e^{–}} \quad {-}0.771
$$
Now add this, term by term, to the Cr half-reaction. The result is
$$
\ce{Fe^{2+} + Cr^{3+} -> Fe^{3+} + Cr^{2+}} \quad {-}1.195
$$
If the potential in the final reaction is negative, the reaction is impossible (more precisely, equilibrium lies to the left --- in the case very far to the left, becase 1 volt goes a long way).
That's the basics.  There is a subtlety involved when the number of electrons exchanged is not the same in the two half-reactions; you'll find it in your textbook.  You could try answering the other 5 questions now and if you get stuck, tell us precisely where.


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Which azo compounds are found in DVD-R and CD-R media?
Does anyone know which specific azo compounds are used in DVD-R/CD-R recordable media?

Hi and welcome to Chemistry.SE!
Please beef up your question a bit with more information: Why do you think azo compounds are used (references) and where are they applied?

Each company in the DVD business has their own proprietary twist on the actual dye they use and they don't go out of their way to reveal the structure of the molecule they use.  Also the actual dye used changes fairly rapidly (several years or so) as technology shifts (e.g. higher density, Blu-ray, etc.).  That said, metal-azo dyes are popular today.  A Mitsubishi patent from a few years ago (WO 2006/104196 A1 (10/03/2006)) says "The recording layer contains a pyridone azo organic dye having a 6-hydroxy-2-pyridone structure as a coupler component and an isoxazole, 1,2,4-triazole or pyrazole as a diazo component, coordinated to a metal, such as nickel, cobalt, iron, zinc, copper or manganese." Here is the structure of 2-pyridone (6-hydroxy would be attached on the other side of the nitrogen) and pyrazole.

Here is a link to a Ciba patent that has more detail. And here is a general structure a 2-pyridone-based metal-azo dye


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Are homemade batteries economically practical?
Every kid who has been to a science fair knows that you can make a battery with a lemon, a potato, a penny and a nail. I've never done this and I'm not entirely clear on how it works beyond a hazy recollection of undergraduate-level chemistry class; but I gather that you do not need to use a lemon and potato as your power source but can in fact use any number of substances willing to donate or accept electrons.
My question is this: are there any home-made batteries that can produce sufficient electrical energy that its cheaper to make them yourself rather than plug your appliance into the wall or use store-bought batteries
I realize that there are a number of variables here that might make a significant difference in whether its cheaper to use established electricity sources or create your own so for arguments sake lets say that I am trying to:
- charge a cell phone/laptop
- run a microwave long enough to make popcorn
- keep a refrigerator cold
Lets also consider the dual possibilities that I:
- live in a city apartment and have to buy all my fuels and materials
- live on my own far where I can grow as much produce as I want to devote to making batteries
If there are other good fuel sources that could be a logistic variable; let me know in the comments so I can edit the question accordingly.

It should be pointed out that the lemon/potato does not function as a power source in such batteries. Instead, it functions as the electrolyte of the battery. The power is instead coming from the oxidation/reduction of the metal in the electrodes. (Copper & zinc in the most common configuration.)

@R.M. Well, oxidation anyways. It's just hydrogen evolution at the copper electrode, not a reaction of the electrode itself. I suppose one could make ethanol out of the potatoes or whatever for use in a fuel cell/generator, but certainly not the most efficient thing.

The second possibility assume you "grow" your electrodes, too, i.e. you have a functioning mine and smelting facility... In other words, it is rather unrealistic. Also, you forgot to discuss environmental issues. Producing energy via batteries will leave you a lot of dead batteries, i.e. most likely toxic metal salts. If you go the DIY way, you should figure out a DIY disposal, too.

No: trying to catch up with 100+ years of technological optimizations in your kitchen won't be an economical enterprise.  Store-brought batteries are standardized and relatively inexpensive.  Power from an outlet is even less expensive (by one or two orders of magnitude).
On the other hand,
in many places you can save on electricity by going solar. Even if you can't install solar panels on your roof because you don't own a house, you could buy solar chargers for small appliances if you have a sunny corner in your apartment...

In general, think of batteries (including chemical nonrechargeable ones) as tools for storing and transporting energy, not as an energy source.
If you want to create energy through chemistry, then the economically efficient way is the way everyone else does it, through exothermic oxidation - burn stuff and use it to power a generator.

A Nickle Iron or iron copper battery,is most practical. A pint jar sized one is similar to a AA in voltage and similar to a lead acid battery in usable current, it's also easy enough to seal since it already has a lid. The self discharge rate is pretty high so you would have to keep your batteries empty or on a trickle charge to keep them usable. The major advantage is the long lifespan (rechargeable for nearly 70 years), and the electrolyte which is common lye. If the battery dies or spills you can neutralize the lye with vinegar,clean the plates with water and then refill it with lye and water. This will fully recharge the battery instantly and restore your battery's ability to hold a charge.  I know there are countless articles online about the dangers of lye but people have been using it for 100's of years to make soap,and it's not particularly harmful to the environment. You still want gloves goggles and vinegar on hand, it eats skin similar to the way household bleach does. (bleach has lye in it)

Bleach doesn't have lye in it, though you can make it by adding elemental chlorine to lye (just as you can make NaCl by adding HCl to lye.) What makes bleach corrosive is the resulting oxidiziing ion (hypochlorite).


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Difference between nickel catalyst and Raney nickel catalyst
Many times I have seen the catalyst as Raney nickel instead of nickel catalyst. Raney nickel catalyst is developed by Murray Raney, and I think Raney nickel catalyst and nickel catalyst are two different things, because various sources use both names.
How do they differ from each other?

Raney nickel is derived from a nickel-aluminum alloy.  The removal of the aluminum results in a higher surface area for the Raney nickel vs regular nickel.

Yes, the Raney nickel catalyst is different from simple, fine nickel powder.  Raney nickel is prepared in two steps.  First, the nickel is alloyed with aluminum and then in second step much of the aluminum is removed through a leaching process with sodium hydroxide. As the aluminum is removed two things happen.  The resulting nickel-aluminum alloy becomes very textured and porous with extremely high surface area (sponge-like).  Also hydrogen gas is liberated during the leaching process and some of it remains adsorbed on the alloy surface.  This is the active catalyst that is ready for use.
$$\ce{2 Al + 2 NaOH + 6 H2O -> 2 Na[Al(OH)4] + 3 H2}$$
If you'd like to read more about the initial alloying process or the activation (leaching) step, see this Wikipedia article.
As an aside, Murray Raney discovered this catalyst (US1628190) in the late 1920's and he was the sole assignee.  I'm guessing that W. R. Grace bought the patent rights from him since they hold the trademark on use of the term "Raney nickel"

As pointed out by ron, a large surface is the key. Another Ni-based catalyst is prepared by 1. reducing a Ni salt with zinc and then treating the resulting mixture with A) acid or B) base. The resulting catalysts are known as Urushibara nickel (type A and B, resp.)

Ron, spot on with the Raney trademark comment. It's as simple as that. Any nickel catalyst prepared in the manner above, not from Grace, is sponge nickel


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Mathematical model of isomerism
Does there exist a mathematical model of isomerism? (a model which is capable of predicting, for given circumstances, the re-arrangement that is going to occur, e.g. Demjanov, Wagner-Meerwein etc.).
The goal is to:

understand how a geometrical structure of a molecule changes for isomers
understand under what conditions certain geometrical structure of one molecule is transformed into a different geometrical structure of an isomer
which geometrical isomer structures are impossible to exist

Try graph theory.

It helps if you are more specific. Technically speaking any quantum chemical theory can predict relative energies of isomers.

@Greg, I clarified my question a bit, so that it is now more descriptive.

It is still very vague. "Any QChem theory predicting energy and PES, qualitatively or quantitatively" is a correct answer. Also note, isomerization is a reaction (geometry change) while isomerism is a relationship between two (several) structures. There are a bunch of different type of isomer relationship (and theory describing them), many of them are not associated with any transformations between them. Also this question technically covers isomerism of organic compounds as well as phase transformations of inorganic solids or nanoparticles.  It is extremely wide range.

@Greg, I am now referring to isomerism instead of isomerization. As for the second part of your comment, doesn't there exist a theory, e.g. one based on group theory / topology, which would cover both organic compounds and inorganic solids (and nanoparticles)? If not, I would be interested to know   the relevant theories for these three cases.


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Adsorption on surface of adsorbent
Why is energy released during the process of adsorption? How is this energy change observed and how is it calculated?

Under normal conditions, atoms will bond to one another and produce a molecule, only if the bond formation process is exothermic (e.g. it releases energy) - this translates into the products (the molecule) being thermodynamically more stable that the starting material (the atoms).  The same is true for the adsorption process.  A gas will adsorb on a surface only if this configuration is more stable than a gas and a free surface.
Typically when a gas is adsorbed on a surface, something akin to bond formation between the surface and the gas takes place and we wind up with a chemisorbed species.  There are many ways to study this process.  One simple way is to measure the amount of gas admitted to the system containing the surface to be studied and measuring the change in pressure over time.  When done in a calorimeter, the heat given of during the process can also be measured.  Combining these two measurements(change in moles of free gas and energy released) allows one to understand the basic energetics of the process. The "isoteric method" is one experimental procedure for accomplishing this in a laboratory.  It relies on measuring moles of gas vs. the pressure in the system (an isotherm) at various temperatures.  From these measurements the enthalpy of adsorption can be determined.
The study of the adsorption of a gas on a surface is complicated.  There is not a surface with a monolayer of gas adsorbed on it with free gas above this.  Rather there is a surface and bulk gas and in between the two is a multilayer structure of adsorbed gas where the gas concentration varies throughout.  Here is a link to a recent Cal Tech thesis on the characterization of hydrogen gas adsorption onto various surfaces; take a look at section 1.5 and the beginning of chapter 3 for a nice overview of the adsorption process and the interfaces involved. The rest of chapter 3 is a discussion of the thermodynamics and various experimental techniques used to study the process.  It sounds like you might also be interested in performing calculations on the adsorption process.  If so, section 5.8 delves into computational methods used on one of the surfaces studied.

Might I add that molecules can still adsorb to a surface even when it's thermodynamically unstable, by becoming kinetically trapped.

You can see it from the Gibbs Helmholtz equation . For the process to be spontaneous the net change in Gibbs free energy of the system has to be negative . But since there is a decrease in entropy the enthalpy change should be more negative so as to make the total Gibbs free energy change negative . Hence adsorption is an exothermic process


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Invariants in chemistry used for model selection
What invariants related to geometrical structures that describe molecular geometry are used in chemistry?
Explanation: Here, Electronic model with highest prediction rate…, one can learn about electronic models. Now, I want to learn which invariants are used in all major models used (and why).
Definition of invariant: Invariants in chemistry used for model selection
Example of invariants in physics of ideal gas (invariance of the geometry of ideal gas with respect to Legendre transformations): http://arxiv.org/pdf/0811.0222v1.pdf

This question could use some context and elaboration.

trb456, I am interested to learn about low-level molecular interaction. Here, http://chemistry.stackexchange.com/questions/14671/electronic-model-with-highest-prediction-rate, one can learn about electronic models. Now, I want to learn which invariants are used in all major models used (and why).

What do you mean by invariant? Do you mean a constant, as in Planck's constant? Or do you mean a variable that is always included in the model? Or do you mean it in one of the senses outlined here?

I use the following definition: http://en.wikipedia.org/wiki/Invariant_(mathematics)

Suddenly, this becomes an incredibly broad question. There are two reasons it will be hard to get a good answer here. First, most chemists aren't also mathematicians, and will have a hard time identifying things as mathematically invariant even if they use them regularly. Secondly, you can find things that are classified as invariant in almost any model in chemistry - from balanced equations to statistical mechanics to quantum matrix mechanics. If you can be more specific about what types of invariants and what types of systems you are thinking about, it would help narrow it down.

thomij, I am mostly interested in invariants related to geometrical structures that describe molecular geometry, see: http://en.wikipedia.org/wiki/Molecular_geometry. In physics, one could mention e.g. invariance of the geometry of ideal gas with respect to Legendre transformations, see: http://arxiv.org/pdf/0811.0222v1.pdf

In that case, maybe you could edit your question to focus on that? I believe that you might be talking about what quantum chemists would call symmetries in molecular orbitals.

thomij, I edited the question to make it more specific. Still, I would be interested to hear about other invariants, too.

various symmetry rules are widely used, mostly in the field of quantum chemistry to simplify calculation models and spectroscopy to select spectroscopic active state transitions.

Martin, I did what you suggested. Now, the question may be clear to non-mathematicians. I also added related links from the comments.

Just a general remark on the functionality of this site: Put an @ in front of the name, to notify this person. See [help] for more.


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Organic chemistry textbook for self-learning?
I am interested in learning organic chemistry in order to solidify my knowledge of molecular biology.
However, there are a lot of options for organic chemistry textbooks, and I plan on simply sampling a few in order to figure out which ones are best, but I thought I would also ask for some advice.
Since I am self-learning, I have some specific items on my textbook wishlist:

The textbook should be insightful - I don't simply want facts presented, I want to see how the facts connect together into a larger whole. Usually, teachers fulfill this task, but in this case, I will need some help from the textbook.
The textbook should have a good section on learning IUPAC nomenclature.
I don't want organic chemistry to be an experience in advanced memorization for me. The previous point helps in addressing this issue, but I would like to reiterate it. I would like to have a textbook that helps me understand why each class of reactions is important (what can one do, that the other can't?) in a chemist (or an organism's) toolbox - I don't want to be memorizing reactions, I want to be putting them into my toolbox.

With such requirements in mind, which textbooks would you recommend?

FYI, I think these sorts of questions are generally considered off-topic, since they're broad and answers are invariably at least somewhat subjective. That said, I could be wrong, and I've provided an answer with my recommendations in any case.

I'm curious why one of your focuses is IUPAC naming. If your true application is molecular biology, I think you'll find that real IUPAC naming will be used little in that context, and instead trivial or other naming systems will be used.

I'll provide a rundown of the general organic chemistry textbooks I've studied in detail:

Organic Chemistry, David R. Klein: A very solid introductory text. Emphasis is on expediency and development of intuition over rigorous theoretical correctness. The coverage of molecular orbital theory is extremely scant, and classic valence bond theory (with associated concept of orbital hybridization) is the major explanatory paradigm. This is probably the most accessible text for students with a minimal background, but it's also somewhat superficial.
Organic Chemistry, Jonathan Clayden, et al.: This is a very thorough text, with a slightly wider array of reactions discussed than Klein's textbook (most notably in the coverage of organometallic reagents). Again, the MO treatment is lacking, but it's invoked more frequently than in Klein. This book tends to delve quite deeply into the mechanisms underlying specific reactions, making it rather lengthy. There's also some superficial discussion of HSAB theory, which is mostly (if not completely) absent in Klein's book.
Advanced Organic Chemistry, Carey & Sundberg. This is something of a seminal classic, but it's not for the faint of heart or for those uninitiated in the fundamentals. Mechanisms are elucidated in great detail (primarily via MO theory, with results from both ab initio and semi-empirical methods), and the exposition of pericyclic reactions is notably excellent. It also contains a great deal of illuminating content on conformational analysis. This is definitely not appropriate for a first textbook, but I would consider it essential reading for advanced undergrads and above.

Ultimately, either of Klein's or Clayden's textbooks should satisfy all of your criteria, but I strongly recommend Carey & Sundberg for subsequent study, if possible.
Beyond those, I've found Fleming's Molecular Orbitals and Organic Chemical Reactions to be a very good introduction to qualitative MO theory, but it's not a general-purpose text. March's Advanced Organic Chemistry is the ultimate, all-encompassing reference manual, and I'd consider it the sine qua non of any good organic chemistry library. As an aside, McQuarrie and Atkins have both published very good physical chemistry textbooks, which in many ways have been more enlightening than any textbook specific to organic chemistry. I'm also a fan of Oxtoby's Principles of Modern Chemistry as a general chemistry textbook; its mathematical and physical rigor is incredibly admirable, even if it does sometimes lack severely in concision. I've also heard unanimous praise for Housecroft & Sharpe's Inorganic Chemistry, but I've yet to read more than a few sections of it.

I find Organic Chemistry by Clayden et al. to be a fine textbook, though I'm not a chemistry major (I'm currently a lowly high school student) and just peruse it for fun.


text:
Reaction mechanism of combustion of hydrogen
After a discussion about the usefulness of hydrogen fuel for cars with a friend of mine, I wondered what the reaction mechanism for combustion of $\ce{H2}$ was. I only study (well almost) organic chemistry, so I have no idea if the mechanisms from organic chemistry can be applied at all to non-organic molecules.
So we have the following reaction:
$$2\ce{H2 + O2 -> 2 H2O}$$
In most textbooks this would be described as a redox-reaction with the two half reactions:
$$2\ce{H+ + 2 e- -> H2}$$
$$\ce{O2 + 4 H+ + 4 e- -> 2 H2O}$$
But I wonder what the real mechanism is. Based on my organic thinking I thought the following (it's probably completely wrong, but I think it is important to state because someone can correct me much better if I express my thinking than if he has no idea where the misconception started):
If the temperature is high enough a long pair of oxygen could break the $\ce{H-H}$ (the lone pair will be donated to the $\sigma^{*}$ bond) bond and form $\ce{2 H-}$ (each H gets one electron because of their equal electronegativity). $\ce{H-}$ is very unstable and one will immediately react with the empty oxygen orbital, the other will break the $\ce{O=O}$ $\pi$ bond. The electrons in the $\pi$ bond will go to the positively charged oxygen atom (the one that donated its electrons in the first step). Now one oxygen atom is neutral and the other has a positive charge (because the electrons of the $\pi$ bond went for the other oxygen), where the $\ce{H-}$ will attack forming hydrogen peroxide ($\ce{HO-OH}$). The previous process will now repeat, but instead, the $\sigma$ bond between the oxygen atoms will be broken, resulting in two water molecules, $\ce{2 H2O}$.

here's an overview: http://arrow.utias.utoronto.ca/~ogulder/ClassNotes3.pdf

Several mechanisms are possible.  It depends on the conditions:  pressure, temperature, presence of catalysts, etc.

This post deals with the mechanism that is observed in the gas phase. It is of course not as simple as the equation might suggest and you did suspect that already.
$$\ce{2H2 + O2 -> 2H2O}$$
This will be divided into many different elementary sub reactions. Any mixture of oxygen and hydrogen is metastable (stable as long as you do not change the conditions). If you provide sufficient energy to overcome the activation barrier, the reaction will take its cause:
$$\ce{H2 <=>[\Delta T] 2 H.}$$
The resulting hydrogen atoms are very unstable and will react with anything in reach, but most importantly with $\ce{O2}$ (which is a triplet biradical), forming hydroxyl radicals. This reaction is endothermic (it requires energy).
$$\ce{H. + O2 -> HO. + O}$$
The resulting oxygen radicals can again react with $\ce{H2}$ to form more hydrogen radicals. This reaction is also endothermic.
$$\ce{O + H2 -> HO. + H.}$$
The hydroxyl radicals can also react with $\ce{H2}$ to form more hydrogen radicals, which is a slightly exothermic (releases energy) reaction.
The net result of those equations leads to a slightly exothermic sum, with high potential:
$$\ce{3H2 + O2 -> 2H2O + 2H.}$$
In principle, the hydrogen radicals react as a catalyst. However, this reaction is a highly branched chain reaction, including a lot of radical reactions. Due to this, more and more hydrogen radicals will be produced.
This scheme will eventually end when the concentrations of $\ce{O2,H2}$ will become lower forcing the excess radicals to react with each other.
$$\ce{HO. + H. -> H2O}$$
Another possibility is forming as a byproduct hydrogen peroxide, which is a very exothermic reaction:
$$\ce{H. + O2 -> HO2.}\\
\ce{2HO2. -> H2O2 +O2}$$
(Also happening but not as important: $\ce{HO. + HO. <=> H2O2}$)
As long as there are hydroxyl and hydrogen radicals present, the peroxide will take part in the chain reaction. However, this reaction also provides the necessary energy to cleave more $\ce{H2}$. peroxide is also easily cleaved again, or reacts with each other:
$$\ce{2H2O2 -> 2H2O + O2}$$
This all results in the main product water.
Please keep in mind, that there are many factors, that influence these reactions. It is strongly dependent on pressure, temperature, and surroundings. Surfaces and/ or catalysts involved in this reaction may change it completely.

Further reading:

How explosive is hydrogen gas?
Eugene P. Dougherty1 and Herschel Rabitz1, J. Chem. Phys. 72, 6571 (1980) most comprehensive (theoretical) study on the $\ce{H2-O2}$ system.
Wiberg, Nils: Lehrbuch der Anorganischen Chemie Kap. 3.2 (german) The mechanism is in part taken from this source.
http://arrow.utias.utoronto.ca/~ogulder/ClassNotes3.pdf Thanks to Silvio Levy

I am sure that the combustion of hydorgen gas in the gas phase is by a radical mechanism, I know that alkyl halides such as halon will modify the combustion

@NuclearChemist I am not sure where you are getting at with your comment as the mechanism I describe is a radical one. Obviously if other species are present, the mechanism will change, but that was hardly the point of the question.

The fact that halon tends to modify hydrogen / oxygen flames conforms that the reaction is free radical. Imagine for a momnet if the H2 / O2 reaction occured by a 2+2 pericylic reaction. Then I would expect that halon would have almost no effect on the rate of reaction. Halon works by intercepting a radical in a combustion mechanism to form a lower energy radical which is unable to continue the reaction. I am sure that a hydrogen atom could abstract a Br from CF3Br to form the CF3 radical which I think will have a lower energy than the H atom.

@NuclearChemist I still don't understand where you are going with your comment. You obviously cannot have a pericyclic reaction, the spins don't match. The whole idea of not having a radical mechanism is problematic, as on species is a already a radical. If you really need to consider halogenated compounds, then it is well known that they are relatively easy to split; I wouldn't be too sure that they intercept existing radicals, or are the source of such.

I gave the pericylic reaction idea as an example of a mechanism which is not operating for hydrogen combustion. CH. Kim, OC Kwon, GM. Faeth, JOURNAL OF PROPULSION AND POWER, 2002, volume 18, page 1059 is all about the effect of halon on hydrogen / air flames.

The scheme starts with the production of OH radicals since Hydrogen dissociation is very endothermic (432 kJ/mole) compared to $\ce{H2 + O_2 \to 2OH\cdot}$ which is the initiation reaction with $\Delta H^{\mathrm{o}}_{298} = 72$ kJ/mole. There is a propagation reaction and  two chain branching reactions and gas phase termination as well as wall termination reactions.
$$\begin{align}\ce{H2 + O_2 &\to 2OH\cdot} \qquad &\text{initiation 72 kJ/mole }\\
\ce{OH\cdot + H2 &\to H2O + H\cdot} \quad &\text{propagation -62 kJ/mole }\\
\ce{H\cdot + O2 & \to OH\cdot + O\cdot} \quad &\text{branching 70 kJ/mole} \\ \ce{O\cdot + H2 &\to OH\cdot + H\cdot} \quad &\text{branching 8 kJ/mole }\\
\ce{H\cdot + O2 + M &\to HO2\cdot + M\cdot} &\quad \text{gas termination -196 kJ/mole }\\
\ce{H\cdot  /  OH\cdot / HO2\cdot} &\to \text{ wall  termination}\\
\end{align} $$
There is strong experimental evidence for $\ce{H\cdot,\, O\cdot,\,OH\cdot}$ and $\ce{\,HO2\cdot}$.
The scheme allows for two explosion limits. The first limit, at low pressure is controlled by surface effects and diffusion to the vessel walls. The second limit, where explosion stops, is determined by the gas phase termination reaction of $\ce{HO2\cdot}$ and is independent of the vessel's size.
Some small amount of peroxide may be formed such as by $\ce{HO2\cdot + H2 \to H2O2 + H\cdot}$ but this acts as an inhibitor $\ce{OH\cdot + H2O2 \to H2O + HO2\cdot}$ because the $\ce{HO2\cdot}$ radical is less reactive that either $\ce{H\cdot}$ or $\ce{OH\cdot}$. Peroxide is perhaps involved in the third explosion limit.

In your branching steps, $\ce{O^.}$ means $\ce{^.O^.}$ ?

By convention it is just half of an oxygen molecule.

The combustion of hydrogen can occur both in the gas phase and on the surface of a metal, it can also occur on the surface of platinum. This is used in the hydrogen recombiners which are present in the containment of many nuclear reactors.
Another example of a flameless combustion on a metal surface is the oxidation of solvents such as acetone on the surface of copper metal. Here is one of the many films on youtube showing the reaction. The oxidation of acetone with air using a copper metal surface is much easier to do than the combustion of hydrogen with a platinum surface.
The hydrogen will adsorb onto the metal surface first by physical adsorption and then by chemisorption. This will result in the platinum surface having hydroegn atoms on the surface.
Oxygen molecules from the air will also adsorb onto the platinum, then on the surface a reaction between hydrogen atoms and the oxygen will occur which forms water.
TPD, XPS and LEED were used by Yuichi Ohno and Tatsuo Matsushima to show that oxygen on a platinum surface can be present both as molecules and as atoms, Surface Science, 1991, volume 241, page 47.
If you want to know more then a good refernece would be Study on chemisorption of H-2, O-2, CO and C2H4 on Pt-Ag/SiO2 catalysts by microcalorimetry and FTIR , X. Wang et. al., JOURNAL OF THERMAL ANALYSIS AND CALORIMETRY, 2005, volume 82, page 103.


text:
"Electrwd" group?
The Amber force field parameters describe hydrogen atoms in various states:
HC 1.008                             H aliph. bond. to C without electrwd.group
H1 1.008                             H aliph. bond. to C with 1 electrwd. group
H2 1.008                             H aliph. bond. to C with 2 electrwd.groups
H3 1.008                             H aliph. bond. to C with 3 eletrwd.groups
HA 1.008                             H arom. bond. to C without elctrwd. groups
H4 1.008                             H arom. bond. to C with 1 electrwd. group
H5 1.008                             H arom. bond. to C with 2 electrwd. groups

But I cannot figure what they mean by 'electrwd group' and it doesn't specify anywhere in the documentation. I'm hoping it will be glaringly obvious to a chemist...
Thank you.

electron-withdrawing (e.g. F in CF3COOH)


text:
When using Gaussian to calculate NMR, what's the default solvent and frequency?
If I choose GIAO HF/STO-3G, is the default solvent $\ce{CD3Cl}$? Is the default frequency for $\ce{^1H-NMR}$ 500 MHz? What about the frequency for $\ce{^13C-NMR}$?

$\ce{CD3Cl}$ is not a solvent (or even a liquid) at ambient temperature and pressure. I have never heard of it being used in NMR. I think you meant deuterochloroform ($\ce{CDCl3}$).

Much faster and in most cases reasonably accurate method is empirical prediction based on group contributions, see e.g. http://www.nmrdb.org/
Unless you are after some geometry dependent stuff, like Karplus parameters, you should first try the empirical methods.

While Greg's answer describes sufficiently what the program does, I would like to extend a little on his remark. (This should have been a comment, but it was too long.)
The HF/STO-3G level of theory is what you call the minimised ab initio approach. It is unreliable in many ways, but it is fast. However, even geometries of well known and/or calculated molecules may be wrong. In short, it fails more often than it is correct. The problems start with HF, since it does not describe correlation sufficiently. (It only describes Pauli correlation, but this is exact.) The next problem is the basis set: It substantially lacks basis functions. It is called the minimal basis for a reason. Using this setup for NMR calculations is worse than guessing.
The first thing to do would be to increase the basis set. Maybe move away from Pople basis all together. A few basis set families come to mind: Dunning's augmented family, aug-cc-pVDZ and larger, Ahlrichs and Weigend's def2 family, def2-SVP and larger, and many others, see EMSL.
Calculating NMR tensors is always an accuracy versus time tradeoff. And it might be complicated to find the best (in the sense of most efficient) setup. The best way is to start a geometry optimisation on quite a low level. For reasonably large molecules a semi-empirical model like pm6 can be used to scan for configurational space.
Increase the level of theory to pure DFT. You can use density fitting here to speed up your calculation: e.g. #P BP86/Def2SVP/W06 DenFit.
Validate your geometries with a higher basis set, e.g. #P BP86/Def2TZVPP/W06 DenFit and different functionals, e.g. TPSS, PBE1PBE, M06, etc.
Validate your geometries with the dispersion correction of Grimme, e.g. BP86 EmpiricalDispersion=GD3BJ, PBE1PBE EmpiricalDispersion=GD3BJ, B2PLYPD3, etc.
Apply solvent corrections for your system, e.g. SCRF=(Solvent=Chloroform). Make sure to run single point calculations first and try optimisations later.
If you have fully converged geometries at DFT level, you might want to check basic properties with ab initio techniques. Most appropriate would be perturbation theory, e.g. MP2.
After you have done all the above you can go ahead and calculate NMR tensors. It may be combined with solvent approximations. It requires a very well converged structure, a local minimum on the potential energy hyper surface. (Check that with a frequency calculation, e.g. FREQ.) You can set it up like this sample input:
%oldchk=filenameOfPreviousOptimization.chk
%chk=newCheckpoitfile.chk
#P MP2/def2TZVPP
SCRF(solvent=chloroform)
NMR(GIAO)
geom=AllCheck guess=read

Be advised that that computing NMR data with only one method can lead to many misinterpretations. It is very important to verify your structures and tensor on a variety of methods. If you cannot afford MP2 with a large basis, then you should at least check 4 or 5 DFT methods. It is also important to know, that including solvent effects is still something where the theory is not exact. If you have a system that changes a lot with the solvent, then you should consider explicit treatment. Then you might want to dive into molecular modelling, molecular dynamics, MM/QM, ONIOM, etc. pp.

Very nice, detailed answer.

Thanks for your information. I've tried DFT and it runs so slow that it seems to be stuck.

@Ath How many atoms are we talking about? What kind of computer are you running these calculations on? Depending on your system these calculations can be quite demanding and may well take more than a day. You can check the logfile to see what your calculation is doing.

@Martin 200 atoms. is it too large? I'm running it on Linux.I keep looking at the output, but it's stuck at "no special actions if energy rises". It's doing convergence and slower than I expected

@Ath Linux is just a name for a range of operating systems. For the scf cycle with about 200 atoms, i would expect about 4-8 hours on a single processor (maybe more). Geometry optimizations should take around a week or two. That depends very much on your guess. It really also depends on the computer you are using, memory, number of processors, speed, ... When talking 200 atoms, one should hsve the right equipment...

@Martin Actually I'm not sure what kind of computer I'm using. The only thing I know is that it's very old and doing everything slowly. I plan to buy a new computer to run Gaussian jobs faster. How large memory do I need? How many processors are ok? I need your advice. Thanks.

@Ath you should by something that has at least 8 processors with at least 8 gb of ram for each processor. Several terabytes of disk space is adviseable. When it comes to computational chemistry, usually more means better. You should check benchmark tests or ask another question on this site. As you have a demanding system you should go big. Access to a supercomputer is also advisable. As an alternative, look for a cooperation with a computational group.

Performing calculations with the "NMR" keyword in Gaussian gives the magnetic shielding tensor in ppm and the spin-spin coupling in Hz. These numbers are independent of instrumental parameters like the frequency of the H-NMR. It doesn't contain any correction for any kind of solvent effect.
One remark: HF/STO-3G is a terrible, terrible low level of theory for ANY kind of calculations.

Thank you, but I think the solvent does influence the result because in different solvent, the solute have different structures.

It may be so, you know your system, you have to work that out. The actual calculation does not contain any solvent correction as far as I know (and why would).

If I choose GIAO HF/STO-3G, is the default solvent $\ce{CD3Cl}$?

No, the default is gas phase as usual. But you can choose whatever solvent is programmed through the scrf=(solvent=...) command. The available solvents are listed at the bottom of the related Gaussian keyword site.

Is the default frequency for $^1$H−NMR $500~\mathrm{MHz}$? What about the frequency for $^{13}$C−NMR?

There is no default frequency as there is no frequency at all. The program calculates the absolute nuclear shielding tensors that you afterwards need to relate to some other substance's nuclear shielding tensor. As In experimental spectra one chooses TMS or the solvent signal, this must be done also by calculating the shielding tensors for TMS on your chosen combination of method and basis set.
BUT as there are sure default values that are chosen if one chooses the "NMR" command, there is no such thing as a "default route" for NMR calculations. I mean, you can simply click "NMR" in GaussView which will turn on NMR=GIAO calculations but there is more to think of, as Martin already stated.

NMR Shieldings
I very much like the approach by Lodewyk, Siebert, Tantillo, Rablen and Bally called CHESHIRE - Chemical Shift Repository, because there you can find many recommended procedures for either G03 or G09.
What they have done, was to calculate the shieldings for many known small substances with various DFT functionals, basis sets and solvents in many combinations to get for each of those combinations a set of scaling factors via linear regression.
For me, I chose to optimize my structures with B3LYP/6-31+G(d,p) and afterwards calculate the NMR shieldings with mPW1PW91/6-311+G(2d,p)/PCM and nmr=giao. Then one simply has to write out the calculated shieldings and modify them by $$\delta=\frac{\sigma-\text{intercept}}{\text{slope}}$$ according to their manual with the appropriate scaling factors.
But as you mentioned that you have 200 atoms, you might consider a smaller approach. Good for you, besides their recommended scaling factors there is also a manual on how to calculate your own scaling factors and a vast table for many solvents and methods with their respective scaling factors.
Another really imporant point for such large systems is that they don't exist as one single configuration! What you have to do is to generate and optimize an enormous amount of possible isomers, calculate the shieldings for all those isomers and Bolzmann-weight them using their energy deviation regarding the isomer with the lowest energy.
$$\delta_i = \sum_j w_j~\delta_{i,j} = \sum_j \frac{\exp\left(\frac{-E_j}{RT}\right)}{\sum_k \exp\left(\frac{-E_k}{RT}\right)}~\delta_{i,j}$$

$^1$H,$^1$H-Coupling Constants
As an $^1$H-NMR doesn't consist of straight peaks but mostly multiplets, there could also be the need to calculate the coupling constants. There have been a few approaches towards accurate J values but one of the easiest is also stated on the CHESHIRE website.

Optimize the geometry with B3LYP/6-31G(d)
Run an NMR single-point calculation with the following route section in GAUSSIAN
#n B3LYP/6-31G(d,p) nmr=(fconly,readatoms) iop(3/10=1100000)
At the end of the molecule specification (separated by a blank line) read in: atoms=H
A sample input file for chloroethane could be like:
\#n B3LYP/6-31G(d,p) nmr=(fconly,readatoms) iop(3/10=1100000)

Chloroethylene CS

0,1
C,0,0.,0.,0.
C,0,0.,0.,1.32727906
Cl,0,1.462671003,0.,-0.9634208566
H,0,-0.8964932057,0.,-0.6096342696
H,0,-0.9458848703,0.,1.8609603221
H,0,0.9144771425,0.,1.9110938644

atoms=H

From the resulting log file, extract the desired Fermi contact J values, and scale them by a factor of 0.9117

Example: Nitrobenzene
According to SDBS, AIST the $^1$H resp. $^{13}$C-NMR spectrum have the following peaks:
$$\begin{array}{ccccccc}
\hline
\ce{Nuc.} & \ce{Calc}. & \ce{Exp}. & & \ce{Nuc}. & \ce{Calc}. & \ce{Exp}. \\
 & \ce{ppm} & \ce{ppm} & & & \ce{ppm} & \ce{ppm} \\
\hline
 & & & & \ce{C_{N}} & 146.47 & 148.30 \\
\ce{H_{A,~A'}} & 8.18 & 8.25 & & \ce{C_{A,~A'}} & 123.25 & 123.46 \\
\ce{H_{B,~B'}} & 7.37 & 7.56 & & \ce{C_{B,~B'}} & 127.75 & 129.43 \\
\ce{H_{C}} & 7.59 & 7.71 & & \ce{C_{C}} & 135.03 & 134.71 \\
\hline
 & \ce{RMSD}  & 0.23 & & & \ce{RMSD} & 2.52\\
\hline
\end{array}$$
The differences are not very big. As the carbon nuclei span a much wider shielding range the average error is considerably smaller when compared to the RMSD of the proton shifts. To get an image out of this pure numbers, the overlay of the calculated versus the experimental $^1$H-NMR spectrum can be seen below. It fits quite good, even if it could be better. Maybe the WP04 functional could have done a better job.

Just curious, how do you translate the corrected J values to plotting them as the above vertically stacked spectra? Did you use a specific software?

Do these "solvents" make sense? Calculating the interaction of solvent and specimen is not as "straightforward" than deriving chemical shifts and couplings from the vacuum electronic stucture that Gaussian has solved. Quite another task, that.

The sample input for chloroethane begins with \#n . Is the first backslash a mistake?

Took some time, but I used a Mathematica-notebook. Don’t know where it is now. ^^" ... @Kurzd Yes ... thanks for highlighting.


text:
Are homemade batteries a practical alternative during a blackout?
If your home lost power without warning; are homemade batteries (such as the lemon/potato variety described in this question Are homemade batteries economically practical?) a practical alternative for powering your appliances while you wait for the electrician or what have you?
Are there other, better, non-produce-dependent methods for producing use-able electricity from household materials and minimal expertise?

Short answer:
Home made batteries are expensive, easy to fail, easy to spoil, potentially harmful to the environment and you need really a lot of them to get out meaningful amount of electricity.
Long answer:
Your question is pretty vague, it is not clear what you mean by "usable electricity".
If you need light, it is easier (cheaper, reliable) to use candles or oil lamps.
If you need electricity to drive electric devices, than you most probably need a predetermined voltage, which will be difficult to reliably produce.
If you need a substantial amount of electricity to drive light and several electric devices, it is very hard to do it with any kind of battery-like equipment: just think it over how many batteries you need the produce a kW electricity.
Other problems:

Batteries per definition contains corrosive in one way or another, and DIY level technology risks that corrosion occurs much less predictable way. You have to figure out how long a battery can be stored reliably before you get everything soaked in rotten citrus juices.
Since you most probably use metals in this battery, it is a good question how poisonous your new and used battery will be and if you can trash it without major environmental damage. It is not nice to spill out large amount of dissolved copper or lead in the garden.
If you do any kind of electricity work, and you are not good at it, especially with potentially corroded equipment, you may have a lot of problem with contact errors and other reliability issues.

... are homemade batteries ... a practical alternative for powering your appliances ...?

No, those type of batteries don't have enough energy to power more than a small light bulb or a transistor radio. If fact the light bulb probably won't stay lit for more than 10 minutes. An LED would be much better.
You need at least 30 car batteries to power your home for a significant amount of time; this is why most people going off the grid have a significant infrastructure cost.

Are there other, better, non-produce-dependent methods for producing use-able electricity from household materials and minimal expertise?

Other than the alkaline batteries you buy in stores, not that I can think of.

And the car batteries would all be DC. Any appliance that plugs into a wall will want AC. And converting DC to AC would require an expensive inverter and incur a pretty large efficiency cost.


text:
Electronic configuration of Ace-Ala-Nme
Below is the structure of Ace-Ala-Nme. I have labelled just four of the carbon atoms (1-4 in blue). I need to identify (i) the hybridization state, and (ii) the number of electron withdrawing groups for each.

I believe the amine (-NH) groups are electron donating while the carbonyls (-CO) are electron withdrawing. I therefore suspect that all four labelled carbon atoms are sp3 with 0 withdrawing groups. But this is assuming that the donating/withdrawing tendency of the amine and carbonyls cancel each other out. Can anyone confirm this to be the case?

This is a neutral, un-ionized (i.e. non-zwitterionic form) form of Ala-Ala correct?  Also, are you asking about the electron configuration of this dipeptide or hybridization of various atoms in this molecule?

@LordStryker It is indeed neutral and I'm interested in the individual atoms. Thank you.

This illustration must be just a fragment of an Alanine chain (polypeptide).  The un-ionized Alanine dipeptide would intially begin with NH2 and terminate with COOH.  Your picture does not indicate this. You may want to provide a proper image, re-label your atoms and then attempt to provide an adjusted answer. (See Figure 3 on page 5 of this article: http://iopscience.iop.org/0004-637X/765/2/111/pdf/0004-637X_765_2_111.pdf)

@LordStryker I have updated my question. I have seen this molecule referred to as "Ace-Ala-Nme" although it's mostly referred to as "alanine dipeptide" in the papers I've read (then again, these are papers by physicists rather than chemists).

I think an amine (-NH) is electron donating but I am not an organic chemist.  EDIT: The consensus in my lab is that -NH is an electron donating group.

I have updated my question in light of this new information.

1: sp3 with 1 electron withdrawing group
2: sp3 with 2 electron withdrawing groups
3: sp3 with 0 electron withdrawing groups
4: sp3 with 1 electron withdrawing group

I would say your earlier analysis was correct.  If you look at the resonance structures that can be drawn for an amide linkage, you find there is considerable positive charge both on carbon and nitrogen.  This should serve to make them both electron withdrawing, relative to an attached carbon, in your structure.

EDIT: To clarify this answer with regards to some of the comments posted above, I'd like to add that typically when an amino nitrogen is situated on a double bond or aromatic ring it can donate electrons through resonance.  In fact, that is what it is doing in the resonance structure on the right - it is donating its lone pair of electrons via resonance.  However, the amino group is also electron withdrawing in an inductive sense relative to carbon.  Further the $\ce{-NH3^+}$ group can no longer interact via resonance and is simply a strong electron withdrawing group.  The resonance structure on the right is analogous to the $\ce{-NH3^+}$ situation and since it is a significant contributor to the overall description of the amide (cf. there is significant double bond character in the amide $\ce{C-N}$ bond) it acts inductively as a strong electron withdrawing group for the attached $\ce{sp^3}$ carbon.