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What reaction is expectable from the atomized reactants?
Let we have a reaction that starts from some reactants to produce some products. When the activation energy is provided to the reactants they will be atomized and produce a gas of individual atoms. The gas contain the whole atoms that can either form again the reactants, or form the products, or instead form the elemental molecules of themselves from which the reactants or products can be produced in other reactions. Now that the gas of all individual atoms is common for all these different reactions my question is what conditions decide which reaction to happen?
Can I say the reaction happens in the direction at which the Gibbs free enthalpy decreases more rapidly? Or do thermodynamic conditions like temperature, pressure and especially concentration of the species also play a role, independent of the variation in the Gibbs function?

The rate for a reaction is dependent upon the rate constant and, depending upon what type of reaction it is, the concentrations of the reactants.  The rate constant can be further broken down into an entropic factor (how likely are the reactants to collide in an orientation that will produce a reaction) and an energy factor (the activation energy).
Assuming for the moment that the reactant concentrations are such that they will not control the direction of reaction, and that the entropic factors don't differ dramatically for one pathway over another (that's usually the case for similar reactions), then the first reaction to happen will be the one involving the reaction with the lowest activation energy.  This will lead to the kinetically controlled product.
If the back (reverse) reaction from the kinetically controlled product also has a low enough activation energy such that it happens rapidly under our reaction conditions, then essentially we are setting up an equilibrium.  In this case, thermodynamics will eventually control the product distribution and we will eventually wind up with the product that has the lowest free energy, the thermodynamically controlled product.
Many reactions first produce the kinetically controlled product, but with time, the thermodynamic product becomes the major product.

thank you @ron, so you mean not the slope of change in the Gibbs free energy but the magnitude of the activation energy is the prime cause that determines the direction at the beginning, but if we wait for a long time eventually it is the magnitude of the Gibbs free energy that determines which product to be the major products? That is, the slope of change in the Gibbs energy is not really very important, neither in the rate of reaction nor in the final result achieved, right?

@topology Yes, that is correct.


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Is the bond in HF ionic while it is covalent in HCl?
Why would a hydrogen atom "donate" to fluorine in an ionic bond but not in $\ce{HCl}$?  Why would $\ce{H}$ and chlorine share instead of $\ce{Cl}$ just stripping it away like $\ce{F}$ does?

Just to clarify, are you stating the HCl bond is covalent and the HF bond is ionic?

http://chemistry.stackexchange.com/questions/17064/why-is-kf-the-most-ionic-compound?rq=1

Bonds can be completely covalent as in $\ce{Cl-Cl}$, $\ce{H-H}$, etc. In these cases the electron density is shared equally between the two atoms.  Bonds can also be ionic as in $\ce{Na^{+} Cl^{-}}$ where much (~80%) of an electron has been transferred from the sodium atom to the chlorine atom.  Between these two extreme cases exist a continuum of bonds that are called polar, covalent bonds.  In such cases the electron density between the two atoms is not shared equally, but neither is the electron transferred extensively from one atom to the other.
Just as we can draw resonance structures for acetone and discuss how the real molecule is a weighted superposition of these two structures, we can draw resonance structures for inorganic compounds and, again, discuss how much the various structures count towards the "real" description of the molecule.  In the diagram below, two resonance structures are shown for sodium chloride.  The structure on the left represents a covalent bond and the structure on the right represents an ionic bond. Just as acetone is a mix of the two resonance structures, so is sodium chloride.  When the weighted mix of the resonance structures lies far to the left, we can say the compound or bond is covalent; when it lies far to the right we can say the compound or bond is ionic.  Everything in between represents a polar, covalent bond.

One can crudely estimate how polar a diatomic covalent bond is by using the following methodology.  The dipole moment of a molecule is given by
$$\ce{\mu~=~Q*d}$$
where the dipole moment is in Debye (D), Q is the partial charge on an atom measured in coulombs and d is the distance between the two atoms (the bond length) in meters.
In the case of $\ce{HF}$, the dipole moment is 1.86 D and the bond length is 91.7 x 10^(-12) m.  Noting that 1 D = 3.3356 × 10^(−30) C·m, solving for Q we find a partial charge on one of the atoms of
$$\mathrm{Q~=~\frac{[1.86 ~x~3.3356 ~x~ 10^{-30} ]}{[91.7~x~10^{-12}]}~=~6.77~x~10^{-20}~C}$$
Now since one electron equals 1.6022 × 10^(−19) C, by taking the ratio
$$\mathrm{\frac{[6.77~x~10^{-20}]}{[1.6022 × 10^{−19}]}~=~ 0.42}$$
we estimate that roughly 42% of an electron has been transferred from the hydrogen atom to the fluorine atom.  This gives us a rough idea of how ionic the polar, covalent bond is in $\ce{HF}$.  Repeating this calculation for $\ce{HCl}$ suggests that roughly only 18% of an electron has been transferred in this case.
Clearly the polar, covalent bond in $\ce{HF}$ has more ionic character than the analogous polar, covalent bond in $\ce{HCl}$.  The reason why the $\ce{HF}$ bond has more ionic character is the greater electronegativity of fluorine (3.98) compared to chlorine (3.16).

Just to clarify, electron density can be donated and quantified in terms of percentages.

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Electron distribution in a C-S bond
My lecturer said that in a $\ce{C-S}$ bond, sulfur is slightly $\delta^{-}$ and the carbon slightly $\delta^{+}$, although they have (almost) the same electronegativity. What is the cause of this phenomena if it is not electronegativity?

I think your lecturer would have been more correct had he said that the carbon-sulfur bond reacts as if the sulfur is slightly $\delta^{-}$ and the carbon slightly $\delta^{+}$.
Sulphur is a larger atom so it has more, loosely held electrons than carbon.  This means that a sulfur atom is more polarizable than a carbon atom.  Although carbon and sulfur have similar electronegativities, the effect of this increased polarizability for sulfur is that sulfur behaves or reacts as if the bond were polar.  Take the case of nucleophilic attack on a $\ce{C-S}$ bond.  As the nucleophile approaches, the $\ce{C-S}$ bond will tend to polarize because of sulfur's polarizability.  And in which direction will the bond polarize?  Electron density will prefer to shift such that it resides on the larger atom (charge to size effect), sulfur.
This is the same argument used in the case of a carbon-iodine bond.  Again the two atoms have similar electronegativities (carbon = 2.55, iodine = 2.66), so although the bond is not polar based on electronegativity considerations, the polarizability of the iodine allows the bond to react as if it were polar when a nucleophile approaches.

Although I think everything you said is correct, I still guess it depends on what kind of bond it is (single, double, triple) and what saturates the carbon. But in the most simplest case you are absolutely right.

This was exactly what I thought, but then why isn't there a $\delta^{-}$ on sulfur if charge can be distributed there better? It seems to me that if the electrons shift towards the sulfur atom, they will become more stable because there is less interaction with the other electrons of the atom on sulfur.

Oh wait, perhaps I get it. The electrons are of course confined to the C-S bonding MO (which is symmetric I guess?). So not until some reaction happens that breaks this bond, can the size of the sulfur atom (and its polarizability) come into play and stabilize these electrons (thus behaving as if it was $\delta^{-}$). Is that right?

Yes, you've got it.  The electronegativities are similar, so the bond is not polar.  But when a reactant approaches, the polarizable bond will polarize such that electron density shifts towards sulfur (or iodine).

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Adjusting PBS with HCl: pH drift after dilution of phosphate buffer
It is some kind of everyday lab science and with a feel of shame I cannot answer this observation.
I noticed a pH drift (~0.5 units) of my 10X phosphate buffer (PBS), when I adjust the pH with $\ce{HCl}$ and dilute the 10X buffer afterwards.
When I prepare another buffer by mixing $\ce{Na2HPO4}$ and $\ce{NaH2PO4}$ to achieve the desired pH value, there is virtually no drift after dilution.
If using carbonate buffer, $\ce{CO2}$ is created if I set the pH with $\ce{HCl}$, but none with the also acidic $\ce{NaH2CO3}$ is used.
I just want to know how the phosphate buffer system is working and why there is a pH drift after dilution if I use $\ce{HCl}$ to set the pH. This might answer the general question, why adjusting the pH with $\ce{HCl}$ in PBS is not suggested.
I feel there could be something wrong with the counter ion maybe?!

Do you know what kind of electrode you're working with?

It's not clear in your question whether you are trying to prepare 1X PBS from  an existing 10X PBS or you are preparing 10X PBS from its constituting salts? Please precise!

"I noticed a pH drift (~0.5 units) of my 10X phosphate buffer (PBS), when I adjust the pH with HCl and dilute the 10X buffer afterwards"
It means exactly what I wrote. I prepare 10X and dilute it to 1X. Then I notice a pH drift or not. Dependent on how I prepared the 10X PBS.

How much HCl are you using?  Could the amount affect ionic strength of the solution?

The differences between PBS and the buffer obtained by mixing $\ce{Na2HPO4}$ and
 $\ce{NaH2PO4}$ are in ionic strength ( because of the presence of $\ce{KCl}$ d $\ce{NaCl}$ in PBS) and in the ratio of [acid]/[conjugated base]. We have to signal that the buffer capacity (buffer effectiveness) is maximum when $\ce{pH=pKa}$.
The best way to dilute 10x PBS into 1PBS (let's assume we want to prepare 1L of 1xPBS from 10xPBs) is to begin by diluting 100 mL of 10xPBS solution with 800mL water, then adjusting the pH and finally completig the volume to 1000mL. In doing so, there will be no drift in pH.


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How much oxygen is in a cup of water?
I know that water is 2 parts hydrogen and 1 part oxygen. Hydrogen and oxygen are quite light gasses, they don't weigh very much compared to water. Water is very heavy and it made me wonder exactly how much oxygen I'd need to make a cup of water.
Please answer in terms of cubic meters of air, assuming the air is 21 % oxygen, and that my small cup contains only 100 ml. I suppose the exact question is: How many cubic meters of air contains the same amount of oxygen as found in 100 ml of water?

2/3 hydrogen, 1/3 oxygen.

@Freddy - The first sentence of my question explains that I already know that. I actually wanted to know how  much that is.

Since the molecular weight of water is $18\ \mathrm{g/mol}$, $100\ \mathrm{ml}$ (density = $1\ \mathrm{g/ml}$, so $100\ \mathrm{ml} \overset{\wedge}{=} 100\ \mathrm{g}$ of water) of water contains
$$\frac{100}{18} = 5.56$$
moles of water, or 5.56 moles of atomic oxygen ($\ce{O}$), or 5.56/2 = 2.78 moles of molecular oxygen ($\ce{O2}$).
At standard temperature and pressure, 1 mole of a gas occupies 22.4 liters.  Therefore, 2.78 moles of molecular oxygen will occupy 62.22 liters.
If air is 21% molecular oxygen, then we would need 62.22/0.21, or 296.3 liters of air to provide that much molecular oxygen.
Since 1 cubic meter equals 1000 liters, this volume converts to 0.3 cubic meters of air.
Therefore, 0.3 cubic meters of air contains the same amount of oxygen as is present in 100 ml of water.

Please note that the value $V_\mathrm m=22.4\ \mathrm{l/mol}$ for the molar volume of an ideal gas corresponds to the old definition of standard temperature and pressure (STP), which is a temperature of $T=273.15\ \mathrm K$ and a pressure of $1\ \mathrm{atm}$. Since 1982, $p=1\ \mathrm{bar}=100,000\ \mathrm{Pa}$ is used as the standard pressure for tabulating thermodynamic data. At this pressure, the molar volume of an ideal gas actually is $V_\mathrm m=22.710,947(13)\ \mathrm{l/mol}$.


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How was Mendeleev able to develop his table?
I asked my general chem instructor this and he didn't know:
Mendeleev was able to group elements and discovered a bunch of them and put them in the correct order in the mid-1800s.  What equipment did he use to do this?  Was he just weighing stuff and guessing?
Specifically, how did he figure out atomic weight?

I don't know a lot of details, but I know part of it was noticing a pattern of similar reactivities, for example, lithium, sodium, and potassium did similar things, magnesium and calcium did similar things, fluorine, chlorine, bromine, and iodine did similar things. However, I don't really know how the concept of atomic mass or atomic number were arrived at, but you'd need some way to order the elements before their similar properties would have a recognizable periodicity.

Mendeleev was not the first, he had predecessors that grouped elements by their similarity, though ignoring their weights. Relative element masses were somewhat known at the time, as $pV=nkT$ relates volume with number of molecules in gas, so it allows to guess relation of moles in two gas volumes. And yes, wiki is your friend https://en.wikipedia.org/wiki/Atomic_theory

This quote from the wikipedia article on History of Chemistry might provide some insight into the determination of atomic mass:

Chemists in different countries developed several different incompatible atomistic systems. A paper that suggested a way out of this difficult situation was published as early as 1811 by the Italian physicist Amedeo Avogadro (1776-1856), who hypothesized that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules, from which it followed that relative molecular weights of any two gases are the same as the ratio of the densities of the two gases under the same conditions of temperature and pressure.

So this implies that you could take equal volumes of different gases at the same temperature and pressure, then weigh them and take the ratios. This would allow you to get relative masses for $\ce{H2}$, $\ce{N2}$, $\ce{O2}$, $\ce{Cl2}$, $\ce{Br2}$, $\ce{I2}$. Other gases could be used, but they wouldn't be as useful here. You could then determine that $\ce{H2}$ is the lightest gas, and that $\ce{O2}$ is 16 times heavier, $\ce{N2}$ is 14 times heavier, and so on. You don't even need to know that these are diatomic gases in order to make these measurements. Once you had these determined you could move on to gases like $\ce{CO2}$, or $\ce{H2S}$. Once you knew the mass of the gas molecule and what elements were in it, you could calculate how much of each element must be there, and by subtracting the mass of the known element, the remaining mass must be the atomic mass for the other element. As you accumulated more data you could approach substances that don't easily form gases, such as $\ce{NaCl}$. If you know the mass for $\ce{Cl}$ you could determine the mass for $\ce{Na}$.
What I suggest might not be the way atomic masses were determined, but it would probably work. Looking at this image of Mendeleev's original table shows they already had pretty good estimates of atomic mass by 1869.


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Why does hand sanitizer leave your skin feeling cool?
I noticed, and perhaps many others have too, that the application of hand sanitizer (mainly ethanol), leaves one's hands feeling rather chilly after application.
What is responsible for this phenomena? Is it the high heat of vaporization of hand sanitizer? However, explanation doesn't hold water; water has a standard heat of vaporization of 40.65 kJ/mol while ethanol has a heat of vaporization of 38.56 kJ/mol.
Could it be the low boiling point of ethanol? Hand sanitizer disappears (vaporizes) within seconds upon rubbing the hands together. Water, however, does not.
Additionally, how does one square a high heat of vaporization with a low boiling point? If it takes a lot of energy to vaporize something, then how can that something have a low boiling point?

I noticed the same thing with acetone. I think it's evaporation related, but that's all I can say.

My best shot at an answer to your last paragraph is here: http://chemistry.stackexchange.com/questions/32504/how-is-it-possible-for-a-substance-to-have-a-high-heat-of-vaporization-but-a-low

As user137 noted, this is from evaporation of the alcohol. Evaporation of a liquid takes the enthalpy of vaporization away with it, making you cooler.  Just like how sweating helps keep you cool in the summer, and how swamp coolers (evaporative coolers) keep houses cool in the southwest.

The above explanations are not incorrect, but they don't address the focus of the original question: the comparison of alcohol with water. The questioner was, in effect, saying they knew about "Evaporative Cooling", but had rejected that explanation due to the higher enthalpy value for water verses alcohol.
I believe the answer to this aspect of the question has to due with the quantity of alcohol vs. water that evaporates per unit time.  That is, although more energy is removed from the skin per gram of water evaporated, it takes a lot longer for each gram of water to evaporate compared to each gram of alcohol. That is, alcohol evaporates so much faster that it ends up being more efficient at removing heat.
The temperature we "feel" is often deceptive because it's usually a difference we detect, having to do with a lot of other things related to our nerves and their recent history, so there could be other factors related to alcohol's stronger interactions with hydrophobic molecules on our skin (especially oils), or the "drying out" effect that alcohol has after it evaporates from our skin.
But none of this rationalization means that much from a scientific perspective unless actual tests are made.  So, I would suggest different solvents be applied to skin and/or suitable skin-like surfaces and the rate of temperature tracked with one of those IR thermometers.  Alcohols of different chain lengths, butanol, propanol, ethanol, methanol (warning, poisonous!) would be perfect, but it'd also be interesting to try things like acetone or even oils. The results of those tests would be the real answer.

Hand sanitizer leaves your hands feeling cool because the particles in the gel which posses the most amount of energy are able to evaporate from the gel and off your hands. This results in only the particles with low energy levels to be left on your hand which therefore reduces the hands overall temperature.


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Which software is suitable to calculate the root mean square deviation between two protein structures?
I need to compute the root mean square deviation (RMSD) between two protein structures: the X-ray determined protein structure and the MOPAC optimized structure. Which software do you suggest?

There is also an interesting python package called ProDy which seems to be able to compute the root mean square deviation.

All of these are freely available packages:
Visual Molecular Dynamics - good visualization tool that can also do quick-and-dirty calculations from a graphical interface. The scripting interface has a learning curve, but yes, it can be quite powerful.
Avogadro - not as powerful as VMD, but can be useful for certain things.
Rasmol - Similar to VMD, although I've never used it.
LAMMPS, Gromacs, or NAMD - If one of these is installed on a computer you have access to, each of them can do this sort of computation and will allow you to do more advanced things if needed (they are actually molecular dynamics simulators). Each of these packages requires some familiarity with running code from the command-line and will have a learning curve.
Alternatively, you could always write your own code. RMSD is a pretty straightforward computation, and it's always nice to have more control over your analysis!

You should add TINKER to the list.  Its free and has a bunch of other things but the (overlay?) package computes RMSD on the fly.

There are a number of command-line utilities that can do this, but I'd suggest Open Babel
For example, you can align the molecules and append or print the RMSD as described in the documentation here.
I'll point out that calculating an RMSD can be difficult if the frame of reference has changed or the order of atoms is different in the calculation. The Open Babel code will ignore such differences.

VMD.  The built-in scripting language is insanely powerful.

Could you elaborate a little bit, like adding a link.

I added the link, but please add a bit more substance as to how the scripting language would suit the OP's purpose, for example.


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Examples of intramolecular hydrogen bonding closing bond angles
Okay, we've established that intramolecular hydrogen bonding can exist in molecules, and doesn't always serve the purpose of closing the bond angle at least relative to other stereoisomers of the same compound.
Consider nitrous acid and its cis and trans configuration; hydrogen bonding is only a possibility in the cis configuration, yet the cis conformation has (rather unexpectedly) a larger bond angle than in the trans conformation. The same goes for nitric acid.
Why does nitrous acid exhibit (unexpectedly) different bond angles?
So, are there actually any examples of molecules in which internal hydrogen bonding serves the purpose of closing up the bond angle relative to other conformations of the same molecule?


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Why is formic acid a stronger acid than acetic acid?
I am told that because the methyl group is electron donating in the conjugate base of acetic acid, this destabilizes the conjugate base by exacerbating the existing negative formal charge on the deprotonated oxygen, while in formic acid the electron donating methyl is absent in lieu of a hydrogen, which is neither withdrawing nor donating.
Is this reasoning correct, and are there any other reasons why formic acid might be stronger than acetic acid?

Yes the reasoning is correct . The more the charge is increased on a molecule the more it gets destabilised.

I don't have an authoritative reference to the effect, but I strongly suspect that its small size is probably a major factor. That potentially results in formation of more efficient hydrogen-bond networks and superior solvation.

Yes, the electron donation effect is pretty small.

@GregE. You may want to look at my answer, which also mentions solvation and identifies a similar effect with oxalate ion.

We are discussing the following equilibrium

We can make the acid a stronger acid by pushing the equilibrium to the right.  To push the equilibrium to the right we can

destabilize the starting acid pictured on the left side of the
equation, and \ or
stabilize the carboxylate anion pictured on the right side of the
equation.

Comparing acetic acid ($\ce{R~ =~ CH3}$) to formic acid ($\ce{R~ =~ H}$), the methyl group is electron releasing compared to hydrogen.  Therefore the methyl group will stabilize the dipolar resonance form of the starting acid where there is a partial positive charge on the carbonyl carbon.  This should stabilize the starting acid. Further, this electron releasing ability of the methyl group will tend to destabilize the resultant carboxylate anion which already has a full unit of negative charge.
Therefore, because the methyl group 1) stabilizes the starting acid and 2) destabilizes the carboxylate anion product, the methyl group will push the equilibrium to the left, compared to the case where the methyl group is replaced by a hydrogen.  Consequently, acetic acid is a weaker acid than formic acid.

I get the point you are trying to get across, but it seems weird that you chose exactly those resonance forms for the free acid and those different ones for the carboxylate. The resonance structure with the positive charge on $\ce{OH}$ should contribute more than the $\ce{C+ -O-}$ one.

@Jan wrote "The resonance structure with the positive charge on $\ce{OH}$ should contribute more than the $\ce{C^{+}-O^{-}}$ one." Jan, why do you suspect that? Carbon is less electronegative than oxygen and I thought the position of the carbonyl carbon in $\ce{^13C}$ nmr suggests a fair amount of positive charge on carbon.

Because all-octet resonance structres as $\ce{^{- }O-C={O+}H}$ generally contribute more to the overall structure than those without. Also, the carboxyl carbon’s NMR shift is about $30$ to $40~\mathrm{ppm}$ lower than that of a ketone, showing that the carbon’s positive partial charge is lower thanks to the neighbouring hydroxyl group.

What's the partial positive charge on the carbonyl carbon in acetone, around 40%?  Reducing that value to account for replacing a methyl group with a hydroxyl would still (IMO) leave significant positive charge on the carbon.

The strength of carboxylic acid can be discussed by positive inductive effect. Here, in ethanoic acid there is an electron releasing inductive effect from the alkyl group. The larger the positive inductive effect, the lesser will be the strength of an acid. Similarly in formic acid there is no alkyl group and hence no inductive effect takes place. So formic acid is a stronger acid than ethanoic acid.

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We may also consider solvation. Formate ion is relatively compact and thus more readily solvated by water molecules than acetate or other carboxylates, which are encumbered by their hydrocarbon groups.
Oxalate ion, $\ce{C2O4^{2-}}$, enjoys a similar solvation advantage, and this table shows that the second dissociation of oxalic acid is stronger than the dissociation of acetic acid.

Yeah, totally this! I'm pretty sure this "electron donating" explanation is incorrect, and solvation is probably the important thing here.

@Mithoron  Formic acid is more acidic than acetic acid in both solution and the gas phase.  Gas phase data can be found here and here

This is because of increasing positive inductive effect of alkyl group attached to the $\ce{-COOH}$ group in acid. In ethanoic acid electron releasing  group ($\ce{CH3}$) increase the electron density on the carboxylic group $\ce{-COOH}$. But, methanoic acid doesn't contain electron releasing group and there is no inductive effect.The larger the positive inductive effect, the lesser will be the strength of an acid. Hence, methanoic acid (formic acid) is stronger than ethanoic acid.

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Ozone resonance structure: differing opinions
I've always thought that the two-form resonance structure of ozone was the scientific concensus (from Wikipedia]:

Then today I found two other opinions. A three-form structure:

And a four-form structure:

Is there a consenus? Are the two and three-form representations just simplifications of the four-form structure? Please include references.
EDIT: The second resonance structure on the four-form image above is drawn incorrectly, as pointed out by @ron. The upper oxygen atom should have a lone pair and a positive charge, just like the second structure of the two-form.

Those are all valid resonance structures for ozone (except the second from the left in the "four-form structure" row, that one has two negative charges, maybe it's a typo).  In fact you can draw even more structures like this one, where this is no single bond between the middle and terminal-right oxygens.

All of these structures contribute to the "real" structure of ozone - however, some resonance structures contribute more and some resonance structures contribute less.  Here is a link to a nice set of rules (see page 2, the "Resonance Contributor Preference Rule" section) that describes how to weight the contribution of various resonance structures.  The most important rule is that each atom in the resonance structure should have an octet of electrons.  Only the structures labeled I and II in your "three-form structure" row meet this requirement, so they are the two most important resonance structures for ozone - but all of the others do contribute, just to a lesser extent.

Covalent bonds suppose to help an atom achieve filled valence shell by sharing electrons. Resonance structures need to have maximum atoms adhering to the octet rule. Look at III, the oxygen atoms on the right and left have only 7 electrons in their valence shall even AFTER forming covalent bonds. That is why it is not one of the resonance structures.


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Does knowing the ratio of masses of the elements in a compound lead to the unique chemical identity of the compound?
I was posed this question from a friend, however I am not necessarily sure. I think: yes, because a molecule of any compound is composed of a whole-number ratio of atoms of its elements, per the law of definite proportions. However, I feel as if I am wrong.

You may wish to read about isomers and the subset of these known as structural isomers.

No, knowing the mass ratios is not sufficient by itself. In the absence of additional information (for example, molar mass) that would only be enough information to determine the empirical formula, which is the formula that contains the smallest integral ratios of atoms. (E.g., the empirical formula of $\ce{C2H6}$ would be $\ce{CH3}$.) There are (theoretically) infinitely many molecules with the same atomic and mass ratios, yet differing in the actual molecular formula.
Further, even given the additional information of molar mass (so that the molecular formula can be determined), the existence of isomers precludes identification of a singular, specific compound without more sophisticated analysis. For example, $\ce{C3H6}$ could be either cyclopropane or propene. $\ce{C3H6O}$ could be as innocuous as acetone, or it could be highly carcinogenic epoxypropane. With more atoms, the number of possible isomers grows very rapidly. There's a nice table giving the number of isomers of alkanes of different lengths in this article. By the time you get to $\ce{C30H62}$, the number is $4111846763$. (Note, this is only considering structural isomers; accounting for stereoisomerism, the number would be even more astronomical.)


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Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?
Does Bent's rule have any utility? I got a vehement earful for "NO." Points raised by my professor:

Coulombic considerations can be used to rationalize bond angles, strengths, and lengths without the use of Bent's rule.
Orbitals are not real so why invoke orbitals when simpler Coulombic considerations are available? (Occam's razor)
The Coulombic effect is also real as opposed to orbitals.

Consider carbon tetrachloride and chloromethane.
Obviously, carbon tetrachloride has the stronger and shorter carbon-halide bond because in chloromethane, the methyl group is slightly electron donating, and this repels the terminal chlorine, which lengthens the carbon-halide bond. No application of Bent's rule is necessary.
Sure, you can justify this difference with Bent's rule; s-character is concentrated in orbitals directed toward more electropositive substituents; the carbon is more electropositive than chlorine; thus, the bond has more s-character around the chlorine and more p-character around the carbon in chloromethane. This greater p-character around the carbon in chloromethane weakens the carbon-halide bond relative to the bond in carbon tetrachloride. And greater p-character in a bond is also correlated with a compression in bond angles.
However, bond angles are also explained nicely through Coulombic effects. The C-H bonds in chloromethane are obviously shorter due to the electron-withdrawing nature of chlorine and the carbon-chlorine bond is obviously expanded due to methyl group/chlorine lone pair repulsion. This shortening of the C-H bond lengths explains the H-C-H bond angle expansion in chloromethane, and the lengthening of the C-X bond explains the X-C-H bond angle expansion.
Is there any case in which other elementary considerations (Coulombic, VSEPR, van der Waals repulsions) fall short in the realm of molecular geometry and Bent's rule can step in to pick up the slack? Anyone got some sort of vindication for Bent's rule?

You write 1) "the methyl group is slightly electron donating, and this repels the terminal chlorine, which lengthens the carbon-halide bond" and 2) "the carbon-chlorine bond is obviously expanded again due to the electron-withdrawing nature of chlorine"  Are you saying that both electron donation and electron withdrawal increase bond length?

Why should the [chlorine lone pair] - [C-Cl bonding electrons] repulsion in $\ce{CH3Cl}$ be different from that in $\ce{CH2Cl2}$ or $\ce{CCl4}$?

In CCl4 the other chlorines isolate electron density from the carbon but in H3CCl the carbon pulls electron density from the hydrogens and the carbon therefore repels the chlorine more

I read Bent's rule as, considering the hybrid orbitals around C, the ones directed toward electropositive substituents (e.g. H) have more s-character and the ones directed toward electronegative substituents have more p-character. I don't know what this statement means: "This greater p-character around the carbon in chloromethane weakens the carbon-halide bond relative to the bond in carbon tetrachloride." As I understand it, C has a certain amount of p-character (i.e. 3x 2p orbitals). How it uses those p-orbitals (invoking hybrids or not) may change depending on the substituents.

Right...p character in the c-cl bond

I think it would be helpful for this discussion to have a table of the actual data that we're trying to explain, perhaps all of the one carbon compounds starting with methane and replacing the H's with Cl's.

Here is a review on the utility of Bent's rule: http://onlinelibrary.wiley.com/doi/10.1002/poc.3382/abstract It provides answers to many questions that "professor" from the original post will not be able to explain. Bent's rule should be in the arsenal of tools for understanding of chemical structure and reactivity

There is a reason for everything.

Does Bent's rule have any utility?

YES! It wouldn't be there if there was not. But I will get back to this at the end of this post.
Let me go through the points raised by your teacher first:

Coulombic considerations can be used to rationalize bond angles, strengths, and lengths without the use of Bent's rule.

This might be true to some extent, but that only means that coulombic considerations give the same result as a complete orbital analysis. Boltzmann, Planck, Einstein, Schrödinger, Pauling, etc. worked the field for a reason - classical mechanics cannot be applied to molecular systems. Any theory that simplifies it beyond quantum mechanics, is a model to understand or comprehend - not a model to predict.

Orbitals are not real so why invoke orbitals when simpler Coulombic considerations are available? (Occam's razor)

Occam's razor would suggest picking molecular orbital or valence bond theory, as it states that from competing hypotheses the one with the fewest assumptions should be selected. Coulombic consideration is purely based on assumptions, it is a rather crude approximation to the correct picture.
The statement, that orbitals are not real is clearly wrong. There are many books that explain how one can comprehend this. Teaching that orbitals are not real is, with all due respect, just plain wrong.

The Coulombic effect is also real as opposed to orbitals.

The coulombic effect is not more real than orbitals. Orbitals should be used to explain the coulombic effect.

But let's talk about Bent's rule. The IUPAC goldbook states:

In a molecule, smaller bond angles are formed between electronegative ligands since the central atom, to which the ligands are attached, tends to direct bonding hybrid orbitals of greater p character towards its more electronegative substituents.

It is important to know, that Bent's rule is, like Lewis rule of two (or eight, known as octet rule), a rule (not a law) based on observation. Hence its character is more or less one of a guideline. The key paper was published in Henry A. Bent, Chem. Rev., 1961, 61 (3), 275–311.
Bent's rule originates from valence bond theory (VB). Early forms of VB theory proposed that a molecule can entirely be described by a set (linear combination) of resonance Lewis structures, i.e. every bonding orbital is a directed localised orbital, which will be occupied by two electrons and the same applies to lone pairs. The advantage of this approach is that it is easily comprehensible. The disadvantage is, that it is quite inflexible. The reason for this was the use of either pure $\ce{s}$ and $\ce{p}$ functions or fixed hybrid orbitals $\ce{sp^3,~sp^2,~sp}$, which only allowed a subset of angles. A direct quote from Bent's original paper states this as follows:

For atoms that satisfy the octet rule, and it is with such atoms (e.g., carbon, nitrogen, oxygen, fluorine) that this review is primarily concerned, the four valence-shell orbitals of an atom may be thought of as compounded from one spherically symmetrical $\ce{2s}$ atomic orbital and three mutually perpendicular dumbbell-shaped $\ce{2p}$ orbitals, designated $\ce{2p_$x$, 2p_$y$, 2p_$z$}$, ($x, y, z$ indicate mutually orthogonal axes). The $\ce{2s}$ orbital is lower in energy than the $\ce{2p}$ orbital; hence the ground state of a carbon atom, for example, is written $\ce{ls^2~2s^2~2p^2}$ (not $\ce{ls^2, 2s~2p^3}$. However, in a molecule such as methane, which is known to have the shape of a regular tetrahedron, the $\ce{2s}$ orbital of carbon may be considered to mix (or “hybridize”) with the three $\ce{2p}$ orbitals to form four equivalent orbitals that point to the corners of a regular tetrahedron and make angles with each other of $109^\circ28’$. These are called hybrid orbitals. Each contains one-quarter $\ce{s}$ character and three-quarters $\ce{p}$ character, abbreviated $\ce{s^{$1/4$}p^{$3/4$}}$ or $\ce{sp^3}$, or $\ce{te}$ (for tetrahedral) (223). In ethylene, on the other hand, the $\ce{2s}$ orbital of carbon may be considered to mix with but two of the three $\ce{2p}$ orbitals to form three equivalent hybrid orbitals that lie in a plane and make angles with each other of $120^\circ$; these three hybrids, which contain 33 per cent $\ce{s}$ character, are abbreviated $\ce{s^{$l/2$}p^{$2/3$}}$ or $\ce{sp^2}$, or $\ce{tr}$ (for trigonal); perpendicular to the plane of these three orbitals is the axis of cylindrical symmetry of the remaining pure $\ce{2p}$ orbital. The geometry of acetylene is often explained by assuming that the carbon $\ce{2s}$ orbital mixes with but one of the three $\ce{2p}$ orbitals, forming two $\ce{di}$ (for digonal) hybrids that point in diametrically opposed directions; these are abbreviated $\ce{s^{$1/2$}p^{$1/2$}}$ or $\ce{sp}$.

The original approach allowed already good results. However, Bent already draw the conclusion, that hybridisation is connected to molecular geometry:

Molecular geometry provides then a direct clue to interorbital angles and
  (table 1) the distribution of $\ce{s}$ character. For example, from the bond angles in ammonia ($106^\circ46'$) (284) and water ($104^\circ27'$) (134, 284), it is inferred that the nitrogen atom of $\ce{NH}$, devotes slightly more $\ce{s}$ character to its bonding orbitals than does the oxygen atom of $\ce{H2O}$.

In conclusion to this (unfortenuately I ran out of time, so this may be extended soon) he suggested to use fractional linear combinations of hybrid orbitals to add more flexibility to the VB approach. Therefore Bent's rule is an extension to VB theory, that refers to the linear combination of atomic orbitals, that eventually form hybrid orbitals.
The formulated rule was later merely a result of the approach to flexibilize this theory. It is therefore a very important part of theoretical chemistry.
Of course there are now many more modifications to VB theory, and it is nowadays considered complementary to molecular orbital theory, but this would certainly exceed the scope of this question.

Very nice answer. I agree that the phrase (and any variation thereof) "orbitals are not real" is very unfortunate. I presume what's meant by that is that they're not observable (as opposed to electron density), being just mathematical constructs, and I think that would've been a much better way of stating the case.

@GregE. It depends on what you mean by "not observable". With photoelectron spectroscopy you can measure the energies of the orbitals.

@martin are there any examples in which bents rule works while other guidelines fail?

@Philipp, I think it might just be an issue of semantics. What I mean to say is that "orbitals" in the most formal sense are not distinct physical entities but a mathematical description, with the "observable" that can be experimentally probed being the electron density itself.

@Philipp, however, your point about PES is well taken. If I understand correctly, Koopmans' theorem relates the first ionization energy to the HOMO and the electron affinity to the LUMO, and in those instances the entire molecule is being probed indirectly. In that sense, I recognize that limiting the observables to the electron density alone is perhaps narrow. (I'm certainly no expert here, so please do correct me if I'm in error, and thank you for the thought-provoking comment.)

@Dissenter Yes, many, almost all. For example, coulombic considerations in the series $\ce{H_{$4-n$}CF_{$n$}}$ would predict longer $\ce{C-F}$ bond lengths with increasing $n$ (repulsion of the negatively charged fluorine atoms), where the opposite is the case. In all other cases you can expect that Bent's rule is correct for the correct reasons, while other, simplified models are heavily relying on error compensation, e.g. VSEPR, therefore being correct accidentally.

@Martin, "From the bond angles in ammonia and water, it is inferred that nitrogen atom of NH devotes slightly more s character to its BOs than does the oxygen atom of H2O" How did you conclude this? Can you provide a source that explains it?

@B.Anshuman That is a direct quote from Bent's paper, which I have cited earlier.


text:
What's the order of oxidising strength of the oxyacids of Chlorine?
What I have been told is that the acidity of the four oxyacids of Chlorine increase in the order $\ce{HOCl} < \ce{HClO2} < \ce{HClO3} < \ce{HClO4}$.
I have also been told that the oxidising strength goes in the reverse order. In my knowledge acidic strength should increase with oxidising strength. Where am I going wrong?
For your reference, the Ka values of the acids are as follows:

$\ce{HOCl} = 2.9 \times 10^{-8}$
$\ce{HClO2} = 1.1 \times 10^{-2}$
$\ce{HClO3} = 5.0 \times 10^2$
$\ce{HClO4} = 1 \times 10^3$

Don't you mean the oxyacids $\ce{HClO}$, $\ce{HClO2}$, $\ce{HClO3}$, and $\ce{HClO4}$?

For many acids, the proton $\ce{H+}$ is the oxidizing agent. For example, in the oxidation of zinc with $\ce{HCl}$, the oxidizing agent is $\ce{H+}$.
$$\ce{Zn + 2H+ -> Zn^{2+} + H2}$$
However, for some acids, the conjugate base (anion) is also an oxidant (and usually a more powerful oxidant than $\ce{H+}$. The oxyacids of chlorine are all oxidizing agents through their anions, as is nitric acid and some others:
$$\ce{Zn + 2H+ + 2OCl- -> Zn^{2+} + H2O + 2Cl-}$$
In the above example, hypochlorite is the oxidizing agent. For the oxyacids of chlorine and others, the strength of the oxidizing agent is independent of the strength of the acid.

Why exactly would the proton often times be a weak oxidizer and why exactly would the conjugate bases sometimes be good oxidizers? Isn't the formation of H2 entropically favorable?

Formation of $\ce{H2}$ is entropically favorable, but so is conversion of $\ce{2HOCl}$ to $\ce{H2O + 2Cl-}$ (three particles from 2). Additionally, you cannot ignore the contribution to spontaneity from enthalpy.

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


text:
When to regard water as a strong base
Question:

Classify each of the following ions according to whether they react with water to give a neutral, acidic, or basic
solution:

As a general rule:

Strong acid + Strong base $\rightarrow$ Neutral solution
Strong acid + weak base $\rightarrow$ acidic solution
Weak acid + Strong base $\rightarrow$ Basic solution

According to my solutions manual.

a) $\ce{F-}$ is the conjugate base of a weak acid, therefore  the solution is basic.
b) $\ce{Br-}$ is the anion of strong acid, therefore the solution is neutral.
c) $\ce{NH4+}$ is the conjugate acid of a weak base, therefore the solution is acidic
d) $\ce{[K(H2O)6]+}$ is the conjugate base of a weak acid, therefore the soltuion is basic.
e) $\ce{SO3^2-}$ is the conjugate base of a weak acid, therefore the solution is basic
f) $\ce{[Cr(H2O)6]^3+}$ is an acidic solution, therefore the solution is acidic.

However, all these answers imply that water is acting as a strong base. How is this possible when water pH is $\sim 7$?

You have to take into consideration the pKa and pKb, not the pH, when making judgments about the acidity or basicity, respectively, of a molecule. In the examples you pulled out, for water to act as a base, it must have a higher propensity to grab a proton or donate electrons than the competing molecule. Thus, it must have a higher pKa or lower pKb than the other molecule to act as a base.

Question:

Classify each of the following ions according to whether they react with water to give a neutral, acidic, or basic
solution:
As a general rule:

Strong acid + Strong base $\rightarrow$ Neutral solution

Strong acid + weak base $\rightarrow$ acidic solution

Weak acid + Strong base $\rightarrow$ Basic solution

That's not true. The only time in which these rules are true is when you have an equimolar concentration of the acid and base in question, and the acids and bases must both be monoprotic and monobasic, or diprotic and dibasic, etc. Otherwise there are too many variables to consider for these guidelines to always ring true. Consider mixing 1.0 molar $\ce{H2SO4}$ with 0.10 molar $\ce{NaOH}$ ... I guarantee you that the pH of that solution is not "neutral."

d) $\ce{K(H2O)6+}$ is the conjugae base of a weak acid, therefore the soltuion is basic.

I didn't know that potassium hydrated to a significant extent in water; even so, what the solution manual has written down is not the conjugate base of hydrated potassium cation - the weak acid.

However, all these answers imply that water is acting as a strong
base. How is this possible when water pH is $\sim 7$?

How do these answers imply that water is acting as a strong base?
Also regarding how water can be a strong base in general - it just depends on your solvent system.

One more thing that you have to keep in mind is that water acts as a strong base ONLY with weak acids.
PECULIAR nature of water is that with weak bases, water acts as an acid.
So, it can be said that it is amphoteric in this regard.

Water is never a strong base, water is always a weak base. The strengths of acids and bases are intrinsic properties of the corresponding molecules/atoms, only dependent on the solution to a lesser extent.


text:
Length of copper wire in 1 lb spool
I was asked the following question in my introductory chemistry class:

The diameter of metal wire is often referred to by its American wire gauge number. A 16-gauge wire has a diameter of $\pu{0.05082 in}$. What length of wire, in meters, is there in a $\pu{1.00 lb}$ spool of 16-gauge copper wire? The density of copper is $\pu{8.92 g/cm^3}$.

First I converted the imperial measurements to metric. $\pu{0.05082 in} = \pu{0.12908 cm}$, $\pu{1 lb} = \pu{0.45359 kg}$, according to Google. So I can use $\pi r^2 h$ to calculate the volume (and thus, using density, calculate the mass) of one meter of wire, I figure the radius to be $\pu{0.06454 cm}$. The equation becomes $V = \pi \times (\pu{0.06454 cm})^2 \cdot \pu{100 cm}$, which equals $\pu{0.41654 cm^3}$. So that's the volume of one meter of wire.
Using the density of copper $(\pu{8.92 g/cm^3})$, I figure the mass of one meter of wire to be $\pu{3.72 g}$. If $\pu{1  lb} = \pu{453.59 g}$, then converting $\pu{3.72 g}$ to lbs gives me $\pu{3.72 g} \cdot \frac{\pu{1  lb}}{\pu{453.59 g}} = \pu{0.01 lbs}$. So assuming I did everything correct, one meter of wire is a hundredth of a pound, so in one pound of wire there is $100$ meters.

Find a function of length depending on mass
Conversion factors:

$2.54 ~\mathrm{\frac{cm}{in}}$
$454 ~\mathrm{\frac{g}{lb}}$

Constants:

Wire diameter $d = 0.129~\mathrm{cm}$
Density of copper $\rho = 8.92~\mathrm{g\, cm^{-3}}$

Thus, our function $l(m)$ for the wire length from the wire mass in pounds reads:
$$ l(m) = m \cdot 454~\mathrm{\frac{g}{lb}} \cdot \frac{1}{\pi d^2 \rho}$$
Evaluate the function
Since I'm too lazy to evaluate something like this by hand (and since I somehow had to record this thought process and did not want to waste the processed remnants of a dead tree) there is a small python script available to do this for you, which you can find here and which, if you run it using the command python3 wires.py, will give you the following output:

Wire length of 1.00 lb of 16-gauge wire: 9.71 m

(Bonus: You can run the command as python3 -i wires.py and will then be dropped into an interactive python shell, where you can call the wirelength function as many times as you wish using different masses for the wire.)


text:
Moist Chlorine & Hydrogen Psychrometric or Mollier Chart
In our electrolysis of brine, $\ce{Cl2}$ and $\ce{H2}$ are produced as gaseous products along with aqueous $\ce{NaOH}$. In this scenario I require what amount of maximum moisture can $\ce{Cl2}$ and $\ce{H2}$ hold up at specific temperature & pressure (precisely 85 celsius & 1.2bar absolute). Two coolers (shell/tube exchanger with chilled water on shell side) are planned to dehumidify these two gases.
I have searched hard but without proper psychrometric data am unable to even estimate preliminary values. Link to data or other alternatives are highly welcome.
regards,

I have searched hard but without proper psychrometric data am unable to even estimate preliminary values.

You have to google for 'vapor pressure' of water and ideal gas model. In cases ideal gas model is close enough to true, the total pressure is divided into partial pressures, attributed to different gases, and the partial pressures are proportional to amount of molecules in the gas mixture. So, in the case the equilibrium partial pressure of water vapors above the liquid can be found in any table of vapor pressures, the rest of the pressure would be other gases. In the case of vapor above saline water, the equilibrium pressure is reduced by the fact that the salt is present. The exact reduction is guarded by Raoult's Law and is linearly proportional to the concentration of the salt.

Two coolers (shell/tube exchanger with chilled water on shell side) are planned to dehumidify these two gases.

To dehumidify these gases effectively, if this is really required, consider bubbling them through layer of concentrated $\ce{H2SO4}$, it is a quite common and easy way to dehumidify gases.

Thanks for replying but I don't know the amount of water vapour being carried by say per kg Cl2 at given conditions. If this is known your method of obtaining water's vapour pressure & hence Cl2 partial pressure is valid, but thats why I need psychrometric data. What we do know is all about Cl2 its temperature & quantity & total absolute pressure (Cl2 + H2O)

@nightcrawler water vapor pressures: http://en.wikipedia.org/wiki/Vapour_pressure_of_water | $pV=\nu RT$ allows to move from the partial pressure to moles of water per liter, and the rest of absolute pressure is the partial pressure chlorine, from which you can get moles of chlorine per liter. From here you can get equilibrium percentage of water vapors in chlorine at any given temperature and total pressure. Seriously, read about partial pressures and ideal gas model, they appear in chemistry regularly.


text:
Is this the correct mechanism of the formation of the nitronium (NO2+) ion from sodium nitrate and sulfuric acid?
I am currently working on a mechanism for forming paracetamol (or N-(4-hydroxyphenyl)-ethanamide). The typical reaction to nitrate the phenol (starting product) is $\ce{NaNo3}$ with a catalyst of sulfuric acid. The sulfuric acid is required as a catalyst to form the nitronium cation - an electrophile - to attack the aromatic phenol. Unlike the nitration of toluene (for TNT) or benzene, it is possible to use $\ce{NaNO3}$ because the oxygen in the hydroxy group is activating.
My proposed mechanism for the formation of the nitronium is this:
$$\ce{NaNO3 + H2SO4 + 2H2O <=> Na+ +NO3- + SO4- + 2H3O+}$$
I think this is a likely equilibrium between the starting ions. Sulfuric acid is a strong acid, and therefore to use a mechanism with it undissociated I think unlikely - the reaction would be slower, due to lower concentrations of the undissociated sulfuric acid. I know that $\ce{NaNO3}$ is fairly soluble, so would dissociate into ions.
I suggest attack of the $\ce{NO3-}$ by $\ce{H3O+}$, protonating one of the oxygens:
$$\ce{NO3- + H3O+ -> HNO3 + H2O}$$
$\ce{HNO3}$ has a Ka of ${24 * 10}$ - 24 moles of $\ce{HNO3-}$ present for every $\ce{NO3}$ we can attribute this to the stability of the $\ce{NO3-}$ ion, but it also is fairly low compared to the Ka of sulfuric acid - ${1*10^3}$. By adding sulfuric acid, a strong acid, it increases the concentration of $\ce{H3O+}$, therefore shifting the equilibrium to the right. This favors the reaction.
In the $\ce{HNO3}$ molecule the oxygen-hydrogen bonds are all different lengths - $\ce{N-O}$ is 119.9pm, or 121.1pm, and the $\ce{N-O-H}$ bond is 140.6pm. The $\ce{N-O-H}$ bond is clearly weaker than the other two bonds. This is attributed to the resonance effect - a valence electron from each oxygen interacts with a lone pair on nitrogen favorably, shortening and strengthening the bonds. Protonating the oxygen in $\ce{NO3-}$ thus weakens the bond. Protonating the oxygen a second time:
$$\ce{HNO3 + H3O+ <=> NO2+ + H2O + H2O}$$
Would probably break that bond.
The hydronium protonates the oxygen already protonated - resulting in an unstable intermediate. The positive charge pulls electron density from the $\ce{N-O}$ sigma bond onto the oxygen (inductive effect). This breaks the bond, and stabilizes the positive charge on the oxygen, and it becomes $\ce{H2O}$
However, the hydronium may protonate one of the other two oxygens. Why is the one already protonated more basic? The other two oxygens have a higher electron density, so would be more nucleophilic - better for the electrophilic $\ce{H3O+}$ to attack. A fraction would surely react in the way I have described, but I would think a much higher proportion would react on other oxygens.
Is this mechanism likely/correct? Could you explain how the second protonation occurs?

The electrons on the two other oxygens are delocalized making them much less basic.

Even if the charge is delocalized, those oxygens should be more basic. For comparison, acetate anion is around a million times more basic than water.

As you described, the reaction depends on formation of the nitronium intermediate. I think you are right that the (partially) negatively charged oxygens are more basic. However, nitronium formation is much much faster by using water as a leaving group rather than hydroxide, which is what would be required otherwise.

So you're saying that while the proportion forming on the already protonated oxygen is lower, but it still proceeds to a significant enough extent that nitronium ion is formed?

Yes, that's right.


text:
Why are DCM and chloroform so resistant towards nucleophilic substitution?
In the book Organic Chemistry by  J. Clayden, N. Greeves, S. Warren, and P. Wothers I found the following reasoning:

You may have wondered why it is that, while methyl chloride (chloromethane) is a reactive electrophile that takes part readily in substitution reactions, dichloromethane (DCM) is so unreactive that it can be used as a solvent in which substitution reactions of,other alkyl halides take place. You may think that this is a steric effect: Indeed, $\ce{Cl}$ is bigger than $\ce{H}$.
  But $\ce{CH2Cl2}$ is much less reactive as an electrophile than ethyl chloride or propyl chloride: there must be more to its unreactivity. And there is: Dichloromethane benefits from a sort of 'permanent anomeric effect.' One lone pair of each chlorine is always anti-periplanar to the other $\ce{C–Cl}$ bond so
  that there is always stabilization from this effect.

So, in MO-terms the situation would look something like this.

The reasoning looks plausible to me. The interaction between the free electron pair on $\ce{Cl}$ and the $\sigma^{*}$ orbital of the neighboring $\ce{C-Cl}$ bond, which would be the LUMO of DCM, lowers the energy of the free-electron-pair-orbital, thus stabilizing the compound and it raises the energy of the LUMO, thus making DCM less reactive towards nucleophiles.
But how important is this anomeric effect actually for explaining the unreactiveness of DCM toward nucleophiles, especially compared to the steric effect? And how important is the steric effect actually: Is the steric hindrance exerted by the second $\ce{Cl}$ in $\ce{CH2Cl2}$ really that much larger than the steric hindrance exerted by the methyl group in $\ce{CH3CH2Cl}$ (which is a good electrophile for $\mathrm{S_N2}$ reactions)?
I'm a little skeptical because if this anomeric effect was very important I would have expected that the $\ce{C-Cl}$ bond length in DCM would be slightly higher than in methyl chloride because there should be some transfer of electron density from the free electron pair into the antibonding $\sigma^{*}$ orbital, which should weaken the $\ce{C-Cl}$ bond. But the actual bond lengths don't show this. They show rather the contrary:
$$\begin{array}{c|c}
\hline
\text{Species} & \text{Average }\ce{C-Cl}\text{ bond length / Å} \\
\hline
\ce{CH3Cl} & 1.783 \\
\ce{CH2Cl2} & 1.772 \\
\ce{CHCl3} & 1.767 \\
\ce{CCl4} & 1.766 \\
\hline
\end{array}
$$
(source: Wikipedia)
Now, I know that the stronger polarization of the $\ce{C}$ atomic orbitals in di-, tri-, and tetrachlorinated methane as compared to methyl chloride (due to the electronegativity of $\ce{Cl}$) should lead to an overall strengthening of the $\ce{C-Cl}$ bonds in those compounds. But I would have expected a trend that would show only a slight decrease (or maybe even an increase) of bond length when going from methyl chloride to DCM and then a more pronounced decrease when going from DCM to chloroform followed by a similar decrease when going from chloroform to tetrachloromethane. But instead the polarization effect seems to only slightly increase on adding more chlorine atoms.

Can you tell which chapter if it's not a problem?

@Marko It's in the first edition, chapter 42, page 1133.

Never mind if one of the starting orbitals was an anti-bonding one. Then electrons aren't put in this anti-bonding orbital, but in the newly formed orbital which has the lowest energy in that system. The new anti-bonding orbital (new LUMO) is very high on the energy scale, but without any electrons in it, no destabilization in the system occurs.

@Marko But the in-phase combination of the $\sigma^{}(\ce{C-Cl})$ orbital and the free-electron-pair-orbital will have to some extent a $\ce{C-Cl}$-antibonding character. The amount of $\sigma^{}(\ce{C-Cl})$-character in the newly formed orbital will of course depend on the energetic difference between the interacting orbitals and their orbital overlap. But if the interaction is strong enough to raise the LUMO enough to make DCM unreactive then I would definitely expect some bond-lengthening effect as well.

@Marko, I don't understand what your objection is. I think the claim is that in-phase combination of $\sigma^{*}\ce{(C-Cl)}$ with $n\ce{(e^{-})}$ is energy-lowering in the reactant ground state. The destabilization would occur in the transition state as the electrons of the nucleophile interact with the LUMO (which has been made more energetic by inclusion of the orbital combination described above).

As the in-phase combination is energy lowering in the reactant ground state, the bond should be shorter. That was my objection.

@Marko The in-phase combination will have a bonding overlap between $\ce{C}$ and the $\ce{Cl}$ atom with the free electron pair but an antibonding overlap between the atoms giving rise to the $\sigma^{*}(\ce{C-Cl})$ orbital. Energy-lowering need not necessarily mean bond-shortening.

@Marko I think I have found an answer to the problem of the mismatching bond lengths. This might also clarify the matter we discussed in the comments here.

Introduction (and Abstract, TLDR)
In very short words you can say, that the anomeric effect is responsible for the lack of reactiveness. The electronic effect may very well be compensating for the the steric effect, that could come from the methyl moiety. In any way, most of the steric effects can often been seen as electronic effects in disguise.
Analysis of Molecular Orbitals
I will analyse the bonding picture based on calculations at the density fitted density functional level of theory, with a fairly large basis set: DF-BP86/def2-TZVPP. As model compounds I have chosen chloromethane, dichloromethane, chloroform and chloroethane.
First of all let me state, that the bond lengths are a little larger at this level, however, the general trend for shortening can also be observed. In this sense, chloroethane behaves like chloromethane. An attempt to explain this will be given at the end of this article.
\begin{array}{lr}\hline
\text{Compound} & \mathbf{d}(\ce{C-Cl})\\\hline
\ce{ClCH3}      &  1.797\\
\ce{Cl2CH2}     &  1.786\\
\ce{Cl3CH}      &  1.783\\\hline
\ce{ClCH2CH3}   &  1.797\\\hline
\end{array}
In the canonical bonding picture, it is fairly obvious, that the electronic effects dominate and are responsible for the lack of reactivity. In other words, the lowest unoccupied molecular orbital is very well delocalised in the dichloromethane and chloroform case. This is effectively leaving no angle to attack the antibonding orbitals.

In the  mono substituted cases there is a large coefficient at the carbon, where a nucleophile can readily attack.
One can also analyse the bonding situation in terms of localised orbitals. Here I make use of a Natural Bond Orbital (NBO) analysis, that transforms the canonical orbitals into hybrid orbitals, which all have an occupation of about two electrons. Due to the nature of the approach, it is no longer possible to speak of HOMO or LUMO, when analysing the orbitals. Due to the nature of the calculations, i.e. there are polarisation functions, the values do not necessarily add up to 100%. The deviation is so small, that it can be omitted.
The following table shows the composition (in $\%$)of the carbon chloro bond and anti bond.
\begin{array}{lrr}\hline
\text{Compound} &\sigma-\ce{C-Cl} & \sigma^*-\ce{C-Cl}\\\hline
\ce{ClCH3}      &  45\ce{C}(21s79p) 55\ce{Cl}(14s85p)
                &  55\ce{C}(21s79p) 45\ce{Cl}(14s85p)\\
\ce{Cl2CH2}     &  46\ce{C}(22s77p) 54\ce{Cl}(14s85p)
                &  54\ce{C}(22s77p) 46\ce{Cl}(14s85p)\\
\ce{Cl3CH}      &  48\ce{C}(24s76p) 52\ce{Cl}(14s86p)
                &  52\ce{C}(24s76p) 48\ce{Cl}(14s86p)\\\hline
\ce{ClCH2CH3}   &  44\ce{C}(19s81p) 56\ce{Cl}(14s85p)
                &  56\ce{C}(19s81p) 44\ce{Cl}(14s85p)\\\hline
\end{array}
As we go from mono- to di- to trisubstituted methane, the carbon contribution increases slightly, along with the percentage of $s$ character. More $s$ character usually means also a stronger bond, which often results in a shorter bond distance. Of course, delocalization will have a similar effect on its own.
The reason, why dichloromethane and chloroform are fairly unreactive versus nucleophiles, has already been pointed out in terms of localised bonding. But we can have a look at these orbitals as well.
In the case of chloromethane, the LUMO has more or less the same scope of the canonical orbital, with the highest contribution from the carbon. If we compare this antibonding orbital to an analogous orbital in dichloromethane or chloroform, we can expect the same form. We soon run into trouble, because of the localised $p$ lone pairs of chlorine. Not necessarily overlapping, but certainly in the way of the "backside" of the bonding orbital. In the case of chloroethane we can observe hyperconjugation. However, this effect is probably less strong, and from the canonical bonding picture we could also assume, that this increases the polarisation of the antibonding orbital in favour of carbon.
In the following pictures, occupied orbitals are coloured red and yellow, while virtual orbitals are coloured purple and orange.

(Note that in chloroform two lone pair orbitals are shown.)
Conclusion
Even though this article does not use the Valence Bond Approach, one can clearly see the qualitative manifestation of Bent's Rule (compare also: Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?). A higher $s$ character means a shorter bond. The lack of reactivity towards nucleophiles can be explained electronically with a delocalised LUMO. In terms of localised bonding, the lone pairs of any additional chlorine atom would provide sufficient electron density, to shield the backside attack on the carbon.

Awesome answer, thank you very much. But since I'm not very familiar with NBO could you maybe explain why you say that the localised $\mathrm{p}$ lone pairs of chlorine are "not necessarily overlapping" with the (anti-)bonding orbital of the $\ce{C-Cl}$ bond? Because looking at the form of the orbitals it look to me that they could overlap quite well, but then again, as I said, I don't know much about NBO. Also would you share your opinion on my own answer which is mostly concerned with solving the apparent contradiction concerning the the bond lengths? Do you think the reasoning is sound?

NBO transforms canonical orbitals into localised orbitals, i.e. it uses linear combinations to form orbitals that are concentrated to a certain spatial area. Symmetry constraints are lifted and these orbitals are no eigenfunctions of the Schrödinger equation. The canonical orbitals are already the optimal solution. With "not necessarily overlapping" I mean, that in the ground state you have two entirely separate orbitals. They do not form new orbitals. They could interact with each other when an external potential would necessitate this.

But the main question remains: Is the anomeric effect the main cause
  for the unreactiveness of DCM towards nucleophiles or is it the steric
  hinderance exerted by the second Cl ?

Assessment of Steric Effects
Bonds, like humans, have a length and a girth.  In some reactions, the reaction is most likely to occur when the attacking group strikes an atom or substituent "head on".  In other reactions, like the $\ce{S_{N}2}$, the attacking group approaches from the back side and slips by the various substituents in a side-ways manner as it approaches the carbon under attack.
There is a nice way to measure this side-ways (girth of a bond, if you will) interference in organic systems.  Axial substituents in a cyclohexane ring are subject to what is termed "A-strain" produced by the interactions depicted in the figure below.
These interactions are very analogous to the gauche interactions

in butane and arise from side-ways interaction of the substituent and the two diaxial hydrogens located two carbons away.  Equatorial substituents avoid these repulsive interactions.  The smaller the substituent, the less repulsive the 1,3 interaction will be and, at equilibrium, more of the substituent should exist in the axial position.  Conversely, the larger the substituent, more of it should reside in the equatorial position.  By measuring the axial/equatorial ratio for various substituents we can meaningfully assess the side-ways steric bulk of the substituent.
For chlorocyclohexane the axial/equatorial ratio is ca. 31/69 ($\ce{\Delta G = 0$.$48 kcal/mol}$) at room temperature.  For methylcyclohexane the axial/equatorial ratio is ca. 5/95 ($\ce{\Delta G = 0$.$1.7 kcal/mol}$) at room temperature.  This analysis clearly suggests that the chloro substituent is "smaller" than a methyl substituent when we use this side-ways steric probe.
For the interested reader, here is a Table of axial/equatorial $\ce{\Delta G's}$ (in kcal/mol) for other substituents.

Since this analysis suggests that a chlorine substituent is smaller than a methyl substituent, in terms of the side-ways interaction expected in a back side $\ce{S_{N}2}$ attack, it would be difficult to use a steric argument to explain the reduced reactivity in the polyhalomethanes towards  $\ce{S_{N}2}$ reaction.  This leaves the anomeric effect as a reasonable explanation for the reduced reactivity of the polyhalomethanes. But...
Other Thoughts

How unreactive are these compounds to $\ce{S_{N}2}$ reaction?  Dichloromethane, chloroform and carbon tetrachloride will all undergo the $\ce{S_{N}2}$ reaction with various nucleophiles, under various conditions.  Before I sign up for the anomeric effect explanation, it would be nice to know exactly what reaction Clayden, et. al. were discussing.  Perhaps some other factor (hydrogen bonding, who knows what) suppressed the nucleophile's reactivity in Clayden's series of compounds.
Bond strengths:  $\ce{C-Cl}$ bond strengths decrease from chloromethane (80 kcal/mol) to carbon tetrachloride (70 kcal/mol).  This would cause us to suspect the chloromethane might react the slowest and carbon tetrachloride the fastest.  That this is not the case is another argument in support of the anomeric effect playing a key role.

Thanks for bringing the A-strain into the discussion and explaining it so nicely. As for Clayden et al.: You can find the passage on p. 1133 of the book (first ed.) if you want to read it. They discuss this anomeric effect in the context of heterocycles and bring in some unusual examples such as that of DCM. The statement about the unreactiveness of DCM is made without reference to specific reactions. But they also write in a small box: "Dichloromethane will react as an
electrophile, but it needs a very
powerful nucleophile and long reaction times" and give the following example reaction...

... $\ce{PhSNa ->[\substack{\ce{CH2Cl2}\ \text{as solvent}} ][\text{several days}] (PhS)2CH2}$

I really dislike the concept of strain. If you put a hydrogen and a chlorine atom in close proximity then this would be primarily an attractive interaction (dipole-dipole and electrostatics and orbitals). The same applies to other substituents. Only if you put them too close the nuclei start to repel each other (but this only starts when it is significantly shorter than the covalent radii). I would expect that hyperconjugation has a much larger effect in stabilising the equatorial position, than stabilising the axial position of the chlorine.

On a completely unrelated matter: Why do you put pictures in the middle of a sentence? I always have the feeling there is missing something...

@Martin As far as I know the origin of the strain is the interaction of filled orbitals on the two neighboring atoms. Only the axial bonds are in the right position to let the bond orbitals overlap appreciably. So, in an axial position of a cyclohexylchloride ring you have the $\sigma(\ce{C-Cl})$ orbital interacting with a neighboring axial $\sigma(\ce{C-H})$ orbital. As both orbitals are filled you get a 2-center-4-electron interaction which will be destabilizing (because the out-of-phase combination will more destabilized than the in-phase combination is stabilized) as it raises the energy.

@Martin Why would you think that "hyperconjugation has a much larger effect in stabilising the equatorial position, than stabilising the axial position of the chlorine"? You could have (nearly) the same degree of hyperconjugation between axial chlorine and the neighboring axial $\ce{C-H}$ bonds that point in the opposite direction as you would get from the hyperconjugation of an equatorial chlorine and the neighboring $\ce{C-C}$ bonds parallel to the $\ce{C-Cl}$ bond within the ring, right?

@Philipp Yes. I think that the interaction of $\sigma^(\ce{C-Cl})$ with $\sigma(\ce{C-C})$ (and $\sigma(\ce{C-Cl})$ with $\sigma^(\ce{C-C})$) is a little bit more stabilising than $\sigma^*(\ce{C-Cl})$ with $\sigma(\ce{C-H})$ (and vice versa). We are talking about a difference of $0.5~\mathrm{kcal/mol}$ here, which is not a lot.

@Martin But wouldn't you think that the 2-center-4-electron interactions (or rather 3-center-6-electrons) account for more destabilization in the axial positions than this stabilization of $0.5 , \mathrm{kcal/mol}$ in the equatorial position and would thus be the more important factor for explaining the preference for an equatorial position? I have actually no idea how big the destabilization of 2-center-4-electron interactions usually is, but maybe you can tell from experience.

@Philipp I do not like the concept of destabilisation, but this might just be a way of phrasing it. Chlorocyclohexane is stable. The conformer where the chlorine is in equatorial position is about half a kcal more stable than the conformer with axial chlorine. So the stabilising effect from the carbon carbon bond is a little higher than from a carbon hydrogen bond. (I was prompted to move to chat, I enjoy this conversation, so I am tempted to accept, but I leave the choice to you.)

Let us continue this discussion in chat.

Thinking about it some more made me realize how the problem with the bond lenghts might be resolved. Let's look at the orbital picture again:

I still think, that the in-phase combination of the $\sigma^{∗}(\ce{C-Cl})$ orbital and the free-electron-pair-orbital $n(\ce{e-})$ will have to some extent - depending on the energetic difference between the interacting orbitals and their orbital overlap - an $\ce{C-Cl}$-antibonding character, thus weakening the blue $\ce{C-Cl}$ bond (a simplified view would be, that on interaction electron density flows from the free electron pair into antibonding $\sigma^{∗}(\ce{C-Cl})$ orbital and more electron density in an antibonding orbital mean a weakening of the corresponding bond). But my mistake was not to also consider the $\pi$-bonding interaction between the red $\ce{Cl}$ and $\ce{C}$ in the in-phase combination. This $\pi$-bonding interaction will of course strengthen the red $\ce{C-Cl}$ bond. So resulting from this one interaction pictured above there should be a bond-shortening of the red $\ce{C-Cl}$ bond and a bond-lengthening of the blue $\ce{C-Cl}$ bond.
But the anomeric effect works not only in one direction in DCM (or Chloroform). The blue $\ce{Cl}$ atom has free electron pairs, too. And there is also a $\sigma^{∗}(\ce{C-Cl})$ orbital for the red bond that can interact with one of those free electron pairs in exactly the same way described before. Now, this interaction will strengthen the blue $\ce{C-Cl}$ bond while weakening the red $\ce{C-Cl}$ bond. So, it counteracts the effects of the interaction presented above and the bond-shortening and bond-lengthening effects should cancel each other out to some extent. But the raising of the LUMO, which makes DCM so unreactive remains unchanged.
Now, the only thing that needs some thinking is: Do the $\pi$-bonding and the $\sigma$-antibonding in the in-phase combination have a comparable magnitude? This question I can't answer in any quantitative way. So I will only list some factors of influence here. The $\ce{C-Cl}$ bond is rather long. This means the $\pi$-overlap won't be optimal. And the  $n(\ce{e-})$ orbital will lie somewhat lower in energy than the $\sigma^{∗}(\ce{C-Cl})$ orbital thus weakening the covalency of the $\pi$-interaction. On the other hand, the same factors will also lower the amount of $\sigma^{∗}(\ce{C-Cl})$-character contained in the in-phase combination MO. All this leaves me with the gut feeling that the $\pi$-bonding might indeed be comparable to or even a little stronger than the $\sigma$-antibonding effect. And this would mean that the observed bond lengths are in accord with the orbital situation and the anomeric effect might very well be quite important for explaining the unreactiveness of DCM towards nucleophiles.
But the main question remains: Is the anomeric effect the main cause for the unreactiveness of DCM towards nucleophiles or is it the steric hinderance exerted by the second $\ce{Cl}$?
Maybe one could go about it by employing the van der Waals radii of a chlorine atom and a methyl group. According to Wikipedia, the radius of a chlorine atom is 175 pm while the radius of a carbon atom (without any hydrogen atoms around) is 170 pm (so it should be safe to say that the radius of a methyl group is higher than that). Here is a picture of the van der Waals surface of methyl chloride created with Avogadro

which seems to support the view that the spacial requirements of a chlorine atom and a methyl group are rather similar (but I don't know how reliable the van der Waals surface calculations by Avogadro actually are).
If those lengths mirror the real spacial requirements of the groups in the context of a nucleophilic substitution reaction then a chlorine atom wouldn't exert a larger steric hinderance than a methyl group and the anomeric effect would clearly be the determining factor when explaining the unreactiveness of DCM. But I'm unsure whether resorting to van der Waals radii is justified.

Your qualitative analysis is sound. A (partial) $\pi$ contribution would strengthen the one bond while weaken the other. But you are interpreting only a concept here, something that is not there. Since you are arguing with orbitals that cannot be assigned energies,  magnitudes would have no meaning. The anomeric effect is therefore a concept, that allows us to understand delocalisation in terms of a much simpler concept - it is not the true reason.

@Martin Thanks for the comment. Your point is well taken. But as for "... something that is not there. Since you are arguing with orbitals that cannot be assigned energies, magnitudes would have no meaning": Are you sure that there is not a little more reality in there than you give it credit. I mean, the procedure of looking at the interaction of the $n(\ce{e-})$ and the $\sigma^{*}(\ce{C-Cl})$ orbitals is basically the usage of perturbational MO-theory on the problem of what happens if I take a $\ce{{}^{+}CH2-Cl}$ fragment and let it interact with a $\ce{Cl-}$ fragment...

...of course this treatment is far from being exact and additional symmetry consideration would have to be made but in principle I think it should be possible to assign energies to the fragment orbitals used here (although I would never go so far as to really do that with any accuracy as MO-theory usually serves only a qualitative purpose).

@Philipp There are many schemes that decompose energy based on orbital interactions, but they all work in the same way, i.e. using the original AO. I guess you can go ahead and do a full Valence Bond Theory approach, to find out which resonance structure contributes how much to the total structure and then you get an idea of how strong the $\pi$ and $\sigma$ interactions are - then this approach would be (at least semi) quantitative. The unfortunate thing is, that you can only observe the total bonding directly and not the singular contributions.

Thinking about it a little more, the EDA-NOCV may give you exactly what you are looking for. (I do not have access to this program, so I cannot help out here.) I would suspect your analysis would be correct. On the other hand, I have grown quite fond of the simplicity of Bent's rule, which explains the shortening of the bonds. So maybe there is no use overanalyzing it. As for many things, there is no definite truth when it comes to interpretation, we always put some of our own to any analysis. Your analysis is well founded and (from my point of view) a valid approach to this problem.

@Martin Thanks again. You're right, to each his own "tool". Since looking at the foundations of perturbational molecular orbital theory I try to use its concepts for everything. But the problem is, as my professor stated it, you can explain everything with this theory in a very elegant and plausible way, even if your explanation is totally wrong :)

The anomeric effect where σ(C−C) orbital overlaps with the σ∗(C−Cl) should be considerably less than the overlap of n(e−) and σ∗(C−Cl). Therefore by comparing the relative rates of substitution of DCM, and a primary alkyl chloride, whose alkyl group is similar in size as the chlorine atom, against MeCl should give us the answer. I would be thankful if someone could provide the data.

Actually, I would say it is even justified to say that there can't be anything like the anomeric effect present in alkylchlorides. A $\sigma(\ce{C-C})$/$\sigma(\ce{C-H})$ orbital is much too low in energy compared to a $\sigma^{*}(\ce{C-Cl})$ orbital to show any appreciable interaction. As for the test you suggested: I think it will be hard to say whether a primary alkyl group/methyl group and a chlorine atoms exert a similar steric effect in the context of a nucleophilic substitution reaction.

Maybe one could go about it by employing the van der Waals radii of a chlorine atom and a methyl group. According to Wikipedia, the radius of a chlorine atom is 175 pm while the radius of a carbon atom (without any hydrogen atoms around) is 170 pm. If those lengths mirror the real spacial requirements of the groups in the context of a nucleophilic substitution reaction then a chlorine atom wouldn't exert a larger steric hinderance compared to a methyl group and the anomeric effect would clearly be the determining factor when explaining the unreactiveness of DCM.

The A-values for methyl and chloro are 1.7 and 0.4 respectively, suggesting that methyl is sterically larger than chloro.

@jerepierre I am not sure if it is valid to apply the A-values to these acyclic molecules, as A-values are derived from cyclohexane conformations.


text:
Why do we need acid when making acetone peroxide?
Acetone peroxide is a very dangerous explosive, easily detonated by mild heating, friction or shock. It appears in the form of small white crystals.

This compound forms from the mixture of hydrogen peroxide, acetone and hydrochloric or sulphuric (and maybe any other) acid. The acid, however, is only a catalyst in the reaction.

I'm curious why the acid is required for the reaction to happen. Is it actually necessary? Does the acid change during the reaction?

May I ask which program you used to generate the first image?  (Great question too!)

@santiago Sorry to disappoint you, but that images come from wikipedia. According to it's metadata, it was created using Adobe ImageReady. I have doubts that this is the actual program that generated the image.

Ah, no disappointment, a treasure found on the wilderness of Wikipedia

Material Studio can be used to generate these type of images

You asked whether the acid changed, well no catalyst means it promotes the reaction and does not change itself

The role of the acid in this reaction (really a polymerization) is two-fold.  First, it protonates the acetone carbonyl oxygen making the acetone carbonyl carbon more electrophilic.  This speeds up the rate of attack by the nucleophilic hydrogen peroxide (or other hydroperoxides formed later in the reaction) on this carbon.  Second, the acid also protonates hydroxyl groups in intermediates formed later in the reaction.  This makes them better leaving groups and allows nucleophiles like hydrogen peroxide or other hydroperoxides formed in the reaction to displace them.
The acid is a catalyst, the more acid the faster the reaction, but the reaction will proceed, albeit more slowly, without the acid.  Any proton acid can be used, but as expected, the stronger the acid, the faster the reaction.
Here are links to two articles on the subject (reference 1, reference 2) along with a diagram of the proposed reaction mechanism.


text:
Is there a systematic way to determine the expected cubic structure of metals?
I've noticed that some groups of metals with similar properties tend to belong to the same cubic structure group, such as $\text{FCC, BCC, or HCP}$. What is a good explanation as to why these metals have similar cubic structures, and how can one determine a metal's expected cubic structure just from knowing its identity (if there is a way)?
For example, copper, silver, and gold all have face centered cubic structures.
This trend seems to break down at some point, I think, because $\frac{2}{3}^{rd}$ of the platinum group have $\text{FCC}$ structures and $\ce{Ru}$ and $\ce{Os}$ have simple cubic structures.

There is no systematic way (besides brute force - that is a system, right?) to predict the crystal structure of the metals.  Even sticking just to the transition metals, as you have noted, there are quite a variety.  The primary ones are as you have noted $fcc$, $bcc$, and $hcp$ (ignoring the rhombohedral Hg).
Several issues arise here. First, there is little difference between $fcc$ and $hcp$ - they are both close packed structures with the only the stacking order changing (ABCABC for $fcc$ and ABABAB for $hcp$). An estimate of the free energy difference between these two crystal structures can be obtained 'SGTE Data for Pure Elements', A.T. Dinsdale, CALPHAD 15(4) 317-425 (1991), the compilation of elemental thermodynamic data. Here one finds that for Ag ($fcc$), the $hcp$ phase is only 300J/mol higher free energy (at T=0K).  Similarly, for Co ($hcp$), the $fcc$ phase is only 427 J/mol higher. So, the free energy differences aren't necessarily that much, making an $a\ priori$ differtiation impossible. (I'll note here that Ru and Os are $hcp$ crystals, not simple cubic as you stated).
The second factor that is difficult to determine is the impact of magnetic free energy on the equilibrium crystal structure (as included in the above reference).  Iron is the most obvious candidate to consider this, and it does not disappoint.  As observed, the equilibrium crystal structure at room temperature is $bcc$.  Heating leads to a change to $fcc$ at $910C$, then back to $bcc$ at $1400C$.  If the magnetic contributions to the free energy are ignored, the observed crystal structures would instead be $hcp$ at low temperatures, a transition to $fcc$ at around $200C$, and a brief appearance of $bcc$ right before melting. I see no way that one could predict these particular changes in phases with temperature.


text:
What additives could make a masonry unit water-repellent but vapor-permeable?
I'm curious about some of the wondrous materials available today are able to repel liquid water but permit water vapor to pass through. I've heard that housewrap-type plastics accomplish this by being perforated with microscopic holes that are slightly too large for liquid water, but large enough for the vaporous form to pass through. I have no idea how silicate mineral paints do it, though.
Housewraps and paints are thin films, though. I'd like to learn about how such properties could be imparted to a masonry unit like a compressed earth block. Could you cast such a block with a mixture of soil and some additive that would grant this kind of water repellency but allow water vapor to pass though?

Is liquid water molecule any larger than a steam molecule? Isn't there rather some trick with hydrogenium bonds?

The answer turned out to be waterglass. That's the carrier for silicate mineral paints.

I appreciate it that you took the time to write an answer yourself, but could you please elaborate so that the next person who had this problem would find a solution here?


text:
Which d orbitals of sulfur take part in the pi bonds of SO3?

In $\ce{SO3}$, 2 $p\pi-d\pi$ bonds are present. But which $d$ orbitals of
  sulfur take part in these $\pi$ bonds?

The answer says $d_{xy}$ and $d_{yz}$. Someone also told me that crystal field theory supports it as while splitting $d_{x^2-y^2}$ and $d_{x^2}$ are raised in energy compared to the other 3, but that theory has nothing to do here.

and the question is?

It is a long-standing myth that $d$ orbitals partake in bonding of most hypervalent compounds in any appreciable amount. There is little to no $d$-orbital participation in the bonding in $\ce{SO3}$ (see here). There are almost only $\mathrm{s}$ and $\mathrm{p}$ orbitals of sulphur involved.

@permeakra Its specified. I just need one reason why only $d_{xy}$ and $d_{yz}$ participate instead of other 3.

@Philipp Maybe you are right, but I read that in "J.D. Lee Concise Inorganic Chemistry" which I think is more trustworthy than the link you provided.

@Hardik I've seen this issue come up several times here on Chemistry.SE. A few other users mentioned that the supposed d-orbital participation in hypervalent molecules is a relic from the past which is contradicted by actual quantum chemical calculations. Maybe people like Martin can chime in an comment on that.

@Hardik Also, have a look here.

If your textbook still needs to use d orbitals to explain the hypercoordinate molecules, then you should get a new one. For the complete molecular orbitals of $\ce{SO3}$ have a look at this question and answer. Apart from this the question is incomplete. In order to determine which orbitals take part in bonding, you have to know the position of the symmetry defining elements. Otherwise the naming of the orbitals is completely interchangeable. (@Philipp thanks for raising awareness.)

@Martin Thanks for giving some support. And once again, nice answer of yours in the link you provided. Really like those orbital pictures of yours.

@Philipp Yes, its specified ".....these models still appear in some beginning-level college texts" . And the book I am following is a also for a beginner in college. I am still learning chemistry, just passed high school. Sorry if the question was not upto the level of CHEM-SE.

@Hardik There is nothing to feel sorry about. It's still a common thing to be told in universities even though it is an outdated view. I've been told it that way too before diving a little deeper into the more theoretical aspects of chemistry. And some of the historical concepts in chemistry still have their uses today, because they give you some easy to grasp guidelines and help you to develop a good intuition, although there might be some limitations and exceptions. But it is always good to be aware of the pitfalls and to know that there are more advanced theories out there.

@Philipp  I just realised that the first link you gave was a bout sulfite, when I answered, I was like... there is something missing there... where is the pi bond... and then i noticed that there are too many electrons ;) Unfortunately, no one published an article about the electronic structure at the basis set limit.

Unfortunately this is an issue that is, even though thoroughly disproved, still taught in many books and schools.
As Philip already pointed out in the comments: There is little to no contribution of the $d$ orbitals to the bonding in $\ce{SO3}$, as modern quantum chemical calculations prove (compare Wikipedia). This bonding scheme is a relic from a time where quantum theory was not yet widespread available. The little contribution of the $d$ orbitals (usually below 1%) we find is due to polarisation functions, that are added for a more accurate description of the molecule.
There are partial answers to the overall bonding in $\ce{SO3}$ in those two question:

Why is the bond order in the SO₃ molecule 1.33 and not 2?
Is the bond enthalpy of S-O stronger in SO₃ or SO₃²⁻?

But I will go a little further here. In the second post linked, you will find the result of a quantum chemical calculation for $\ce{SO3}$ on the BP86/cc-pVTZ level of theory. Here is the part of the output, containing the major contributions to the molecular orbitals (including the LUMO):
 Atomic contributions to Alpha molecular orbitals:
 Alpha occ 1 OE=-88.394 is S1-s=1.00
 Alpha occ 2 OE=-18.855 is O2-s=0.67 O3-s=0.17 O4-s=0.17
 Alpha occ 3 OE=-18.855 is O3-s=0.50 O4-s=0.50
 Alpha occ 4 OE=-18.855 is O3-s=0.33 O4-s=0.33 O2-s=0.33
 Alpha occ 5 OE=-7.922 is S1-s=1.00
 Alpha occ 6 OE=-5.945 is S1-p=1.00
 Alpha occ 7 OE=-5.941 is S1-p=1.00
 Alpha occ 8 OE=-5.941 is S1-p=1.00
 Alpha occ 9 OE=-1.113 is S1-s=0.39 O2-s=0.17 O3-s=0.17 O4-s=0.17
 Alpha occ 10 OE=-0.993 is O3-s=0.37 O4-s=0.37 S1-p=0.18
 Alpha occ 11 OE=-0.993 is O2-s=0.49 S1-p=0.18 O4-s=0.12 O3-s=0.12
 Alpha occ 12 OE=-0.585 is S1-s=0.24 O3-s=0.14 O4-s=0.14 O2-s=0.14 O2-p=0.11 O4-p=0.11 O3-p=0.11
 Alpha occ 13 OE=-0.502 is S1-p=0.55 O3-p=0.14 O4-p=0.14 O2-p=0.14
 Alpha occ 14 OE=-0.482 is O2-p=0.31 S1-p=0.20 O4-p=0.15 O3-p=0.15 O2-s=0.12
 Alpha occ 15 OE=-0.482 is O4-p=0.26 O3-p=0.26 S1-p=0.20
 Alpha occ 16 OE=-0.374 is O4-p=0.36 O3-p=0.36 O2-p=0.13
 Alpha occ 17 OE=-0.374 is O2-p=0.43 O3-p=0.21 O4-p=0.21
 Alpha occ 18 OE=-0.358 is O2-p=0.60 O4-p=0.15 O3-p=0.15
 Alpha occ 19 OE=-0.358 is O3-p=0.45 O4-p=0.45
 Alpha occ 20 OE=-0.305 is O4-p=0.32 O3-p=0.32 O2-p=0.32
 Alpha vir 21 OE=-0.141 is S1-p=0.45 O2-p=0.18 O4-p=0.18 O3-p=0.18

Now it is important to realise, that these numbers have to add up to 1 (accounting for rounding issues 0.99-1.01). So the reminding contributions are those of polarisation functions. Most notably missing is in orbital 15 the contribution of the O2-p orbital. The missing parts in 16-18 can be attributed to the diffuse and polarisation functions, i.e. higher $p$ and $d$-functions, 0.15; 0.15; 0.10; 0.10. As you can see, the involvement of $d$ orbitals in the total bonding is very little, almost negligible.
If you, however, consider $d$ orbitals to play a role in the bonding, then you have to consider, that all will participate.

authors: عبدالوهاب دفع الله


text:
Explaining Ba(OH)2 as a base?
It is well known fact that Barium Hydroxide is a base. How does the Brønsted-Lowry concept explain $\ce{Ba(OH)2}$  as a base?
Is it that $\ce{Ba(OH)2}$ first accepts a proton then leaves behind $\ce{H2O}$?

"Is it that Ba(OH)2 first accepts a proton then leaves behind H2O?" You know that's a really interesting statement; if you want to explore mechanisms of acidity, try checking out this thread: http://chemistry.stackexchange.com/questions/14165/the-acidity-of-boric-acid

Barium hydroxide – or more exactly, the hydroxide ion – is a Brønsted-Lowry base, because in aqueous solution, it is completely dissociated into $\ce{Ba^2+}$ and $\ce{OH-}$ ions:
$$\ce{Ba(OH)2 -> Ba^2+ + 2OH-}$$
$\ce{OH-}$ is a proton acceptor, forming $\ce{H2O}$. In aqueous solution, the following reaction takes place:
$$\ce{OH- + H3O+ ->~ 2H2O}$$
Barium hydroxide thus reacts with acids in a neutralization reaction, with water and the barium salt of the respective acid as the products. For example, with hydrochloric acid:
$$\ce{Ba(OH)2 + 2HCl ->~ BaCl2 + 2H2O}$$

But here it does \n't follow thw basic ideo of bronsted lowry base

It does if you envision it dissociating first, which it will do in aqueous media.


text:
Density of an object
To complete homework I am working on, I need help finding the density of a penny. I'm not given the mass or the volume. All I know about this penny is it's zinc coated in copper, and I am given the percentage of copper in the penny: 3.0% (by mass). We are to use these values for the density of copper and zinc:
Copper = 8.93, Zinc = 7.14
My friends did this:
$$\mathrm{8.98 \times 0.03 + 7.14 \times 0.97 = average~density~of~a~penny}$$
I told them "I think you're wrong, you can only calculate the average density of something if you use the volume", but I couldn't explain it any better.
Here's what I did:
Let $M$ be the mass of this penny in grams.
$$\frac{0.03M~\mathrm{g~mL^{-1}}}{8.98~\mathrm{g}} = \frac{0.03}{8.98}M~\mathrm{ml}$$
$$\frac{0.97M~\mathrm{g~mL^{-1}}}{7.14~\mathrm{g}} = \frac{0.973}{7.14}M~\mathrm{ml}$$
Adding them together we get $7.1952M$.
I'm not sure what this number represents but we can use it to find the ratios of volume.
$$\frac{\frac{0.03}{8.98 M}}{7.1952 M} = 0.0374$$ $$\frac{\frac{0.97}{7.14 M}}{7.1952 M} = 0.963$$
Using these numbers, we do what my friends did:
$$8.98 \times 0.0374 + 7.14 \times 0.963 = \mathrm{average~density~of~a~penny}$$
Am I correct?
If I am, is there a formula or any other simpler way to do this.
This is my first question on this site, so let me know if I did anything wrong.

For a new penny, the mass is 2.5 g. The composition is slightly different, but acceptable. http://en.wikipedia.org/wiki/Penny_(United_States_coin)

The volume of a solution (or alloy) is not quite additive either.

You have the right idea, but somewhere your math is incorrect.
Let the mass of the penny be $M$.
The mass of the copper is $\pu{0.03M}$ and the mass of the zinc is $\pu{0.97M}$.
The volume of copper is $\pu{0.03M}/(\pu{8.93g/cm^3})$ and the volume of the zinc is $\pu{0.97M}/(\pu{7.14g/cm^3})$.
The total volume is
$$\pu{0.03M}/(\pu{8.93g/cm^3}) + \pu{0.97M}/(\pu{7.14g/cm^3})$$
The density of the penny is
$$\frac {M}{\pu{0.03M}/(\pu{8.93g/cm^3}) + \pu{0.97M}/(\pu{7.14g/cm^3})}$$
Factoring out M gives
$$\frac {1}{0.03/(\pu{8.93g/cm^3}) + 0.97/(\pu{7.14g/cm^3})}$$
So the density of a mixture works out to be:
$$D=\frac {1}{(\%_1/D_1)+(\%_2/D_2)}$$