text:
How do you calculate the power (in watts) required per mole of product for or an electrochemical reduction?
Is it the transfer current density of the product multiplied by the overpotential?

Faraday’s law is usually summarized as
$$m=\frac{M\cdot Q}{z\cdot F}\tag{1}$$
where
$m$ is mass,
$M$ is molar mass,
$Q$ is electric charge,
$z$ is charge number, and
$F=96\,485.332\,89(59)\ \mathrm{C\ mol^{-1}}$ is the Faraday constant.
It essentially consists of two parts:
$$Q=n\cdot z\cdot F\tag{2}$$
and the definition of molar mass $M$
$$m=M\cdot n\tag{3}$$
Since you want to calculate values per amount of substance $n$ and not per mass $m$, you only need equation $\text(2)$.
In case of electrolysis at a constant current, the electric charge $Q$ is given by
$$Q=I\cdot t\tag{4}$$
where
$I$ is electric current, and
$t$ is time.
The power $P$ is given by
$$P=U\cdot I\tag{5}$$
where
$U$ is voltage.
Inserting $\text(4)$ into $\text(5)$ yields
$$P=U\cdot\frac Qt\tag{6}$$
And inserting $\text(2)$ into $\text(6)$ yields
$$P=U\cdot\frac nt\cdot z\cdot F\tag{7}$$
Now you can see why your question doesn’t make sense. The power $P$ (e.g. in watt) does not correspond to the deposited amount of substance $n$ (e.g. in mol); it actually corresponds to a deposition rate $n/t$ (e.g. in mol per second).
However, you could calculate the energy $E$ (e.g. in joule, watt seconds, or kilowatt hours) per amount of substance $n$ since
$$P=\frac Et\tag{8}$$
and thus
$$\begin{align}
E&=P\cdot t\\[6pt]
&=U\cdot n\cdot z\cdot F\tag{9}
\end{align}$$


text:
What does "n" stand for in n-akyldiols, n-alkanes, n-dicholoroalkanes?
I have read two conflicting answers upon a google search:

" 'n' in this context stands for normal or the latin equivalent. That is why, like other Latin and Greek abbreviations, it is in italics. It refers to straight chain alkanes. If we weren't dealing with n-alkanes we would have branched chain isomers to consider and potentially very different answers. It is old nomenclature from the days when adjacent carbons in a molecule were designated alpha,beta,gamma, etc and when acidity was given in 'N' units of normality."

and

" 'n' in science is like "x". It is just a reminder that a number needs to be there. You need to know that numbers can be replaced by x, x/y, n, or any variable and that groups can be replaced by their generic names."

"n" means a straight chain of carbon atoms.

The first one is relevant here - the nomenclature dates to at least 1844 (Gerhardt). From Wikipedia: "Straight-chain alkanes are sometimes indicated by the prefix "n-" (for normal) where a non-linear isomer exists. Although this is not strictly necessary, the usage is still common in cases where there is an important difference in properties between the straight-chain and branched-chain isomers, e.g., n-hexane or 2- or 3-methylpentane." The second one is identifying a generic case where the letter n is used as a variable.

The "n" stands for "normal", which in this context means "straight-chain".  It is not italisized, since it is not Latin, but English.  In contrast, the prefixes "i" or "a" for "iso" and "anteiso" are Latin and are properly italisized, but rarely are.
Your second choice, "n is like x", can be found in the subscripts of chemical formulae, where it does denote a variable amount.

But the n IS italicised for n-alkyldiols. Doesn't that mean it is Latin and not English? And therefore does not mean it is "a straight chain of carbon atoms"?

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text:
Why is the most probable KE for a monoatomic ideal gas exactly 1/3 of its average KE?
The most probable value of the kinetic energy for a monoatomic ideal gas can be calculated by setting the first derivative of the probablity density function equal to $0$, this turns out to be $\frac{1}{2}kT$ which is $\frac{1}{3}$ of the average value, $\frac{3}{2}kT$ what is the physical reason behind this?

There is no physical reason. It's bare mathematics. The probability distribution in terms of energy is asymmetric, for it is bounded by 0 from below and not bounded at all from above. Asymmetric distributions are known to have (generally speaking) different values for mean, mode, and median.
Think of the exponential distribution: its average value is some positive number, but the mode (the most probable value) is $\bf 0$.
Moreover, 3 is not even some intrinsic characteristic of the system. If you rewrite the distribution in terms of velocity (the way it is commonly done), you'll have an altogether different situation. All of a sudden, the ratio of mean to mode is... $2\over\sqrt\pi$, how would you like that?

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text:
Does 1-bromo-4-chlorobenzene have a permanent dipole moment?
My first thought was there would be a dipole moment as chlorine and bromine are unequal in electronegativity. I have since looked up the electronegativity of both chlorine and bromine through the Pauling scale. The difference between the electronegativity is 0.20 according to the table. For reference, the difference in electronegativity between carbon and hydrogen is 0.25 on the same table.
Since this is the case, I am very confused about the dipole moment of this molecule. Since the carbon-hydrogen bond is so negligible, why should this difference be significant?
To add to my confusion I found a source stating Dipole moment = 0.
And Wikipedia only states: Dipole moment = D.
However, my textbook questions lists it as having a dipole moment with the simple explanation that Br and Cl have differing electronegativities (question 1.61).

In my thought the dipole moment of 1-bromo-4-chlorobenzene is 0, there is not that much difference in the electronegativity of  Chlorine(3.0) and bromine (2.8).

Well, the correct answer to the way you phrased your question is: of course it does have a dipole moment. Whether the dipole moment is significant enough for us to consider the molecule to be polar, is a different matter entirely. But that is not the question asked. The question is "does it have a dipole moment" and the answer is "yes".

Sorry for the confusion, I meant a permanent dipole moment. I will adjust the question to clarify this.

There was no confusion at all; when you say "dipole moment", you do mean "permanent dipole moment", that's the default assumption. Yes, by the reasons listed above this molecule does have some dipole moment, but I would not be surprised if it is 0.00 when rounded to 2 decimal digits.

As others have stated before in the comments, the molecule itself has a dipole moment, but it is indeed very, very small, to the point where you would say it is negligible. To illustrate this, I have performed a few quick calculations, rounded to two decimal places.
Remember that the dipole is defined from negative to positive pole, i.e. $\displaystyle \ominus \xrightarrow{~\mathbf{q}~}\oplus$.
\begin{array}{lr}\hline
\text{Method}  & \mathbf{q}(\ce{Br-C4H4-Cl})~/~\mathrm{D}\\\hline
\text{BP86/def2-SVP}           &     0.07\\
\text{BP86/def2-TZVPP}         &     0.11\\
\text{B3LYP/def2-TZVPP}        &     0.09\\
\text{B3LYP+D3(BJ)/def2-TZVPP} &     0.09\\
\text{PBE0/def2-TZVPP}         &     0.10\\
\text{M11/def2-TZVPP}          &     0.10\\
\text{B2PLYP/def2-TZVPP}       &     0.09\\
\text{MP2/def2-TZVPP}          &     0.06\\
%\text{CCD//MP2/def2-TZVPP} &     ----\\
\hline\end{array}
In this case the dipole is directed towards the chlorine. Here are the NBO (natural bond orbital) charges for the CCSD(T)//MP2/def-TZVPP level of theory:

We can see that the chlorine (light green) is essentially neutral, while the bromine (dark green) is slightly positive polarised. From this we would assume the dipole moment to be directed towards the bromine. This is also what you would expect from the electronegativity differences.
Slightly different values are obtained from an QTAIM (quantum theory of atoms in molecules) analysis on the MP2/def2-TZVPP level of theory, where the volume is cut off at an electron density smaller than 0.001.
 Normalization factor of the integral of electron density is    1.000001
 The atomic charges after normalization and atomic volumes:
      1 (C )    Charge:    0.071729     Volume:    68.316 Bohr^3
      2 (C )    Charge:    0.078474     Volume:    79.386 Bohr^3
      3 (C )    Charge:    0.088508     Volume:    78.816 Bohr^3
      4 (C )    Charge:   -0.122884     Volume:    73.483 Bohr^3
      5 (C )    Charge:    0.088513     Volume:    78.816 Bohr^3
      6 (C )    Charge:    0.078474     Volume:    79.386 Bohr^3
      7 (H )    Charge:    0.026701     Volume:    46.501 Bohr^3
      8 (H )    Charge:    0.025890     Volume:    46.475 Bohr^3
      9 (Br)    Charge:   -0.097472     Volume:   260.461 Bohr^3
     10 (H )    Charge:    0.025886     Volume:    46.475 Bohr^3
     11 (H )    Charge:    0.026702     Volume:    46.501 Bohr^3
     12 (Cl)    Charge:   -0.290521     Volume:   214.364 Bohr^3

However, the calculated dipole moment points in the opposite direction. Reasons for this are similar to the case of carbon monoxide (see here). In short: it is basically because bromine is larger and more diffuse than chlorine. The σ-lone pairs, which are mainly responsible for the direction of the dipole moment, are further away from the atomic centres of the molecule for bromine compared to chlorine.
This causes a larger charge difference than the polarisation of the bonding structure.

Above graphic corresponds to the QTAIM analysis, the bold blue line represents the electron density contour of 0.001.

There are two aspects to dipole moments:
The first is the amount of charge transfer between the linked atoms - this is determined by the electronegativity difference.
The second is the bond length.
So, the greater difference in electronegativity between Cl and C vz Br and C is partially compensated by the longer Br-C bond vz Cl-C.
Many dipole moment are determined in solution and between different solvents they may differ by more than 0.2 units. I.e. 0 < μ < 0.2


text:
Can aldehydes be reduced to alcohols efficiently using LiAlH4 (excluding cost effectiveness)?
In my textbook (NCERT 12th class Chemistry, Part 2, India), $\ce{H2}$ with $\ce{Pd}$ catalyst is the only method mentioned.

Yes they certainly can, although I don't know why anyone would want to use LiAlH4 when you can just use NaBH4.

Yes, $\ce{LiAlH4}$ can be used for the reduction of an aldehyde to an alcohol. Compared to $\ce{NaBH4}$ it is a stronger reducing agent due to the electronegativity differences (Pauling electronegativity of $\ce{Al}$ is 1.61 & $\ce{B}$ is 2.04) making the metal-H bond more polar in case of $\ce{LiAlH4}$. Therefore, it reacts very violently with water and other acidic group containing compounds, and careful addition of $\ce{LiAlH4}$ is required.
The mechanism is rather straightforward, with alkoxide formation and protonation steps to get the final product.

In the lithium aluminum hydride reduction, water is usually added in a second step, whereas in the sodium borohydride reduction the methanol solvent system achieves this protonation immediately.

It really depends on the molecule itself. If you're talking about a molecule where the aldehyde is virtually the only functional group present, then yes $\ce{LiAlH4}$ will be very efficient. However, in the presence of other sensitive functional groups such as nitriles which can also be reduced with $\ce{LiAlH4}$, using milder reagents such as $\ce{NaBH4}$ makes more sense since nitriles cannot be reduced by it.

I would suggest you to kindly refer page no. 360 of Chemistry Part II NCERT of class 12.
The paragraph having title Reduction reads:

Reduction to alcohols: Aldehydes and ketones are reduced to primary and secondary alcohols respectively by sodium borohydride $\ce{NaBH4}$ or lithium aluminium hydride $\ce{LiAlH4}$ as well as by catalytic hydrogenation.


text:
What exactly does κ stand for in this equation connecting solubility and molar conductivity?
I came across the following equation while studying electrochemistry ("Physical Chemistry" by Wallwork and Grant):

The solubility $s$ of a sparingly soluble salt can be determined from conductivity measurements, provided that the molar conductivities of the ions at the temperature of these measurements are known to enable $\Lambda_0$ to be calculated. Since the saturated solution is dilute, $\Lambda_0 \approx \Lambda = \kappa/c$, where $c = s~(\pu{mol m-3})$. The conductivity $\kappa$ is the conductivity of a saturated solution of the salt in conductance water minus that of the water alone. The solubility $s$ is calculated from:
  $$s = \kappa/\Lambda_0 \tag{8.36}$$

It's mentioned that κ is the conductivity of the saturated solution of the salt in conductance water, minus that of the water alone. Now the thing is, from what I've read, conductance water has NO (or negligible) conductivity.
In that case, what's the point of having to subtract the conductivity of conductance water? Or is there something I've misunderstood here?

Let's clear up one thing first. There is no such thing as conductance water. It is a totally weird phrase.  I can only assume that pure water was the intended phrase. I'd suspect that the book is a translation, or written by an author in what is his second language.

The specific conductance, $\kappa$ (kappa) is the reciprocal of the specific resistance.

I'm working on an answer. The author has written this in a slipshod manner. This will never make sense unless the whole thing gets explained using the right terminology. I'm going to lunch. If someone else wants to jump in fine. In the meantime you can look at the wikipedia article on conductivity https://en.wikipedia.org/wiki/Conductivity_(electrolytic)

Nomenclature changes and authors don't always follow the rules.  The more typical use now is $\Lambda_m$ for the molar conductivity and $\Lambda_m^0$ for the limiting molar conductivity.
The molar conductivity, $\Lambda_m$, is given by the equation
$\Lambda_m = \dfrac{\kappa}{c}\tag{1}$
where:

$\kappa$ is the specific conductance (which is what you measure...)
$c$  is the concentration (which you know because of how the solution was made)

Now typically the specific conductance, $\kappa$, is plotted against concentration. For dilute solutions, there will be a linear relationship where the slope of the line is the limiting molar conductivity, $\Lambda_m^0$.  However at high concentrations the relationship becomes non-linear and the actual conductance is less than what the linear relationship would predict. That is because of the formation of cation and anion clusters. The local effect is thus to make the cluster uncharged (or charged less is the charges don't balance to zero).
Now the author of your book specified a sparingly soluble salt, so at low concentrations $\Lambda_m \approx \Lambda_m^0$, and we can substitute for $\Lambda_m$ to get:
$\Lambda_m^0 = \dfrac{\kappa}{c}\tag{2}$
which can be rearranged to give:
$c = \dfrac{\kappa}{\Lambda_m^0}\tag{3}$
If the conductivity of the solution is much greater than that of pure water then this is fine. The measured specific conductance will be directly proportional to the concentration since $1/\Lambda_m^0$ is a constant.
However the author is talking about a salt of limited solubility, so the conductivity of pure water is not negligible. Going back to the plot of specific conductance, $\kappa$, against concentration, $c$, we can determine that at 0 concentration there is some conductance which I'll call $\kappa_w$ which is the conductance of pure water. So for a salt with limited solubility the equation is:
$c = \dfrac{\kappa - \kappa_w}{\Lambda_m^0}\tag{4}$
If you look back at equation (2) you can see why equation (3) doesn't work for solutions of very low conductivity. When $c=0$ equation (2) blows up since you can't divide by $0$.
So the author used $\kappa$ in two different ways which is very poor writing. We could define $\kappa'= \kappa - \kappa_w\tag{5}$ then $c = \dfrac{\kappa'}{\Lambda_m^0}\tag{6}$

The conductivity $\kappa$ of pure water is very low. The value for highly purified water at a temperature of $T=25\ \mathrm{^\circ C}$ is only $\kappa=5.50\times10^{-6}\ \mathrm{S\ m^{-1}}$. However, the conductivity of water is sensitive to dissolved impurities. The conductivity of water that is saturated with atmospheric $\ce{CO2}$ is already increased to $\kappa=1.10\times10^{-4}\ \mathrm{S\ m^{-1}}$. Therefore, if necessary, the actual conductivity of the water that is used in the experiment should be measured and subtracted from the measured values for the concerned salt solutions.
Certainly, when measuring the conductivity of strong electrolyte solutions at high concentrations, the contribution of water is not significant and may be neglected. For example, the conductivity of a potassium chloride solution with a molality of $b=1.0\ \mathrm{mol\ kg^{-1}}$ is about $\kappa\approx10.9\ \mathrm{S\ m^{-1}}$.
However, if you want to measure the conductivity of ions in solutions of sparingly soluble salts (for example in order to calculate the solubility of such salts as mentioned in your book), the contribution of water can be significant. For example, the solubility product constant for silver chloride $(\ce{AgCl})$ is $K_\mathrm{sp}=1.77\times10^{-10}$. Ignoring activity coefficients, this solubility corresponds to a concentration of about $c=1.3\times10^{-5}\ \mathrm{mol\ l^{-1}}$ for $\ce{Ag+}$ as well as $\ce{Cl-}$. Note that this value is not much larger than the concentration of $\ce{H+}$ and $\ce{OH-}$ in pure water, and that the ions $\ce{H+}$ and $\ce{OH-}$ have a relatively high molar conductivity $\lambda$.


text:
How to select a schematic of the coordinate vs energy representing a rearrangement reaction?
Here's the question

For a reaction to take place the net energy should go down, and thus option (d) can be discarded.
We are now left with options (a), (b), and (c).
A $3^{\circ}$ carbocation is more stable than a $2^{\circ}$ carbocation, therefore the 3rd dip in the graph should be lower than the 2nd dip as more stable means lower energy.
Thus now we are left with options (b) and (c).
How do you decide between (b) and (c)?

The answer is (b).  The initial step of the reaction should take energy.  Otherwise the reactant molecule itself is unstable.  The very initial portion of the curve in (c) is downward-sloped, which doesn't make sense.  If the reacting alkene could relax to a different conformation without any activation barrier, then it would do so.  But the fact that the reactant is a stable, isolable molecule with a well-defined chemical structure indicates that no such barrier-less relaxation is possible.
Thus, (b) is a better answer than (c).

no, the answer is c

Not according to me it isn't.

well, the answer key could be wrong, but your answer doesnt really convince me much

Feel free to downvote.

I would agree with you on the initial downward bit, which makes 0 sense. However, (b) also doesn't make sense to me, with a carbocation intermediate that is more stable than the starting alkene. Even (c) says that the tertiary carbocation is lower energy than the alkene. I would prefer to believe that carbocations are high-energy intermediates, and that if you added, say, perchloric acid to the double bond, and let the system reach equilibrium, you would not obtain a salt of the carbocation with perchlorate counteranion, because that is what is implied. I don't like any of these answers.

One might like to think that because a carbocation is tertiary, it is really stable because of all the wonderful hyperconjugation. That neglects the fact that it is still a carbocation, featuring positively charged, octet-deficient carbon, which is inherently more unstable than any neutral molecule from which it is derived. What do you think the pKb of an alkene is? I don't think it's anywhere near significant.

Well with the data in place that the alkene is more stable than the carbocation, it is quiet evident that the answer is c and not b. As alkene is more stable than carbocation thus there is a deeper dip initially than later on. As the carbocation is less stable than the alkene the dip isn't as great as the dip around the alkene.

As Curt said and as I said c doesn't make sense either. You are basically choosing the least bad answer, and between b and c, which answer is least bad is really quite subjective. I would argue for this. http://i.stack.imgur.com/SbKkb.png

I'm pretty sure the answer is c. As the carbocation would have higher energy than the alkene. Which is the case in (c) but not in (b). In (b) it can be seen clearly that carbocation is at a lesser energy than alkene which certainly isn't true!

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text:
If a transition metal loses electrons so easily, how does it maintain its integrity?
If a transition metal can lose electrons so easily, then after it has been used for a conductor for x amount of time, why doesn't it fall apart?  If electricity can flow through it because of loose electrons, why doesn't the wired disintegrate.  Are there that many electrons that can be spared in a piece of metal?

The metal is not being used as a source of electrons, it is being used as a means to transport electrons from one place to another. When you hook a battery up to a circuit, the potential difference across the wire (created by the chemicals in the battery) causes the electrons to flow from the negative end of the battery and into the wire. As these electrons enter the wire, electrons are simultaneously being pushed out the other end of the wire, and into the positive end of the battery. The wire is always neutrally charged.

Isn't it the means because it has so many electrons that are delocalised?

@johnny it is because the electrons are delocalized, yes, though it doesn't have anything to do with how many there are. Conductivity is an intrinsic property.


text:
Lewis structure of ions
In class, we learned about the Lewis structure of some elements such as carbon, but we didn't learn how to do the Lewis structure of cations and anions.
Do we just draw the number of electrons in the ion, or do we do something different?
So for example, what would be the Lewis structure of $\ce{Li+}$ and $\ce{Li-}$? How would you draw it out?

The charge on a centre in Lewis structures is not of special relevance, except that it helps to provide the information which is useful: the number of electrons on that centre.
A positive charge of $X$ means $X$ electrons were lost, while a negative charge, $-Y$ means $Y$ electrons gained.
From this you can either simply remove/add the corresponding number of dots from/to the diagram, or calculate how many valence electrons are now present, and then draw the diagram ab initio from this information.
A "normal" lithium centre has 1 valence electron, so the positive ion just loses it (leaving no dots) while the negative ion gains one more (and thus has two dots on the diagram).

@Nij Okay, this may be a somewhat different question, but what would the atom do if it had the same number of electrons added compared to electrons removed to get the $8$ valence electrons? For example, Carbon. It can add $4$ electrons to get $8$ valence electrons, or remove $4$ to get the same number of valence electrons as Helium. Which does it do?

@frank It tends not to do either. This is where you get the difference between covalent and ionic bonding.


text:
How many significant figures?
In Brown, Theodore E.; LeMay, H. Eugene; Bursten, Bruce E.; Murphy, Catherine; Woodward, Patrick; Stoltzfus, Matthew E.. Chemistry: The Central Science (Page 35). Pearson Education. Kindle Edition, an assignment is:

Make the following conversions: $\pu{2500^\circ F}$ to $\pu{K}$

The answer given is $\pu{1644 K}$ but I say it's $\pu{1600 K}$. What is the correct answer?

It is ambiguous. 2500 can either be 2.5E3 or 2.500E3. The number 2506 would clearly be 2.506E3. Changes of last two digits being zeros randomly is 1%. If this is a home question or on a test then work the answer to 4 figures state that the problems seems ambiguous and that your assumption is that there are only two significant figures in the given temperature and round your answer to 1.6E3.
I'll point out that your answer has the same ambiguity. Is 1600 really 1.6E3 or 1.600E3?

I thought that the only time we should use the zeroes is if there is a period... But you say its better to calculate as if the period was there for all problems?

has four sig figs while 1600 has only two. I thought.

does indicate 4 significant figures, but 1600 is ambiguous. It actually could be 1.6E3 or 1.60E3 or 1.600E3.


text:
What are two superimposable mirror images called?
I know that enantiomers are mirror images which are not superimposable on each other.
What about mirror images which are superimposable? Are they necessarily the same molecule? Is there a particular term for them?

Well, if you can superimpose them then they are identical. What is there to ask about?

Mirror images are mirror images. I suspect that what you mean is ‘what is a molecule called, which is superimposable on its mirror image?’

If two molecules are superimposable, they would by definition be the exact same molecule. There is no chemical test you could perform to differentiate them. This is true for all achiral compounds.
The most interesting case of this would be a class of molecules called meso compounds. They contain multiple stereogenic centers, but due to an internal plane of symmetry (or certain other symmetry elements), reflection of these molecules across a mirror plane results in superposable mirror images.


text:
Why do polysaccharides consisting of alpha-glucose have helical structures?
Why do the polysaccharides consisting of alpha-glucose such as starch and glycogen have helical structure, while the one such as cellulose consisting of beta glucose don't display such structures?
The textbook said that is because of the angle of 1-4 linkage, but what kind of angle is it that makes helical structure?

https://chemistry.stackexchange.com/questions/33105/why-starch-amylose-and-cotton-cellulose-are-so-different

Note: This is an off the top of my head answer (I might be horribly wrong), but it has prompted me to do some research. I will add more details, and/or make corrections once I am done. In the meantime, I would love to hear what others have to say on this.
Pictured below is the 2D structure of amylose

This primarily composed of $\alpha$-(1-to-4)-D-glucose units. Although the $\alpha$-(1-to-4) links are capable of relatively free rotation about their torsional angles, hydrogen bonding between the $\ce{O}$3 and $\ce{O}$2 encourages helicogenesis.
Single helical amylose has hydrogen-bonding between $\ce{O}$2 and $\ce{O}$6 atoms on outside surface of the helix and the ring oxygen pointing inwards.
Contrat this with cellulose which is a linear polymer of $\beta$-(1-to-4)-D-glucopyranose units.
The fully equatorial conformation of $\beta$-linked glucopyranose residues stabilises the chair structure, minimising its flexibility.
Pictured below, the 2D structure of cellulose and displaying the intrastrand hydrogen bonding

Interactive 3D visualisation for amylose and cellulose.

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text:
Relation between constant-pressure and constant-volume heat capacities: Cp - Cv = nR
For an ideal gas, we have
$$C_p - C_V = nR$$
where
$C_p$ is heat capacity at constant pressure,
$C_V$ is heat capacity at constant volume,
$n$ is amount of substance, and
$R=N_\mathrm A\cdot k_\mathrm B=8.314\,462\,618\,153\,24\ \mathrm{J\ mol^{-1}\ K^{-1}}$[source] is the molar gas constant.
How can I prove this?

The heat capacities are defined as
$$C_p = \left(\frac{\partial H}{\partial T}\right)_{\!p} \qquad \qquad C_V = \left(\frac{\partial U}{\partial T}\right)_{\!V} \tag{1}$$
and since $H = U + pV$, we have
$$\begin{align}
C_p - C_V &= \left(\frac{\partial H}{\partial T}\right)_{\!p} - \left(\frac{\partial U}{\partial T}\right)_{\!V} \tag{2} \\
&= \left(\frac{\partial U}{\partial T}\right)_{\!p} + \left(\frac{\partial (pV)}{\partial T}\right)_{\!p} - \left(\frac{\partial U}{\partial T}\right)_{\!V} \tag{3}
\end{align}$$
Ordinarily, we have $\mathrm{d}(pV) = p\,\mathrm{d}V + V\,\mathrm{d}p$, but under conditions of constant pressure $\mathrm{d}p = 0$ and so we can write
$$C_p - C_V = \left(\frac{\partial U}{\partial T}\right)_{\!p} + p\left(\frac{\partial V}{\partial T}\right)_{\!p} - \left(\frac{\partial U}{\partial T}\right)_{\!V} \tag{4}$$
For an ideal gas, $V = nRT/p$ ($n$ being constant throughout this whole discussion - or else $nR$ in the original equation makes no sense!) and so
$$\left(\frac{\partial V}{\partial T}\right)_{\!p} = \frac{nR}{p} \tag{5}$$
Substituting $(5)$ into $(4)$ gives
$$C_p - C_V = \left(\frac{\partial U}{\partial T}\right)_{\!p} - \left(\frac{\partial U}{\partial T}\right)_{\!V} + nR \tag{6}$$
At this stage, you have two choices.

The easy way
From the equipartition theorem, the internal energy of an ideal gas is given by
$$U = \left(\frac{\text{Degrees of freedom}}{2}\right)nRT \tag{7}$$
The number of degrees of freedom doesn't matter, because it is a constant for any given gas. The important point is that $U = U(T)$, i.e. internal energy is a function of only temperature. Therefore,
$$\left(\frac{\partial U}{\partial T}\right)_{\!p} = \left(\frac{\partial U}{\partial T}\right)_{\!V} = \frac{\mathrm{d}U}{\mathrm{d}T} \tag{8}$$
and the desired result, $C_p - C_V = nR$, immediately follows.
The "easy way" is described in a large number of textbooks, or web pages. I was even told to use it in one of my tutorials, at Oxford no less. It is not incorrect. However, I don't like the easy way, because it invokes the equipartition theorem, which (in my opinion) makes it only half a proof, unless one proves the equipartition theorem along with it.
In fact, you don't need to use equipartition to derive this result. The only thing that we need to use is $pV = nRT$. If you are not convinced: read on!

The hard way
Without invoking the equipartition theorem, we don't know that $U = U(T)$. In general, we would expect $U$ to be a function of all three variables: $U = U(V, T, p)$. However, for an ideal gas, $p$ itself is a function of $(V,T)$: $p(V,T) = nRT/V$; so, we can eliminate the $p$-dependence of $U$, since the $p$-dependence is adequately described by the $(V,T)$ dependence. The total differential for $U = U(V,T)$ is
$$\mathrm{d}U = \left(\frac{\partial U}{\partial V}\right)_{\!T}\,\mathrm{d}V + \left(\frac{\partial U}{\partial T}\right)_{\!V}\,\mathrm{d}T \tag{9}$$
Using a similar argument, we could also treat $U$ as a function of $p$ and $T$ instead. The total differential for $U = U(p,T)$ is
$$\mathrm{d}U = \left(\frac{\partial U}{\partial p}\right)_{\!T}\,\mathrm{d}p + \left(\frac{\partial U}{\partial T}\right)_{\!p}\,\mathrm{d}T \tag{10}$$
And the third total differential we need is that of $V = V(p,T)$:
$$\mathrm{d}V = \left(\frac{\partial V}{\partial p}\right)_{\!T}\,\mathrm{d}p + \left(\frac{\partial V}{\partial T}\right)_{\!p}\,\mathrm{d}T \tag{11}$$
Substituting $(11)$ into $(9)$, we have
$$\begin{align}
\mathrm{d}U &= \left(\frac{\partial U}{\partial V}\right)_{\!T} \left[\left(\frac{\partial V}{\partial p}\right)_{\!T}\,\mathrm{d}p + \left(\frac{\partial V}{\partial T}\right)_{\!p}\,\mathrm{d}T \right] + \left(\frac{\partial U}{\partial T}\right)_{\!V}\,\mathrm{d}T \tag{12} \\
&= \left[\left(\frac{\partial U}{\partial V}\right)_{\!T}\left(\frac{\partial V}{\partial p}\right)_{\!T}\right]\mathrm{d}p + \left[\left(\frac{\partial U}{\partial V}\right)_{\!T}\left(\frac{\partial V}{\partial T}\right)_{\!p} + \left(\frac{\partial U}{\partial T}\right)_{\!V}\right]\mathrm{d}T \tag{13}
\end{align}$$
Equations $(10)$ and $(13)$ have very similar forms: they both look like $\mathrm{d}U = a\,\mathrm{d}p + b\,\mathrm{d}T$. We can equate the coefficients of $\mathrm{d}T$ in both equations to get
$$\left(\frac{\partial U}{\partial T}\right)_{\!p} = \left(\frac{\partial U}{\partial V}\right)_{\!T}\left(\frac{\partial V}{\partial T}\right)_{\!p} + \left(\frac{\partial U}{\partial T}\right)_{\!V} \tag{14}$$
and therefore, looking back to equation $(6)$ and substituting equation $(14)$ in, we see that
$$\begin{align}
C_p - C_V &= \left(\frac{\partial U}{\partial T}\right)_{\!p} - \left(\frac{\partial U}{\partial T}\right)_{\!V} + nR \tag{6} \\
&= \left(\frac{\partial U}{\partial V}\right)_{\!T}\left(\frac{\partial V}{\partial T}\right)_{\!p} + nR \tag{15}
\end{align}$$
The final step is that, for an ideal gas,
$$\left(\frac{\partial U}{\partial V}\right)_{\!T} = 0 \tag{16}$$
and the proof of this can be found in this question: Internal pressure of ideal gas. This yields
$$C_p - C_V = nR \tag{17}$$
as desired.

I'd just add that this only works for a monoatomic gas. The vibrations of the atoms adds more degrees of freedom for a polyatomic gas. This leads down a rabbit hole to a discussion about perfect gases and ideal gases. ;-)

No, it applies for all ideal gases. The exact value of $C_p$ and $C_V$ differs from monoatomic to diatomic to polyatomic, but the relation $C_p - C_V = nR$ holds true for all.

Strictly speaking the perfect (ideal) gas does not have any internal molecular nature, but is just a mass point that does not interact with others. The total energy is just its kinetic energy and there is no potential energy arising from interaction between them.  What ever the gas the difference between heat capacities is caused by expansion work done at constant pressure :)

@orthocresol Does it apply to all ideal gases or perfect gas? With ideal gas the Cp and Cv are sensitive to temperature. And I have read that Cp and Cv changes at a different rate with respect to temperature. If that is true, then the Meyer's relation will not be valid at higher temperatures. With perfect gas since there is no temperature dependence of Cp and Cv the Meyer's relation will hold at all temperature.

@AbhishekPG sorry, please ask a new question.

Preliminaries
Consider $U = U(V,T, p)$. However, assuming that it is possible to write an equation of state of the form $p = f(V,T)$,  I don't  have to explicitly address  the $p$ dependence of $U$, and I can write the following differential:
$$\mathrm{d}U = \underbrace{\left ( \frac{\partial U}{\partial V} \right)_T}_{\pi_T} \mathrm{d}V  + \underbrace{\left ( \frac{\partial U}{\partial T} \right)_V}_{C_v} \mathrm{d}T \tag{1}$$
so one writes,
$$ \mathrm{d}U = \pi_T \mathrm{d}V + C_v\mathrm{d}T \tag{2}$$
also, for an ideal gas, the internal pressure ($\pi_T$) $= 0$
Additionally for changes in internal energy at constant pressure,
$$ \left (\frac{\partial U}{\partial T}\right)_p = \pi_T \left (\frac{\partial V}{\partial T}\right)_p +C_v \tag{3}$$
Here, I take the opportunity to define two new quantities, namely $\alpha$ and $\kappa_T$ (the expansion coefficient and the isothermal compressibility respectively)
$$\alpha = \frac{1}{V}\left (\frac{\partial V}{\partial T}\right)_p$$
and, $$\kappa_T = \frac{-1}{V}\left (\frac{\partial V}{\partial p}\right)_T$$
Rewriting (3) using $\alpha$
$$ \left (\frac{\partial U}{\partial T}\right)_p = \alpha \pi_T V +C_v \tag{4} $$
Now, although the constant volume heat-capacity is defined as $C_v = \left (\frac{\partial U}{\partial T} \right)_v$, if we let $\pi_T = 0$ in equation (4), we get $C_v = \left (\frac{\partial U}{\partial T} \right)_p$
This holds true for a perfect gas, and one can quickly obtain the desired relation at this stage.

Derivation: Difference between constant volume and constant pressure heat capacities for a perfect gas
Consider, $$C_p - C_v  = \overbrace{\left (\frac{\partial H}{\partial T} \right)_p}^{\text{definition of} \ C_p} - \overbrace{\left (\frac{\partial U}{\partial T} \right)_v}^{\text{definition of} \ C_v} \tag{5}$$
Introducing, $H = U+ pV = U+nRT$, and exploiting $C_v = \left (\frac{\partial U}{\partial T} \right)_v =  \left (\frac{\partial U}{\partial T} \right)_p$, equation (5) yields:
$$ C_p - C_v = \left (\frac{\partial (U+nRT)}{\partial T} \right )_p - \left (\frac{\partial U}{\partial T} \right )_p = nR$$

Derivation: Difference between constant volume and constant pressure heat capacities (general case)
However, as my contribution to this discussion I would like to derive a relation between heat capacities that is universally true for any substance, not just a perfect gas. So let's return to equation (5):
$$C_p - C_v  = \overbrace{\left (\frac{\partial H}{\partial T} \right)_p}^{\text{definition of} \ C_p} - \overbrace{\left (\frac{\partial U}{\partial T} \right)_v}^{\text{definition of} \ C_v} \tag{5}$$
Here, we substitute $H = U + pV $ and obtain,
$$ C_p -C_v = \overbrace{\left( \frac{\partial U}{\partial T} \right)_p}^{\text{evaluated in} \ (4)}+ \left( \frac{\partial (pV)}{\partial T} \right)_p - C_v \tag{6}$$
The first partial derivative was already taken care of in equation (4). For the second one, since the derivative is to be evaluated at constant pressure, we can do the following
$$\left( \frac{\partial (pV)}{\partial T} \right)_p = p \overbrace{\left( \frac{\partial V}{\partial T} \right)_p}^{\alpha V}$$
Putting all of this together, one obtains
$$C_p -C_v = \alpha \pi_T V + \alpha pV = \alpha(p+ \pi_T)V \tag{7}$$
At this stage, I will make use of the following relation (derived in additional comments) $$\pi_T = T \left (\frac{\partial p}{\partial T}\right )_v - p$$ After substituting this into (7) we get:
$$ C_p -C_v = \alpha T V \left( \frac{\partial p}{\partial T} \right)_V \tag{8}$$
I wish to transform the last remaining partial derivative, and to do so I consider $ V = V(T,p) $ which yields the following differential
$$ \mathrm{d}V = \left( \frac{\partial V}{\partial T} \right)_p \mathrm{d}T + \left( \frac{\partial V}{\partial p} \right)_T \mathrm{d}p $$
At constant volume, $\mathrm{d}V = 0$ so one gets,
$$\left( \frac{\partial V}{\partial T} \right)_p \mathrm{d}T  = - \left( \frac{\partial V}{\partial p} \right)_T \mathrm{d}p$$
$$\overbrace{ \left( \frac{\partial V}{\partial T} \right)_p}^{\alpha V} = \overbrace{-\left( \frac{\partial V}{\partial p} \right)_T}^{\kappa_T V} \left( \frac{\partial p}{\partial T} \right)_V$$
Note: One can avoid all of this work, and simply invoke the Euler Chain Rule
Rearranging,  $$\left( \frac{\partial p}{\partial T} \right)_V = \frac{\alpha}{\kappa_T}$$
We can finally substitute this into (8) to get
$$C_p -C_v = \frac{\alpha^2 TV}{\kappa_T} \tag{9}$$
This is true for any substance, not just a perfect gas. Now, for a perfect gas $ pV = nRT$ holds true, and thus $\alpha = \frac{1}{T}$ and $\kappa_T = \frac{1}{p}$. Making these substitutions into (9), we get our desired result
$$C_p -C_v = nR $$

Additional Comments

This might seem like an unnecessarily complex, and not to mention convoluted way to get to the desired result (especially, in light of a much simpler method presented by @orthocresol), however, I think the deriving the expression for a general case first, and then reducing it to the special case is illuminating. Moreover, in spirit and approach, it is not that far from what @orthocresol did.
Physical Significance of certain terms/quantities

$\pi_T$ is called the internal pressure (it has the dimensions of pressure) and is a consequence of the interactions between molecules. For an ideal gas it is necessarily zero.
$\alpha$ i.e the expansion coefficient is the fractional change in volume that accompanies a rise in temperature. A large volume of $\alpha$ implies that the sample responds very strongly to changes in temperature.
Similarly, $\kappa_T$ is a measures of the response to a change in pressure. The negative sign insures that $\kappa_T$ is a positive quantity, because a pressure increase causes a decrease in volume ($\mathrm{d}V$ is negative)
Since equation (9) holds true for any substance, for solids and liquids one might be tempted to say $C_p \approxeq C_V$ because $\alpha$ is small for solids and liquids. However, one must be careful because $\kappa_T$ can be small as well, which makes the fraction $\frac{\alpha^2}{\kappa_T}$ large. In other words, even though a little work has to be done to push back the atmosphere when a solid expands, a great deal of work will go into pulling the atoms apart.

Supplementary Derivation: For a system where, $N$ doesn't change the fundamental equation of thermodynamics is:

$$\mathrm{d}U = T\mathrm{d}S -p\mathrm{d}V$$
This seems to suggest that, $U = U(S,V) $. Thus, one can write the following differential and after comparing to the one above can equate $T$ and $-p$ (as indicated by annotations) with the partial derivatives given below:
$$\mathrm{d}U = \underbrace{\left ( \frac{\partial U}{\partial S}\right)_V}_{T} \mathrm{d}S + \underbrace{\left ( \frac{\partial U}{\partial V}\right)_S}_{-p}\mathrm{d}V$$
Moreover, dividing both sides of the fundamental equation by $\mathrm{d}V$ (yeah, I know), and imposing constraint of constant temperature) we can manipulate it into the following form:
$$\overbrace{\left( \frac{\partial U}{\partial V} \right)_T}^{\pi_T} = \overbrace{\left ( \frac{\partial U}{\partial S}\right)_V}^{T}\left ( \frac{\partial S}{\partial V}\right)_T - \overbrace{\left ( \frac{\partial U}{\partial V}\right)_S}^{-p}$$
Thus, we have $$ \pi_T = T\left ( \frac{\partial S}{\partial V}\right)_T - p$$
Invoking, the Maxwell Relation
$$ \left ( \frac{\partial S}{\partial V}\right)_T = \left ( \frac{\partial p}{\partial T}\right)_V$$
one finally gets, $$\pi_T = T \left (\frac{\partial p}{\partial T}\right )_v - p$$

Some elegant descriptions have already been given but it is useful to consider in words what is actually happening. When a system absorbs heat q its temperature is raised and the average heat capacity between two temperature is $q/\Delta T$. As the temperature difference becomes infinitesimally small the actual heat capacity at that temperature is produced.
However, the heat needed to raise a substance by a given temperature depends on the circumstances, namely if this is done at constant volume or pressure, for if it is done at the latter, then some work has to be done against the prevailing pressure and the heat absorbed differs from the increase in internal energy by the amount of work done or $q=dU+pdV$.
If the heat absorbed is equal to the increase in internal energy then this has been done at constant volume and no work is involved, thus
$$ C_V=\left ( \frac{\partial U}{\partial T}\right )_V$$
This equation is quite general and applies to any system.
At constant pressure work is done and as $q=dU+pdV$
$$ C_p = \left ( \frac{\partial U}{\partial T}\right )_p + p\left ( \frac{\partial V}{\partial T}\right )_P$$
Next we can study the effect of heating from one temperature to another at the same pressure in two parts. The first part is that the increase in energy is the same as it would be if the substance was heated from the first temperature to the second at constant volume, and the second the change as the system is brought to the original pressure at constant temperature. This is
$$ \left ( \frac{\partial U}{\partial T}\right )_p= \left ( \frac{\partial U}{\partial T}\right )_V + \left ( \frac{\partial U}{\partial V}\right )_T \left ( \frac{\partial V}{\partial T}\right )_p$$
(This is obtained by letting $dU= \left ( \frac{\partial U}{\partial T}\right )_VdT + \left ( \frac{\partial U}{\partial V}\right )_TdV$ and then differentiating by T.
The last two equations may now be combined by substituting $\left ( \partial U/\partial T\right )_p$ into $C_p$;
$$C_p=C_V + \left( p+ \left ( \frac{\partial U}{\partial V}\right )_T   \right) \left ( \frac{\partial V}{\partial T}\right )_p $$
and for a perfect gas this has a simple form as $(\partial U/\partial V)_T = 0$ and $(\partial V/\partial T)_p=R/p$ which produces
$$ C_p  = C_V+R$$


text:
Why can't one make C60 with a combination of hexagons and heptagons?
$\ce{C60}$ has a pentagon-hexagon system. Why can't one make it from a hexagon-heptagon system, or a pentagon-heptagon system for that matter?

It is possible to build carbon skeletons of other geometric forms (though they usually contain hydrogen as well). But they are not exactly easy to make. Dodecahedrane, for example, has a carbon skeleton with just pentagons, derivatives of cubane (just squares) are known. And they don't have 60 carbons.

Well, let's see. This is not quite a chemical question, though I wouldn't recommend moving it to Math.SE either, because they might have hard time recognizing what's $\ce{C60}$.
Now, let's imagine we're considering locally planar conjugated carbon
structures. Each carbon has to reserve one $p$ orbital for $\pi$-bonding,
which leaves it with 3 bonds. Now, the bonds tend to go in cycles; let's
confine ourselves to 5-, 6-, and 7-membered cycles, because otherwise we'll
have too much of angular strain (not that other cycles are totally impossible, though). So in effect, we are considering graphs
consisting of pentagons, hexagons, and heptagons, with 3 edges meeting at
each vertex. Let's say the numbers of these n-gons are $n_5, n_6$, and $n_7$,
correspondingly.
That's when old Euler kicks in. Polyhedra must meet a certain relation
between the number of faces (F), edges (E), and vertices (V):
$$V - E + F = 2$$
OK, what is the number of faces? Why, that's quite simple: $F=n_5+n_6+n_7$.
Then what is the number of vertices? Apparently, each pentagonal face has
five of them, and so on, so it would be $5n_5+6n_6+7n_7$. But wait, we've
counted each atom three times (for it belongs to three polygons), hence we must divide by 3: $V=(5n_5+6n_6+7n_7)/3$. By similar reasoning, the number of edges would be $5n_5+6n_6+7n_7$, if not for the fact that each edge belongs to two faces and thus is counted twice, hence $E=(5n_5+6n_6+7n_7)/2$.
All in all,
$$
V - E + F = 2\\
{5n_5+6n_6+7n_7\over3} - {5n_5+6n_6+7n_7\over2} + n_5+n_6+n_7 = 2\\
-{5n_5+6n_6+7n_7\over6} + n_5+n_6+n_7 = 2\\
{n_5-n_7\over6} = 2\\
n_5-n_7 = 12
$$
That's pretty much the size of it. You can do more or less whatever you want, as long as you have twelve more pentagons than heptagons. There is no going around it.
Now chemistry kicks in, too, and brings its own limitations. An $sp^2$ carbon wants to have its bonds in one plane, at $120^\circ$ angles. You can convince it to budge, but only so much. If you strain it more, the system becomes unstable. It turns out that two pentagons meeting at one carbon is already too much (not for an $sp^3$ carbon, though; see dodecahedrane). That's why you can't have constructions made of pentagons alone (or pentagons and heptagons, for that matter). Geometry would allow that, but chemistry wouldn't.
Let's see how that translates to actual shapes. Several fused hexagons give us coronene; it's planar.

(It is tilted sideways to give it a 3D feel. There are hydrogens attached on the outer edge; I omitted them for clarity.)
Extend it further, and you'll have an infinite flat sheet; that would be graphene.
Throw in a pentagon, and here's corannulene; it's bowl-shaped.

Extend that pattern, and you'll have fullerene.

Use a heptagon, and get circulene:

See that? It's bent indeed, but it's bent the wrong way. It's a
saddle-shaped bend. An example of negative Gaussian curvature, so to say. You can't wrap it to a sphere, no matter how hard you try.
Can we extend this further? Well, we kinda can (on paper, that is).
Moreover, it can be done in a number of ways, some leading to magnificent structures (purely hypothetical, of course).

Upd. It was brought to my attention that according to some studies, so-called carbon nanofoam might be a real example of structures of the latter kind. I'm not quite convinced (to me, until you have an X-ray diffraction study, everything is just hand-waving), but let's mention that anyway.

I was wondering whether anyone would look at this from a geometrical standpoint, and your answer is even more pleasant to read than I expected! I shall briefly point out Stone-Wales defects and carbon nanofoam as chemically-relevant examples of fused heptagons in carbon structures.

Thought of including the Stone-Wales thing at some point or another, but it didn't quite fit anywhere, and the answer is long enough already (by my standards, that is), so let's save it for another day.

Really, really great answer!


text:
Why does chemisorption increase with rise in temperature, even though it is an exothermic process?

Chemisorption is exothermic process, but still it increases with increase in temperature. Why?

I know that there needs to be a minimum activation energy to be achieved in order to initiate chemisorption. Also physiosorption can change to chemisorption in some cases on merely rising the temperature.
But why does it increase with an increase in temperature? How is this related to Le Chatelier's principle?

I am not sure if you are asking about the rate of chemisorption or the amount of bonds that are adhered at equilibrium. What do you mean by "it" in "But it still increase in temperature".

If you studied thermodyanamics you should know $\Delta G =\Delta H -T \Delta S$.Read the Wiki article.I hope you'll undertand after reading it. https://en.wikipedia.org/wiki/Chemisorption#Adsorption_Kinetics

@ZOZ The entropy is negative in case of sorption. Increasing temperature means $\Delta G$ to tend to 'positiveness'.

@ReeshabhRanjan You are correct "The entropy is negative in case of sorption. Increasing temperature means $\Delta G$ to tend to 'positiveness." But your mistake was that you though chemisorption increases with temperature which is incorrect.

You might find my answer to this question helpful: Is adsorption exothermic, and if so, why?

To answer that first think about a reaction like combustion reaction. Even though combustion reactions are exothermic you need to provide heat to the system by increasing its temperature. See, every reaction or chemical change happening requires a certain minimum amount energy called the "activation energy" for the reaction to start. This energy is required to break the bonds of the reactants so that they can react to form product. That's why we say that at low temperatures, kinetics controls the reaction rather than thermodynamics. Now, once you are at a temperature which is able to provide the sufficient activation energy to the reactants, the thermodynamics comes into play which essentially says in this case that if a reaction is exothermic, increasing the temperature will shift the reaction in backward direction.
Now coming to chemisorption, in this case too you need to increase the temperature to provide sufficient activation energy so that the reaction can start. So increasing the temperature increases the rate of reaction. After a certain temperature when thermodynamics become more important, the reaction shifts backward.
Hope it helps!

This behavior is expected because, like all chemical reactions, some activation energy is required for chemisorption. At  low temperature, x/m is small.  As temperature is increased, the molecules of the adsorbate gain energy and become equal to activation energy so that proper bonds are formed with the adsorbent molecules. Therefore, initially, the amount of gas adsorbed increases with rise in temperature. Further increase of temperature will increase the energy of the molecules which have already been adsorbed. This would increase the rate of desorption, thereby decreasing the extent of adsorption

No doubt, chemisorption is (mostly) exothermic. So if we increase the temperature, Le Chatelier tells us that the extent of adsorption decreases, i.e., equilibrium will lie less towards the adsorbed side.
But kinetics tells us that the temperature is a "universal catalyst", hence will increase the rate.
So our equilibrium favouring less adsorption will be achieved in less time.
It is the interplay of both the factors that gives the familiar hump-shaped plot of chemisorption with temperature. But I've never come across a graph online which labels the y-axis accurately. It must be the /extent of chemisorption achieved in a "feasible" time scale/, instead of just extent of chemisorption.

We know that in case of chemical adsorption, the forces acting are chemical bonds (ionic or covalent). We also know that temperature is directly proportional to activation energy.
So, when increasing the temperature, the activation energy is increased, so more bonds are formed. Thus, the magnitude of adsorption increases on increasing the temperature.

The increase in temperature is required to provide sufficient activation energy so that the reaction can start . So increasing the temperature increases rate of reaction initially . Then after a certain temperature by Le-Chatelier's principle further increase will shift the equilibrium to the left side and thus desorption will prevail. Moreover with further increase the energy of the molecules of adsorbate and in turn increases desorption.

Chemisorption first increase on increasing in temp. then decrease because initially breaking of old bond takes place the formation of new bond will be there on the surface of adsorbent after that these bond also start at very high temperature.


text:
Steady State Approximation
Is there anyone who can explain the steady state approximation to me? We’re given a series of reaction steps, and we need to deduce the rate law for the formation of the product.
It would be super helpful if someone could help me out with an example!
Have a nice day :)

Could you be a little more specific please? Also, welcome to chemistry.SE
Feel free to take the tour to learn more about the site. cheers!

http://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Kinetics/Reaction_Mechanisms/Steady-State_Approximation

I'm sure a simple google search would yield some results. I have, nevertheless, posted an answer, which might prove useful to you, and perhaps to future visitors.

The steady state approximation posits that after an initial induction periods, an interval during which the concentration of reaction intermediates rises from zero, and during major parts of the reaction, the rates of change of concentration of all reaction intermediates ($I$) are nearly equal to zero.
$$\frac{\mathrm{d}[\ce{I}]}{\mathrm{d}t} \approx 0 $$

Illustrative example
Consider the following sequence of consecutive elementary reactions:
$$\ce{A} \ce{->[k_a]} \ce{I}  \ce{->[k_b]} \ce{P} $$
One can write the rate of change of $\ce{I}$ as follows:
$$\frac{\mathrm{d}[\ce{I}]}{\mathrm{d}t} = k_a[\ce{A}] -k_b[\ce{I}] \approx 0$$
Thus, one obtains $\ce{[I]} = \frac{k_a}{k_b}[\ce{A}]$
The rate of formation of product, $$\frac{\mathrm{d}[\ce{P}]}{\mathrm{d}t} = k_b[\ce{I}]$$
However, using the previously obtained result one obtains $$\frac{\mathrm{d}[\ce{P}]}{\mathrm{d}t} \approx k_a[\ce{A}]$$
This brief example, I believe captures the essence of the steady state approximation. Intuitively, this would be a reasonable approximation to make if $k_b > k_a$. Which means that the first step is the bottleneck in the process, because once the intermediate appears, it immediately decomposes to yield products.

A more concrete example
Consider the following reaction:
$$ 2 \ce{N2O5}(g) \ce{->} 4\ce{NO2}(g) + \ce{O2}(g) $$
With the proposed mechanism,
$$  \ce{N2O5}(g) \ce{<=>[k_a][k'_a]} \ce{NO2}(g) + \ce{NO3}(g) \tag{i, reversible}$$
$$\ce{NO2}(g) + \ce{NO3}(g) \ce{->[k_b]} \ce{NO2}(g) + \ce{O2}(g) + \ce{NO}(g) \tag{ii}$$
$$\ce{NO}(g)+ \ce{N2O5}(g) \ce{->[k_c]} \ce{NO2}(g)+ \ce{NO2}(g)+\ce{NO2}(g) \tag{iii}$$
First step would be to identify the intermediates, i.e anything that is not a reactant or a product. In this case the intermediates would be $\ce{NO}$ and $\ce{NO3}$ because they are produced and consumed during the course of the reaction.
Second, write down the rates of change of concentration of intermediates and set them equal to zero.
$$\frac{\mathrm{d}[\ce{NO}]}{\mathrm{d}t} = \overbrace{k_b[\ce{NO2}][NO3]}^{\text{produces intermediate, step ii }} - \overbrace{k_c[NO][N2O5]}^{\text{consumes intermediate, step iii}} \approx 0 \tag{1}$$
$$\frac{\mathrm{d}[\ce{NO3}]}{\mathrm{d}t} = \overbrace{k_a[\ce{N2O5}]}^{\text{produces intermediate, step i }} -
\overbrace{k_a'[\ce{NO2}][\ce{NO3}]}^{\text{consumes intermediate, step i, rev}} -
\overbrace{k_b[\ce{NO2}][\ce{NO3}]}^{\text{consumes intermediate, step ii}} \approx 0 \tag{2}$$
Solving (1) and (2) one can obtains
$$[\ce{NO3}] = \frac{k_a[\ce{N2O5}]}{(k_a'+k_b)[\ce{NO2}]} \tag{3}$$
and $$ [\ce{NO}] = \frac{k_b[\ce{NO2}][\ce{NO3}]}{k_c[\ce{N2O5}]} = \frac{k_ak_b}{(k_a'+k_b)k_c} \tag{4}$$
Thus, the rate of change of $\ce{N2O5}$ is
$$\frac{\mathrm{d}[\ce{N2O5}]}{\mathrm{d}t} = -k_a[\ce{N2O5}] + k_a[\ce{NO2}][\ce{NO3}] -k_c[\ce{N2O5}][\ce{NO}] \tag{5}$$
Substituting the results from (3) and (4) into (5)
$$\frac{\mathrm{d}[\ce{N2O5}]}{\mathrm{d}t} = -k_a[\ce{N2O5}] + \frac{k_a'k_a[\ce{N2O5}]}{(k_a'+k_b)} - \frac{k_ak_b[\ce{N2O5}]}{(k_a'+k_b)} $$
Which simplifies, into a first order expression with respect to $\ce{N2O5}$
$$\frac{\mathrm{d}[\ce{N2O5}]}{\mathrm{d}t} = - \underbrace{\left( \frac{2k_ak_b}{k_a'+k_b} \right)}_{k_r}[\ce{N2O5}]$$
$k_r$ can be thought of as a "combined" rate constant for the first order decay of $\ce{N2O5}$ which depends on $k_a$, $k_a'$, $k_b$ but not on $k_c$


text:
Is it possible that for a reaction, it's activation energy equals total enthalpy of reaction?
Is there any reaction for which activation energy equals total change in enthalpy for the reaction . Is it possible in the first place ? Are there any interesting examples relating this concept ?

That would imply zero barrier for the reverse reaction, BTW.

What can we say about homolytic fission of... say chlorine molecule ?

Yeah, maybe something like that.

So we can say that activation energy for homolytic fission is same as  enthalpy change for overall reaction ? It's confusing

@getafix any help ?

Your question is confusing. What do you mean by total change in enthalpy?

Sorry for that ambiguity. I meant the heat of reaction . Or total enthalpy of reaction .

As Ivan said I guess $\ce{2Cl. -> Cl2}$ would fit the bill. Reverse reaction pretty much has no activation energy.

@orthocresol Did you mean $\ce{Cl2 -> 2Cl*}$?

Yes, I did. $\ce{}$

According to DavePhD, the following reaction satisfies the requirement:
$$\ce{O2N-NO2 -> 2NO2}$$


text:
A comprehensive list of theoretical approximations that are used in computational chemistry
I am having trouble recognizing all the approximations that are used in computational chemistry. I would like to start an answer list (similar as to the list of resources for learning chemistry) that addresses this question.
I will try to answer this question myself. I am hoping that anyone who is knowledgeable in the topic will correct and/or contribute. I am planning to start from general (e.g. Born Oppenheimer, LCAO) to specific (e.g. pseudopotential, functional). I am also planning to include why this approximation is necessary.

Asking for a comprehensive list makes this question impossible to answer between reasonable limits, as it effectively asks for a list that includes all the tricks used from integration, basis set approximations, through diagonalization to whatever that is included every single QChem software that meant for production and not just simple demonstrations.

@Greg I will try to avoid the math. I know that it is important and that it is sometimes difficult to separate the physics/chemistry from math. Otherwise it is too large, as you say. For example, I will not delve into how Frank Boys figured out that STO could be represented with GTO. Nor will I explain the different types of diagonalization (conjugate gradient, Davidson, etc). I will try not go to basis sets either unless there is a physical/chemical reasoning. Additionally, the reason why some basis sets are better or worse can be illustrated by first showing what has been sheared away first.

Overlooked in answers here is the reality of classical simulation, encompassing huge fields of monte carlo and molecular dynamics, which seeks to define mathematical fits to highly accurate quantum phenomena, and thereby runs a lot faster than solving the Schrodinger equation. There is no fundamental theory that proves this can work in all cases, but it has doubtless been useful: https://github.com/khavernathy/mcmd

@khaverim I suspect that this bias is primarily due to the site's users; most of us use or develop only the electronic structure theory side of things, so we have far fewer experts in MM/MD/MC, which means many related questions go completely unanswered.

@khaverim if you have anything to add, you can always post it as a new answer. Even if you don't think you have a full answer, you could start a community wiki answer that could help this post become a more comprehensive reference for other users.

The goal of computational chemistry is to obtain the properties of a system. This is done by solving Dirac's Equations.
Treating particles as point particles with mass
In most computational software, particles are treated as points with some mass. Neutrons and protons may be lumped into a nucleus. However, this is not true in all cases as pointed out in this question: Does computational chemistry include neutrons in ground state calculations? .
Electrons, protons, and neutrons are not simply point particles with some mass and charge. The theory and forces can get quite complex. I haven't been able to find journals that address the effect of this approximation. But I will try to post something if I come across it.
Neglecting Relativistic Effects
Wien2k has a nice summary of relativistic effects that need to be considered [8]. The relativistic effects that need to be included are:

Mass-velocity correction
The Darwin term
Spin-orbit coupling
Indirect relativistic effect

However, relativistic effects usually become important for elements further down the fifth row in the periodic table. This is because relativistic effects are dependent on nuclear charge. The velocity of the electron as well as spin orbit coupling increases as the nuclear charge increases.
This does not mean that relativistic effects do not affect "light" atoms. A good example that shows this is the sodium doublet which is a result of spin-orbit coupling. Nevertheless, solving Schrodinger's equation is enough in many cases.
Most computational chemistry software have the option of scalar relativistic calculations. The scalar relativistic technique was developed by Koelling and Harmon. These are calculations that include the Darwin term and the mass velocity correction. According to Koelling and Harmon's paper:

We present a technique for the reduction of the Dirac
equation which initially omits the spin-orbit interaction (thus keeping spin as a good quantum number), but retains all other relativistic kinematic effects such as mass-velocity, Darwin,and higher order terms.[9]

Some offer fully relativistic calculations, which include all (or most) relativistic effects. But these are rarer to find, and are only available for certain cases.
So why are relativistic effects neglected? Well, relativistic effects are small for light molecules and relativistic calculations are expensive

This is so because relativistic calculations need self-consistent solutions of about twice as many states as non-relativistic ones. [10]

Thus, scalar relativistic calculations offer a nice middle ground between efficiency and accuracy.
The Born-Oppenheimer Approximation
I am not sure what motivated Born and Oppenheimer to use this approximation. The seminal paper is in German.[1] It appears that the motivation was to simplify the Hamiltonian. According to the introductory course at MIT,[2]

it allows one to compute the electronic structure of a molecule
without saying anything about the quantum mechanics of the nuclei

And according to Wikipedia,[3] it reduces the amount of computations:

For example, the benzene molecule consists of 12 nuclei and 42
electrons. The time independent Schrödinger equation, which must be
solved to obtain the energy and wavefunction of this molecule, is a
partial differential eigenvalue equation in 162 variables—the spatial
coordinates of the electrons and the nuclei.

As many approximations, it does not hold true in all scenarios. Cases where the Born-Oppenheimer approximation fail are:[4]

ion-molecule, charge transfer, and other reactions involving obvious
electronic curve crossings

Qualitatively, the Born-Oppenheimer approximation says that the nuclei are so slow moving that we can assume them to be fixed when describing the behavior of electrons. Mathematically(?), the Born-Oppenheimer approximation allows to treat the electrons and protons independently. This does not imply that the nuclei and electrons are independent of each other. In other words, it does not mean that the nuclei are not influenced by the motion of electrons. The nuclei still feel the motion of the electrons. In addition, the Born-Oppenheimer approximation does not say that the nuclei does not move. It only means that when describing the motion of electrons, we assume that the nuclei are fixed.
No Analytical Solution to Dirac/Schrodinger Equation
Unfortunately there is no analytic solution to the Dirac equation for any atom that has more than one electron even after the Born-Oppenheimer approximation (a list of quantum-mechanical problems that have an analytical solution is available on Wikipedia[5]). Many texts state that the reason as to why the Schrödinger equation is not exactly solvable for more than one electron is due to the Coulomb repulsion between electrons.[6]
However, this is not entirely true. A counterargument is Hooke's atom. The Hamiltonian for Hooke's atom has an Coulomb electron-electron repulsion term. However, it has an exact solution for more than one electron under certain circumstatnces.[7]
The true reason as to why the Schrödinger equation is not solvable for multi-electron atoms is due to the fact that the motion of electrons cannot be decoupled from each other. In other words, the Hamiltonian is not separable for a multi electron system. If we were to get rid of the electron-electron Coulomb repulsion, the motion of the electrons can be decoupled. This may be the reason as to why the electron-electron Coulomb repulsion (a.k.a. electron correlation) is used as the reason why the Schrödinger equation is not exactly solvable.
From non-interacting to the real thing
Since the Dirac equation cannot be solved analytically, we must make  models that are solvable and add approximations to it. These approximations are further refined to get more accurate results.
The most simple model (and the foundation for computational chemistry) is the system of non-interacting electrons. As the name suggests, the electrons do not interact with other electrons. This allows us to write the Hamiltonian for all electrons as the sum of one-electron Hamiltonians.
The one electron Hamiltonian consists of a kinetic energy term and a potential energy term.
$$\mathcal{H}=\sum^N_i \left(\frac{\hbar^2}{2m_i\nabla^2 }+V_i\right)$$
The solutions for the non-interacting system of electrons are analytical. And they give a starting point for other calculations. Hartree-Fock, DFT, and solid state physics use this simplified model.
Note: it appears that there are several terms that say something similar to this. There is the independent electron approximation and the central field approximation. I'll dig into literature to see what are the differences between these three terms.
Variations of the non-interacting system of electrons
There are several variations of non-interacting electron systems. In all these, the electrons do not interact with each other. What sets them apart from each other is the potential energy term of the Hamiltonian $V_i$. This potential energy term is often called the effective potential.
Free electron model
This model is used as the starting point in solid state physics. The free electron model describes the behavior of valence electrons in a metal or semiconductor. Here, the potential $V_i$ is equal to 0 for all electrons. It does not feel the effect of other electrons or nuclei. The wavefunctions of this model are plane waves.
Nearly free electron model
The nearly free electron includes a weak periodic potential. In other words $$V_i(r)=V_i(r+k)$$ It is used to describe periodic systems such as ideal crystals. The wavefunctions of the nearly free electron model are Bloch waves. Bloch waves are plane waves that are multiplied by a periodic function.
Hartree method
In the Hartree method, the potential considers repulsion between two electrons and attraction to the nucleus. In this model,

Hartree assumed that each electron moves in the averaged potential of
the electrostatic interactions with surrounding electrons [11]

Hartree replaced the electron-electron interaction with an effective potential that only depended on the coordinates of the $i$th electron. This effective potential describes an electron interacting with an electron cloud.
From one-electron orbitals to the multielectron orbital
As mentioned above, the total Hamiltonian can be approximated as the sum of one-electron Hamiltonians for non-interacting electrons. However, the wavefunction has to satisfy antisymmetry.
Exact exchange
Electron correlation
Some define electron correlation as everything that the Hartree-Fock method leaves out.
Electron correlation: DFT edition
Electron correlation: Post-Hartree-Fock edition
Periodic systems and pseudopotentials

References

M. Born and R. Oppenheimer, Ann. Phys. 1927, 389, 457–484.
doi: 10.1002/andp.19273892002
Born Oppenheimer Approximation. Open Courseware MIT:Introductory Quantum Mechanics. Fall 2005. Section 12 Lecture. (https://ocw.mit.edu/courses/chemistry/5-73-introductory-quantum-mechanics-i-fall-2005/lecture-notes/sec12.pdf)
(https://ocw.mit.edu/courses/chemistry/5-73-introductory-quantum-mechanics-i-fall-2005/)
Born-Oppenheimer approximation. Wikipedia (https://en.wikipedia.org/wiki/Born%E2%80%93Oppenheimer_approximation)
L. J. Butler, Annu. Rev. Phys. Chem. 1998, 49, 125-71.
PMID: 15012427
doi: 10.1146/annurev.physchem.49.1.125
List of quantum-mechanical systems with analytical solutions. Wikipedia
LibreTexts: 9.1: The Schrödinger Equation For Multi-Electron Atoms
Hooke's atom. Wikipedia
Summary of relativistic effects. Wien2k  (http://www.wien2k.at/reg_user/textbooks/WIEN2k_lecture-notes_2013/Relativity-NCM.pdf)
A technique for relativistic spin-polarised calculations. Journal of Physics C: Solid State Physics, Volume 10, Number 16
(http://iopscience.iop.org/article/10.1088/0022-3719/10/16/019/meta)
The Scalar Relativistic Approximation. Takeda, T. Z Physik B (1978) 32: 43. doi:10.1007/BF01322185 (http://link.springer.com/article/10.1007/BF01322185)
Density Functional Theory in Quantum Chemistry. Tsuneda, T. 2014. ISBN: 978-4-431-54824-9. Page 36.

The reason is due to the Coloumbic Repulsion between electrons.  Hooke atom also has Coloumbic Repulsion between electrons.

@Rodriguez I will look into that. Also, this is an ongoing answer. It is still not complete! Nevertheless I would appreciate any feedback.

Schroedinger equation is approximation, better go with Dirac to have relativity (and spin).

Born-Oppenheimer is two-step process. First is separation of total wavefunction into product of nuclear and electronic part, second is setting the nuclear kinetic energy to zero. (see https://en.wikipedia.org/wiki/Born%E2%80%93Oppenheimer_approximation)

Please use manual markup for links, i.e. [link](http://...). It looks better, is easier readable, and will not lead into nirvana if interpreted wrong by the page. Just for funs, you can check what I mean by clicking the links in revision 8.

As @ssavec already mentioned, the first usual approximation is neglecting relativistic effects. 2. Big mistake in the very first sentence: the goal is to obtain properties of a molecular system. Then, depending on the formalism used (MM, WFT, DFT, ...)  you are looking for a mathematical entity that can somehow spits out the values of the properties. And this entity is not always the wave function. 3. Don't get me wrong, but if you don't quite understand the purpose of arguably the central approximation in present day QC (BO one), it might be a bit to early for you to answer the question.

@Wildcat I agree to some degree. Part of the purpose for this question is to verify what I know since I haven't taken a formal course of computational chemistry at Uni. In addition, I'm currently not in academia...so it's hard to find someone to ask. Nevertheless, I am already finding this question useful due to feedback from the community. Could you tell me any text that explicitly mentions the motivation of BO. I haven't been able to find one so far. Also thanks to everyone for feedback.

@QuantumAMERICCINO You want a chapter about RI-approximations? I can write one if the need arises.

@Fl.pf. That would be awesome!  Wikipedia doesn't even have an article about it. Could you add it as a new answer ?

Where is the term "multielectron orbital" coming from? I have never heard it before, and it is actually contradictory to the definition of an orbital being an "one-electron wave function".

@QuantumAMERICCINO There is some stuff I would explain differently. I think I will write something about the Variational Principle and (Post)-Hartree-Fock as a new answer. Maybe even some DFT. Then we can see how to connect it best to your content.

I have several issues with this answer. For instance, not all the properties of a system can be obtained from the Dirac equation. Also, while protons and neutrons have structure, electrons are point particles.

Speeding up two-electron integral evaluation through approximate methods
Remarks on Notation (using Mulliken Notation)

$\mu,\nu,\kappa,\lambda:$ atomic orbital (AO) basis functions
$P,Q,R,S:$ auxiliary basis functions
$(\mu\nu|\kappa\lambda) = \int\int\phi_\mu^*(r_1)\phi_\nu(r_1)\frac{1}{r_{12}} \phi_\kappa^*(r_2) \phi_\lambda(r_2)d\tau_1d\tau_2 = \left<\mu\kappa|\nu\lambda \right>$
$(\mu\nu|1|P) = (\mu\nu P)$
$(\mu\nu|g_{12}|P) = (\mu\nu|P)$ where $g_{12}$ is the two-electron interaction, almost always $= \frac{1}{r_{12}}$
$S_{PQ} = (PQ)$
$V_{PQ} = (P|Q)$

Resolution of the Identity approximations
Computation of 2-electron 4-centre Integrals $(\mu\nu|\lambda\sigma)$ can be a significant bottleneck in electronic structure calculations.
$\rightarrow$ Idea of RI is to avoid such integrals
Proposed solution: Reexpansion of pair products $|\mu\nu)$ with
$$
|\mu\nu) \approx |\widetilde{\mu\nu}) = \sum_P C_{\mu\nu}^P |P)
$$
Where we choose the auxiliary basis set $\{P\}$ to be $\sum_P |P)(P| \approx 1$  (Hence the name Resolution of the Identity)
If we approximate the unity operator $\hat{1}$ in the Coulomb metric
$\hat{1} \approx \sum_{PQ} |\chi_P)(P|Q)^{-1} (\chi_Q|$
The integral $(\mu\nu|\kappa\lambda)$ can then be approximated as
$(\mu\nu|\kappa\lambda) \approx \sum_{PQ}(\mu\nu|P)(P|Q)^{-1}(Q|\kappa\lambda)$.
The same formulation is obtained if the the integral error $\Delta (\mu\nu - \widetilde{\mu\nu}|\kappa\lambda - \widetilde{\kappa\lambda})$ is minimized with respect to the coefficients $C_{\mu\nu}^P$.
This leads to a reduction of computation cost since we need only to compute three-index integrals. These are much faster to compute than four-index integrals. Additionally, a lot fewer three-index index integrals have to be computed than four-index integrals would be.
The RI approximation is used to accelerate the calculation of the Coulomb operator $\hat{J}$ in the RI-J approximation Here the electron density is expanded in the auxiliary set ($D_{\kappa\lambda}$ is the density matrix):
$$
J_{\mu\nu} = (\mu\nu|\rho) \approx \sum_{PQ} \sum_{\kappa\lambda} (\mu\nu|P)(P|Q)^{-1}(Q|\kappa\lambda)D_{\kappa\lambda}
$$
If you implement this in your program in such a way that you calculate all the quantities strictly from right to left, the scaling is brought down from $N^4$ to $N^3$.
It can also be used to approximate the Exchange operator $\hat{K}$. Here no reduction of the scaling is possible, but the prefactor is reduced.
RI approximations are essential also for MP2/CC2-R12/F12 methods. There you need the robust expression $(\mu\nu|\kappa\lambda) \approx (\widetilde{\mu\nu|\kappa\lambda}) = (\widetilde{\mu\nu}|\kappa\lambda) + (\mu\nu|\widetilde{\kappa\lambda}) - (\widetilde{\mu\nu}|\widetilde{\kappa\lambda})$
for example for the operators $r_{12}$ and $[\hat{T}_{12},r_{12}]$.
The auxiliary basis sets can be used freely, but they have to be fitted for their purpose (meaning there are individual auxiliary basis sets for every operator, often denoted $jbas$ (for pair products $(ij|$), $cbas$ (for pair products $(ia|$), $i,j$ being occupied, $a$ being virtual orbitals. I know of no aux basis sets for doubly occupied pair products $(ab|$, so there are certain integrals where RI/DF can't be used afaik.
Cholesky Decomposition
The two-electron integrals $(\mu\nu|\kappa\lambda)$ can be expressed in a positive definite Hermitian $V_{\mu\nu,\kappa\lambda}$.
Since this matrix is positive definite it can be decomposed via a Cholesky decomposition $\mathbf{V} = \mathbf{LL}^{\dagger}$.
For a detailed explanation read the original publication by Beebe and Linderberg.
In this approximation the basis set $\{P\}$ is conceived via the orbital basis, so no generation of a basis set is needed.
While CD is computationally more demanding than RI, it is numerically robust and can be used for more precise results. Since the only constricting value is the threshhold at which the expansion is stopped, the precision of the CD in theory can be chosen arbitrarily and systematically.
References

P. Merlot, T. Kjaergaard, T. Helgaker, R. Lindh,  F. Aquilante, S. Reine, T. B. Pedersen, J. Comp. Chem. 2013, 34, 1486 - 1496
O. Vahtras, J. Almloef. Chem. Phys. Letters 1993, 213, 514-518
F. Weigend, M. Kattanek, R. Ahlrichs, J. Chem. Phys., 2009, 130, 164106-1
N. H. F. Beebe, J. Linderberg, Int. J. Q. Chem., 1977, 12, 683-705

Variational Principle
The Variational Principle states that any approximate solution to the true wave function $\Psi$ will be higher in its total energy. The total energy is calculated as the expectation value of the Hamiltonian $E=\langle\Psi|\hat H|\Psi\rangle$. Thus we can set up a trial wave function $\tilde\Psi$ and vary its parameters until its corresponding energy $\tilde E=\langle\tilde\Psi|\hat H|\tilde\Psi\rangle$ reaches a minimum. This solution provides an upper limit for the true (exact) energy ($\tilde E \le E$) as well as the best possible approximation to the wave function.
When comparing variational methods we can directly judge which result is better by comparing their total energy. Examples for variational methods are Hartree-Fock and Configuration Interaction.
Non-variational methods are Coupled Cluster, Density Functional Theory and Pertubation Theory.
Note that the total energy by itself has not much physical meaning and depends on different numerical parameters. Of practical interest are always difference between total energies, for example binding energies.
Hartree-Fock
The main issue of solving the Schrödinger is the Coulomb interaction between electrons. There is no general analytic solution for more than 1 electron. However, for the non-interacting system,  which ignore electron-electron interaction completely, one can employ separation of variables for the electronic coordinates. This allows us to express the total electronic wave function as a product of one-electron wave functions (orbitals). This is called the Hartree-Product and when additionally accounting for the anti-symmetry as required by the Pauli principle, it leads to the Slater-Determinant.
In the Hartree-Fock method the Slater-Determinant, which is only exact for the non-interacting system, is combined with the Hamilonian including the electron-electron interation term. Further applying the Variational Principle leads to the Hartree-Fock equations, which may be solved numerically. The physical interpretation of this approximation is, that the electrons feel each other only in an averaged way, hence the name mean-field approximation.
For the numerical solution one needs to diagonalize the Fock matrix resulting in its eigenvalues and eigenvectors. Since the Fock matrix depends on its own eigenvectors, this is solved iteratively: After choosing an initial guess the result is used to update the Fock matrix, which is then diagonalized again. After each iteration an improved result is obtained. This is known as self-consistent field method.
The SCF algorithm scales with $M^3$, where $M$ is the number of basis sets. However, the preceding calculation of all required one- and two-electron integrals is usually more time consuming than the actual HF calculation.
Electron Correlation & Post-Hartree-Fock
The correlation energy is defined as the difference between the Hartree-Fock energy and the exact energy. Electron Correlation is the error made in the Hartree-Fock approximation, since the Ansatz made for the electronic wave function does not reflect its true form.
The numerically exact solution can be obtained with the Full Configuration Interaction method.
Configuration Interaction
The electronic wave function, like any other wave function in quantum mechanics, can be expanded in an arbitrary basis set. In Configuration Interaction it is recognized that the electron configurations, which can be created based on the molecular orbitals obtained from a Hartree-Fock calculation, can be used as such a basis.
Note that this basis of electron configurations is a different one, than the one-electron basis set which is used to to represent the molecular orbitals.
A Configuration Interaction calculation is thus preceded by a Hartree-Fock calculation. The Slater-Determinant used in the Hartree-Fock calculation is the first, and usually most important, configuration. Further excited determinants (configurations) are generated by promoting electrons from occupied orbitals to virtual (in the HF configuration unoccupied) ones. Applying again the Variational Principle one can calculate that linear combination of such configurations that minimized the electronic energy.
Computationally one needs to set up the matrix representation of the Hamiltonian (in the basis of the chosen basis of the configurations) and diagonalize it. In contrast to the Fock matrix it does not depend on its own solution, so no iterative procedure is required here. However, this is commonly speed up by iterative approaches which are much less time consuming than an exact diagoanlization.
In total $\binom{M}{N}$ configurations can be generated, where $M$ is the number of orbitals and $N$ is the number of electrons. Since this number increases exponentially, only the smallest molecules can be calculated even on the fastest super computers (of course this also depends on various other numerical parameters). Therefore different truncation schemes exist to only include some of the configurations.

Full CI includes all possible configurations and is the numerically exact limit.
Hartree-Fock from the perpective of CI it is known the one-determinant approximation.
CIS includes only configrations where 1 electron is excited with respect to the HF configuration (Single excitations). Due to Brillouin's theorem, no improvement of the HF ground state energy is made, but rough approximations for excited states can be obtained.
CID includes only configrations where 2 electrons are excited with respect to the HF configuration (Double excitations). First improvements to the ground state energy.
CISD since Single excitation couple with Double and those in turn with the HF ground state, this is an improvement over CID. Furthermore, there are much less Single excitations than Double excitations, which means computationally CISD has no considerable additional efford over CID.
CISDT... The higher the excitation degree the smaller the correction to the electronic energy, but the higher the computational demand.
CASCI stands for Complete Active Space CI. The active space is a chosen set of orbitals within which all possible configurations are considered. Orbitals with energy below the active space orbitals are (doubly) occupied for all configurations, orbitals above in energy are always left empty.

Multi-configurational self-consistent field
The MCSCF approach combines a CI calculation with the HF method in the sense that the coefficients of the chosen configurations (CI optimization) and the molecular orbital coefficient (SCF optimization) are optimized simultaneously. A common example is the CASSCF method.
Coupled Cluster
Coupled Cluster is a reparametrization of the electronic wave function. In the CI approach the wave function can be written as an excitation operator $\hat T$ acting directly on the HF determinant and creating all possible excited configurations. The CC wave function is obtained by having the exponential operator $\exp(\hat T)$ acting on the HF determinant.
The effect is that for example CCSD will not only include the Single and Double excitations, but certain Triple and Quadruple excitations as well, making the method size consitent.
Unfortunately the arising Coupled Cluster equations cannot be solved by employing the Variational Principle, hence the resulting total energy may be lower than the exact FCI energy. This also means that a CC calculation including all possible configurations is not equivalent to FCI. Although CCSD seems to be an exception here.
CC is computationally more expensive than a CI calculation of the same excitation operator, but results are more accurate.
CCSD(T) is considered the gold standard of quantum chemistry and includes pertubative estimates of the Triple excitations. It's overall scaling is in the order of $N^7$.
Multi Reference Methods
Instead of creating excitated configurations only based on the HF configuration, one can also have multiple reference configurations. Commonly one starts with a MCSCF wave function and its configurationsi as reference. In a second step one can do for example do a MRCI or MRCC calculation.
This approach is required for strongly correlated systems, where other methods (DFT, CCSD(T)) are failing.
References

A. Szabo and N. S. Ostlund. Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory (Dover Books on Chemistry). Dover Publications, 1996.
T. Helgaker, P. Jorgensen, and J. Olsen. Molecular electronic-structure theory. 1. Auflage. New York: Wiley, 2000.


text:
What household compound will neutralize naphthalene-stained washer (for clothes)?
Please help me tag this post.
Googling provided no answers except for vinegar, which did not work.
I washed this blanket for the first time that was purchased off of ebay.  There was a smell emitting from the washer as it was washing clothes.
This is no ordinary "mothball smell".  It is extremely stinky, like 30 magic markers were opened in the wash.  It's almost as if the blanket was DIPPED in naphthalene.
I tried running it empty with all the vinegar we had which was about 6-10 cups, together with powdered laundry soap, and both hot and cold wash cycles.  Will it help if I added some throw-away cloth in it to clean it out?  I even put it on a 2-hour wash.  The smell is as potent as when we started!
Will airing it out reduce the smell?  Ideally, I would like to completely remove the naphthalene, not just the smell.
Is cold better than hot water for this?  And lastly, what household compound can I buy that will neutralize the naphthalene?
Thanks.

Again, this is no ordinary mothball-like smell.  The smell was so intense, that I thought I almost poisoned myself.  It fills the whole house without the windows open.

Going to the dry cleaners may be an option; both their experience as well as proper equipment (including using not only water, but organic solvents to remove to clean).  Because washing a large tissue like a blanket in vodka or the non-drinkable Everclear's denaturated / rectified spirit may be a bit expensive.

Air-drying may do the trick. I.e. leave it open in warm wind for a week or two. Other than that the best bet would be either organic solvent, like tetrachloroethylene used in dry cleaning, or a professional detergent. By 'professional' I mean something industrial-grade, made to handle heavy machine oil stains.

Napthalene is not water soluble, so the temperature doesn't matter as much as the detergent. What makes you think it's napthalene, though? It could be some organic compound in which case you need a good de-greaser (Dawn is the easiest to get, but there are better industrial ones like what cleaning crews may use). If what you're dealing with is an aromatic hydrocarbon (which napthalene is), airing it out will go a long way. I would see how far you can get with a fan blowing it out before you wash the rest out with a good de-greaser.


text:
Why is DMSO more polar than other solvents, yet it dissolves ionic compounds worse?
I'm working right now in the solvent importance in ionic character, and I noticed how a binary salt ($\ce{CuCl2}$) dissolved better in methanol than in dimethylsulfoxide, and in contrast, a coordination compound (Cu[salen]) dissolved better in DMSO that in the aforementioned solvent.
It doesn't make a lot of sense to me considering these properties I found:
DMSO:

Dipole moment: 3.96
Dielectric constant: 47

Methanol:

Dipole moment: 1.71
Dielectric constant: 33

I would have expected the binary salt to dissolve better in DMSO rather than Methanol, and the inverse in the coordination compound yet it didn't happen like that. Is there a intermolecular detail I'm not considering? An hydrophobic-hydrophilic portion in the structure? Thanks everyone.

Steric hinderance?

I was thinking more of the aprotic nature of the DMSO, and its inability to form hydrogen bonds. What were your thoughts?

Having large dipole moments and dielectric constants, as DMSO does, does not imply that it has the ability to solvate ionic compounds like $\ce{CuCl2}$.
Although both methanol and DMSO have oxygen atoms with their lone pair of electrons to help solvate cations, there is no corresponding functionality in DMSO to solvate a small, inorganic anion like chloride.  Methanol on the other hand can solvate chloride via hydrogen bonding in the same way that water does.
These properties are not necessarily applicable to a coordination compound, so there is no reason to expect methanol to act as a better solvent than DMSO in this case.


text:
Can flames be colored black?
Most of us here would know that flames can be colored by adding an appropriate salt to the 'fuel' or by simply introducing it into a flame. Boron and Barium salts give the flame a green tinge, Strontium colors it red, Sodium does gold, copper a bluish-green, etc, etc.
Now from my (not-so-in-depth) high-schooler level of understanding, the energy released by whatever is being burnt to produce the flame in the first place (as is usually the case), excites electrons in metal ions in the salt to higher energy levels. And when these electrons 'de-excite' and drop back to the original energy levels, they emit that extra energy as photons (whose frequencies are determined by the Planck equation $E = h\nu$ ) and these photons are what we perceive as the colored flame.
If I'm not mistaken the book Harry Potter and the Philosopher's stone mentions a 'black flame' of sorts, being one of the obstacles Harry has to circumvent in order to reach the Stone. (Artist's depiction)

Now, as I understand it, for something to be 'black', it must absorb all ( or at least a fairly significant amount of) visible light, encompassing a wide range of frequencies.
Going back to colored flames, I mentioned that the ions must emit light energy, not absorb it, in order to color the flame.
Armed with all this information, would anyone know if it really is possible to create 'black colored flames'? Has it ever been achieved?
[I tried thinking it out myself.....just gave me a headache]
Edit 1: And yes, I've made note of what @Mithoron mentioned, but what I mentioned was just conventional logic, it isn't universal though.
...Newtonian mechanics was 'conventional logic', but we all saw how that failed in relativistic scenarios. So I didn't want to take the risk, hence I've asked If it actually is possible...
Edit 2: Just because someone mentioned otherwise in an answer... I emphasise that by black flames I don't mean invisible flames.

You basicly answered this - "black flame" is nonsense.

Thinking outside the box here, vaporization of solid iodine may be the closest you can get to a "black flame". There is no chemical reaction involved, and the process consumes heat rather than releases it (so it cannot be self-sustaining), but it would generate dark purple wisps.

Off course: it is called smoke

The flame might of course give off radiation in a range not visible to the eye; one night even suppose that an exothermic reaction that heats the air around it and makes the air rise thereby makes a "flame" of sorts from the rising air.  But that is invisible as such, not black.

In a way, all flames are colored black.

If black means that the flame does not emit light that is visible to the human eye then there is a certain chance that black flames might exist.
Since hot flames emit black body radiation and the spectrum of visible light reaches from about $\pu{400nm}$ to $\pu{800nm}$, a black flame had to be cool enough to not emit too high intensities of light below $\pu{800nm}$. So we're searching for a cool flame.
While cool flames are hardly visible in daylight, it might be difficult to find one that is invisible in a dark environment. Here is a report of a cool flame that could only made visible in the dark by long photographic lighting.

This video shows how a black flame is achieved. If you illuminate the fire with a monochromatic light source (sodium vapor lamp) and introduce a species in the fire that absorbs that wavelength (sodium ions) then the fire will in fact appear black under the illumination. A screen shot is shown below:

Click image for video.

This is an exceptionally interesting video, thanks very much for sharing it. I do feel, however, that a more proper description for that flame is "anti-yellow", or to be extremely precise, "anti-sodium D line", rather than black. It is not my intention to demote your contribution, nor to be needlessly nitpicky. I merely wish to state that there is a subtle distinction, which provides interesting food for thought and a nice stepping stone into further topics, such as atomic absorption spectroscopy.

When you think of light, white is a mixture of all the colors, but paint and pigment black is a mixture of all the colors. Black to get in light is either the absence of light, or a white shade of black when compared to a white background. The parlor trick is maybe the only way to achieve the black flame.

Nonetheless it does change the proprities of the plasma. Where fire is a light source, the experiment also shows the differences between the flame compared to a lighter with an ordinary flame. The results depicted the lighter with no shadow and the black flame with a shadow. Like you said this does make an interesting question. I would also add that there are many theroys that darkness is faster then the speed of light. I’m sure I have a well appreciate video to correspond to that as well.

Prelude:  I do not know this part of the story you refer to.
The picture you display here, however, contradicts the idea of "black flame", because if there were a black flame, it should absorb the light, and attenuate the sight across it.  There is little evidence, however, that one of the seven containers is source of a "black flame" as the background is (still) clearly visible, and the blur likely more a question of "out of focus".
As @ZOZ correctly indicated, "black flames" you / the story refer(s) to do not exist as such.  The black of a flame occasionally seen often is result of incomplete combustion in shortage of oxygen. To provide an example, starting a Diesel engine may generate soot.
A third cent.  Your "Going back to colored flames, the ions must emit light energy, not absorb it, in order to color the flame." needs a revision.  True, ions present in the flame may emit light due to their de-excitation; similar to additive color mixing. To enlarge the picture, however, absorption may still occur -- similar to subtractive color mixing. This is element specific (Fraunhofer lines in the solar spectrum, for example) and is used in analytical chemistry atomic absorption spectroscopy, too.  The likelihood of the two processes is different, however.  Hence to be efficient, AAS works on temperatures largely exceeding the ones provided by a normal fire, a candle, a medieval torch.

There is no way to create black fire because when it comes to light black is the absence of light not the mixture of all the colors light can produce. Although pigments which refers to paint, mixing them would create black paint but there is no way to make black fire sorry.

Using an (I think) hydrogen vapor lamp and salt water it is possible to make the fire absorb all of the light at the particular wavelength that the vapor lamp emits.

This answer has already been given.