text:
Why is BCl3 a monomer whereas AlCl3 exists as a dimer?
What makes dimerization possible in $\ce{AlCl3}$? Are there 3c-2e bonds in $\ce{Al2Cl6}$ as there are in $\ce{B2H6}$?

Possible duplicate of Why does AlCl3 dimerise to Al2Cl6 at low temperatures?

I think the question is related, but I think it is distinctively different also. On the other hand, the answer at the other question does not really explain why the boron compound does not dimerise.

Extra links in decreasing order of relevance: 1, 2, 3, 4, 5, 6.

Introduction
The bonding situation in $\ce{(AlCl3)2}$ and $\ce{(BCl3)2}$ is nothing trivial and the reason why aluminium chloride forms dimers, while boron trichloride does not, cannot only be attributed to size.
In order to understand this phenomenon we need to look at both, the monomers and the dimers, and compare them to each other.
Understanding the respective bonding situation of the monomers, is key to understand which deficiencies lead to dimerisations.
Computational details
Since I was unable to find any compelling literature on the subject, I ran some calculations of my own. I used the DF-M06L/def2-TZVPP for geometry optimisations. Each structure has been optimised to a local minimum in their respective symmetry restrictions, i.e. $D_\mathrm{3h}$ for the monomers and $C_\mathrm{2v}$ for the dimers.
Analyses with the Natural Bond Orbital model (NBO6 program) and the Quantum Theory of Atoms in Molecules (QTAIM, MultiWFN) have been run on single point energy calculations at the M06/def2-QZVPP//DF-M06-L/def2-TZVPP level of theory.
A rudimentary energy decomposition analysis has been done on that level, too.
Energy decomposition analysis
The dissociation energy of the dimers $\ce{(XY3)2}$ to the monomers $\ce{XY3}$ is defined as the difference of the energy of the dimer $E_\mathrm{opt}[\ce{(XY3)2}]$ and double the energy of the monomer $E_\mathrm{opt}[\ce{XY3}]$ at their optimised (relaxed) geometries $\eqref{e-diss-def}$.
The interaction energy is defined as the difference of energy of the relaxed dimer and double the energy of the monomers in the geometry of the dimer $E_\mathrm{frag}[\ce{(XY3)^{\neq}}]$ $\eqref{e-int-def}$. That basically means breaking the molecule in two parts, but keeping these fragments in the same geometry.
The deformation energy (or preparation energy) is defined as the difference of the energy of the optimised and the non-optimised monomer $\eqref{e-def-def}$. This is the energy required to distort the monomer (in its ground state) to the configuration it will have in the dimer.
$$\begin{align}
E_\mathrm{diss} &=
  E_\mathrm{opt}[\ce{(XY3)2}] - 2E_\mathrm{opt}[\ce{XY3}]
  \tag1\label{e-diss-def}\\
E_\mathrm{int} &=
  E_\mathrm{opt}[\ce{(XY3)2}] - 2E_\mathrm{frag}[\ce{(XY3)^{\neq}}]
  %\ddag not implemented
  \tag2\label{e-int-def}\\
E_\mathrm{def} &=
  E_\mathrm{frag}[\ce{(XY3)^{\neq}}] - E_\mathrm{opt}[\ce{XY3}]
  \tag3\label{e-def-def}\\
E_\mathrm{diss} &=
  E_\mathrm{int} + 2E_\mathrm{def}\tag{1'}
\end{align}$$
Results & Discussion
The Monomers $\ce{XCl3; X{=}\{B,Al\}}$.
Let's just get the obvious out of the way: Boron is (vdW-radius 205 pm) smaller than aluminium (vdW-radius 240 pm). For comparison chlorine has a vdW-radius of 205 pm, too. That is pretty much reflected in the bond lengths and the chlorine-chlorine distance.
\begin{array}{llrrr}\hline
&\ce{X{=}}& \ce{Al} &\ce{B} &\ce{Cl}\\\hline
\mathbf{d}(\ce{X-Cl})&/\pu{pm} & 206.0 &173.6&--\\
\mathbf{d}(\ce{Cl\bond{~}Cl'})&/\pu{pm} & 356.8 & 300.6 & --\\\hline
\mathbf{r}_\mathrm{vdW}&/\pu{pm} & 240   & 205   & 205\\
\mathbf{r}_\mathrm{sing}&/\pu{pm} & 126   & 85   & 99\\
\mathbf{r}_\mathrm{doub}&/\pu{pm} & 113   & 78   & 95\\\hline
\end{array}
From this data we can draw certain conclusions without further looking. The boron monomer is much more compact than the aluminium monomer. When we compare the bond lengths to the covalent radii (Pyykkö and Atsumi) we find that the boron chloride bond is about the length that we would expect from a double bond ($\mathbf{r}_\mathrm{doub}(\ce{B}) + \mathbf{r}_\mathrm{doub}(\ce{Cl}) = 173~\pu{pm}$). While the aluminium chloride bond is still significantly shorter than a single bond ($\mathbf{r}_\mathrm{sing}(\ce{Al}) + \mathbf{r}_\mathrm{sing}(\ce{Cl}) = 225~\pu{pm}$), it is still also much longer than a double bond ($\mathbf{r}_\mathrm{doub}(\ce{Al}) + \mathbf{r}_\mathrm{doub}(\ce{Cl}) = 191~\pu{pm}$).
This itself offers compelling evidence, that there is more π-backbonding in $\ce{BCl3}$ than in $\ce{AlCl3}$. Molecular orbital theory offers more evidence for this. In both compounds is a doubly occupied π orbital. The following pictures are for a contour value of 0.05; aluminium (left/top) and boron (right/bottom)

In numbers, the main contributions are as follows (this is just a representation, not the actual formula):
$$\begin{align}
\pi(\ce{BCl3}) &=
   21\%~\ce{p_{$z$}-B} + \sum_{i=1}^3 26\%~\ce{p_{$z$}-Cl^{$(i)$}}\\
\pi(\ce{AlCl3}) &=
   13\%~\ce{p_{$z$}-Al} + \sum_{i=1}^3 29\%~\ce{p_{$z$}-Cl^{$(i)$}}
\end{align}$$
There is still some more evidence. The natural atomic charges (NPA of NBO6) fairly well agree with that assesment; aluminium is far more positive than boron.
$$\begin{array}{lrr}
                    & \ce{AlCl3} & \ce{BCl3}\\\hline
\mathbf{q}(\ce{X})~\text{[NPA]}  & +1.4       & +0.3     \\
\mathbf{q}(\ce{Cl})~\text{[NPA]} & -0.5       & -0.1     \\\hline
%\mathbf{q}(\ce{X})~\text{[QTAIM]}  & +2.4       & +2.0    \\
%\mathbf{q}(\ce{Cl})~\text{[QTAIM]} & -0.8       & -0.7     \\\hline
\end{array}$$
The analysis in terms of QTAIM also shows that the bonds in $\ce{AlCl3}$ they are predominantly ionic (left/top) while in $\ce{BCl3}$ are predominantly covalent (right/bottom).

One final thought on the bonding can be supplied with a natural resonance theory analysis (NBO6). I have chosen the following starting configurations and let the program calculate their contribution.

The overall structures in terms of resonance are the same for both cases, that is if you force resonance treatment of the aluminium monomer. Structure A does not contribute, while the others contribute to about 31%. However, when not forced into resonance, structure A is the best approximation of the bonding situation for $\ce{AlCl3}$. In the case of $\ce{BCl3}$ the algorithm finds a hyperbond between the chlorine atoms, a strongly delocalised bond between multiple centres. In this case these are 3-centre-4-electron bonds between the chlorine atoms, resulting from the higher lying degenerated π orbitals.

This all is quite good evidence that the monomer of boron chloride should be more stable towards dimerisation than the monomer of aluminium.
The Dimers $\ce{(XCl3)2; X{=}\{B,Al\}}$.
The obvious change is that the co-ordination of the central elements goes from trigonal planar to distorted tertrahedral. A look at the geometries will give us something to talk about.
\begin{array}{llrrr}\hline
&\ce{X{=}}& \ce{Al} &\ce{B} &\ce{Cl}\\\hline
\mathbf{d}(\ce{X-Cl})&/\pu{pm} & 206.7 &175.9&--\\
\mathbf{d}(\ce{X-{\mu}Cl})&/\pu{pm} & 226.1 &198.7&--\\
\mathbf{d}(\ce{Cl\bond{~}{\mu}Cl})&/\pu{pm} & 354.1 & 308.0 & --\\
\mathbf{d}(\ce{{\mu}Cl\bond{~}{\mu}Cl'})&/\pu{pm} & 323.6 & 287.3 & --\\
\mathbf{d}(\ce{B\bond{~}B'})&/\pu{pm} & 315.7 & 274.7 & --\\\hline
\mathbf{r}_\mathrm{vdW}&/\pu{pm} & 240   & 205   & 205\\
\mathbf{r}_\mathrm{sing}&/\pu{pm} & 126   & 85   & 99\\
\mathbf{r}_\mathrm{doub}&/\pu{pm} & 113   & 78   & 95\\\hline
\end{array}
In principle nothing much changes other than the expected elongation of the bonds that are now bridging. In case of aluminium the stretch is just below 10% and for boron it is slightly above 14%, having a bit more impact. In the boron dimer also the terminal bonds are slightly (> +1%) affected, while for aluminium there is almost no change.
The charges are not really a reliable tool, especially when they are that close to zero as they are for boron. In both cases one can see that charge density is transferred from the bridging chlorine to the central $\ce{X}$.
$$\begin{array}{lrr}
                    & \ce{(AlCl3)2} & \ce{(BCl3)2}\\\hline
\mathbf{q}(\ce{X})~\text{[NPA]}  & +1.3       & +0.2     \\
\mathbf{q}(\ce{Cl})~\text{[NPA]} & -0.5       & -0.1     \\\hline
\mathbf{q}(\ce{{\mu}Cl})~\text{[NPA]} & -0.4       & +0.1     \\\hline
\end{array}$$
A look at the central four-membered-ring of in terms of QTAIM offers that the overall bonding does not change. In aluminium they get a little more ionic, while in boron they stay largely covalent.

The NBO analysis offers a maybe quite surprising result. There are no hyperbonds in any of the dimers. While a description in these terms is certainly possible, after all it is just an interpretation tool, it is completely unnecessary. So after all we have two kinds of bonds in the dimers four terminal $\ce{X-Cl}$ and four bridging $\ce{X-{\mu}Cl}$ bonds. Therefore the most accurate description is with formal charges (also the simplest). The notation with the arrows is not wrong, but it does not represent the fact that the bonds are equal for symmetry reasons alone.

To make this straight: There are no hyperbonds in $\ce{(XCl3)2; X{=}\{B,Al\}}$; this includes three-centre-two-electron bonds, and three-centre-four-electron bonds. And deeper insight to those will be offered on another day.
The differentiation between a dative bond and some other for of bond does not make sense, as the bonds are equal and only introduced by a deficiency of the used description model.
A natural resonance theory for $\ce{(BCl3)2}$ gives us a overall contribution of the main (all single bonds) structure of 46%; while all other structure do contribute, there are too many and their contribution is too little (> 5%). I did not run this analysis for the aluminium case as I did not expect any more insight and I did not want to waste calculation time.
Dimerisation - yes or no
The energies offer us a clear trend. Aluminium likes to dimerise, boron not. However, there are still some things to discuss. I am going to argue for the reaction
$$\begin{align}
\ce{2XCl3 &-> (XCl3)2}&
\Delta E-\mathrm{diss}/E_\mathrm{o}/H/G&.
\end{align};$$
therefore if reaction energies are negative the dimerisation is favoured.
The following table includes all calculated energies, including the energy decomposition analysis mentioned at the beginning. All energies are given in $\pu{kJ mol^-1}$.
\begin{array}{lrcrcrcrr}
 \Delta & E_\mathrm{diss} &(& E_\mathrm{int} &+2\times& E_\mathrm{def}&)& E_\mathrm{o} &H &G\\\hline
\ce{Al} & -113.5 &(& -224.2 &+2\times& 55.4&)& -114.7 & -60.4 & -230.4\\
\ce{B}  &   76.4 &(& -111.2 &+2\times& 93.8&)&   82.6 & -47.1 &  152.5\\\hline
\end{array}
The result is fairly obvious at first. The association for aluminium is strongly exergonic, while for boron it is strongly endergonic. While both reactions should be exothermic, stronger for aluminium, the trend for the observed electronic energies ($E_\mathrm{o}$ including the zero-point energy correction) and the (electronic) dissociation energies reflect the overall trend for the Gibbs enthalpies.
While it is fairly surprising how strongly entropy favours association of $\ce{AlCl3}$, it is also surprising how it strongly disfavours it for $\ce{BCl3}$.
A look at the decomposed electronic energy offers great insight into the reasons why one dimer is stable and the other not (at room temperature).
The interaction energy of the fragments is double for aluminium than it is for boron. This can be traced back to the very large difference in the atomic partial charges. One could expect that the electrostatic energy is a lot more attractive for aluminium than it is for boron.
The deformation energy on the other hand clearly reflects the changes in the geometry discussed above. For aluminium there is a smaller penalty resulting from the elongation of the $\ce{Al-Cl}$ bond and pyramidalisation. For boron on the other hand this has a 1.5 times larger effect. The distortion also weakens the π-backbonding, which the additional bonding would need to compensate.
The four-membered-ring is certainly not an ideal geometry and the bridging chlorine atoms come dangerously close.
Conclusion, Summary and TL;DR:
The distortion of the geometry of the monomer $\ce{BCl3}$ cannot be compensated by the additional bonding between the two fragments. Therefore the monomers are more stable than the dimer. Additionally entropy considerations at room temperature favour the monomer, too.
On the other hand, the distortion of the molecular geometry in $\ce{AlCl3}$ is less severe. The gain in interaction energy of the two fragments well overcompensates for the change. Entropy also favours the dimerisation.
While size of the central atom is certainly the distinguishing factor, its impact is only severe on  the electronic structure. Steric crowding would not be a problem when the interaction energy would compensate for that. This is quite evident because $\ce{BCl3}$ is still a very good Lewis acid and forms stable compounds with much larger moieties than itself.
References
The used van der Waals radii were taken from S. S. Batsanov Inorg. Mat. 2001, 37 (9), 871-885. And the covalent radii have been taken from P. Pyykkö and M. Atsumi Chem. Eur. J. 2009, 15, 12770-12779.
Computations have been carried out using Gaussian 09 rev D.01 with NBO 6.0. Additional analyses have been performed with MultiWFN 3.3.8. Orbital pictures were generated with the incredible ChemCraft.

Nice. Although the entropy does intrigue me - that $-T\Delta S$ contribution is huge. I find it counter-intuitive that the entropy change can be so drastically different for two reactions that are fundamentally the same: $\ce{2A <=> A2}$. (Also would have expected the dimers to have $D_\mathrm{2h}$ symmetry.)

Martin, It would be good to perform similar analisys for boron trifluoride. It also is monomeric, by fluorine is oh so smaller than chlorine, so other factors should dominate.

@ortho (D2h is correct, don't know where my brains went wrong - it's a long post.) There seems to be a slight numerical error in the calculation which causes (AlCl3)2 to have a tiny imaginary mode. While this usually has no effect, it can lead to wrong corrections and this is more predominant in such small systems. Gaussian has a paper on how they do thermochemistry. I'll look into this further when I get the chance next week, maybe lifting the symmetry restriction will run into the minimum.

@permeakra This is indeed an interesting question, and when I get the chance I'll have a look. I think the reasoning will be very similar as the pi bonding should also increase with a smaller bond length. But we'll see.

@orthocresol entropy is hugely influenced by vibrational zero-energy and it falls rapidly with lowered mass of the atoms involved.

TL;DR

$\ce{BCl3}$ does not dimerize to $\ce{B2Cl6}$ due to a conflict between the short $\ce{B-Cl}$ bond length in $\ce{BCl3}$ $(1.74~\mathring{\mathrm{A}}$ in my calculations$)$ and the long $\ce{B-Cl}$ bond lengths $(\sim 2~\mathring{\mathrm{A}})$ that would be required for the $\ce{B2Cl2}$ core of a $\ce{B2Cl6}$ dimer.
$\ce{B2Cl6}$ does appear to have a positive (albeit small) dissociation energy, however, forming a van der Waals complex at a $\ce{B-B}$ distance of $3.63~\mathring{\mathrm{A}}$. Thus, this weakly-bound dimeric form may have the potential to form at very low temperatures.
$\ce{Al2Cl6}$ does not possess 2e-3c $\ce{Al-Cl-Al}$ bonds, in contrast to $\ce{B2H6}$. Instead, each of the four $\ce{Al-Cl}$ bonds in the $\ce{Al2Cl2}$ core is moderately 'sub-valent,' carrying $\sim1.5~e^–$.
The valence orbital localization and orbital shapes of an optimized $\ce{B2Cl6}$ system with one $\ce{B-Cl}$ bond on each boron constrained to $2.08~\mathring{\mathrm{A}}$ are very similar to those of the fully optimized $\ce{Al2Cl6}$, in agreement with Martin’s results.

These conclusions are based upon ORCA v3.0.3 DFT calculations and subsequent MultiWFN 3.3.7 electron density and ELF topology analysis on the three dimeric systems: $\ce{B2H6}$, $\ce{B2Cl6}$, and $\ce{Al2Cl6}$. Where used, "$\mathrm{E_h}$" indicates the Hartree energy unit.

As a first entry into the question, I performed relaxed surface scans on each dimer system, adjusting the $\ce{B-B}$/$\ce{Al-Al}$ distance from $9.0~\mathring{\mathrm{A}}$ down to $2.0~\mathring{\mathrm{A}}$ ($1.5~\mathring{\mathrm{A}}$ for $\ce{B2H6}$). A representative ORCA input file for these scans is:
! RKS PBE0 def2-SVP def2-SVP/J RIJCOSX D3BJ
! GRID4 GRIDX5 TIGHTOPT
! PRINTBASIS

%output  PrintLevel Small  end

%geom
   Scan
      B 0 4 = 9.0, 1.5, 16
      end
   end

* xyz 0 1
  B -4.91130   1.95714  -6.10613
  H -3.80330   1.95714  -6.10613
  H -5.46530   2.39076  -5.25014
  H -5.46530   1.52351  -6.96212
  B -2.06337  -4.02491   0.00000
  H -0.95537  -4.02491   0.00000
  H -2.61737  -4.79680  -0.57003
  H -2.61737  -3.25303   0.57003
*

Note that unlike in Martin’s calculations, I did not impose any symmetry constraints.
The figure below shows the potential energy curves obtained for these scans (scale increased in the right panel to show $\ce{B2Cl6}$ more clearly; click to enlarge):

The solid curves are the best-fit Morse potentials, determined by minimizing the sum-squared deviation of the fit from the data:
$$
\begin{array}{ccccc}
\hline
 & D_e~(\mathrm{E_h}) & D_e~\left(\mathrm{kcal\over mol}\right)
   & a~\left(\mathring{\mathrm{A}}^{-1}\right)
   & r_e~\left(\mathring{\mathrm{A}}\right) \\
\hline
\ce{B2H6}   &  0.0867  &  54.41  &  2.3304  &  1.7025 \\
\ce{B2Cl6}  &  0.00599 &  3.76   &  0.79832 & 3.9632 \\
\ce{Al2Cl6} &  0.0532  &  33.36  &  0.97947 & 3.0259 \\
\hline
\end{array}
$$
To note, these data likely do not run along any particular internal reaction coordinate. The images below are of the optimized geometries (left: $\ce{B2H6}$; center: $\ce{B2Cl6}$; right: $\ce{Al2Cl6}$); click each to display an animated GIF of the optimized geometries at each step of the respective scan, which show the irregularities in the progression of the scans (all GIFs pause at the step closest to the optimized geometry):

Despite these irregularities, the energy profiles seem qualitatively reasonable.  Of particular interest is the fact that $\ce{B2Cl6}$ may actually form a van der Waals complex, enabling it to be bound at a sufficiently low temperature (plotted at the $0.5$ isosurface of the reduced density gradient):

So, these computations agree with the experimental observation that $\ce{BCl3}$ does not dimerize in the way that $\ce{BH3}$ and $\ce{AlCl3}$ do.

In the course of exploring the $\ce{B2Cl6}$ system, I performed a constrained optimization where I locked one $\ce{B-Cl}$ bond on each boron atom to be considerably longer than its equilibrium value, starting from the above optimized geometry. I arbitrarily chose $2.08~\mathring{\mathrm{A}}$, which is the length of the terminal $\ce{Al-Cl}$ bonds in $\ce{Al2Cl6}$.  Once the energetic cost is paid to stretch the two $\ce{B-Cl}$ bonds, an appreciable minimum in the energy profile is exhibited:

The data in red is the same $\ce{B2Cl6}$ data from above; that in yellow is with the two $\ce{B-Cl}$ bonds constrained at $2.08~\mathring{\mathrm{A}}$. All energies are relative to the relaxed system at a $9~\mathring{\mathrm{A}}$ $\ce{B-B}$ separation.
The resulting optimized structure under this constraint is below; as above, click the image to view an animated GIF of the progression of the constrained optimization:

I suspect that the symmetry-constrained $\ce{B2Cl6}$ species studied by Martin is very similar to this constrained-optimized structure, and if he had relaxed the symmetry constraint he would have obtained the van der Waals complex above.  Based on his data, though, it appears my choice of $2.08~\mathring{\mathrm{A}}$ for the stretched $\ce{B-Cl}$ bonds was too large by $0.1~\mathring{\mathrm{A}}$ or so.
In any event, the fictional "pseudo-dissociation curve" for this "extended $\ce{B-Cl}$" system follows much the same form as that for $\ce{Al2Cl6}$:

The '$\ce{B2Cl6}$-extended' data in the above chart was shifted vertically so that $E[9~\mathring{\mathrm{A}}] = 0$, in order to illustrate its qualitative similarity to the $\ce{Al2Cl6}$ data.
So, simply stretching one $\ce{B-Cl}$ bond at each boron atom is sufficient to induce $\ce{B2Cl6}$ to behave in substantially the same fashion as $\ce{Al2Cl6}$.

$\ce{B2H6}$ does show the expected 2e-3c bonding behavior.
ORCA localization of the $\ce{B2H6}$ valence orbitals (omitting the $\ce{B}$ core orbitals) indicates their presence (the bridging hydrogens are 1H and 5H):
----------------------------------------
LOCALIZED MOLECULAR ORBITAL COMPOSITIONS
----------------------------------------
The Mulliken populations for each LMO on each atom are computed
The LMO`s will be ordered according to atom index and type
  (A) Strongly localized MO`s have populations of >=0.950 on one atom
  (B) Two center bond orbitals have populations of >=0.850 on two atoms
  (C) Other MO`s are considered to be `delocalized`
FOUND  -   0 strongly local MO`s
       -   4 two center bond MO`s
       -   2 significantly delocalized MO`s
Bond-like localized orbitals:
MO   5:   7H  -   0.494574  and   4B  -   0.525955
MO   4:   6H  -   0.494586  and   4B  -   0.525936
MO   3:   3H  -   0.494588  and   0B  -   0.525939
MO   2:   2H  -   0.494573  and   0B  -   0.525953
More delocalized orbitals:
MO   7:  0B - 0.269 1H - 0.478 4B - 0.269
MO   6:  0B - 0.269 4B - 0.269 5H - 0.478

Both the localized orbitals and the QTAIM LCP paths are consistent with the 2e-3c bond paradigm, and ELF basin analysis finds attractors at the bridging hydrogen atoms, whose basins are calculated to contain $1.94~e^-$ each, very close to the expected value of $2.0$.

Orbital plots here and below are at the $\phi = \pm 0.02$ isosurfaces.
On the other hand, $\ce{Al2Cl6}$ does not show multicenter bonding upon localization of the valence orbitals (the bridging atoms here are 7Cl and 1Cl):
FOUND  -   8 strongly local MO`s
       -  16 two center bond MO`s
       -   0 significantly delocalized MO`s
Rather strongly localized orbitals:
MO  47:   7Cl -   0.955366
MO  46:   7Cl -   1.001974
MO  45:   6Cl -   1.007096
MO  44:   5Cl -   1.007111
MO  43:   3Cl -   1.007113
MO  42:   2Cl -   1.007094
MO  41:   1Cl -   0.955367
MO  40:   1Cl -   1.001973
Bond-like localized orbitals:
MO  63:   7Cl -   0.844289  and   4Al -   0.149451
MO  62:   7Cl -   0.843809  and   0Al -   0.150019
MO  61:   6Cl -   0.932028  and   4Al -   0.074001
MO  60:   6Cl -   0.938027  and   4Al -   0.063923
MO  59:   6Cl -   0.746625  and   4Al -   0.250592
MO  58:   5Cl -   0.931694  and   4Al -   0.074421
MO  57:   5Cl -   0.938034  and   4Al -   0.063923
MO  56:   5Cl -   0.746451  and   4Al -   0.250727
MO  55:   3Cl -   0.931694  and   0Al -   0.074422
MO  54:   3Cl -   0.938034  and   0Al -   0.063923
MO  53:   3Cl -   0.746450  and   0Al -   0.250728
MO  52:   2Cl -   0.938028  and   0Al -   0.063922
MO  51:   2Cl -   0.932027  and   0Al -   0.074001
MO  50:   2Cl -   0.746627  and   0Al -   0.250589
MO  49:   1Cl -   0.843808  and   0Al -   0.006200
MO  48:   1Cl -   0.844290  and   0Al -   0.149450

The shape of the bond orbitals in the $\ce{Al2Cl2}$ core is consistent with strictly two-center bonding:

Interestingly, the valence orbital localization pattern and core bond orbital structures in the 'extended $\ce{B-Cl}$' system above are very similar to $\ce{Al2Cl6}$:
FOUND  -   8 strongly local MO`s
       -   16 two center bond MO`s
       -   0 significantly delocalized MO`s
Rather strongly localized orbitals:
MO  39:   7Cl -   0.967482
MO  38:   7Cl -   1.006838
MO  37:   6Cl -   1.008398
MO  36:   5Cl -   1.008336
MO  35:   3Cl -   0.967481
MO  34:   3Cl -   1.006837
MO  33:   2Cl -   1.008334
MO  32:   1Cl -   1.008398
Bond-like localized orbitals:
MO  55:   7Cl -   0.830908  and   4B  -   0.171645
MO  54:   7Cl -   0.769240  and   0B  -   0.232367
MO  53:   6Cl -   0.941563  and   4B  -   0.066433
MO  52:   6Cl -   0.931032  and   4B  -   0.073365
MO  51:   6Cl -   0.658532  and   4B  -   0.343433
MO  50:   5Cl -   0.941702  and   4B  -   0.066259
MO  49:   5Cl -   0.930994  and   4B  -   0.073395
MO  48:   5Cl -   0.658437  and   4B  -   0.343525
MO  47:   4B  -   0.232364  and   3Cl -   0.769241
MO  46:   3Cl -   0.830907  and   0B  -   0.171646
MO  45:   2Cl -   0.941703  and   0B  -   0.066256
MO  44:   2Cl -   0.930995  and   0B  -   0.073394
MO  43:   2Cl -   0.658441  and   0B  -   0.343521
MO  42:   1Cl -   0.931031  and   0B  -   0.073367
MO  41:   1Cl -   0.941565  and   0B  -   0.066432
MO  40:   1Cl -   0.658533  and   0B  -   0.343432

Of particular interest, in both chlorine systems, all of the ELF basins with attractors lying along the $\ce{B-Cl}$ / $\ce{Al-Cl}$ bonds each have an integrated electron density of $\sim 1.5~e^-$. Thus, taken together, the four bonds in each rectangular core comprise six electrons in total.  While I agree with Martin that there are no multicenter bonds here, there does appear to be an interesting sort of resonance phenomenon at play.
In sum: $\ce{B2H6}$ does display 2e-3c bonding; $\ce{Al2Cl6}$ does not.

"Once the energetic cost is paid to stretch the two B−Cl bonds, an appreciable binding energy is exhibited." At no point in that plot is the extended $\ce{(BCl3)2}$ bound (it has a positive energy at all points), yet in the next plot you show it as being bound.

@pentavalentcarbon Mm, did I not emphasize enough that the dissociation energy curve for the "extended" system is fictional? I can add that....

I get what you're saying, that there is a minimum along that coordinate, but to call it bound would be wrong. Fixing the wording still doesn't correct that either the first plot or second plot is wrong in some way (look at the energy scale, then look at the "extended" curves).

What makes dimerization possible in AlCl3?

Aluminium trichloride is an electron deficient molecule and it tends to get stability by accepting a lone pair of electron from Cl atom of another AlCl3 molecule so that it now has an octet in outer shell.

Are there 3c-2e bonds in Al2Cl6 as there are in B2H6?

No, the Cl forms dative bond.

Why is BCl3 a monomer whereas AlCl3 exists as a dimer?

STABILITY, As it is seen in the following figure

The bond length and atomic size of boron in BCl3 are smaller than in AlCl3 so this leads to interatomic repulsion so BCl3 could not accept another BCl3 via a dative bond because it leads to more repulsion and finally breaking of dimer. Also some degree of pi backbonding is present which may make Cl unable to form dative bond.
One may argue that BCl4- exists but in this cases size of 4th entity which binds to Boron will be smaller than a BCl3 monomer.

I find your last sentence hard to believe. The immediate surrounding of the boron atom is identical in $\ce{BCl4-}$ and $\ce{B2Cl6}$.  The extra boron atom on the far side doesn’t interact with the first.

@Jan but that would prevent dimerisation by straining the bonds.

I think your answers to the first two questions are spot on, from Cotton and Wilkinson "formation of dimers is attributable to the tendency of the metal atoms to complete their octets". AlCl3 also exists as a monomer in the gas phase near the b.p., the dH for disassociation is between 46-63 kJ/mol. For the third question, I think the MO diagram of B2H4 vs AlCl3 would help explain why the MO levels of B and H allow 3C-2E bonds while Al and Cl do not.

It is said that $\ce{AlCl_3}$ exist as tetra-coordinate aluminium when as dimer, meaning the aluminium has an $sp^3$ like structure with a coordinate bond provided by a $\ce{Cl}$ atom.
The $\ce{BX3}$ halides attain an octet by dative $\pi_{p-p}$ interaction, forming a four-centre $\pi$ molecular orbital covering all four atoms which contains two bonding electrons. This is not feasible in case of other elements in the group. They have larger atoms and cannot get effective $\pi$ overlap, so they polymerize to remedy the electron deficiency.
In case of $\ce{B2H6}$, 3 centred-2 electron bonds are formed because boron as well as hydrogen do not have enough electrons to form a classic single bond. But here, chlorine does. So there is no 3-c 2-e bonds and instead there are coordinate covalent bonds.
As dimer, they exist as 3 centred-4 electron non-hypervalent bond, forming a delocalized dative bond as shown:

Also note that similar structure is shown by $\ce{(BeCl2)_n}$ polymer, with beryllium atom having $sp^3$ hybridisation.
This type of 3c - 4e bonds are not possible in case of boron because boron is too small and the chlorine atoms will not fit in between them to form the bridge. Thus boron handles it through back bonding.

I dont think the AlCl3 dimer is considered a 3C-4E bond? I think itbis more dative or backbonding of non bonding Cl electrons to fill an Al oxtet, the multi center bond requires a MO orbital with equal distribution? Do you have a source for that?

Bond length of $\ce{Al-Cl}$ bridges are equal to $2.3a$ with $2e^-$ in each bond. They all have the same bond orders (Mulliken), $\approx 0.66$. Visit Schupf Computational Chemistry Lab

The definition of a multicenter multielectron bond is 3-center orbitals one bonding, one non bonding, and antibonding. An open 3 center bond is called 3 center 4 electron, where in addition to uniform electron distribution in the bonding orbital, there are a pair of electrons in the non bonding orbitals, ENTIRELY concentrated on the end atoms, giving a relatively polar bond. I do not think Al2Cl3 meets those requirements. (From Cotton&Wilkerson verbatim)

In Al2Cl6 you have 2 electrons in 2 Al-Cl bridges for a total of 4 electrons, but those 4 electrons are not distributed relative to eachother in a 3 centered orbital with 2 electrons in a bonding, and the other two removed from the center orbital in non bonding orbitals.

I found a reference for the dH association for borane to diborane giving a value of -142 kJ/mol vs dH for AlCl3 dimer formation is ~ 52 kJ/mol. The first value for diborane makes sense as a covalent bond is being formed (multicenter multielectron) and the smaller positive dH for AlCl3 is consistent with a lewis-acid lewis base dative bond interaction.

Sorry, the -142 kJ/mol is for association, and the +52 kJ/mol is for disassociation, so the bond enthalpy for diborane is +142 kJ/mol.

@EframGoldberg The dative bond is continuously being switched. That is why I believe they are distributed. So you are saying they exist as Structures 1 and 2 only and not 3, in the above picture?

@EframGoldberg, what do you say in case of $\ce{(BeCl2)_n}$? I'm pretty sure that it has a 3c-4e bond.

@Efram The Al-µCl-Al are 3c-4e-bonds, because symmetry dictates so. Cotton and Wilkerson are oversimplifying and the concept of the dative bond is just a book keeping tool.

So are the only requirements for multicenter multielectron bonding just symmetry and equal distribution? Wouldnt all dibridged dinuclear complexes be considered the same?

@EframGoldberg According to my text book, they are considered same. But notice one thing, the structures we both refer to are in fact the same. Only difference lies in how we define multi centred bonds. I refer to Inorganic Chemistry by J D LEE for these structures.

In $\ce{AlCl3}$, $\ce{Al}$ has 6 valence $\ce{e-}$ now to complete its octet it needs 2 more $\ce{e-}$. So it accepts a pair of electron from the $\ce{Cl}$ of the neighboring $\ce{AlCl3}$. This donation of electron results in the formation of dimer.

But in $\ce{BCl3}$, due to the small size and the extent of π-bonding in $\ce{B}$ results in the short bond length, which also prevents $\ce{Cl}$ to form coordinate bond. And due to this short bond length when they dimerises (like in the above structure) chlorine atoms are closer than that in $\ce{AlCl3}$ and it breaks so it exist as monomer

$\ce{BCl4^-}$ does exist, so the size argument is void.

The second boron atom in the ring would strain the bonds, so a direct comparison to BCl4- cannot really be made.

A planar four membered ring with equal sides should form in both cases. The optimum bond angles would be 90 degrees within the ring. (This is independent of the bond lengths.)
Both the B and the Al atoms are four coordinate and thus would prefer a tetrahedral arrangement (109.5 degrees).
90 degrees is not too bad for the Cl-Al-Cl angle, (see H2S).
However the ring is too strained in the case of boron, the Cl-B-Cl angle is too strained.
It is another question why B and C atom show much higher strain compared to heavier counterparts when distorted from the tetrahedral arrangement.

Actual bond angles in Al2Cl6 are 93, 87, 110, and 122 far from tetrahedral.

110 is not far actually from tetrahedral, but the argument was about the angles in the ring. These are close to 90 in Al2Cl6, which is tolerable for elements not in the 2nd period, but the same would cause too much strain in B2Cl6.

BH3 and AlCl3 each form "dimers" but by different processes.

AlCl3 is trigonal planar but in vapor phase at certain temperatures, it can dimerize to Al2Cl6 with a dH of ~ 46-63 kJ/mol. A few degrees above the boiling point, and the monomer is predominant. This happens by change of geometry from trigonal planar to tetrahedral and backbonding of Cl pi electrons to fill the octet of Al giving a dimer. This is possible due to the low relative energy difference between Cl and Al valence orbitals in the MO diagram. See http://pages.swcp.com/~jmw-mcw/Parsing%20the%20Bonding%20in%20the%20Aluminum%20Trichloride%20Dimer.htm for a good overview of bond angles and lengths.
BH3 is tetrahedral and forms a dimer  B2H6 but through a different process. The bonding is described by Cotton&Wilkerson as: "each boron atom uses two electrons and two roughly sp3 orbitals to form 2 center 2 electron (covalent) bonds to two hydrogen atoms. The boron atom in each BH2 group still has one electron and two hybrid orbitals for use in further bonding. The plane of the two remaining orbitals is perpendicular to the BH2 plane. Thus when two BH2 radicals approach eachother with the with the hydrogen atoms also in the plane, a B-H-B 3 center 2 electron bond is formed." The MO orbital diagram for the 3C2E bond requires close energy of the three atomic orbitals which is met between boron and hydrogen.
BCl3 does not form dimers as others have mentioned above, the difference in atomic radii between B and Cl explain the instability of the potential dimer. Additionally, larger difference in orbital energy would make such bonding unfavorable.

This is as wrong as most of the textbooks on this topic. Your cited source in 1 calls three-centre bonds an aberration, clearly an indicator that they have not understood the first thing about bonding. They try to explain their reasoning by using the hybridisation model completely and utterly wrong. The dimerisation process (the bonding?) is pretty much the same for BH3 and AlCl3. The size argument is only a small part of a more complicated matter. All in all, this answer just repeats what most of the others already stated.

The website you cited is borderline pseudoscience. The author published that web page (which cites - amongst other articles by himself - Wikipedia) in a non-peer-reviewed online "journal" and is most "famously" known for arguing against the spdf model of atomic orbitals in favour of his own construction. One downvote doesn't do justice to any answer that uses him as a reference. cc @Martin

@orthocresol I was only citing the website for the bond angles, I did not read the rest of it. I actually use Cotton & Wilkerson's definition for multicenter multielectron bonding. I think my answer adds some thermodynamic values to help show the difference and I think is the only answer to mention MO theory. Either way, if the other responses are adequate, then should the question be considered answered?


text:
Which is the correct name: 2,3-dimethyl cyclopentene or 1,5-dimethyl cyclopentene?
According to the lowest locant rule, 2,3-dimethyl cyclopentene should be the correct IUPAC name as the sum of the locants of the substituents(2+3=5) of 2,3-dimethyl cyclopentene is lower than the sum of the locants of the substituents(1+5=6) of 1,5-dimethyl cyclopentene.
But my book says that the correct name of the compound is 1,5-dimethyl cyclopentene. I guess this is due to the fact that methyl gets the 1st locant position. But I am doubtful about this.
What is the correct name?

The lowest sum of locants rule does not exist in the IUPAC recommendations.

The correct name is 1,5-dimethylcyclopent-1-ene. The double bond has the highest priority and so these two carbons must be numbered 1 and 2 if possible (yes in this case because the molecule is cyclic). We then have to choose between numbering the methyl substituents 1,5 or 2,3. The lowest locant sum rule that you mentioned doesn't actually exist, even though it is often taught. It is a agglomeration of several rules which works most, but not all of the time. The actual rule we are interested in is the first point of difference rule which states:

2.2 - The longest chain is numbered from one end to the other by Arabic numerals, the direction being so chosen as to give the lowest numbers possible to the side chains. When series of locants containing the same number of terms are compared term by term, that series is "lowest" which contains the lowest number on the occasion of the first difference. This principle is applied irrespective of the nature of the substituents.

Therefore the locant set 1,5 is lower than the locant set 2,3 since at the first point of difference, 1 is lower than 2.

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text:
What alcohol concentration makes bathroom cleaner "disinfecting"?
There are many do-it-yourself cleaner solutions out there, using a combination of vinegar, alcohol (in a "cheap vodka" concentration), dish soap, baking soda and essential oils. I like these recipes for two reasons, first for cost and second for health (sometimes my toddler helps me clean his bathtub, and I wouldn't want him to use too harsh stuff).
However, many of these recipes claim to be "disinfecting", and as far as my understanding goes, alcohol has that effect, as has stuff that offsets the pH-level (such as the acidity of vinegar). But given that this stuff is just sprayed onto the surface, can a mixture of 2 parts water 1 part vodka really be concentrated enough to be "disinfecting"? And if you mix baking soda and vinegar, won't that bring you closer to a neutral pH, i.e. offset any "disinfecting powers"?
Is there some sort of guideline as to how much alcohol/other agent should be in the final solution for it to be of good use in cleaning a bathroom?

see also http://www.cdc.gov/hicpac/Disinfection_Sterilization/6_0disinfection.html#a1 and
http://cmr.asm.org/content/12/1/147.full

I like how vinegar and baking soda are each magic reagents in the household, so if you put them together, you'll get benefits from both!

Thanks @Loong. I can't vote up your comment (not enough reputation points :)) but it's really helpful. From reading it, I think it would probably be better to just spray some concentrated alcohol on dirty areas (eg. toilet bowl) and apply regular cleaner after that has already done its magic. (I'm not aiming for clinical sterilization :) .) That would be better than mix the two and apply it in one go, right?

According to this Center for Disease Control document:

Ethyl alcohol (70%) is a powerful broad-spectrum germicide and is considered generally superior to isopropyl alcohol. Alcohol is often used to disinfect small surfaces (e.g. rubber stoppers of multiple-dosemedication vials, and thermometers) and occasionally external surfaces of equipment (e.g. stethoscopes and ventilators).

Although 70% alcohol is ideal, another CDC article states that it can be effective at 50%.  According to this article:

In the healthcare setting, "alcohol" refers to two water-soluble chemical compounds—ethyl alcohol and isopropyl alcohol—that have generally underrated germicidal characteristics.  FDA has not cleared any liquid chemical sterilant or high-level disinfectant with alcohol as the main active ingredient. These alcohols are rapidly bactericidal rather than bacteriostatic against vegetative forms of bacteria; they also are tuberculocidal, fungicidal, and virucidal but do not destroy bacterial spores. Their cidal activity drops sharply when diluted below 50% concentration, and the optimum bactericidal concentration is 60%–90% solutions in water (volume/volume).

So, alcohol has certainly been shown to be a powerful, broad-spectrum antimicrobial agent, though not at the dillution you proposed. I would just use the straight vodka if it were greater than 50% alcohol, or probably beter (cheaper) yet would be denatured alcohol if that's available where you live.

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text:
How would I find an apporpriate dopant element for a lattice
I am trying to figure how would I go about finding a dopant element that would fit in a copper lattice? I already have the data for the atomic dimensions and structure (available below), however how would I find a dopant element that would make it a P-Type material? (Relatively new to chemistry, so bear with me if I ask a stupid question)
Atomic dimensions and structure
Atomic Radius 145 pm
Covalent Radius 138 pm
Van der Waals Radius 140 pm
Crystal Structure Face Centered Cubic
Lattice Angles
π/2, π/2, π/2
Lattice Constants
361.49, 361.49, 361.49 pm
Space Group Name Fm_ 3m
Space Group Number 225

Just how do you intend to make Cu, a metal, "p-type"? What do you mean by that?

Copper chalcocite, for example, is a degenerate p-type. To clarify, how would I find a material(s) that would fit within the copper lattice, however a material(s) that leave holes within the valence band.

Sure, but your example is a compound and not fcc I believe. So, look for other similar transition metal compounds and look at their lattice constants.

So would Copper and Nickel work? They are both fcc I believe, thus would it still leave a valence hole?


text:
Why are the reactants in the solubility product constant multiplied by their coefficients and raised to the power of their coefficients?
It is probably easiest for me to explain the question in an example. Consider the following chemical reaction:
$\ce{MgF2(s) -> Mg^{2+} (aq) + 2F- (aq)}$
Plugging this into the formula for the solubility product constant, $K_{sp}=[M^{y+}]^{x}[A^{x-}]^{y}$, we get:
$K_{sp}=[\ce{Mg^{2+}}][\ce{F-}]^{2}$.
The solubility product constant for magnesium fluoride is $5.16\times10^{-11}$. To determine molar solubility, the equation to solve for magnesium and fluoride is $5.16\times10^{-11}=X\times (2X)^{2}$.
$X$ is magnesium and $(2X)^{2}$ is fluoride. This is because there are two moles of fluoride for every one mole of $\ce{MgFl_{2}}$, and one mole of magnesium for every one mole of $\ce{MgFl_{2}}$. (Please correct me if I'm wrong anywhere.)
But here's whats confusing to me. Why is fluoride denoted as $(2X)^{2}$? I know that there are two moles of fluoride per one mole of magnesium fluoride, but why does it need to be denoted by being both multiplied to its coefficient and raised to the power of that coefficient?
Sources: Magnesium Fluoride Solubility product constant

The reason why in the equilibrium constant expression, the concentration is to the power of coefficient is because the above equation can be written like this:
$$\ce{MgF_2(s) -> Mg^{2+} (aq) + F^- + F^- (aq)}$$
Hence when you plug this into the equilibrium constant expression, you get this:
$$K_{sp}=[\ce{Mg^{2+}}][\ce{F-}][\ce{F-}]$$ which becomes the equivalent to:$$K_{sp}=[\ce{Mg^{2+}}][\ce{F-}]^{2}$$ So really the equilibrium constant is proportional to the square of the concentration of fluoride. So that is the reason for the square. However, the concentration of fluoride is twice as much as the concentration of Mg, hence that is the reason for the multiplying the concentration of fluoride by 2. So that is why fluoride is denoted as $(2X)^2$.

How is the fluoride concentration twice as much? I understand why the equilibrium constant expression squares, but why would it multiply? It seems redundant.

The reason the fluoride concentration is squared is due to the equilibrium constant. The reason why the fluoride concentration is also multiplied by two is because of its relationship with Mg

Say you have $X$ moles of magnesium fluoride powder that you dissolve in water.  After it dissolves, how many moles of fluoride anion will there be in solution?

We'd get 2X moles of fluoride anion. Thanks for both answers, that clears it up. Raising it to the square has nothing to do with its relationship with Mg, it is just part of the equation.


text:
Can we determine the amplitude of vibration of O-H bonds in water?
This is going to be a long shot, and I might be wrong in any and all of my assumptions, be warned (... and please correct me). I might also be trivially right, I have no idea.
At equilibrium, each degree of freedom has the kinetic energy:
$$E = \frac{1}{2}k_\mathrm{B}T$$
So, I have water at, say, $300\ \mathrm K$. Water has three atoms, so 9 degrees of freedom (3 per atom, because space is three-dimensional). One of them is for a $\ce{O-H}$ stretching. I know that usually, you would need to divide into modes of oscillations, but let's say we combine half the energy of the symmetric and antisymmetric stretchings, giving that one $\ce{O-H}$ stretching has a full $\frac{1}{2}k_\mathrm{B}T$ in it. So we have an average, roll of drums,
$$ E = \frac{1}{2}(1.3806488 \times 10^{-23})(300) =  2.071\times 10^{-21}\ \mathrm J$$
in a $\ce{O-H}$ stretching.
NOW, let's say this stretching behaves like an harmonic oscillator, its' energy is given by the amplitude of oscillation:
$$ E = \frac{1}{2}kA^2 $$
where $k$ is the stiffness constant of the chemical bond.
That's nice, but we have two unknowns, $k$ and $A$. So let's go further. If we know the absorption band's wavenumber, we can say that the amplitude corresponds to this wavelength, $2.734\ \mathrm{\mu m}$ (source = Wikipedia). This would mean that we have a stiffness constant:
$$ k = 5.5412 \times 10^{-10}\ \mathrm{N/m}$$
I know that, although it was fun to blindly math around, this result is wrong. Just the fact that the stiffness constant would depend on temperature doesn't sound right. Neither does the bit about wavelength = amplitude.
My question is: is it possible to know the amplitude of oscillation/stiffness constant of a $\ce{O-H}$ bond?

My question is: is it possible to know the amplitude of
  oscillation/stiffness constant of a O−H bond?

Yes.
Your approach is along the right lines but -
The Hooke's Law formula you used
$$E = \frac{1}{2}kA^2$$
is for the stretching of a spring.  Hooke's law changes if the spring has mas(ses) attached at the end(s), now it is a harmonic oscillator.  In this case Hooke's Law becomes
$$E = \frac {h}{2 \pi} \sqrt{\frac{k}{\mu}}$$
where k is the force constant you are interested in and $\mathrm{\mu}$ is the reduced mass of the system.
Since
$$E=h\nu$$
The frequency, which can be experimentally measured, is given as
$$\mathrm{frequency} = \frac {1}{2 \pi} \sqrt{\frac{k}{\mu}}$$
The frequency is independent of the amplitude, affected only by the masses and the stiffness of the spring.
If the masses of the atoms at the ends of the spring are given as A and B, then
$$\mu =\frac{A \cdot B}{A+B}$$
So for an $\ce{O-H}$ bond $\mathrm{\mu=\frac{1 \cdot 16}{1+16}}\ \mathrm{u}$. If we measure the frequency of the $\ce{O-H}$ stretch absorption, then we can calculate the force constant.
Here is a link to a one-page example that shows the actual computation and explains the specific units to be used very nicely.
You also ask about the amplitude of oscillation.  If, by this you mean the bond length, then yes this can also be calculated from spectra (relatively easy in simple diatiomic cases, but more difficult in complex polyatomic cases.  The methodology involves assuming a rigid rotor and then using the Schrodinger equation to solve for the energy states and extracting the bond length.  A nice example using $\ce{HCl}$ can be found here, look down the page for the section entitled "Bond Length of HCl".

What I meant with "amplitude" is the deviation from the mean bond length. Your document is very helpful,  just to make it clear, though, they deduce the "spring constant" from the wavenumber of the stretching mode's absorption band and the bond length from the wavenumber of a rotational mode's absorption band. Can we return to my idea with HCl at 300K and find the amplitude   $A = \sqrt{\frac{2E}{k}} = \sqrt{\frac{2(2.071\times 10^{-21}J)}{481N/m}} = 2.93fm $? Which would seem to fit with the  $127fm$ bond length of HCl.

Finally, I have a day off to read that question and response. Half-way reading it I was already a bit suspicious mostly because of the half-jolly and half-formal style OP used. :) Don't get me wrong, in principle I have nothing against such style. But the thing is that usually when such style is used some nonsense is eventually introduced at some point, and this nonsense is not so easily detectable since it is buried under all these jolly informalities.
That is exactly what happened this time too.

we can say that the amplitude corresponds to this wavelength

Wait, what? Amplitude has no relation whatsoever to wavelength.

My question is: is it possible to know the amplitude of
  oscillation/stiffness constant of a $\ce{O-H}$ bond?

Yes, you can calculate the force constants of molecular vibrations (as well as frequencies). These calculations are routinely done in the domain of quantum chemistry these days.

Perhaps you could expand a bit upon both of your responses. At the moment they are more or less unqualified statements with no explanation.


text:
Is the absolute temperature concept 100% viable?
Maybe I'm wrong about it but I wanted to know whether the concept of absolute temperature is real or not.
As every element behaves differently at a particular temperatures how could we assign a common temperature to all these elements where they possess zero point energy? If they will do so then what will happen if we find an new element which ground state energy is at $\mathrm{-10\ K}$ theoretically?

The concept of absolute temperature is real. Temperature can be thought of as thee measure of the concentration of heat energy in a substance and isn't effected by different elements or compounds. Absolute temperature which is −273.15 degrees Celsius (0 kelvin) is the temperature where there is no heat energy in the substance at all. That means that all the molecules are existing in their lowest possible energy state. So conventionally is thought that it is impossible to get lower than absolute zero as to reduce the temperature, you have to remove heat energy but you can't remove heat energy in a substance which doesn't have any heat energy in the first place.
However theoretically, the Boltzmann equation does allow for substance to go minus Kelvin and in fact it actually lies above infinity Kelvin. Recently there have been several claims that this has been achieved. If you search this up on Google I am sure that you will be able to find more information on this.

Negative temperatures are easily achieved: it implies a population inversion, where the lowest energy states are unpopulated. A laser works because of that inversion; see https://en.wikipedia.org/wiki/Negative_temperature#Lasers.


text:
Effect of quantum tunnelling on chemical reactions at very low temperature
Can anybody give me a schematic representation of how quantum tunneling allows chemical reactions to occur at very low temperatures like in the interstellar medium.

What exactly are you looking for? Do you understand how quantum tunneling works?

Yes I do. But now Scientists are thinking only quantum tunnelling can say how molecules are formed in ISM at a very low temperature...

I am aware of that but what part of it do you not understand?

Sir, actually I can't visualize this process. I am unable to imagine it. For example if you say about activation energy, I could imagine this process how bonds are breaking and making at that energy level. But in this case I only know the mathematical explanation in which I can only imagine electron has a tendency to exist across a potential barrier through a tunnel but can't understand how reactions occur through this process.


text:
Why does equilibrium exist?
I've read an article that equilibrium exists when $G$ (Gibbs free energy) becomes minimum. If formation of products decreases Gibbs free energy, why won't the reaction proceed to completion so that Gibbs energy becomes minimum.

This is because the Gibbs free energy also depends on the concentration of the reactants and products. Hence, for a reaction, if the energy change is negative it moves forwards, forming more product, which also changes the concentrations of the species involved. The Gibbs energy change at this new concentration will be less negative than that in the initial case, but the reaction still proceeds since the reaction is accompanied by a negative change of free energy.
A point comes, where the concentrations of the reactions and the products is such, that the Gibbs energy change now becomes zero for the particular reaction. This is the equilibrium condition (minimum free energy, maximum entropy). If the energy change is positive, the reverse reaction becomes favorable, again inching towards the zero energy change condition, i.e. equilibrium. So, the reaction is already at the minimum Gibbs energy when it is at equilibrium. Further proceeding of the reaction will be accompanied by a positive change, and hence a rise in energy.
For how it depends on the concentrations, see these questions

Why does the reaction quotient use the products (multiplications) of reactants and products, rather than their respective sums?
Derivation of relationship between equilibrium constant and Gibbs free energy change

Just because the Gibbs energy change of reaction ($\Delta _r G$) is negative, doesn't mean that it reaches a minimum when all reactants have been converted to products.
Consider the reaction: $$\ce{A <=> B}$$
We can plot the total Gibbs energy of the system against the percentage conversion from $\ce{A}$ to $\ce{B}$ and we get a graph like this:

Technically the curve should be a parabola but it was much easier to draw an arc in the program I used. As you can see, the minimum Gibbs energy is reached when there is a mixture of products and reactants. This is in fact true for all chemical reactions: the minimum on the Gibbs energy graph cannot be at the point where only reactants or products exist. It can be extremely close to that point, in which case we tend to call the reaction irreversible (if the minimum is close to 100% products) or we say that the reaction doesn't happen (if the minimum is close to 100% reactants).
It turns out that there is a relationship between $\Delta _r G$ and the equilibrium constant which is expressed as: $$\Delta _r G = -RT\ln K$$

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text:
Are common names of substituents accepted by IUPAC?
Are names of substituents like isopropyl and sec-butyl accepted by IUPAC?
Or do we need to name them according to standard IUPAC conventions?
(Ex- 1-methylethyl for isopropyl)
My book uses common names for substituents at a lot of places but I am doubtful about this.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) various prefixes are retained for use in general (IUPAC) nomenclature. However, many prefixes are no longer recommended.

Trivial, common, and traditional prefixes have always been an integral part of organic nomenclature. However, as systematic nomenclature develops and becomes widely used, many of these prefixes fall by the wayside. Accordingly, each set of IUPAC recommendations contains fewer of these traditional prefixes.

The prefix “isopropyl” for $\ce{(CH3)2CH-{}}$ is retained only for use in general (IUPAC) nomenclature.
For the preferred IUPAC name (PIN), the preferred prefix is “propan-2-yl”.
The prefix “1-methylethyl” may be used in general (IUPAC) nomenclature.
The prefix “sec-butyl” for $\ce{CH3-CH2-CH(CH3)-{}}$ was still contained in the 1993 recommendations but is no longer recommended as approved prefix.
For the preferred IUPAC name (PIN), the preferred prefix is “butan-2-yl”.
The prefix “1-methylpropyl” may be used in general (IUPAC) nomenclature.


text:
Calculating Entropy

Atkin's Physical Chemistry Book; Chapter 3; question 3.3(a):
Question
  Calculate $\Delta S$ when the state of $\mathrm{3~mol}$ of perfect gas atoms, for which $C_{p,m}$ = $\frac{5}{2}R$, is changed from $\mathrm{298~K}$ and $\mathrm{1~atm}$ to $\mathrm{398~K}$ and $\mathrm{5~atm}$. How do you rationalize the sign of $\Delta S$?

Attempt
Since this is at a constant pressure, $dq = dH = C_pdT$. Then I used the equation $$dS = \frac{C_pdT}{T}$$
I integrated both sides to get the following equation: $$\Delta S = C_p   \ln\frac{398}{298}$$
Subbing in the values I got that $\Delta S =  \mathrm{+18.3~J}$. However, the answer in the book is $\mathrm{-21~J}$. Shouldn't that be definitely wrong because the temperature of the system increased and hence the total entropy should have increased?

Notice that your assumption that you are at constant pressure is incorrect. You are given the constant pressure heat capacity, but the pressure increases from $\text{1 atm}$ to $\text{5 atm}$.

So how should I attempt this problem?

Your energy change involves PV work. Try calculating the volume change with IGL, then incorporate it into your calculation of DeltaS

The pressure changes, so we cannot use the simplified equation you use. From the perfect gas law, we can calculate the volume change, but we don't need to go there. The big problem is that in addition to the heat transfer, pressure-volume work is done. Work is easy to determine at constant pressure. When pressure changes also, it gets harder. However, let's use enthalpy to get rid of work (and thus the change in volume), leaving us with $dS=f\left(\frac{dT}{T},\frac{dP}{P}\right)$, which is the data we are given.

Here's the derivation.

Second Law of Thermodynamics:
$$dS = \frac{\delta Q}{T}$$
First Law of Thermodynamics:
$$dU = \delta Q + \delta W = \delta Q - PdV_m$$
Definition of Enthalpy:
$$dH = dU + d(PV_m) = dU +PdV_m + V_mdP$$
Substitutions:
$$dH = \delta Q + PdV_m + V_mdP - PdV_m = \delta Q +V_mdP$$
$$\delta Q = dH - V_mdP$$
Perfect gas equation (notice that I am using molar volume $V_m$, so we will be calculating molar entropy change):
$$PV_m = RT$$
$$V_m = \frac{RT}{P}$$
At constant pressure (Yes, I know pressure changes, bear with me. We'll put that back in).
$$dH = C_p dT$$
Substitute to get rid of $V_m$, since we don't have volume data.
$$dQ = C_p dT - \frac{RT}{P}dP$$
Substitute into the second law:
$$dS = C_p \frac{dT}{T} - R\frac{dP}{P}$$
Integrate (now we account for change in pressure:
$$\int_{S_1}^{S_2} dS = C_p \int_{T_1}^{T_2} \frac{dT}{T} - R \int_{P_1}^{P_2} \frac{dP}{P}  $$
$$\Delta S = C_p \ln \left(\frac{T_2}{T_1}\right) - R \ln \left(\frac{P_2}{P_1} \right)$$
Using this method, I get the correct answer after I multiply the molar entropy change by the number of moles.
Note, that if you are given volume change instead, you can use a similar derivation to reach a similar equation:
$$\Delta S = C_v \ln \left(\frac{T_2}{T_1}\right) + R \ln \left(\frac{V_{m,2}}{V_{m,1}} \right)$$
Reference:

Nasa Glenn Research Center

Since the change in entropy does not depend on path, you can do it as a two step process.  First you heat it at constant pressure to the final temperature, then you compress it at constant temperature to the final pressure.  The change in entropy for each of these steps is easy to determine.  You already got the result for the first step.


text:
Determining percentage of copper in brass

Source: an IB Chemistry past paper.
I looked it up and found out that brass is an alloy of copper and zinc - however, I don't see zinc in any of the equations, and therefore any brass being separated into copper and zinc, which I had originally thought would be the method to determine the percentage of copper in brass (by looking at the coefficients of the compounds in the resulting chemical equation).
My question is how do these reactions help determine the percentage of copper in brass?

As this is a homework question, please share your thoughts and attempts towards the solution. It'll make us certain that we aren't doing your homework for you.

Well the question would had more to it. The question would have probably gone something like this. A sample of brass was dissolved with excess nitric acid. The $\ce{Cu^{2+}}$ ions are then reduced with excess iodide ions. Iodine is then titrated with thiosulfate (giving its concentration and volume used). Find the percentage of copper in brass.
To do this question, calculate the amount of moles of thiosulfate used, using its concentration and volume. Then using equation 3, you can see the number of moles of iodine is half the number moles of thiosulfate. Then using equation 2, you can see that the number of moles of $\ce{Cu^{2+}}$ ions are twice as the number of moles of iodine and hence are equal to the number of moles of iodine. Finally, using equation 1, the number of moles of copper is equal to the number of moles of $\ce{Cu^{2+}}$ ions and hence are equal to the number of moles of thiosulfate.
Now that we know the number of moles of copper, you can multiply by its molar mass to get its mass and then find the percentage of copper in brass.


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pH of sodium metasilicate anhydrous vs pentahydrate when dissolved in water
I'm not a chemist and don't remember much chemistry from school, so please forgive a naive question (I've read the Wikipedia article on sodium silicate though).
I'm researching this topic to make a positive photo resist developer for making PCBs.

Will dissolving $\ce{Na2SiO3}$ (the anhydrous form) in water produce essentially the same solution as dissolving $\ce{Na2SiO3*5H2O}$ but in different concentration or are these intrinsically different?
Will $\ce{Na2SiO3}$, if left in contact with open air, eventually hydrate to $\ce{Na2SiO3*5H2O}$?
Also, the Wikipedia article talks about the $\ce{SiO2:Na2O}$ ratio affecting the pH of the water solution. Is the information about this ratio already present in the chemical formula so if I buy $\ce{Na2SiO3*5H2O}$ it is always the same ratio or is this a different parameter that is specified in addition to the chemical formula? (I understand that different concentrations of the solution will produce different pH, but consider equal concentrations)

I believe I need the developer to be an alkaline solution so I need to understand what will affect the pH when dissolving sodium metasilicate in water (e.g. if I need $\ce{Na2SiO3}$ or $\ce{Na2SiO3*5H2O}$, if I need to consider other specifications besides the chemical formula, if I need to consider the shelf life of the chemical if it changes with time e.g. gets hydrated from $\ce{Na2SiO3*5H2O}$ into $\ce{Na2SiO3*9H2O}$).
 Searching electronics forums on the internet I haven't found a clear answer. For some people using this sodium metasilicate as a developer works for some it doesn't develop at all, some add $\ce{NaOH}$ to it but there are places that advertise their sodium metasilicate developer as sodium hydroxide free.

1)In the short run they may be different. They key point is that anhydrous sodium metasilicate contains chains, while hydrated sodium metasilicate is actually dihydro-orthosilicate. I guess, the first one will transform into the second one over time in solution, but it will take some time.

As explained in the comment by permeakra, due to the structural differences between anhydrous and hydrated Na2SiO3, their solutions will also be intrinsically different, at least for a while.
Whether anhydrous salt will turn to hydrate or vice versa depends on your ambient humidity and temperature. Anyway, I'd not recommend leaving either of them in contact with open air, for fear they might react with CO2. Silicic acid is so weak that it can be displaced even by carbonic acid.
Well, theoretically, the information about the SiO2:Na2O ratio is already present in the chemical formula Na2SiO3 (with or without water). However, the industrial-grade product may not quite match that formula. In fact, it may be a wild mixture of different silicates, where each individual compound cannot and need not be characterized. For many large-scale applications it suffices to know just the said ratio. Not sure about your application, though.


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What is the parent chain in a hydrocarbon with two different rings?

There is cyclopentyl attached to one terminal carbon atom and cyclohexyl attached to the other terminal carbon atom and there are three carbon atoms linking them. Since the cyclohexyl ring has the greatest number of carbon atoms, it should be the parent chain. So the name should be 3-cyclopentylpropyl cyclohexane.
But my book says that the middle propyl chain should be the parent chain, the cyclohexyl and cyclopentyl groups should be the substituents, and the correct answer is 1-cyclohexyl-3-cyclopently propane.
What is the correct name. And if my book is correct, why is it so?

http://www.chemspider.com/Chemical-Structure.206342.html My guts say you're right. || I imagine some say it's possible to imagine this as a propane substituted with a cyclopentyl and cyclohexyl; but that surely won't be PIN.

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)
Two methods are recognized to name systems composed of rings and chains
  (exclusive of linear phanes).
(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.
(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

So far, we have three possible names:
Method (1): “(3-cyclopentylpropyl)cyclohexane” or “(3-cyclohexylpropyl)cyclopentane” (ring preferred to chain)
Method (2): “1-cyclohexyl-3-cyclopentylpropane” (chain preferred to ring)

(…) For selection of a preferred IUPAC name, see P-52.2.8.

P-52.2.8 Selection between a ring and a chain as parent hydride
Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

This leaves us with two possible preferred IUPAC names:
“(3-cyclopentylpropyl)cyclohexane” or “(3-cyclohexylpropyl)cyclopentane”

P-44.1.3 Seniority order only for rings and ring systems. Criteria that apply only when the choice for parent structure is between two or more rings or ring systems are given in P-44.2.

P-44.2 SENIORITY ORDER ONLY FOR RINGS AND RING SYSTEMS
P-44.2.1 If application of P-44.1 does not effect a choice, general criteria for determining ring seniority given below are applied successively until there are no alternatives remaining. (…)
  The senior ring or ring system:
  (…)
  (e) has the greater number of skeletal atoms;
  (…)

Therefore, the preferred IUPAC name is “(3-cyclopentylpropyl)cyclohexane” (not “(3-cyclohexylpropyl)cyclopentane”) since cyclohexane has more skeletal atoms than cyclopentane.

What are linear phanes?


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How to stop goggles from fogging?
I wear corrective eyeglasses, and because of this, I frequently experience problems with fogging when I wear safety goggles. How can I prevent or reduce fogging?
The safety goggles I use are the standard-issue goggles supplied by the campus store, but I can use other goggles rated at least Z87+D3 ("+"=impact, "D3"=splash) if I want.
One method of reducing the fogging I have tried is wedging a folded paper towel under the top edge of the goggles. It works, but I don't feel comfortable with it because it compromises the splash protection of the goggles.
I have also tried placing desiccant packets (of the sort found in various packaging) in the goggles. This has helped somewhat but doesn't solve the problem completely.
One possible solution would be to get safety goggle with prescription corrective lenses built-in (making them Z87-2D3 instead of Z87D3), but I have not been able to find a supplier online for such eyewear.
What can I do to reduce or eliminate fogging?

I have no idea if this works on safety goggles, but with diving masks spitting on the inside of the mask, rubbing it over the clear area and washing off any excess with water is very effective at preventing fogging.

FWIW - I do not wear corrective eyeglasses, and my goggles still fog. I recently bought an expensive (ca. 40 USD) pair of 'fog-resistant' safety goggles from a company that also makes skiing and snowboarding goggles. They take longer to fog up, but they still do.

Why would eyeglasses affect whether goggles fog up?

@DavidRicherby Not sure, but putting my eyeglasses outside the goggles results in no fogging. Although for a general solution that won't work because it is not easy to balance them there.

None of these options actually work. I've been a lab tech for over 10 years, and I've tried everything out there, and they simply do not work. I'm reduced to having my lab assistant pull my goggles out and wipe them for me every 2-3 minutes or so. AND FYI, please stop rubbing spit inside of your goggles, it's completely ineffective and could compromise any lab projects you are working on, not to mention you look like an ass doing so.

Be careful with desiccant: silica may create tiny dust particles that could scratch the cornea, and hygroscopic salts may liquefy, possibly irritating.
A traditional anti-fog treatment is to rub the interior of the lens with bar soap (or a drop of mild detergent) and polish it. The soap decreases the surface tension of water, causing droplets to make a smoother, more transparent film. Wikipedia suggests some other anti-fog treatments.
There are also vented anti-fog goggles (the vents should have foam or other filter to make them splash resistant.
My personal choice for lab work or mechanical work is a full-face splash shield, which may offer more protection and does not fog.

There are sprayable hydrophobic coatings available such as NeverWet, though you should check compatibility with the reagents you're using.

NASA had this problem in the late 1960s, on the Moon.  They developed a very effective solution.  Sub-Aquatic Systems (SAS) licensed their formula and sold it to sport SCUBA divers.  (I used it on my dive mask and on my motorcycle face shield.)
Since then, there have been MANY anti-fogging solutions formulated for diving.  A former dive instructor of mine swore by no-tears baby shampoo.  (You OBVIOUSLY want something that won't irritate your eyes.)  I personally used Sea Drops (tm), when I dropped back into diving after being away for many years and learning that SAS was no longer in business.

I used to have a brown wax crayon that I would rub onto my glasses and polish in order to prevent fogging. After a quick google search I couldn't find the same product (mind you, my experience was almost 20 years ago). I found something that might be similar, albeit with a interesting name, here.

My experience on this is also at least 20 years old, but I used to have a small stick type anti-fogging material that worked well.  It resembled a chapstick and was rubbed onto the surface.  I don't remember the name, but some of the other answers sound similar.
A Google search for "anti fog for glasses" or for "anti fog for safety glasses" yields a LOT of hits, including sprays, home made methods, and anti fog glasses.


text:
Solution for storing a glass electrode
What kind of storage solution is required for a glass electrode?
Is 3 M $\ce{KCl}$ sufficient for this purpose? If so, what is the approximate pH of that solution (the solution I use presently has a pH $\approx$ 5)?
Can filling solutions be used as storage solutions for glass electrodes?

A glass electrode is usually stored in a 3 or 4M KCl solution, depending on whatever the manufacturer advices you to do.
But the pH that you measure from this solution is absolutely not the pH. It's the measured potential of this concentrated KCl solution that is (in this case wrongly) converted into the pH scale.

The normal pH measurements with glass electrodes cannot be done without the calibration of your device through references solutions, which are common buffer solutions with acidic, neutral and basic pH values like, e.g., pH 4, 7, 9.
In order to calibrate the pH electrode, you or actually your device saves the measured potential for each of these known buffer solutions and calculates a regression function that should give you something near: $$E = 414.12 - 59.16~pH~~~~~(\text{at}~25~^\circ C)$$
(The time when I had to do this in practical courses gave us $E = 401.93 - 56.711~pH$ ... I have no idea what the temperature was :D )
For further informations on all this you can read Mettler-Toledo's pH guide.

Now, as your KCl solution has a certain potential the device converts it through this regression equation into a pH value. But as I said, that is wrong - I guess that pH indicators like the universal indicator would show no big difference from pH 7 (I've never done this actually, so this is seriously a guess).
Your actual question should rather be about what the potential of a 3M or 4M KCl solution should be. And that, as can be found here, is:

Thanks for the Mettler-Toledo's pH guide link - I've included it in this question.

Most pH electrode storage solutions are 4 M KCl with a pH about 4. The solution you have should be fine if you can change the internal solution on a monthly basis (more often if you use it continuously).
The internal solution is also saturated with silver chloride. I would only use it for the external storage solution if I couldn't make the external solution, but it should not be a problem for a short term.


text:
Is hydroboration a pericyclic reaction?
According to Ahn.N.T "Frontier Orbitals" hydroboration isn't a pericyclic reaction because boron uses two AOs and not one. The same applies for cheletropic reactions, which aren't pericyclic either. But the IUPAC goldbook says different: http://goldbook.iupac.org/P04491.html - that cheletropic reactions are pericyclic. So is hydroboration, too?

Based on the IUPAC Goldbook definition, the only two requirements for a pericyclic reaction are:

Concerted mechanism
Cyclic transition state

Hydroboration reactions (and similar reactions) are pericylic reactions by these criteria, as are all cheletropic reactions.
The language about fully conjugated transition states can be used to explain some pericyclic reactions, but not all, and thus IUPAC uses the phrasing may be viewed as.... Hydride shifts, (and all sigmatropic rearrangements ene reactions) involve non-$\pi$ orbitals (even so, the transition states do tend to follow the Hückel rule).

What type of pericyclic reaction is hydroboration?

It's similar to an ene-reaction, but I am not saying that it is an ene reaction. Does it need to fit easily into a subcategory?

I guess not... .


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Why does lactose-free milk last so much longer than regular milk?
I buy lactose-free milk. The sell-by date is usually as much as eight weeks from the date of purchase, which is in the neighborhood of twice that of regular milk. This seems puzzling; lactose-free milk, which still has the same mass of sugar, has twice as many sugar molecules as regular milk, since all the lactose molecules have been split in to galactose and glucose. With twice as many sugar molecules per unit volume, one would think lactose-free milk would last half as long as regular milk, not twice as long.

You have regular milk with a four-week shelf-life?  The milk I buy (UK) typically has a shelf-life of about five or six days.

@DavidRicherby, since I don't buy regular milk, I am really not sure of the shelf-life; that was just a guess. I know in Spain, and probably much of Europe, irradiated milk is available that has a room-temperature shelf-life of something like six months. We Americans are so afraid of anything "irradiated" that it would never sell here. We'd rather drink spoiled milk or throw it out. [shrug]

Fair enough. (I've not seen irradiated milk in the UK but we do have some that's been put through an ultra-fine filter to remove bacteria. As I recall, that has a shelf-life of a couplefew weeks.)

Apparently the process used by manufacturers (ultra-pasteurization) to produce lactose-free milk is responsible for the longer shelf life.
One source reports:

Lactose-free milk is pasteurized at a higher temperature than regular milk. The process, known as ultra-pasteurization, is designed to remove the bacteria content entirely, giving lactose-free milk a refrigerated shelf-life of 60-90 days, compared with regular pasteurized milk, which retains some bacteria. It has a shelf life of 1-3 weeks.

A producer (in the United States) of organic dairy products (Organic Valley) describes ultra-pasteurization and the extended shelf-life here as:

Ultra Pasteurization (UP), or Ultra High Temperature (UHT) pasteurization, is the process of heating milk to approximately  280 °F for just 2 seconds and then chilling it back down rapidly.  The result is milk that's 99.9% free from bacteria.  This pasteurization process creates an extended shelf-life for milk products of up to three times the length as HTST pasteurization; while providing the same wholesome, quality dairy product.  This allows us to distribute UP milk products regionally as well as to other areas of the country that might not have access to our dairy products. This is referred to as 'ultra pasteurized' on the milk package.  Organic Valley offers its quarts and some of its half gallons of milk with Ultra Pasteurization (UP).

That first source tells very dubious things... Lactase is an enzyme, as indicated by the -ase suffix, while lactose is a sugar (-ose). The enzyme breaks down lactose, a disaccharide (two base sugars), to glucose and galactose, which taste sweeter than lactose. A lack of the enzyme causes lactose to go through as is until bacteria come into play to ferment it. Correct me if I'm wrong. Source: "milk sugar lactase, while lactose-free milk does not. Lactase, an enzyme produced by the stomach, breaks down the milk sugar galactose"

I believe this answer is correct, it might also be worth mentioning that you can buy (at least in some areas, the UK being one) regular milk which has been UHT pasteurised. It tastes awful and is best avoided unless you really need the long unrefrigerated shelf life, eg for a camping trip.


text:
Why is the cathode reaction required in a fuel cell?
So, at the anode of a fuel cell hydrogen is split into a positively charged proton and an electron. Then, there is the membrane which can let only protons through to the cathode, whereas electrons need to follow external circuit (effectively powering it).
Question: why do we need any other reaction on the cathode? Isn't the fact that positively charged ions travel through the membrane to the cathode enough to create the EMF?
Trying to solve it I assume that the reaction is there to actually apply some kind of force to move the ions though the membrane to the cathode, otherwise they'd just stay and reunite with electrons on the anode. What is that force and how does it work?

Why do we need two half reactions, one at the anode and another at the cathode?

The whole thing eventually grinds to a halt without the other reaction. The electrons generated at the anode travel through the circuit to the cathode. What happens then? Something needs to happen to the electron.
Additionally, what happens to that proton? As the concentration of protons increases, what will balance the charge that is produced in the cell, given that the electron is elsewhere?
We can imagine that we can set up a cathode in which the reverse reaction of the anode occurs:

Cathode: $\ce{H2 -> 2H+ + 2e-}$
Anode: $\ce{2H+ +2e- -> H2}$

However, in this case, there is overall no net reaction, which means that $\Delta G =0$ and since $\Delta G = nFE$, then the cell potential $E=0$. We need a different reaction for the cathode to consume electrons.
For example:
$$\ce{O2 + 2H2O +4e- -> 4OH-}$$
Since this reaction produces hydroxide anions, then we have something for the proton to do:
$$\ce{H+ + OH- -> H2O}$$
Now we have the following combination of reactions:

Anode: $\ce{H2 -> 2H+ + 2e-}$
Cathode: $\ce{O2 + 2H2O +4e- -> 4OH-}$
Neutralization $\ce{H+ + OH- -> H2O}$
Net: $\ce{2H2 + O2 -> 2H2O}$

This net reaction is exergonic, and thus produces a potential.

Wow thanks, this answer takes me to the next level. However I still have that dangling drawback: what makes the proton to actually travel through the membrane? The fact that on the cathode anions are produced? On the other hand, shouldn't the anode be positively charged for the circuit to work?

@LeonidUsov electrons always flow into the cell at the cathode and out at the anode. So when a cell is being charged the + plate is the anode & the - plate is the cathode, but when a cell is being discharged (as here) the + plate is the cathode & the - plate is the anode. So it's the cathode that's positively charged: electrons flow out of the anode, round the circuit back to the cathode . For current to flow it's essential for a charged species (in this case H+) to pass through the membrane, but really it's the sum of all reaction steps that provide the energy that drives the current.

@steveverrill I just noticed that I made a mistake in my comment above saying that anode should be positively charged, but I meant cathode of course. The paradox that I was trying to describe is how a positively charged cathode attracts ions+ through the membrane. There should be some non-electrostatic force that makes them come. Apparently this force is somehow a result of the cathode reaction...

@LeonidUsov at the membrane itself, there is no paradox: electrostatic forces pull the H+ ions through, where they combine with the OH- ions. (This is relevant to the need to have two half reactions. If there were only one half reaction, static electricity would build up very quickly and block the current.) At the surface of the electrodes, what we see is a jump in electrical potential, due to the energy released by the half reactions. This is analogous to the jump in electrical potential at the junction of two dissimilar metals in a thermocouple. Does that seem paradoxical too?


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Reverse-Engineer a material
Let's say I am seeking to find an alloy with certain properties, is there any method or process that enables one to backtrack from desired properties to actual structure and composition of a material?

Try MatWeb, where you can search by physical properties, material type, alloy composition, and up to 3 physical/chemical properties (a very extensive list); and NIST's Material Measurement Laboratory, as well as NIST's Standard Reference Data. These can at least get you started with a name of something from which you can look into the structure and composition.

In general no.
What you're describing is known as inverse design, that is, designing something with particular properties.
I agree completely with the comment above, that you can find many things from databases. But when you mix together multiple components, into new inter-metallic compounds or alloys, the results do not necessarily match the individual components.
Consider, for example, "Galinstan", the various alloys of gallium, indium, and tin. While each of these have low melting points, the eutectic point for the ternary alloy is $\:\mathrm{-19 ^{\circ} C}$.
There are many other examples, but the point remains that without careful experiments or theory for materials design, it's tricky to do inverse design.

Now if you really want to do it anyways, there are multiple experimental or computational strategies to get it done. For metals or inorganic systems, so-called "alchemical DFT" methods let you optimize the atomic number of a system for a particular property.


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Fermi Level of an alloy
How could one calculate the Fermi level of a copper-nickel alloy? They are both structurally face-centered cubic, however how would this structure affect the Fermi level of the individual atom or the entirety of the material?

I'm unclear on what you mean when you say the Fermi level of an individual atom. Since the Fermi level is defined in a statistical sense, it applies to the entirety of the solid.  And, while Cu and Ni are both fcc metals, the binary phase diagram exhibits a miscibility gap up to about 400C. above that they are mutually soluble across the entire composition range.  What information do you want to get to using the Fermi level? That might perhaps be an easier question to answer.


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Curious precipitate from substitution reaction
Two solutions were prepared.
'Solution A' being $\ce {Na2S4}$ prepared from aqueous $\ce {NaOH}$ and elemental sulfur, the mixture heated and additional sulfur added until sulfur remained as a precipitate.
'Solution B' being a solution of  $\ce {FeSO4}$ and $\ce {Fe2(SO4)2}$ prepared from reacting steel wool with $\ce{H2SO4}$, adding steel wool until reaction ceased. The liquid portion drained off and collected for use.
Solution A was added to Solution B via squeeze pipette in intervals allowing precipitate to settle between additions. This was continued until the net solution began to turn yellow from a rising conventration of  $\ce{Na2S4}$.
The solution was washed drained and the precipitate collected, however its character is not as expected from micro particulate $\ce{FeS2}$ but rather a black oily colloidal mass was the collected product. Did I somehow manage to acquire $\ce{Fe2S4}$?

from WIKI :which decays at a temperature over 20°C into FeS and sulphur.[2]
$\ce {Fe2S3 → 2 FeS + S↓}$

Seems I made Iron III sulfide, the sample has decayed during the day becoming stiff and grainy with a sizable rotten egg odor in the house despite the sample being in a closed container.......


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Relationship between dipole moment of a molecule and electronegativities of constituent atoms
I want to know if I can calculate the dipole moment of a molecule given the electronegativity values for the atoms comprising it.
I think that dipole moments determine the polarity of molecules, so if one molecule has a larger dipole moment than another, the one with the larger dipole moment is more polar than the other. Is this correct?
Can the magnitude of how polar a molecule is be calculated from the difference in electronegativities among the elements comprising the molecule?

Yes you can. By looking at the different electronegativities of each element in a molecule, you are able to determine the dipole moment for each bond. Then by adding the dipole moments for each bond, you are able to determine whether a molecule is non-polar or polar and if it is polar, how polar it is.
For example, in $\ce{CH4}$, the carbon atom is bonded to four hydrogen atoms. The electronegativity of carbon is 2.55, while the electronegativity of hydrogen is 2.2 This means that for each $\ce{C-H}$ bond, there will be a dipole moment from the hydrogen to carbon atom. However, since methane has a perfect tetrahedral structure, the four dipole moments cancel each other out so that there is no net dipole moment. Hence, methane despite having polar bonds, is actually not a polar molecule, shown below.

Another example that we can consider is ammonia, $$\ce{NH3}$$. Ammonia consists of 3 $\ce{N-H}$ bonds. The electronegativity of nitrogen is 3.04. That means that ammonia consists of 3 polar $\ce{N-H}$ bonds. The molecular structure of ammonia is trigonal pyrimadal and hence unlike methane, when you add up the 3 dipole moments, they don't cancel each other out. Instead, they result in a net upward dipole moment. Hence, the molecule itself is also polar, shown below

Is there a mathematical equation between the dipole moment and electronegativity difference?

What about ozone ($\ce{O3}$)? Does your explanation account for why it is polar even though the atoms (and thus electronegativities) on both sides of each bond are the same?

@user102008 ozone is slightly more complicated. In short, since O3actually exists in resonance, the two end O atoms have a slight delta negative charge. Since O3 is bent, these dipoles dont cancel each other out, hence O3 is polar.

@Nanoputian: I know, but it doesn't seem to be explained by the method you described in your answer, i.e. determining the dipole moment of each bond by looking at the electronegativities of the atoms.

OP asked whether one can calculate the dipole moment from electronegativity. Hence, is there atleast an approximate equation for dipole moment given electronegativity?

I don't know any to my knowledge and I doubt any reasonable formula can be given. The ozone example above demonstrates that the dipole moment not only depends on electronegativity but also other equally important factors such as resonance structure and molecular structure.


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Baking Soda + Citric Acid + Water = CO2, how much though?
Baking soda (sodium bicarbonate) + citric acid react when mixed with water to form some amount of carbon dioxide gas ($\ce{CO2}$). I am trying to determine what percentage of the products is carbon dioxide: in other words, how much does reaction make?

It makes a lot of CO2.

OK, snark aside, the answer depends on how much citric acid and baking soda you use.  What does "a lot" mean to you?  "A little"?

I mean the percentage, a lot would be upwards of 90% of the reaction, a little would be like %10

I edited, hope I was right about your intentions

I'm not sure what you mean by 90% and 10%. If you mean 90% yield (i.e. 90% or more of the bicarbonate ions in sodium bicarbonate will become carbon dioxide), then yes, that will happen if there is enough acid. How much is enough acid? We'll get to that in a minute. If you mean something else, like 90% of the mass of the components will become carbon dioxide, then that probably will not happen.

Let's look at why 90% of the mass of products is not carbon dioxide.

Bicarbonate anions react with protons from acid to produce carbonic acid, which decomposes to form carbon dioxide:
$$\ce{H+ + HCO3- -> H2CO3 -> H2O + CO2}$$
Let's start with a gram of sodium bicarbonate. How many bicarbonate ions are present? The molar mass of sodium bicarbonate is $84.0066\mathrm{\ g/mol}$, where a mole (mol) is a quantity equal to the number of carbon-12 atoms in 12 grams of carbon-12 or an Avogadro's number $(6.022\times 10^{23})$ of things. Atoms and molecules are very small, and the numbers of them in macro measurements are large.
So we have
$$\mathrm{1.00\ g \times \dfrac{mol}{84.0066\ g}= 0.0119\ mol}$$
$$\mathrm{0.0119\ mol \times \dfrac{6.022\times 10^{23}}{mol}=7.17\times 10^{21}\ molecules}$$
Let's stick with the mole unit since it gives us easier to conceptualize numbers.
The molar mass of carbon dioxide is $\mathrm{44.01\ g/mol}$, so if all of the bicarbonate ion in sodium bicarbonate is converted into carbon dioxide, we will get $\mathrm{0.0119\ mol}$ of carbon dioxide which is
$$\mathrm{0.0119\ mol \times \dfrac{44.01\ g}{mol}=0.524 \ g}$$
So 1.00 gram of sodium bicarbonate will product 0.524 grams of carbon dioxide. We have not even accounted for the citric acid yet. Citric acid is $\ce{H3C6H5O7}$, and it is tempting to think that some of those carbon and oxygen atoms in that molecule might become molecules of carbon dioxide. They can, but not in a reaction with sodium bicarbonate (acid-base). If you react citric acid with oxygen (combustion), then you can produce carbon dioxide.

How much acid?

Citric acid serves as a proton source - in fact it has three acidic protons. That means in theory, each molecule of citric acid can donate three protons to convert three bicarbonate anions into carbon dioxide:
$$\ce{H3C6H5O7 + 3HCO3- -> C6H5O7^3- + 3H2O + 3CO2}$$
However, those three acidic protons are not equal (and the reason for that would make a great follow-up question). Two of the protons are more acidic than carbonic acid, and the third one is slightly less acidic than carbonic acid. In theory, since carbonic acid decomposes into carbon dioxide, which escapes from solution, we can count on Le Châtelier's principle to pull even that third proton. However, in practice, we might not want to wait around. Let's use one equivalent of citric acid per two equivalents of sodium bicarbonate:
$$\ce{H3C6H5O7 + 2HCO3- -> HC6H5O7^2- + 2H2O + 2CO2}$$
So, to figure out how much citric acid (molar mass $\mathrm{192.12\ g/mol}$) we need, we have to do some calculations. Remember that 1 gram of sodium bicarboncate is 0.0119 moles, and we need 0.5 moles of citric acid for every mole of bicarbonate:
$$\mathrm{0.0119 \ mol\ bicarb\times \dfrac{1\ mol\ CA}{2\ mol\ bicarb}\times\dfrac{192.12\ g\ CA}{1\ mol\ CA}=1.14\ g\ citric\ acid}$$

Thanks Ben, that's all the chemistry (plus some) I forgot from 30 plus years ago. I sure lost it from lack of using it. Given your careful calculations, I go back to the kitchen from whence the question arose and find roughly two teaspoons of Citric Acid to a bit over three quarters of a teaspoon of Bicarbonate of Soda (Baking Soda) to get to your ratio of 1.20 g of Citric acid to 0.55 grams of Baking Soda. 2 teaspoons of Citric Acid weigh close to 8.5 grams. A little over 3/4 teaspoon of Baking Soda weighs in at 4.25 grams.

@FredRock, good real-world use, except you grabbed the wrong number from the baking soda equation.  1g is the amount of baking soda input. 0.55g is the amount of CO2 is produced.
TL:DR, Combine 1.20g citric acid with 1g baking soda


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Calculating Gibbs Free Energy
This is a question from Atkin's Physical Chemistry Book.
Question
Suppose that 3mmol $\mathrm{N_2\,(g)}$ occupies $\mathrm{36~cm^3}$ at $\mathrm{300~K}$ and expands to $\mathrm{60~cm^3}$. Calculate $\Delta G$ for the process.
My Attempt
I decided to split this process up into two separate process (I am not sure if this allowed. Please correct me if this wrong). The first process is an reversible isothermal expansion of the gas to $\mathrm{60~cm^3}$. Then the second step is an adiabatic cooling of the gas. I was planning to calculate $\Delta G$ for each step then add them up to get the answer. For the first step, $\Delta U = 0$, therefore:
$$q = -w = n~R~T~\ln\frac{60}{36}$$
$$\Delta S = \frac{q}{T} = n~R~\ln\frac{60}{36}$$
Since pressure is constant, $$\Delta H = q = n~R~T~\ln\frac{60}{36}$$
Therefore: $$\Delta G = \Delta H - T\Delta S = 0$$
I stopped her because I am sure that this is incorrect as Gibbs Free Energy shouldn't be equal to $0$.
Can someone please clarify if the above procedure is correct and perhaps a better way to attempt this question?

I'm quite certain about the following, because it's from the related (german) woorkbook to the 4th (german) edition. The beneath numbering is from the linked online version, which should be the 9th english edition.
Your way to the solution is part of subsection 3.9 Properties of the Gibbs energy and subsubsection 3.9 c) The variation of the Gibbs energy with pressure.
I seriously entrust you to read this subsubsection and try to solve it with the given information. If this does not help, the following "hidden" section reveals what you need to solve it.

 You need to insert the equation for the molar volume of an ideal gas $V_m = R~T/p$ into (3-57b) $$G_m(p_f) = G_m(p_i)+\int_{p_i}^{p_f}V_m~\mathrm{d}p \tag{3-57.b}$$ which yields one of the two equations that you need: $$G_m(p_f)=G_m(p_i)+R~T\int_{p_i}^{p_f}\frac{\mathrm{d}p}{p}=G_m(p_i)+R~T\ln\frac{p_f}{p_i}\tag{3-59}$$ The other one is the Boyle's law.

Good luck.

Thanks for that. I missed that section when I was reading the chapter before. That makes the question much easier. However would it still be possible to the question from my method? Also was my result for $\Delta G$ being equal to $0$ correct?

For your first question ... maybe someone knows that but I am not qualified enough. To your second question ... calculate $\Delta G$ with this "new" way and you will see.

For an ideal gas, $\Delta H = 0$ for a constant temperature change between two thermodynamic equilibrium states of a system.  So, for your system, $\Delta G=T\Delta S$.


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Are decomposition and dissociation reactions the same?
I have seen these terms being used almost interchangebly, but are decomposition and dissociation reactions the same ?

I would say they are different because I wouldn't say an acid dissociating in to its ions (in an aqueous solution) is the decomposition of the acid. However I may be proven otherwise.

Could you give us some of those examples that you mention?

Decomposition and disassociation refer to similar processes, however they are different. The main difference between them is that decomposition reactions are usually irreversible while disassociation reactions are reversible and exist in equilibrium.
A decomposition reaction is where a compound breaks up into smaller elements or compounds. For example: $$\ce{2H2O -> 2H2 + O2}$$ is an example of a decomposition reaction. Note that in this case that this is actually a reversible reaction.
A disassociation reaction is the general process where molecules split into smaller particles which can include elements, compounds, ions and radicals. For example: $$\ce{NaCl(s) -> Na+ (aq) + Cl- (aq)}$$ is a disassociation reaction.

they partially overlap - decomposition can proceed via dissotiation

Your example of a "decomposition" reaction is not very "irreversible", in fact the opposite:  the reverse reaction (hydrogen combustion) that is irreversible.  It isn't really accurate to classify reactions by their reversibility.

Thanks for pointing that out. I have edited my answer now.