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Integrated rate law graphs
Why does the graph of $\ln[\ce A]$ vs $t$ produce a straight line for the first order rate law?
Also why does the graph of $t$ vs $\frac{1}{B}$ produce a straight line for the second order rate law?

The short answer is that the equations you describe as graphs are the solutions of the differential equations that define the rate laws in question.  If you don't understand what a differential equation is, any sensible answer is going to be problematic.
Assuming you do know what a differential equation is, we proceed!
For a first order rate law, the differential equation is something like:
$$-\frac{\mathrm d[\ce{A}]}{\mathrm dt}=k[\ce{A}]$$
Rearrange the concentration and time to get:
$$-\frac{\mathrm d[\ce{A}]}{[\ce{A}]}=k\,\mathrm dt$$
You now integrate both sides, with $[\ce{A}]$ as a function of $t$. This is a well-known integral, which is:
$$\ln [\ce{A}] = \ln [\ce{A}]_0 - kt$$
where $[\ce{A}]_0$ is an initial value for $[\ce{A}]$ at the start of the reaction.
Similarly, the differential equation for the second order rate law is:
$$-\frac{\mathrm d[\ce{A}]}{\mathrm dt} = k[\ce{A}]^2$$
Rearrange the concentration and time to get:
$$-\frac{\mathrm d[\ce{A}]}{[\ce{A}]^2} = k\,\mathrm dt$$
And again integrate to get:
$$\frac{1}{[\ce{A}]}=kt+\frac{1}{[\ce{A}]_0}$$
where $[\ce{A}]_0$ is an initial value for $[\ce{A}]$ at the start of the reaction.
This all comes from the following Wikipedia page, which has a lot more details: Rate equation

]Wikipedia got it wrong for the 2nd order differential rate equation.
It should be this:
-d[A] / dt = k[A]^2 for a second order reaction, where there is no 2 by the k constant, and the negative sign denotes the disappearance of the concentration of A with respect to time, and where d[A] / dt means the derivative of the concentration of [A] with respect to time (dt).
Divide both [A]^2 and multiply by negative dt:
{1 / [A]^2} d[A] = -k dt
[A]^(-2) d[A] = -k dt ... This is a separable equation.(Actually, 0 order to third order and even three-halves order are separable differential equations.)
Integrate both sides with definite bounds.
(from [A]0 to [A]t ∫[A]^(-2) d[A] = -k (from 0 to t) ∫(dt),
use power rule of integration (only the first law is special because if you use power rule of integration on 1 / [A] you get a divide by zero error.) So in power rule of integration, you add 1 to the existing power which is -2 and then divide by the new power which is -2 + 1 = -1, so that gives us the integration [A]^(-1) / -1 or -1 / [A]. And when you integrate dt, you are really integrating t^0 dt, so you get t, and "k" is a constant (non variable) so it is not integrated ever in all the rate laws.
-(1 / [A]) {[from [A]0 to [A]t} = -kt {from t = 0 to t}
Since both sides are negative, divide both sides of the equation by -1, and everything's positive.
(1 / [A]) {[from [A]0 to [A]t} = kt {from t = 0 to t}
(1 / [A]t) - (1 / [A]0) = k(t - 0) = k(t)
Then some algebra, adding 1 / [A]0 to both sides and tada!!
(1 / [A]t) = (1 / [A]0) + k(t)
Now how do we graph this linear? think about it. If we substitute 1 / [A]t as "y" and (1 / [A]0) as "b" and k(t) is m(x), that's basically the equation of a line: y = m(x) + b.
So you redefine the axes with the vertical axis as 1 / [A], and the horizontal axis as time (t), and you expect a straight line if your chemical reaction is second order.
Theoretically, you should be able tabulate your data and calculate logarithms and inverses and make tables, and compare how well the scatter plots form a line to figure out which model is the best model. But, if you do not follow the reaction over enough time, then all the models could have good fits which is undesirable.
Not showing the work, but for the 1st order reaction, the integrated rate law is this:
ln[A]t = ln[A]0 - k(t),
where remember [A]0 is the conc of reactant A at time 0, and [A]t is the concentration of reactant A at any given time. So if I just replace the ln[A] as the units of the vertical axis, and the "t" as the units of the horizontal axis, then I get another y = m(x) + b line.
Once you get your axes right in a table, you just format, you just do a linear regression in Microsoft Excel, QtiPlot, R, etc...  to get the equation of your line to fit the model.
I hope I helped you understand this.

Here is how you integrate the first order rate law:
https://www.youtube.com/watch?v=smcyZ_r0A88


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Substance of higher latent heat of melting than water, similar melting point
I saw ingeniously simple "cooler box" yesterday - a double-walled plastic box with some liquid between the walls. You freeze it, and it keeps temperature near the freezing point of that liquid, until the liquid melts, while you can store dry foods inside.
Now, the efficiency of such a box will depend on volume of the liquid, to a lesser degree - how far below its melting point it was frozen, and on latent heat of its melting. (and insulation of course, but that's an entirely different matter.)
While I suspect the liquid used in the box I saw was just water with a little coloring added, I wonder if I could do better than that. Is there a substance that can be reasonably obtained by a DIY tinkerer, that has a latent heat of melting significantly higher than water, and a melting point at normal pressure somewhere "within reason" - achievable by normal kitchen freezer, and not above "food inside will spoil"? (bonus points for it not being overly dangerous...) Or maybe is there some way to increase latent heat of melting water ice by adding something to it?

There are not that many (common) chemicals with a latent heat of fusion above that of water. Check out this list for example: http://www.engineeringtoolbox.com/latent-heat-melting-solids-d_96.html ---- therefore it will be very hard (maybe not possible) to find a liquid with the melting point roughly equal to water and a higher latent heat of fusion.

If you find one, give me a call! I definitely want to be in on the kind of business we could make out of that. But i think what your asking for is some kind of free lunch. And the basic rule is: There's no such thing.

Good answer by @Jerry, and comments about the uniquely high latent heat of fusion of water. Depending on the plastic, you could also use a mixture of dry ice chunks/powder and isopropanol. Isopropanol freezes about 10C below the sublimation point of dry ice, so you aren't likely to freeze it. Iit will just be a very cold bath until the dry ice sublimates and the solution heats up. Extra insulation, inside and out, would be in order to keep things from getting too cold and to make it last much longer. But step one would be testing that the plastic is OK with isopropanol. Then reuse the isoprop.

Indeed, it will be hard to find a substitute for water. You might want to look into how ice packs work.
Some of the things that are added to water in such packs include substances to prevent bacterial growth and to prevent the water to become fully liquid. The water thus become a gel.

Gel packs may be made by adding hydroxyethyl cellulose (Cellusize) or vinyl-coated silica gel.

One factor that you might consider, is that if there is more mass (for example, dissolving something in the water) will provide more substance for the any heat to go to, so that the temperature rise will be slower.
Another, is hydrogen bonds, which are very present in water molecules and which explains its high specific heats (latent heat and heat capacity).
Unfortunately, adding in stuff to the water will tend to disrupt the hydrogen bonds. I haven't tried it, nor can find something to support my next sentence, but I would expect ammonia to be a possible additive. It has a higher specific heat capacity than water and dissolves easily as well since it does not disrupt hydrogen bonds that much due to its molecular structure.
Glucose is also a potential additive since it doesn't disrupt the hydrogen bonds much, but its specific heat is apparently a bit lower than ammonia. I found a site where I could input some parameters and it outputs the heat capacity of the sugar water and the outputs often say that the specific heat capacity is lower than that of ammonia alone (from the same site that michielm mentioned, ammonia has a specific heat capacity of a bit less than 4.6 at $0^oC$).

The problem with ammonia (besides its smell and toxicity) is its melting point is -77C, so while I'd be increasing specific heat of the mix, I think I'd be reducing its latent heat at "home freezer temperature" - only after crossing -77C I'd get summarily more "cold stored" (heat removed) than I would in case of pure water.

@SF the ammonia will be dissolved in the water, so its melting point is not relevant. And since it will be dissolved, the smell will be significantly less (about 40 times less according to a pdf doc. The latent heat of water should not change by much since ammonia bonds well with water. And on the other side ammonia will absorb the heat as well all the way through the temperature changes. Meaning that when water will melt, more heat will be required to melt it since ammonia is appropriating part of that supplied heat.

Yes, but the volume is my limit. To make room for adding 1cc of pure liquid ammonia that will only absorb heat proportional to its specific heat, I have to remove 1cc of water, which would at the same time go through phase change and absorb massive amounts of heat proportional to its latent heat, on top of its similar (if lesser) heat absorption. Unless the bonding somehow reduces the volume of the mix... does it?

@SF You don't have to remove the 1 cc of water. You can add 1200 cc of pure ammonia and the 1 cc of water will remain the same. Or alternatively, if you have pure liquid ammonia, replacing 1 cc of water by 1 cc of it will actually equate to adding back 1 cc of water together with 1200 cc of ammonia (1200 volumes of ammonia dissolves in 1 cc of water at 0C). There is enough intermolecular space to fit ammonia between water molecules. I looked for some actual experiments where heat absorbed and temperature change were plotted for water+ammonia and water alone, but without success. :(

Wait wait wait. I have the dual-walled box with 100cc of room for coolant liquid. I fill it up with 99cc of water. How much pure liquid ammonia can I add before the mixture spills over?

@SF Add 1 cc pure liquid ammonia. It already contains as much ammonia as it can.

but then I'd gain more in means of thermal capacity if I added 1cc of water instead filling it to 100cc of pure water (because on the range between the, say, -20C and +5C 1cc of water has more thermal capacity (latent+normal) than 1cc of ammonia (only normal)).

@SF "Pure liquid ammonia" contains water and ammonia. Ammonia is a gas at 0 C. 1 cc of pure liquid ammonia contains 1 cc of water and 12 cc of (gaseous) ammonia dissolved in the water.

Oh, I think we've confused saturated ammonia water vs pure liquid ammonia below its boiling point. Ammonia is 0.86 kg/m^3 at boiling point, 681.9 kg/m^3 as liquid; 783 times more dense. 1188cc of gas soluble in 99cc of water is 1.5cc of pure liquid ammonia... Still doesn't seem it will fit ;)

@SF Okay, I didn't think about ammonia in its liquid phase because water would be ice at that temperature for mixing! And if that was possible, the ammonia will go into the water without adding the volumes (i.e. the net volume would stay at 99 cc).

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Chemical properties of isotopes
I've always heard that different isotopes of the same element have exactly the same chemical  properties.
But for example, I've read that some compounds which have hydrogen atoms interchanged by deuterium become odorless (this is one of the arguments for the vibrational theory of olfaction).

A 2001 study by Haffenden et al. showed humans able to distinguish benzaldehyde from its deuterated version. In addition, tests with animals have shown fish and insects able to distinguish isotopes by smell

So, are there any situations in which  neutrons do matter (besides nuclear decays)?
Addendum 2013/10/28
Another curious effect is that water is slightly blue, due to vibration. But heavy water is not.

Heavy water is colorless because all of its corresponding vibrational transitions are shifted to lower energy by the increase in isotope mass. For example the H2O band at 760 nm is shifted to approximately 1000 nm in D2O.

(Left: tube if filled with (light) water. Right: empty tube.)

Short answer: nuclear neutrons greatly affect chemistry involving hydrogen.
Longer answer
Hydrogen is the lightest element, and deuterium is just less than twice as heavy. This makes a pretty massive (pun intended) difference to any vibrational modes involving moving hydrogen/deuterium. This means that, for example, the deuterated equivalents of weak acids are less acidic, and any reaction involving the loss of a proton will be significantly slower with the deuterated version.
Since helium is pretty unreactive, and by the time you're at lithium an extra neutron is only increasing the mass by about 14%, the effects on reaction energetics are much, much smaller for anything else.
I'd suggest looking up the "Kinetic Isotope Effect" for more information and more complex effects the mass change on substituting isotopes can cause to chemical activity.
As Eric Brown said in his answer, different isotopes may also have very different spectroscopic properties if the spectroscopic technique depends either on vibrational or nuclear properties.
As a minor curiosity, I'm given to understand this also means that heavy water is mildly hazardous to your health, since deuterated water and deuterated analogues of various vital chemicals have a nasty tendency to react mostly irreversibly or very slowly with vital enzymes that are "tuned" for hydrogen.

I think isotopic analogues have mostly the same bulk physical properties.

There are some striking differences, e.g. water and heavy water (D2O) have perceptibly different boiling points.
There are many spectroscopies where isotopic substitution has a dramatic effect on the spectrum: including vibrational spectroscopies (IR/Raman), and nuclear spin (NMR, EPR, Mossbauer?).
Rates of reaction can be substantially altered by using isotopic substitution.

But I think there is some confusion here, as far as smell.  It is when you change chiral centers from $\ce{R -> S}$ or $\ce{S->R}$, thereby generating enantiomers and/or diastereomers, that you begin to affect things like smell.  It is profound in, e.g. menthol.
Diastereomers can have drastically different properties.  Enantiomers may interact differently with olfactory sensors, because our bodies (amino acids, sugars) have an intrinsic chirality.

My first idea when I read this question was: the masses are different, so everything that is affected by the mass of the ion or molecule must be different.
Diffusion and dissociation also come to mind. I did a quick search and found this: http://www.mendeley.com/research/secondary-deuterium-isotope-effects-acidity-glycine/

This is correct.  Any property depending on the mass of the substituent atomes will differ with the isotope used.  This is most apparent with hydrogen/deuterium because the mass is twice as much.

Lithium 6 and 7 have different electrolysis properties such that given an appropriate membrane - Lithium 7 can be enriched. Common is the mercury process.
Alternatively this process.
http://jolisfukyu.tokai-sc.jaea.go.jp/fukyu/mirai-en/2012/4_10.html

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Do proteins bind to carbon? If so how?
Can someone explain me whether a protein (like an antibody) would bind to carbon? The carbon arises due to electron beam induced deposition of Polyethylglycol (PEG). The carbon molecules that arise are either due to the PEG that dissociates or due to molecules, that are present in the microscope chamber as contamination, that fall apart. I am thus not sure about what the carbon molecule is specifically. You can look in this experiment to see more of the procedure.
And why? (what bond?)

@aaab, rather than clarifying your question in the comments, you should edit the question itself. I have edited the content of your two clarifying comments back into your question.

Ok thanks, didn't know it was possible. Does anyone have some suggestions on answering my question?

Various carbons arising from thermal decomposition of organics have complex and different structures. While there are amorphous carbons, containing unordered sp3 mesh of carbons with various organics, there are also graphitizing carbons, containing packets of graphite layers with unordered interconnects and non graphitizing carbons, containing interwoven graphite layers. The article does no clarify, what kind of carbon is produced in the experiment, while all types of carbons have different adsorption properties. My bet is that proteins sorbs on oxygen species terminating graphite layers.

@Ben When you insert comments into the body, which is very much appreciated, custom flag the ones you made obsolete with that action so that they don't clutter this part.

I can't tell you for sure what is happening, but I can tell you what could be happening.
Physical Adsorption
It is possible that the proteins are not chemically adsorbed, but rather physically adsorbed. The article you cited implies that this is the case.
Many proteins in the presence of nano-scale hydrophobic "features" will partially denature, and the hydrophobic core will physically adsorb (attach via van der Waals forces). This effect is amplified if water is present, since the hydrophilic parts of the protein will be stabilized in water, allowing more of the hydrophobic parts to attach to the carbon feature.
Now you are probably wondering, "Why doesn't it attach to the PEG surface as well, since that is hydrophobic?" There are two possible reasons.
PEG is not as hydrophobic as you think

The PEG chains have oxygen in each monomer, which would reduce the non-polar character of the chains. It is possible that the difference is enough to make adsorption of non-polar amino acids on PEG thermodynamically unfavorable, while on pure-carbon structures it would be favorable.
Shape matters
Another possibility is that the actual shape of the features makes the difference. In order for a protein to denature and "attach" to carbon feature via hydrophobic interactions, it has to first expose the hydrophobic groups in the core. For a flat surface, the amount of initial denaturization (maybe not a real word) would need to be higher than for a "sharp" feature. Sort of like how a needle can slip into your skin without doing much damage. So, the smaller the characteristic length of the features, the more likely the protein would adsorb. There are also protein-protein effects on the surface, which are again reduced by smaller-scale features. In this scenario, you could think of the proteins as kinetically trapped in solution when near a regular PEG surface, while the activation energy for adsorption to a small feature is smaller.
I can think of two ways to test this idea:

Use different proteins on the same type of surface features - the ones with less exposed hydrophobic groups should adsorb less easily
Use different sizes of features - there should be a feature size that gives a maximum adsorption for a given protein.

Chemical Bonding
The other possibility is chemical bonding between the carbon features and the protein. It is unlikely that this will happen if the carbon features are really pure carbon (graphene, for example). However, if they are creating them by ionizing PEG with an electron beam, then it is highly likely that they are functionalized with something. Carbonyl groups would be my guess, but depending on the conditions there could be nitro groups as well. In this case, there is a possibility that hydrogen bonds would form, and if the surface is reactive enough, chemical bonds might form as well.
To test this idea, I would do the same thing as before - only this time, if we are right:

Proteins with more exposed hydrophobic groups should be less likely to adsorb
Feature size should not matter at all

If this turned out to be the case, you could then systematically test polypeptides made up of different amino acids to verify it and determine whether hydrogen bonding or covalent bonding was more likely.


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Round Bottom Flask with Inner Coating
I am using a rotary evaporator to separate DCM (dichloromethane) from some nasty tars.
Before the evaporation is completed, I stop the process and collect the remaining slurry.
The slurry is then positioned in a constant temperature oven, for a secondary evaporation. After cooling in a desiccator, the tar is measured.
The problem is that some of the tars remain deposited on the flask's wall. I can easily remove them with a soft brush, however this is a problem for 2 reasons:

Some of the tar is lost, retained on the brush itself ( I require a very high precision since I work with samples of 100 or so mg)
The amount of DCM for secondary evaporation is increased (I am trying to reduce that amount as much as possible)

Is there a round bottomed flask that has a TEFLON like coating on the inside, that can prevent the tar deposition on the walls?
There seem to be a lot of products that are referred as " Flask with Plastic Safety Coating". On which side is the coating applied (I assume is the outside)

I've seen a couple of companies making flasks entirely from PTFE or PFA, which might be suitable. They may not hold up to heating with internal vacuum though, for example, check out the notes here.
And yes, the typical safety coatings are on the outside to hold the glass together on fracture.

it says that the maximum temperature under vacuum is 50 C.
Luckily for me, I only need 40.

Removing solvent from goo is very difficult.  I had an oil that was a solid melting above 100 C.  I pumped it under good vacuum as a thin film while doing all the usual tricks with ice then Dry Ice.  It finally crystallized - and bubbled while doing it.
Siliconize the glass flask's inside.  Cheap and easy is Rain-X for car windshields.  Spray some inside the clean dry (air dry not oven dry) flask.  Swirl for a couple of minutes, drain, dry.  Couple of quick rinses with methylene chloride, set coating by baking at 110 C for 15 minutes.  The coating is stripped by alkali and nucleophiles like NaOH or KOH in water, a base bath, aqueous ammonia, aqueous fluoride.  Renew as needed.
If that is better, make it a formal procedure using Gelest Aquaphobe® CM or Aquaphobe® CF.  Rain-X is $\ce{(RO)3Si}-$terminated polydimethylsiloxane oligomer in alcohol.  It contains some silicone oil.  The Gelest products are $\ce{Cl3Si}-$terminated oligomers.  You do the 8% dilution in dry methylene chloride or toluene.
http://www.gelest.com/gelest/forms/GeneralPages/Applications/hydrophobic_materials.aspx
If you want to spend money, there are medical siliconizers for treating blood-handling glassware.


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acyclically stereocontrolled reaction: organolithium and 2-chiral compound
Aoyagi et al. (J. Am. Chem. Soc., Vol. 115, No. 24, p 11393) reported that the addition of 2-lithio-1,3-dithiane to the alpha-chiral (S)-2-Acetylpiperidine gives the (2 S,3 S)-configured product. It is stated that the diastereoselectivity of this 1,2-addition (dr = 6.2) is predicted by Cram's model involving complexation of the Li-ion by the piperidine-N. I tried to reproduce the diastereoselectivity in the following scheme:

As you can see I come up with the wrong product by applying Cram's rule with a O,N-complexed Li-ion (right side of the scheme). By applying the non-chelate Cram rule (left side; biggest substituent oriented anti to the carbonyl, attack via smallest group) I get the correctly configured (2 S,3 S)-product.
Where is the mistake?

I have just had a look at the paper and actually they reported that their main product was S,R:
"...treated with excess 2-lithio-1,3-dithiane in THF at -50°C; this treatment generated the tertiary alcohols 22a and 22b in 72% combined yield with 6.2:1 diastereoselectivity as predicted by Cram's model based on participation of the piperidine nitrogen atom, giving a rigid cyclic complex with the lithium compound".
Compound 22a: (2S)-2-[(R)-l-(1,3-Dithian-2-yl)-l-hydroxyethyl]piperidine
Compound 22b: (2S)-2-[(S)-1-(1,3-Dithian-2-yl)-l-hydroxyethyl]piperidine
I do not find any mistake.


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Tetrahedron is minimum energy conformation for atom with 4 electron domains in VSEPR theory
Background
A student asked me to prove that the regular tetrahedron is the minimum energy geometry available to describe the locations of the electron domains in three dimensions – the prediction of valence shell electron pair repulsion (VSEPR) theory for a central atom with 4 mutually repelling electron domains. With the tetrahedron demonstrated, it could then be shown using vector multiplication that the angles between all substituent atoms is $109.5^{\circ}$.

Some ideas
I can think of a few ways to start this proof:

As mentioned above, we could define a regular tetrahedron and work backwards to see if the distance between all vertices is equal and maximized. Start with the vertices: A(0, 0, 0), B(1, 1, 0), C(1, 0, 1) and D(0, 1, 1), and a central point O($\frac{1}{2}$, $\frac{1}{2}$, $\frac{1}{2}$). With these, we can get vectors that point to each vertex. Call them $\vec{r_1}$ through $\vec{r_4}$. We remember that we can calculate the angle between any two vectors using the geometric and algebraic definitions of the dot product. I won't bore you with the details of this calculation but if anyone wishes to see it, I would be happy to share it. If you do it you will see that all the angles are equal and satisfy: $\cos^{-1}\left(-\frac{1}{3}\right) = \theta_i \approx 109.47122^{\circ}$. From here we would need to calculate the six distances bewteen the four vertices to show that the distances are all equal.
Using multivariable calculus, we could treat this as an optimization problem subject to the constraint matrix, $g$; the sum square distance function, $f$; and use the optimization formalism described by Lagrange. This results in a system of 13 unknowns – the 4 points with three coordinates each plus $\lambda$ – and 13 equations – the four constraint equations plus the nine partial derivative equations from the Lagrange method:

$$g = 1 = \left\{\begin{array}{c}g_1(x_1,y_1,z_1) = {x_1}^2 + {y_1}^2 + {z_1}^2\\
g_2(x_2,y_2,z_2) = {x_2}^2 + {y_2}^2 + {z_2}^2\\
g_3(x_3,y_3,z_3) = {x_3}^2 + {y_3}^2 + {z_3}^2\\
g_2(x_4,y_4,z_4) = {x_4}^2 + {y_4}^2 + {z_4}^2\\
\end{array}\right.$$
$$f = \mathop{\sum_{i=1}^{3}\sum_{j=1}^{3}}_{i<j}\left[\left({x_i}-{x_j}\right)^2+\left({y_i}-{y_j}\right)^2+\left({z_i}-{x_j}\right)^2\right] $$
$$\nabla f=-\lambda \nabla g$$

An approach similar to the one outlined above could be used with spherical coordinates instead. This version reduces the set of variables from 13 to 8 (the four sets of two angles for each point $\theta$'s and $\phi$'s; note the spherical radius could be set to 1 or any other bond distance, $a$, so is not a variable). These could be further reduced using symmetry arguments. For example, if we take the first vertex at the point $(0, 0, 1)$, we know based on symmetry that if we look down the z axis at the xy-plane, the angle between the three other vectors must have an angle of $120^{\circ}$ with respect to one another. Let one of the other vertices have the point $(x, 0, -\sqrt{1 - x^2})$ – satisfying the radius of 1 unit constraint we have imposed. Now based on symmetry, we know that the other two points must be $(x\cos(\frac{2\pi}{3}),x\sin(\frac{2\pi}{3}),-\sqrt{1-x^2})$ and $(x\cos(\frac{4\pi}{3}),x\sin(\frac{4\pi}{3}),-\sqrt{1-x^2})$ using spherical symmetry. From here, we can write a formula for the total sum square distance between all the points and set the derivative with respect to $x$ to zero to find the maximum sum square distance. Notice we have reduced the problem with these symmetry arguments from a multivariable calculus problem to a calculus BC problem. Alternatively, we could use some other guess and check method to solve for $x$. Note, we should find the same answer using this method as the others above (and below).

Sketch of a Simpler Solution
There is one remaining way that is simpler than all of the above, and it has to do with circumscribing a tetrahedron with a cube.
I will leave it to the reader to work on this most beautiful version. The properties of the cube ensure that the distances are maximized, and some more geometry will show that the inscribed shape is indeed a regular tetrahedron. For a hint, check out this excellent answer to this question on the math stack exchange site.


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Surface Functional Groups/Notation
I'm reasonably unfamiliar with chemical notation, and recently I've come across something I'm not sure how to interpret.  In this article, the authors refer to the "surface groups" denoted by $\ce{#Ti-F}$ and $\ce{#Ti-OH}$.
Similarly, this article refers to surface functional groups $\ce{#Si-O-}$ and $\ce{#Si-OH}$.
To my (limited) knowledge, "$\ce{#}$" denotes a triple bond.  If this is the case, what am I to take from this?  Are they referring to a (triple) dangling bond, or is this some notation I'm not familiar with to denote surface species?  The first page of the first article does state that "$\ce{#Ti-X}$" refers to surface species, but I'm finding it difficult to believe the notation was chosen arbitrarily.
I appreciate any explanation.  As I've said, I am a novice with chemistry and chemical notation.

Better representation would be e.g. ⚞Si−OH, but I did not see any official material discussing this topic.

The notation used does not necessarily represent a triple bond - for example, using one of the examples that you have provided:

$\equiv\text{Si}-\text{OH}$

Which is for the functional group silanol, that is defined by the $\ce{Si-O-H}$ connectivity, it's chemical structure is

Image source
Using this example, the hydrogens represented by the  ≡ symbol can be replaced by other chemicals forming other members of the silanol functional group, for example, Trimethylsilanol:

To further clarify, that $\ce{#}$ indicates that the $\ce{Si}$ atom has three bonds sub-surface things: it may be three single bonds to three individual atoms or groups as indicated here. It may be one single and one double bond to two atoms or groups. It may be a triple bond to a single atom or group. The $\ce{#Si-OH}$ notation indicates that since we are only interested in surface groups, we don't care about the sub-surface structure and connectivity.


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What happens when you mix alum with soap?
I was handling alum one day and accidentally I touched toilet soap after that. They reacted with each other and formed a rubberlike sticky substance (like dried glue on palm). I am curious what was the chemical reaction and the substance which was formed. I re-tried the same thing with detergent and got similar results.

What type of alum? Potassium Alum, Chrome alum, Ammonium alum, soda alum, a type of selenate alum?

In common household applications, "alum" as one might purchase is usually potassium alum, $\ce{KAl(SO4)2\cdot 12H2O}$

You could've formed an aluminum carboxylate gel.

I think you’ve made a so called “aluminium soap”. It is a compound of the form $\ce{(RCOO)3Al}$ where $\ce{RCOO-}$ is a fatty acid chain. The $\ce{Al^3+}$ ions from the alum you handled would have displaced the $\ce{Na+}$ ions in soap to form the thick gel like aluminium version of soap.
This is similar to how $\ce{Mg^2+}$ or $\ce{Ca^2+}$ ions form an insoluble scum (which consists of $\ce{(RCOO)2Mg}$ or $\ce{(RCOO)2Ca}$) when soap is used in hard water.
I assume here that you’re talking about potash alum $\ce{KAl(SO4)2.12H2O}$ as @BenNorris has mentioned in his comment above.
Any other aluminium containing alum will also have the same effect. But chrome alums (containing chromium instead of aluminium) might have a different effect on soap.
These “aluminium soaps” are used as a thickening agent in oils and greases and also in napalm (as Georg has mentioned in his comment above).

Hi Kaliaden, maybe you are right as the substance formed was sticky in nature. And yes, I used household Alum only which we generally use for cleansing of water, shaving, disinfectant etc.

This reminds me of two things in my professional life, I have been involved both in solvent extraction chemistry and coordination polymers.
The carboxylic acid salts in a typical bar of soap would be able to react with the aluminium ions to form the aluminium carboxylate (aluminium soap). In solvent extraction what you can attempt to do is to have a mixture of the extractant (such as DEHPA or a carboxylic acid) in an inert organic solvent (diluent) such as aliphatic kerosene.
Normally aluminium forms a series of water soluble complexes or hydroxides, these are normally very insoluble in non polar solvents. However the complexes of aluminium with liophilic anions such as deprotonated DEHPA or long chain carboxylates are normally soluble in non polar diluents.
If you look at figure 9 in this cogent chemistry paper you will see how the distribution ratio of aluminium (and the early lanthanides) follows the same shaped line when plotted against the sulfuric acid content of the lower phase. We can write the following equation for the distribution ratio of aluminium (first approximation).
D = [DEHPA]^3[H+]^-3
Now if the solvent extraction experiment is conducted without adding sufficent organic solvent (diluent) to dissolve the metal complex, and where the extractant is added either as a pure liquid or as an aqueous solution then it is possible for the metal organic ligand complex to separate from the water as either a greasy (lipophilic) liquid or as a crystaline solid. Conisder for a moment what happens if you add acetylacetone (H-ACAC) to a solution of copper sulfate in aqueous ammonia, What will happen is that the complex [Cu(ACAC)2] will form a insoluble solid.
If a dilute solution of ACAC is used then it is possible to use it to extract metals from aqueous solutions with modest pH values into organic solvents. For example one of my coworkers did part of their PhD by doing an experiment on the extraction of thorium with ACAC or a similar ligand.
I also see that what can happen is that the metal and the carboxylate can form a polymeric or polynuclear coordination compound. These can be very low in solubility and can separate out from the water.


text:
How to find a transition state for an electrophilic addition with Gaussian and map the reaction pathway?
I aim for two things with this post – on the one hand to produce a manual for other people to use, but also to verify what I obtained via searching, asking a more experienced user as well as receiving a reply from Gaussian support.
I will assume that the user is familiar with the general basic idea of the format of a Gaussian input file.
The objective is to find a transition state for an electrophilic aromatic substitution reaction in decane as  a solvent.
The ideal initial method to obtain a transition state is a QST2 calculation which requires a geometry for the reactants as well as the products. If it fails, a QST3 calculation which also provides a guess for the geometry of the transition state may also be attempted or as a last resort, a transition state may be manually built and optimized.
Ideally the user should have optimized geometries for the reactants as well as the product(s) in a reaction, however it is possible to build the geometry “on the spot”.
In the example I provide the input geometries for m-toludine and acetone:
In this case the oxygen atom  has been shifted out of the way – however the hydrogen has not been (should be though) tweaked on the m-toluidine. If bonds are broken, it is advisable to lengthen the bonds slightly as for example singe bonds will be longer than double bonds.
In the product, the initial molecules have been “rebounded” for the product.
“Imaginary bonds”, bonds from the product in the reactants or bonds from the reactants in the product are shows as dashed lines).
%nprocshared=4
%rwf=m-toluidine-acetone-2.rwf
%nosave
%mem=750MB
%chk=m-toluidine-acetone-2.chk
#p opt=(calcall,qst2) b3lyp/6-31++g(d,p) scrf=(solvent=n-decane) nosymm
geom=connectivity

Title Card Required

0 1
 C                 -0.66738150    0.29867781   -0.01472477
 C                  0.72777850    0.29867781   -0.01472477
 C                  1.42531650    1.50642881   -0.01472477
 C                  0.72766250    2.71493781   -0.01592377
 C                 -0.66716250    2.71485981   -0.01640277
 C                 -1.36476350    1.50665381   -0.01540677
 H                 -1.10194426   -0.62306930    0.00047259
 H                  1.27728650   -0.65383519   -0.01340977
 H                  1.27786250    3.66708081   -0.01598277
 H                 -2.46436750    1.50683681   -0.01558677
 N                 -1.40248531    3.98773009   -0.01767660
 H                 -1.16147100    4.51273462    0.79858460
 H                 -1.16060228    4.51160224   -0.83440824
 C                  2.96531624    1.50654085   -0.01383691
 H                  3.32245021    2.00928088   -0.88825546
 H                  3.32144259    2.01265639    0.85904400
 H                  3.32205530    0.49776307   -0.01168239
 C                 -2.94426822   -1.38737908   -0.21119110
 C                 -2.43095250   -2.83931123   -0.21119110
 H                 -2.78762331   -3.34370870    0.66246165
 H                 -2.78762737   -3.34371014   -1.08484136
 H                 -1.36095250   -2.83932442   -0.21119359
 C                 -4.48426822   -1.38736010   -0.21119110
 H                 -4.84094078   -1.89192796    0.66236254
 H                 -4.84092213   -0.37854994   -0.21099580
 H                 -4.84094128   -1.89158925   -1.08494009
 O                 -2.63797974   -0.83944966    0.94169704

 1 2 1.5 6 1.5 7 1.0 18 0.5
 2 3 1.5 8 1.0
 3 4 1.5 14 1.0
 4 5 1.5 9 1.0
 5 6 1.5 11 1.0
 6 10 1.0
 7 27 0.5
 8
 9
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 11 12 1.0 13 1.0
 12
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 14 15 1.0 16 1.0 17 1.0
 15
 16
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 18 19 1.0 23 1.0 27 2.0
 19 20 1.0 21 1.0 22 1.0
 20
 21
 22
 23 24 1.0 25 1.0 26 1.0
 24
 25
 26
 27

Title Card Required

0 1
 C                 -1.10949687    0.47711117    1.88331619
 C                  0.28566313    0.47711117    1.88331619
 C                  0.98320113    1.68486217    1.88331619
 C                  0.28554713    2.89337117    1.88211719
 C                 -1.10927787    2.89329317    1.88163819
 C                 -1.80687887    1.68508717    1.88263419
 H                 -1.73208670   -0.29950945    3.46519163
 H                  0.83517113   -0.47540183    1.88463119
 H                  0.83574713    3.84551417    1.88205819
 H                 -2.90648287    1.68527017    1.88245419
 N                 -1.84460068    4.16616345    1.88036435
 H                 -1.60358637    4.69116797    2.69662556
 H                 -1.60271765    4.69003560    1.06363271
 C                  2.52320087    1.68497420    1.88420404
 H                  2.88033484    2.18771423    1.00978550
 H                  2.87932722    2.19108974    2.75708496
 H                  2.87993993    0.67619643    1.88635857
 C                 -2.60557153   -0.76810572    1.76628847
 C                 -2.09225582   -2.22003788    1.76628847
 H                 -2.44892663   -2.72443535    2.63994122
 H                 -2.44893069   -2.72443679    0.89263821
 H                 -1.02225582   -2.22005106    1.76628598
 C                 -4.14557153   -0.76808675    1.76628847
 H                 -4.50224410   -1.27265460    2.63984211
 H                 -4.50222545    0.24072341    1.76648378
 H                 -4.50224460   -1.27231589    0.89253948
 O                 -2.55774834   -0.63242367    2.73721431

 1 2 1.5 6 1.5 7 0.5 18 1.0
 2 3 1.5 8 1.0
 3 4 1.5 14 1.0
 4 5 1.5 9 1.0
 5 6 1.5 11 1.0
 6 10 1.0
 7 27 1.0
 8
 9
 10
 11 12 1.0 13 1.0
 12
 13
 14 15 1.0 16 1.0 17 1.0
 15
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 18 19 1.0 23 1.0 27 1.0
 19 20 1.0 21 1.0 22 1.0
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 23 24 1.0 25 1.0 26 1.0
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The calculation can then be started.
Once it has successfully finished, it is advisable to verify that the molecule exhibits one imaginary frequency only and that the motion of the atoms or molecules under that frequency agree with the expected reaction. If this is the case, one should follow up with an IRC calculation to follow the reaction path to verify one has indeed found the desired transition state.
(The geometry below is the transition state from the calculation above, set up for an IRC calculation.)
%nprocshared=4
%rwf=m-toluidine-acetone-2.rwf
%nosave
%mem=750MB
%chk=m-toluidine-acetone-2.chk
#p irc=(maxpoints=200,recalc=5,calcfc) b3lyp/6-31++g(d,p)
scrf=(solvent=n-decane) nosymm geom=connectivity

Title Card Required

0 1
 C                 -1.20983500    0.26428000    0.39208600
 C                  0.14314000    0.04829600    0.07611800
 C                  0.97119600    1.12420400   -0.26335000
 C                  0.42483200    2.41568200   -0.28932800
 C                 -0.92232100    2.65929200    0.02889700
 C                 -1.73598700    1.56297000    0.36139500
 H                 -1.80370000   -0.35728900    1.39005500
 H                  0.56171000   -0.95292700    0.09719200
 H                  1.06141800    3.25561300   -0.56193000
 H                 -2.77794400    1.73200600    0.62392300
 N                 -1.45199400    3.94909800   -0.04588200
 H                 -2.29620600    4.10853800    0.48772900
 H                 -0.78926200    4.70638200    0.05450600
 C                  2.43332600    0.91594500   -0.58932400
 H                  2.68752500    1.33957100   -1.56739900
 H                  3.07670000    1.40280600    0.15294900
 H                  2.68804800   -0.14719300   -0.60548500
 C                 -2.49137900   -1.22149000    0.43327000
 C                 -1.66737300   -2.42548800    0.01309500
 H                 -2.34686200   -3.28416400    0.08629100
 H                 -1.29025700   -2.36043500   -1.01081300
 H                 -0.84776200   -2.60357900    0.71204300
 C                 -3.57286200   -0.83131000   -0.55877300
 H                 -4.30815500   -1.64705100   -0.52976700
 H                 -4.08297000    0.08110500   -0.24873300
 H                 -3.20072400   -0.72720100   -1.58114900
 O                 -2.83178700   -1.17441600    1.73871300

 1 2 1.5 6 1.5
 2 3 1.5 8 1.0
 3 4 1.5 14 1.0
 4 5 1.5 9 1.0
 5 6 1.5 11 1.0
 6 10 1.0
 7
 8
 9
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 11 12 1.0 13 1.0
 12
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 14 15 1.0 16 1.0 17 1.0
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 18 19 1.0 23 1.0 27 1.0
 19 20 1.0 21 1.0 22 1.0
 20
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 23 24 1.0 25 1.0 26 1.0
 24
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So:

I am aware that the initial reactants guess is not great in this calculation as it dates back to me starting to work with Gaussian.
Is there anything I am missing that could have a significant impact on the obtained transition state (and the resulting energy barrier for the reaction)?

Why are you putting diffuse functions on the hydrogen atoms?  Why are you computing the force constants in your QST2 job at every step (CalcAll)?  Also, if you are running an IRC computation on a TS, you don't have to recalculate the force constants if you verified your TS beforehand via a harmonic vibrational frequency computation.  You can read in the force constants from a checkpoint file.

I found that Gaussian tends to be more reliable with CalcAll over CalcFC. But it is a valid point, often trying with just "CalcFC" first is enough. Regarding the Force Constants in the transition state: I tend to create a new file copying the geometry across from the QST2 calculation.

You should never have to do a CalcAll except in extreme circumstances.  It is much too expensive and inefficient.  For the IRC, you have to verify your structure as a TS first by doing the freq calculation.  This means that you can just read in your force constants (RCFC) without having to recalculate them.  Also, QST2 and QST3 jobs generally fair better when creating your own Z-matrix as opposed to allowing the program to create its own internal coords. Also, your post deals more along the lines of mapping out a reaction pathway, not just finding a TS.  Your title should reflect this.

@LordStryker I think problems might arise in QST2/3 calculations when employing solvent approximations. It massively slows down the calculation as they usually tend to oscillate. That's often also a problem for the updating of the force constants, hence leading to convergence issues. Since these type of calculations are preliminaries at best, one can simply omit the solvent and include it after having obtained a well converged TS.

I strongly disagree on "as a last resort, a transition state may be manually built and optimized." If you have well enough insight into the mechanism, this should be the first choice. QST methods are expensive, even a scan can be much more helpful and less costly.

Mapping a Reaction Pathway: A Simple Guide
There are various ways to go about determining a reaction pathway using electronic structure theory.  This post is to serve as a brief guide, highlighting a few useful techniques that may make this process more efficient.  I will restrict this guide to a simple reaction mentioned below.
Preliminaries
Suppose you have a reaction of interest such as A + B -> C.  In terms of relative energies we could define the electronic energies of an isolated A and an isolated B (i.e. (A)+(B)) as the separated reactants (SR) energy.  Furthermore, we could define the electronic energy of C as the product (prod) energy.  So now we have determined two electronic energies that can be set relative to each other in order to determine the deltaE of reaction which I will call dE which is simply (E(prod)-E(SR)).  However, when it comes to reaction pathways, we are also interested in the activation energy (Ea) of the reaction.  This means, for our basic reaction example, that there is one other electronic energy which we must consider, which leads us to the transition state (TS).  Resolving this structure will be discussed from this point on.
Technique #1 - Finding a TS Using the QST2 & QST3 Methods
The QST2 and QST3 methods allow the user to specify two minimum energy structures lying on each 'side' of a TS, and have an algorithm determine a TS which connects the two.  In our example, SR and Prod are these two structures.  Now, for our simple reaction example, using QST2 is unwise as our SR is simply two molecules, A and B, separated at infinite distance.  This is not a meaningful geometry in terms of a QST method.  Our Prod structure would work quite well, however.  The QST methods work well if both of your input geometries represent coherent molecular systems.  So if you were resolving a more complicated reaction pathway with one or more intermediates, you could use the QST method to determine TS structures connecting an intermediate to another intermediate or an intermediate to a product.
The difference between QST2 and QST3 is simply that the latter allows the user to input a TS guess.  If you have fully characterized both the low-lying minimum energy structures which you wish to find a connecting TS structure, you may use your own intuition and guess what the TS may or should look like.  This can help speed up the process of locating a TS with QST3.  However, if you do not know what the TS would look like, you can revert to using QST2 and not provide a TS guess.  In my experience with these methods, it is usually more efficient to provide the program with your own user-defined internal coordinates rather than allowing the program to create its own. If you are lucky enough to have this type of job complete successfully, you will want to optimize the resulting geometry and perform a frequency analysis to determine the nature of the stationary point.  Since we are looking for a TS, we would have to end up with one imaginary mode.  If you don't get this, you need to go back try finding another TS.

QST2 only requires two fully characterized minimum energy structures.  These must be coherent chemical systems.
QST3 allows the user to provide a TS guess
Consider using your own internal coordinates rather than allowing the program to create its own
Fully characterize the resulting structure by performing an optimization and a corresponding harmonic vibrational frequency analysis.

Technique #2 - Finding a TS via Scanning
We have already stated that SR is simply (A)+(B) where A and B are separated at infinite distance.  We can look at our Prod and gain some insight as to how A and B come together to form Prod.  So rather than leaving our TS guess up to some program to decide, we can be smart about the way we go about finding it by using our own, well-informed intuition.  Perhaps by observing SR and Prod we make a reasonable guess that the distance between some atom on molecule A and some atom on molecule B decreases as we go from SR to Prod.  Looking at the fully characterized Prod, we know the final distance between these two atoms.  In SR, this distance is infinite.  What we do is we take our Prod and scan across this distance, partially optimizing the geometry as we go.  So if the distance between the two atoms is 1 Angstrom in the Prod, we would gradually increase this distance in increments of say 0.2 Angstroms a total of, perhaps, 10 times.  Our final distance would then be 3.0 Angstroms of separation.  From this can, we can look at the electronic energies and, if we guessed a good coordinate to scan over, we should have a smooth energy curve.  From this we can yank the geometry with the highest electronic energy from the scan and use this as our guess to the TS.
Take your guess TS from the scan and insert it into a TS optimization job.  With a bit of luck, your guess will be close to a real TS.  Again, perform a frequency analysis and determine the nature of the stationary point.  We must end up with one imaginary mode for our structure to be a TS.

Scanning requires a good guess as to which coordinate to scan across; you will have to specify your own internal coordinates
The maximum energy structure from your scan is used for a TS optimization
This manual approach maximizes user intuition

Verifying the TS - The IRC Computation
Just because we've now characterized a TS does not mean that the TS connects, in our case, SR and Prod.  We must perform an intrinsic reaction coordinate (IRC) computation.  Simply, we give a program our fully characterized TS and we follow the reaction path in the forward and reverse directions.  Be sure to give your IRC calculation a large number of steps.  IRC calculations require force constants on your TS.  Since you've already fully characterized your TS via a frequency analysis, you already have the force constants.  Just feed them into your IRC job, eliminating the need to recalculate them.
Once your IRC calculation has finished, you must optimize the final structures from both the forward and reverse directions.  The resulting structures should have the same energy and same geometry as the structures you used to find the TS.  Of course in our simple reaction case, you would only be comparing the Prod as SR is molecules A and B at infinite distance.  For this, just ensure that the IRC proceeded in such a way that A and B are separating from each other across the entire IRC pathway.

Perform an IRC on the TS you have characterized in both the forward and reverse directions
Optimize the final structures from the IRC computation and compare to the fully characterized structures you used to initially find your TS.  If they match both energetically and geometrically, then you have found a plausible TS which connects both structures.

What do you mean when you say "Consider using your own internal coordinates ..."? Will you specify every coordinate with 'modredundant'/'connectivity' or do you simply use a z-matrix (and force the program to stick with it)?

@Martin I was referring to using your own defined z-matrix but obviously this only works well for very small systems.  Even systems with a dozen atoms can be a pain to create a z-matrix when dealing with high symmetry (dummy atoms are your best friend but knowing where to put them can be difficult).

Absolutely true, also I guessed that already. Although I have had quite good experiences using qst3 with cartesian coordinates. But sometimes you just need luck ;) Btw. Happy new year!

@LordStryker For large systems, I usually open them in Molden and edit their z-matrix from there when I want to use opt(z-matrix) in Gaussian. That way I only have to build the parts of the system wherein I need to change the way parameters are defined, for example, in order to link things.

I would like to build on LordStryker's well-written guide by providing some elementary examples of input file formatting for these calculations.
Now that I've re-read the question, I see that my answer is a little too general for the desired guide. I will leave it here, but you can skip to the end for the practical example. I did one with an electrophilic aromatic substiution reaction. I guess this isn't exactly what was asked for, but it might still be useful.
Part 1: The Basics
Formatting & Optimization
Before using we can find a transition state, we will usually need optimized structures corresponding to minima on the reaction coordinate. If I want to explore a complex reaction with multiple intermediates, I find that it can be very useful to start by obtaining optimized geometries for all minima before I start looking for transition states, that way I can get a good preliminary understanding of the reaction's thermodynamics.
If you are studying a reaction from experimental chemical literature, geometries of intermediates may have been located by x-ray crystallography. If so, these structures will usually be provided in the form of an XYZ coordinate system (.xyz).
.xyz files are formatted as follows:
3
ABC
A x y z
B x y z
C x y z

where 3 is the number of atoms. XYZ files can easily be converted into the format required by Gaussian input files (.gjf) by removing the first line (n atoms) and the second line (title card) and replacing them with the charge & multiplicity card ("q m", i.e.: "0 1"). In a gaussian file, the above coordinate system would look like
ABC

0 1
A x y z
B x y z
C x y z

All this will come after a blank line, following the input section. By default, Gaussian also includes a connectivity section (after the redundant coordinates), therefore a file from GaussView will usually have Geom=connectivity in the input line. Connectivity is important for force field methods or molecular mechanics, but it is not required in DFT calculations. I will leave connectivity out of my examples, but it can always be replaced by leaving Geom=connectivity in the input line.
In order to optimize this geometry; I would use the following input:
%chk=ABC.chk
%nproc=8
%mem=12907MB
#P functional/basis-set opt

ABC

0 1
A x y z
B x y z
C x y z

Of course, there are a lot of other commands that can be used here. For large organic molecules with many tetrahedral centers, I would use an input line like this:
#P functional/basis-set opt int=grid=ultrafine nosymmetry

I use nosymmetry if this is a large organic molecule with low symmetry. Sometimes Gaussian's attempt to introduce symmetry will affect the calculated energy. I use UltraFineGrid if the molecule has a lot of methyl groups or flexible, saturated chains. Keep in mind that it is necessary to be consistent with the inputs in order to have comparable energies for a given reaction path, so if you use these commands once, you must use them for all related calculations.
Another useful technique is constrained optimization. For QST2 calculations, for example, it may be important to find a starting structure wherein two seperated products come together to form a complex by noncovalent interactions of some sort, before reacting to form the desired product. In some cases, this can be found using a constrained optimization on some coordinates of the system (this might move other groups to make the sterics more favourable), then I would re-optimize the system to see if I can get my complex. An input for a constrained optimization at 1.97 angstroms between atoms B and C would look like this:
#P f/b opt(modred)

AB-CD

0 1
A x y z
B x y z
C x y z
D x y z

B 2 3 1.97 F

omitting the %chk, %nproc and %mem lines and abbreviating "functional/basis-set" to f/b (as I will do from here on out). 2 is for atom B, 3 is for atom C and 1.97 is for the distance.
QST2 & QST3
LordStryker has explained these methods, so I will just provide an example of the input format:
#P f/b freq=noraman opt=(qst3,calcfc)

Reactants

q m
<cartesian>

Products

q m
<cartesian>

Transition Structure Guess

q m
<cartesian>

for QST2, you would remove the lines after the product cartesian input.
In all my TS calculations I use opt(calcfc) so that Gaussian calculates the force constants at the initial geometry, helping ensure that the the TS search goes in the right direction. CalcAll is also used, but this is much more expensive. opt(addredundant) and opt(redundant) may also be used here for some cricumstances, and I will link some guides that discuss them:

Transition State Search (QST2 & QST3) and IRC with Gaussian09, Dr. Joaquin Barroso's blog
Technical Note: Locating Transition States (Gaussian 09 via the Internet Archive

I always use freq when searching for transition states because it is necessary for IRC calculations. I use freq=noraman because these Raman frequency calculations are expensive, and unnecessary if I'm just looking for the TS.
Scanning
Scanning is a very useful method of finding starting structures for transition states. Much of the time, I use the following procedure to find a TS:

Conformation Search (beyond the scope of this guide)
optimize interesting conformers
perform loose scans to identify the ones most likely to lead to a maximum
perform a full scan
create a .gjf from the scan's maximum energy geometry (and sometimes one or two other similar points)
perform a TS search

For simple cases, it is only necessary to do the last two steps. I will provide an example of a loose scan:
#P f/b opt(modred,maxcyc=4) scf=conver=6

Title

q m
<cartesian>

B 1 2 S 15 0.1

Reducing the number of optimization cycles from the default using opt(MaxCycles=N) will make the computation cheaper, and so will limiting the SCF convergence criterion using SCF(conver=N). Another option is to use opt(loose). All of these commands should speed of the initial scan. Afterwards, they can be removed and we can do a scan using the default input parameters, or we can just start searching for a TS if the maximum looks close enough to the expected TS structure.
The modredundant line B 1 2 S 15 0.1 corresponds to scanning along an increase in the distance between atoms 1 and 2, in 15 steps, by 0.1 Angstroms. If you wanted to decrease this distance instead, you would use "-0.1". For things like proton transfers, I find that I have much more success scanning an increase in the distance between the proton and the atom it's bonded to, as opposed to a decrease in distance between the proton and the atom we're trying to transfer it to.
Transition State Search
To find a transition state, all we've got to do is save cartesian coordinates of a geometry corresponding to a scan maximum and use the commands opt(ts,calcfc) and freq. For a large organic system, I might use the following input line:
#P f/b opt(ts,calcfc) nosymmetry freq=noraman int=grid=ultrafine

Of course, it is necessary that only one negative imaginary frequency be observed in order to confirm the TS is a first order saddle point. It makes me very happy to click on this frequency and see a nice bouncy TS.
IRC Calculations
The final step necessary to generate a true energy profile for a reaction is to compute intrinsic reaction coordinate (IRC) calculations from the identified transition state, confirming that the TS indeed corresponds to the expected reactant(s) and product(s). I would like to stress that it is very important to take good care of you TS search or QST2/3 checkpoint file. I often use several links (--link1--) in my Gaussian inputs in order to calculate things like solvent corrections, but you cannot do this on a TS search input that you want to link to an IRC calculation. This is because IRC calculations require a TS checkpoint file ending with the frequency calculations. Usually, I will duplicate this checkpoint file and save it in a new directory, along with my IRC inputs. In order to compute both directions along the intrinsic reaction coordinate, we will create two IRC inputs:
Forward:
%chk=TS.chk
%nproc=8
%mem=12970MB

#P f/b geom=allcheck irc(rcfc,forward,MaxPoints=1000)

Title

and Reverse:
...
#P f/b geom=allcheck irc(rcfc,reverse,maxpoints=1000)

The final structures resulting from these calculations can then be optimized.
Once we have all these data, we can compare their energies and create an energy profile. I like to use kcal/mol, so the Heartree energy values are multiplied by 627.5. It's important to note that we have not discussed methods of calculating free energy, so this profile will not take into account things like entropy. Here is an example of an energy profile for a reaction that involves two seperated molecules coming together to form a complex then reacting via a transition state to form a covalently-bound product:

In order to create the above diagram, the following series of steps was performed:

optimization of A & B
constrained optimization
optimization of the complex A...B
scanning
TS search
IRC (forward & reverse)
optimization of IRC structure outputs

It is important to note that the above diagram does not factor in entropy or solvent effects. The second species would probably be higher in energy than the reactant state if we had done free energy calculations. Also, all these commands are subject to interpretation. It depends what you're trying to do. For example, you might not use maxpoints=1000 for IRC calculations or UltraFineGrid for optimizations.
Part 2: A Practical Example: SEAR
For a practical example of SEAR (electrophilic aromatic substitution), I will study the reaction of benzene with $\ce{AlCl3}$ and chloroethane. This is a simple Friedel-Crafts alkylation, which is, of course, well understood. Generally, theoretical methods would be more applicable to studying the selectivity of such reactions, but I think this simple reaction of unsubstituted benzene is hard enough to start with. I will study the actual substitution step corresponding to the first two states shown in this diagram:

First, I created a reactant state and a product state and optimized them. The inputs looked like this:
%chk=FCrafts-r.chk
%nproc=8
%mem=12907MB
#P b3lyp/6-311g* opt

FCrafts Reactant

0 1
 Cl                -1.01985389    0.67072426   -1.99433146
 C                 -2.76984013    0.75358073   -1.82615605
 H                 -3.23193423    0.33394909   -2.69522355
 H                 -3.07145579    0.20199133   -0.96031707
 C                 -3.20190703    2.22391229   -1.67435589
 H                 -2.73981293    2.64354394   -0.80528838
 H                 -2.90029137    2.77550169   -2.54019486
 Al                -0.05247932    1.54920546   -0.17497519
 Cl                -1.22820352    0.97949494    1.64455842
 Cl                 2.03778367    0.76578567    0.01118083
 Cl                 0.00035598    3.78081654   -0.36130926
 C                  0.36368686   -1.97969745   -5.12095489
 C                  0.28443706   -1.06813823   -4.06774519
 C                  1.34622657   -0.94305300   -3.17210102
 C                  2.48747174   -1.73060169   -3.32876331
 C                  2.56643674   -2.64225875   -4.38146600
 C                  1.50464059   -2.76655759   -5.27787474
 H                 -0.47337036   -2.07784492   -5.82724457
 H                 -0.61488949   -0.44739067   -3.94475736
 H                  1.28409161   -0.22412673   -2.34229335
 H                  3.32450411   -1.63184247   -2.62242078
 H                  3.46573369   -3.26308847   -4.50519829
 H                  1.56719555   -3.48518579   -6.10780812
 H                 -4.26581912    2.27428526   -1.57211288

E=-2395.13072206
%chk=FCrafts-p.chk
%nproc=4
%mem=6453MB
#P b3lyp/6-311g* opt

FCrafts Product

0 1
 Cl                -1.01985423    0.67072404   -1.99433140
 Al                -0.05247942    1.54920537   -0.17497533
 Cl                -1.22820366    0.97949483    1.64455823
 Cl                 2.03778376    0.76578615    0.01118094
 Cl                 0.00035647    3.78081647   -0.36130909
 C                  0.40151149   -2.23919085   -4.59712077
 C                  0.43310980   -1.30289194   -3.53217462
 C                  1.74514857   -1.02252407   -3.07205236
 C                  2.89870053   -1.60733521   -3.61195409
 C                  2.80190378   -2.51471733   -4.65239437
 C                  1.54785721   -2.83052726   -5.14520180
 H                 -0.55032526   -2.50900028   -5.00468633
 H                  1.86063959   -0.32586549   -2.26816887
 H                  3.85968351   -1.35082712   -3.21747303
 H                  3.67987963   -2.96356162   -5.06784259
 H                  1.45293337   -3.53014547   -5.94920667
 C                 -0.37963392   -1.87799793   -2.35731261
 H                 -0.39123491   -1.17283060   -1.55263599
 H                 -1.38232175   -2.07026225   -2.67754931
 C                  0.26743991   -3.19130077   -1.87968188
 H                  1.27012774   -2.99903644   -1.55944518
 H                 -0.29725865   -3.59088739   -1.06338165
 H                  0.27904090   -3.89646809   -2.68435849
 H                 -0.00888163   -0.38943615   -3.87150056

E=-2395.1196219
Next, I attempted to use the optimized structures in a QST2 calculation. Since QST2 requires identical atom ordering, I opened the outputs in Gaussview and used Edit>Connection to manually reorder the atoms. Ultimately, none of my initial QST2 calculations worked, probably because the optimized reactant and product states are too different. Generally, I only use QST2 for intramolecular reactions like proton transfers, or reactions between molecules that can form strong complexes through non-covalent interactions.
After this, I decided to attempt to find the transition state using scanning techniques. I took the optimized reactant structure and attempted three scans: two decreasing distances, and an increasing distance between the electrophilic carbon of chloroethane and the chlorine that I want to move to $\ce{AlCl4-}$ in my product state. As I mentioned before, scanning bonds you want to break seems to work better. Indeed, this was the successful scan. The input looked like this:
%chk=scan3.chk
%nproc=8
%mem=8596MB
#P opt(modred) b3lyp/6-311g*

scan3 (C2 <-> Cl1)

0 1
Cl    -0.401371     2.016139    -1.511245
 C    -1.304629     0.538355    -2.237885
 H    -0.506575     0.002707    -2.743614
 H    -1.640303    -0.022236    -1.369005
 C    -2.418105     0.993950    -3.145472
 H    -3.173833     1.568296    -2.609155
 H    -2.042249     1.585674    -3.981376
Al    -1.287380     2.319602     0.728099
Cl    -0.886318     0.423395     1.584647
Cl    -0.090841     3.946397     1.332537
Cl    -3.336028     2.695133     0.340344
 C     1.252751    -0.767160    -4.974849
 C     2.089419    -0.531302    -3.883505
 C     2.055121    -1.385353    -2.780838
 C     1.184354    -2.475005    -2.769452
 C     0.347941    -2.711078    -3.860667
 C     0.382053    -1.857277    -4.963396
 H     1.285388    -0.107349    -5.836093
 H     2.770412     0.313582    -3.893742
 H     2.706979    -1.203076    -1.932781
 H     1.160512    -3.140482    -1.912631
 H    -0.324246    -3.563240    -3.854463
 H    -0.262249    -2.046121    -5.816288
 H    -2.898095     0.099454    -3.556324

B 1 2 S 15 0.1

This scan resulted in a nice bump in energy as the Cl-C bond broke and the ethyl group was transferred to the benzene, forming $\ce{AlCl4-}$ and $\ce{C6H6CH2CH3+}$. The scan eventually terminated in an error, but that's okay. I chose two reasonable looking transition state structures; the first one, ts1, was higher in energy, and the second one, ts2, better resembled the desired TS geometry. Here is the input for my TS search using ts2:
%chk=ts2.chk
%nproc=8
%mem=13107MB
#P b3lyp/6-311g* opt(ts,calcfc,noeigentest) freq=noraman

FCrafts-ts2 -2395.093367 TS search

0 1
Cl    -0.445964     2.845220    -1.106683
 C    -0.865174     0.452262    -2.650118
 H     0.070071     0.786614    -3.073869
 H    -0.824619     0.021555    -1.658487
 C    -2.151432     0.661827    -3.315090
 H    -2.697128     1.419092    -2.730528
 H    -2.065131     0.994322    -4.347280
Al    -1.548552     2.054891     0.676390
Cl    -0.708408     0.079254     0.994127
Cl    -1.277194     3.382724     2.305605
Cl    -3.584672     1.842336    -0.001997
 C     1.261152    -1.065729    -5.126851
 C     2.356845    -1.021440    -4.263905
 C     2.227297    -1.419535    -2.929948
 C     1.000215    -1.863721    -2.453899
 C    -0.114196    -1.887487    -3.310358
 C     0.026820    -1.503333    -4.654853
 H     1.372280    -0.765799    -6.163090
 H     3.318060    -0.677928    -4.631351
 H     3.084711    -1.384062    -2.267323
 H     0.892596    -2.173516    -1.420312
 H    -1.052428    -2.303629    -2.958824
 H    -0.826107    -1.555052    -5.323327
 H    -2.786334    -0.225861    -3.235183

As you will see if you open this input in a molecular visualizer, this structure looks like the ethyl group's C2 is being transfered between Cl1 and benzene's C16, with Cl1-C2 bond distance of 2.88 Angstroms and C2-C15 bond distance of 2.54 Angstroms. Also, C2 has become planar.
This TS search resulted in a nice TS wit the one negative imaginary frequency.
freq=-200.09
E=-2395.094514

Compared to my optimized reactant state, the activation energy is calculated as follows: (-2395.094514-(-2395.13072206))*627.5=~22.72 kcal/mol
In order to confirm this transition state, I will perform IRC calculations in both directions and optimize their output. I duplicate the checkpoint file ts2.chk and upload it to a new directory, which includes both IRC input files. The forward-direction input looks like this:
%chk=ts2.chk
%nproc=8
%mem=17210MB
#P b3lyp/6-311g* Geom=allcheck irc(rcfc,forward,MaxPoints=1000)

ts2-ircf

This provided the following structures (XYZ):
   24
scf done: -2395.130704
Cl     0.272888     1.333254    -0.513417
 C    -0.571943    -0.192957    -1.211006
 H     0.199323    -0.614633    -1.849153
 H    -0.726304    -0.830340    -0.344614
 C    -1.838977     0.185473    -1.933990
 H    -2.568763     0.644706    -1.266937
 H    -1.643268     0.858586    -2.769987
Al    -0.397497     1.490022     1.814452
Cl     0.175277    -0.420412     2.532179
Cl     0.779882     3.144882     2.379307
Cl    -2.489389     1.781502     1.653827
 C     2.101401    -1.455592    -3.878180
 C     2.856218    -1.287190    -2.716820
 C     2.702148    -2.172652    -1.650117
 C     1.793294    -3.226627    -1.744938
 C     1.039189    -3.396159    -2.906443
 C     1.192936    -2.510279    -3.972827
 H     2.226950    -0.770703    -4.710731
 H     3.567332    -0.470486    -2.645085
 H     3.289784    -2.042184    -0.747393
 H     1.675906    -3.916748    -0.915847
 H     0.337147    -4.220667    -2.983047
 H     0.611785    -2.646294    -4.879442
 H    -2.278034    -0.732958    -2.337378

   24
scf done: -2395.109858
Cl     1.581951     0.967535    -0.168204
 C    -0.907511    -0.785408    -1.885356
 H    -0.626761     0.267058    -1.933744
 H    -1.181027    -0.976669    -0.846209
 C    -2.097517    -1.082433    -2.794246
 H    -2.950908    -0.480214    -2.476948
 H    -1.908820    -0.835611    -3.842724
Al     0.286518     0.980850     1.647730
Cl     0.162609    -1.114876     2.229327
Cl     1.210097     2.186737     3.137784
Cl    -1.648423     1.669197     1.018079
 C     2.280567    -1.555648    -3.768606
 C     3.161897    -1.771764    -2.675531
 C     2.711522    -1.851341    -1.367252
 C     1.355427    -1.710821    -1.103867
 C     0.366817    -1.633133    -2.195560
 C     0.940921    -1.472293    -3.548161
 H     2.686377    -1.459115    -4.769022
 H     4.225121    -1.864580    -2.874879
 H     3.409722    -1.972777    -0.548364
 H     0.980262    -1.815841    -0.091354
 H    -0.007797    -2.688333    -2.194489
 H     0.257734    -1.322751    -4.376589
 H    -2.401431    -2.132583    -2.743008

This gives an activation energy of 22.71 kcal/mol and an energy change of +13.01 kcal/mol, which makes sense since this product is a tetrahedral intermediate. Once all this is complete, I will re-optimize everything with scrf=solvent=n-decane and provide a graph, similar to the SPE graph shown in part 1, and maybe try to study an actual electrophilic addition reaction.

I am surprised B 2 3 1.97 F actually works as input for modred (I have not tested it). I only knew you can freeze a bond or activate it...

*And for sanity's sake, don't run compound jobs! (That means opt and freq together.) It is also completely unnecessary to do that for TS/QST jobs, as you are already calculating the force constants you need (remember you specified CalcFC). And don't overwrite your checkpointfile from the TS! (From rev. D.01 on you can specify %oldchk and it will create a copy for you.)


text:
Solvation effect on cyclohexane
I ran molecular dynamics simulation with a molecule of methylcyclohexane soaked into a water box. On the other hand I ran Monte Carlo with the molecule alone (no soaking). I performed this for the equatorial and axial conformers.
The axial conformer with MD stayed axial (the methyl group). The equatorial with MD stayed equatorial. So far no visual change. Is there a way to compute the average energy in VMD for these systems (I have around 250 frames).
On the other hand, the axial conformer alone with MC was transformed into equatorial. While the equatorial stayed the same (visually speaking).
I am trying to understand what's happening here. What I read is that always the equatorial configuration is more stable as it has less steric hindrance (BTW I am computer scientist learning this), so this explains the interconversion in Monte Carlo simulation from axial to equatorial.
On the other hand, I would expect also that in MD the axial conformer turn into equatorial but it didn't, what I believe here is that the water molecules are having some "Solvation effect" on the methylcyclohexane that prevents the interconversion.
So I would like to know if my thought are good and if you could give some references to get more info about solvation effects.

The difference in behavior between the two methods is likely how much "wiggle" the molecule is allowed to have, or how much additional kinetic energy you allow the molecule to have when searching for the minimum energy conformation. The equatorial conformation is the global minimum conformation, but the axial conformation is a local minimum. Additionally, between the two conformations, there is another minimum called the twist-boat. You can read about them on this Wikipedia article or you could check out an introductory organic chemistry book from a library. Molecular dynamics tend to stop when a minimum is found. In other words, if a small alteration to the geometry increases the potential energy, the calculation stops. Monte Carlo calculations on molecules are designed to examine the evolution of conformation with time, and so geometry moves out of energy minima are allowed.
Unsubstituted cyclohexane has four different conformations: Chair (relative energy 0 kJ/mol), half chair (+45 kJ/mol), twist boat (+23 kJ/mol), and boat (+30 kJ/mol). A molecular dynamics calculation may not allow large enough steps to get over the activation barrier to get out of a chair conformation. The following energy diagram shows the conformational space of cyclohexane.

The energy difference between equatorial and axial methylcyclohexane is 7.6 kJ/mol (1.6 kcal/mol on this table), which is small compared to the difference between the chair and the half-chair conformation. The energy diagram below approximates the energy steps between equatorial and axial methylcyclohexane.

Any computational method that allows potential energy steps large enough to get to the half-chair will certainly isomerize the axial conformer to the equatorial conformer. Presence or absence of solvent molecules in the simulation may be irrelevant.

Thank you everything starts to fitting well. But then the role of water molecules is completely irrelevant to the interconversion between axial en equatorial? There are no solvation effects at all?

I'm sure the solvent affects both the ratio of the two conformers (about 95:5 equatorial to axial in pure methylcyclohexane at room temperature) and the rate of interconversion, but the polarity and/or presence/absence of solvent neither causes nor prevents this phenomenon.


text:
Regioselectivity in Kharasch addition of bromotrichloromethane
When hex-1-ene is treated with bromotrichloromethane in the presence of a peroxide initiator (e.g. dibenzoyl peroxide), what is the regioselectivity of the addition?
I know that the radical addition of HBr to an alkene is considered anti-Markovnikov, but is this reaction considered a Markovnikov or anti-Markovnikov addition?

This looks like a typical Kharasch (chain) reaction. The reaction is initiated by the formation of peroxide radicals (by homolytic cleavage of the weak O–O bond). These peroxide radicals then go on to abstract Br• from the starting material:

Note that the alternative generation of Br• is unfeasible, as radicals tend to attack at less hindered sites: hence the phenyl radical attacks the bromine which only has one attached atom, and not the carbon which has four flanking halogens. The weaker C–Br bond is also preferentially broken over the other three C–Cl bonds, so there is no chance of generating a CCl2Br• radical.
The trichloromethyl radical then adds to C-1 of the alkene, resulting in the more stable secondary radical. This can now abstract a bromine atom from unreacted bromotrichloromethane to give the product and a new Cl3C• radical.

Unfortunately, with additions of X–Y across a C=C double bond (as opposed to additions of H–X), the designations Markovnikov and anti-Markovnikov lose their meaning. A Markovnikov addition of H–X would be one where the hydrogen adds to the less substituted carbon. However, here it is not clear whether to define "Markovnikov" as the regioselectivity where X adds to the less substituted carbon or where Y adds to the less substituted carbon.

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


text:
In which cases does the 3d subshell take precedence over the 4s subshell?
I've learnt in Chemistry that for example in Chromium, the 4s subshell only has one electron in favor of making the 3d subshell half-full. Also, for Copper the same happens.
What other circumstances does this happen under? Do ions obey the same pattern, e.g. Mn+?

It's a bit of a complicated matter. Though it can be explained qualitatively for the lighter elements with some finer arguments than merely comparing orbital energy levels via the aufbau principle, things start to get messy when $f$ (and eventually $g$) sublevels and strong relativistic effects are present. If I find the time, I may write up an answer, but for now I shall link this NIST webpage, which has very recent data on ground state electronic configurations of atoms and ions. Just type in an element and play around! Lots to see.

Do note however that the listed data uses a slightly different nomenclature than usual in chemistry for atoms/ions. For example, neutral iron ($\ce{Fe^0}$) is listed as "Fe I", while the $\ce{Fe^3+}$ ion, often called iron(III), is listed as "Fe IV".

If we take the definition of "3d subshell takes precedence over the 4s subshell" as being that the configuration has zero 4s electrons, yet at least one 3d electron, then all such cases are ions.
+1
V: 3d^4
Cr: 3d^5
Co: 3d^8
Ni: 3d^9
Cu: 3d^10
2+
All elements from Sc to Zn
3+
All elements from Ti to Ga
Etc.

I agree with @Nicolau Saker Neto that it's complicated. You might want to look at these two papers, if you're really interested in more of the details:

R. N. Keller, J. Chem. Ed. 39(6), 1962. PDF available here
Robin M. Hochstrasser, J. Chem. Ed. 42(3), 1965. PDF available here

The links might be paywalled. Sorry. :-/

Welcome to Chemistry.SE! Unfortunatlely, we usually don't allow answers that are little more than a link to another resource. Please quote relevant portions of the papers and/or explain them in brief. Thank you :)

So would it have been better as a comment? It was more of a "here's where to learn more" than a full-out explanation. It is, in fact, complicated, and a little post here, even attempting to summarized the papers, would hardly do it justice.

No one had "answered" the question, so I was attempting to note that @Nicolau Saker Neto basically had answered it, and provide further resources to back up the explanation. How should I approach this in the future?

Well, in the future try to leave a comment OR add some meat to the answer. Even a partial summary is good. :) Unfortunately, you need 50 reputation to comment, which you don't have.

Rules for filling up atomic orbitals can be understood by considering the governing principles:
Aufbau principle: electrons fill from lowest energy level to highest (attraction of positive nucleus on electrons: the farther they are from nucleus, higher is their energy or shallower potential well)
Pauli exclusion principle: no two electrons can have all quantum numbers the same; in particular, only two electrons with opposite spins can occupy an orbital. Pairing up of electrons with opposite spin raises its energy. Electrons with parallel spins have a repulsive exchange interaction between themselves.
Hund's rule: electrons are stabilized if they have parallel spins spatially distributed among different orbitals of a subshell. They'll get paired up only after each subshell has one electron, all having parallel spins.
Formation of half filled or fully filled orbitals thus has a stabilizing effect. Energy difference between $n$s and $(n-1)$d is not so high. In case of some special elements, transition of an outer electron from $n$s to $(n-1)$d is energetically favorable if it effects net stabilization due to maximization of electrons with parallel spins in d-shell, thereby over-riding the aufbau principle.
Nb, Mo, Pd, and Ag are some examples of special configurations. These principles should hold true for ions as well if we consider the remaining valence electrons in the system.


text:
Is there anything in the literature that supports revolutionary claims for this "new" sulfuric-acid battery?
I've got the following from a supposed new energy-storage technologist. It's supposed to be a fancy new cheap, high-powered, highly scalable battery. I want to know if this is known chemistry, and if there's anything genuinely revolutionary about it:

"Basic materials in MEG's E-SOURCE™ power cells are lead metal, absorptive glass mat paper, and sulfuric acid solution. Materials unique to the MEG advanced power cell are S2 glass fiber yarn; sodium sulfate; Teflon emulsion, glass-filled polypropylene; bituminous, petroleum-based wax; and metal-free nano-oxides. "

The lead is apparently in the form of lead oxide.
The claim is that this is a radical new cheap energy-storage system - but someone makes that claim for yet another technology every week or two. Is there anything in the recent literature that would give credibility to this one?

There is nothing in the chemistry of what they describe that is different in any way to a conventional Lead-Acid battery. What they really seem to claim is improvement in materials in the battery not to change the electrochemistry but to improve other aspects of battery operation. For example:

MEG's E-SOURCE™ power cell technology is based on a lead-fiberglass composite material that is one-third the weight and 50 times the strength of metal alloys used in conventional batteries. To date, more than $100 million has been invested in research, development, manufacturing, and market testing of this technology in the United States and several international markets.

This suggests they have built a better lead-acid battery rather than something more radical. This is not ridiculous. They claim the following advantages:

Extraordinary manufacturing and user cost savings
Significant weight reduction compared to conventional lead-acid    batteries
Exceptionally long service life
Dramatically fast recharge speeds at all operating temperatures
Safe, simple, and low-cost maintenance

It is not impossible to imagine clever overall design and improved microstructural components and electrode materials could deliver these sorts of improvements and it is certainly far easier than inventing a whole new battery electrochemistry. For example, charging speed is often dominated by the physical material properties of the electrodes and not the underlying electrode chemistry, so clever design can mitigate the physical limits on charging speed such as by increasing the effective surface area.

Sodium sulfate isn't in a normal lead-acid battery, is it?

@EnergyNumbers True, but there is plenty of sulphate as this is essential to the reactions. So sodium sulphate won't change the electrochemistry.


text:
Why does heat solidify cake batter?
Increased vibration causes the atoms in an object to spread out farther away. Thus, with added heat, and object goes form a solid to a liquid to a gas (and then to plasma, with enough heat). Why is it then, that when a heat something like cake batter, it solidifies?

It is caused by physical and chemical reactions in the cake batter of two types:
1) When heated and mixed with water, starches present in the flour will undergo a process called gelatinization in which water is absorbed in the structure. This will cause the batter to 'set' from a liquid into a solid.
2) Additionally, reactions of the proteins from the egg that you added can also work to 'set' the batter. At first the proteins will unfold (denature) after which they start to form bigger proteins through cross-linking reactions that turn the liquid into a solid.
The baking soda in the batter turns into sodium carbonate, water and carbon dioxide. The latter will make your cake nice and fluffy.

Mostly wrong! 1st there are numerous doughs which do not contain eggs, nevertheless they set on heating. For that the swelling of starch with water and heat is resposible. But both processes are physics, not chemistry!

@Georg Thanks for the remark. I was indeed still wondering how it was possible for cake batter without eggs to work out and starch gelatinization is indeed the answer. I'll edit it in

@Georg wrong, the swelling of starch doesn't set the dough. It thickens it, but it doesn't make it "solid" - cook a pudding to see what happens. The eggless doughs set by denaturing of the gluten network, which arises when two wheat proteins bind in the present of water. As for cake, both the egg and the gluten are at work, and depending on the recipe, the egg may prevail.


text:
CI without Hartree-Fock SCF
Usually textbooks on CI takes the SCF Hartree-Fock orbitals as a starting point. Can one use the configuration interaction method starting directly from the single-particle Slater determinants avoiding the iterative procedure. This question concerns mainly small molecules like the hydrogen molecule. Do I get a high inaccuracy in this case? If yes, why? Could this inaccuracy be decreased taking into account more configurations?

ummmm… why would you want to do that?  :-)  I guess I'm wondering if you are doing singles and doubles or toward full CI?  I think you would get toward the Full CI limit faster (with fewer excitations) if you start with more reasonable orbitals.

Thank you for the reply. That is actually what I am asking. How faster do I get convergence in the case of SCF orbitals. How many additional excitations I have to take into account in the case of avoiding preliminary HF computations.

@freude: I don't think it's really possible to make an accurate but general statement about it: it'll vary a lot by system, and will be less important in simple systems where the difference between cheap approximations and accurate models is smaller.

Szabo and Ostlund describes CI calculations using natural orbitals where the convergence is faster than with Hartree-Fock orbitals. You should read that book for more information.

"How faster do I get convergence in the case of SCF orbitals?"  Start with a better initial guess.

If you were to perform Full CI (within a finite basis) then I don't think that it matters whether you start with SCF orbitals or some other choice.  This solution will have the lowest conceivable energy no matter what.
But if you are doing truncated (less than full) CI, then you are very much at the mercy of your starting orbitals. In general, one would want to start with orbitals that resemble those of the actual state that you are trying to describe via CI.
Otherwise, your precious excitations (singles and doubles) may be consumed correcting the "slop" that you provided, rather than moving toward the full CI limit.

(As an example, I know some people using CI-related approaches (CASSCF and CASPT2, specifically) on heavier elements who prefer to start with DFT orbitals because they produce a better result than starting with HF.)

Right. I had a little trouble answering this question, because practically speaking, there is always optimization of the inital vectors. It's sort of a "textbook" question. Very interesting about DFT orbitals for metals. I've never done MCSCF on metals--they probably need all the help they can get :-)

Isn't there a theory that the exact natural orbitals (and approximations to them) should give you the quickest convergence, though? I believe that's the idea behind Iterative Natural Orbital approaches, for example.

I think that is how Natural Orbitals are defined: orbital set that puts the most "electron" into the smallest number of MOs (in an aufbau sense).  Then one could just use the occupied orbitals.  I think the problem is that one has to solve the CI problem to determine the natural orbitals -- sort of a "no free lunch" problem.

@EricBrown: Natural orbitals are the eigenfunctions of the 1st order reduced density matrix of a wavefunction. If you take n strongly orthogonal wavefunctions and mix them, your leading NOs will not make up the Slater determinant with the largest prefactor. Of course that's contrived, since the wavefunctions also are orthogonal in the energy product.

If by "some other choice" you mean a basis that can be obtained from the other one by unitary rotation then of course you are right. Otherwise it is not true that the Full CI will give you the same result. Energy minimization by Full CI only depends on your basis.

Adding to the answer and all the related comments.
Using the HF reference as starting point provides a "clean" interpretation for the different contribution to the total energy. Indeed, the difference between HF and CI energy is, by definition, the electron correlation energy. One can then furthermore break down the correlation into single, double, etc contributions and consider a step-wise convergence to a full CI.
When the reference used for the GS already contains different "effects" (by using NOs or DFT orbitals) the lower energy obtained after performing a CI is not easily interpretable.

"lead to a final lower full CI energy". Certainly not - full CI is full CI, no matter the starting orbitals (as long as they have the correct number of electrons and similar basic conditions). I have not run such calculations, but I would wager that if one starts with orbitals other than canonical SCF orbitals, a HF-like energy is obtained, thus not the DFT energy is used.

Thank you for your comment. I have difficulty to agree with your statement "full CI is full CI, no matter the starting orbitals". In the present state of my understanding, this will only be valid when considering a infinite basis. As soon as you truncated the basis, you also impose a constrain on the way the space is spanned. The calculation should thus become orbital depended (for basis sets of the same size). Isn't it ? I think that some literature search would be needed.

I assume that the same AO basis is used. It then does not matter how the reference determinant is constructed because "all" determinants will be used in the FCI. The result is the lowest possible energy for that system in that AO basis.

Agreed. It is the underlying AO basis that defines the space that can be spannend, not the way the MO basis is constructed. The same full CI will always lead to the same energy, whatever the kind of orthogonal MO basis (HF or other). Thanks for the clarification.


text:
How can potassium disappear after cooking?
I read this at this webpage: http://www.whfoods.com/genpage.php?dbid=90&tname=nutrient#impactcookingstorageprocessing

How do cooking, storage, or processing affect potassium?
Potassium losses from cooking of high-potassium foods can be
  significant. In the case of spinach for example, potassium levels have
  been shown to drop from 6.9 to 3.0 grams in 3 and 1/2 ounces of
  spinach after blanching for several minutes (a loss of about 56%).

I don't understand this. How can potassium disappear after cooking ?

As Eric has mentioned in his answer, the potassium present in foods will leach out into water when you cook them.
Potassium is present in food in the form of potassium ions $(\ce{K+})$. These $\ce{K+}$ ions happen to be extremely soluble in water.
When food is boiled, the cells break down and release their contents more easily which makes the $\ce{K+}$ ions free to move into the water. The $\ce{K+}$ ions remain in the water since they are so highly soluble in it.
This happens to all water-soluble nutrients like the B vitamins, vitamin C, sodium, etc. Some ways to avoid the loss of all these nutrients would be to steam such foods instead of boiling them since the food wouldn’t come into contact with water.
Side note: The same problem occurs when you deep fry foods containing fat-soluble nutrients like vitamins A and E. These leach out easily into oil and so all those nutrients would also be lost in the process.

Water does contact food when steamed, otherwise it wouldn't get heated. Also, the food drips water. You do loose less nutrients since less water comes in contact with the food.

Side note: The constant addition of heat will destroy most of the vitamins, esp. when deep frying. Minerals however will be watered out.

@Martin What is the process by which heat "destroys" vitamins?

@RobJeffries It might be worthwhile asking this as a separate question on the site, it is probably too complex for a comment and I am not an expert on this. I believe they undergo denaturation and often will be deactivated due to oxidation with readily available oxygen from air. But there are certainly other factors.

The potassium is leeched out into the water used to cook it, and unfortunately poured down the drain -- vitamins and all.


text:
How is it determined how many electrons are transferred in redox reactions?
For example, iron and tin(II) sulfate will react to produce iron(II) sulfate and tin:
$\ce{Fe + SnSO4 -> FeSO4 + Sn}$
With the following half equations:
$\ce{Fe + SO4^{2-} -> FeSO4 + 2e-}$
and
$\ce{Sn^{2+} + 2e- -> Sn}$
but where does the "2" come from? Why can't it be:
$\ce{Fe + SO4^{5-} -> FeSO4 + 5e-}$
and
$\ce{Sn^{5+} + 5e- -> Sn}$

Can you clarify whether you're asking about how the number of electrons transferred is determined empirically/experimentally, or merely how that's calculated on paper given skeletal or half equations? Also, how familiar are you with the basics of chemical bonding, electron configurations, etc.? I think most answers would have to presuppose an understanding of those things, and based on the phrasing of your question, I suspect you might be best served by just studying a general chemistry textbook (Oxtoby's is probably the most rigorous I've come across).

Anybody know why the TeX formatting doesn't seem to be working? The format strings look good, but they're showing verbatim in my Chrome browser.

@Cameron Your first half equation isn't really a half equation... the half equation should have been $\ce{Fe -> Fe^{2+} + 2e^-}$

In short, the species you mention ($\ce{Sn^{5+}}$ and $\ce{SO_4^{5-}}$) do not exist in any chemical system at anywhere near ordinary conditions. The number of electrons shuffled in the reaction is not chosen arbitrarily, but is based on the initial and final oxidation numbers of the elements in the reaction, after the equations are balanced. The only stable compound with formula $\ce{SnSO_4}$ is made of $\ce{Sn^{2+}}$ and $\ce{SO_4^{2-}}$ ions. Hence in your reaction, $\ce{Sn}$ atoms start with oxidation number +2 and end with 0 (therefore gaining two electrons) and $\ce{Fe}$ atoms start with oxidation number 0 and end with +2 (therefore losing two electrons).
As another example of electron counting, here's the reaction between $\ce{Ba}$ and $\ce{Al^3+}$:
Oxidation: $\ce{Ba^0 -> Ba^{2+} + 2e^{-}}$
Reduction: $\ce{Al^{3+} + 3e^{-} -> Al^0}$
To balance the equations, the oxidation half-reaction must be multiplied by 3 and the reduction half-reaction must be multiplied by two. Therefore:
Oxidation: $\ce{3Ba^0 -> 3Ba^{2+} + 6e^{-}}$
Reduction: $\ce{2Al^{3+} + 6e^{-} -> 2Al^0}$
Global: $\ce{3Ba^0 + 2Al^{3+} -> 3Ba^{2+} + 2Al^0}$
In the global reaction, six electrons are involved.

After you calculate everything and balance the $\ce{e-}$ too, then you will know that one side loses a certain number of $\ce{e-}$ and the other gains this certain number of $\ce{e-}$.
There isn't any other way apart from balancing the equation and figuring it out from there.
Check here and here for more info.