text:
Why does the energy gap for π - π* transitions shrink with the size of the pi-conjugated system?
Quoting from this site:

As conjugated pi systems become larger, the  energy gap for a π - π* transition becomes increasingly narrow, and the wavelength of light absorbed correspondingly becomes longer.   The absorbance due to the π - π* transition in 1,3,5-hexatriene, for example, occurs at 258 nm, corresponding to a ΔE of 111 kcal/mol.

It's also quite easy to see that this is true glancing at various dye absorption tables.
But why is this true?  I can understand why the optical cross-section would increase, implying a higher quantum efficiency at some maximum absorbance wavelength (is this generally true?), but why would the energy gap between the ground state and the excited state correspondingly shrink?

For those interested in the matter: please also see this follow-up question.

Klaus Warzecha's answer pretty much answers your question. But I know that this subject is easier to understand if supported by some pictures. That's why I will take the same route as Klaus at explaining the concept behind why the absorption in conjugated systems is shifted to higher wavelengths but I will provide some pictures on the way.
In a conjugated carbon chain or ring system you can think of the $\ce{C}$ atoms as $\text{sp}^{2}$-hybridized. So, each carbon has 3 $\text{sp}^{2}$ orbitals which it uses to form $\sigma$ bonds and 1 $\text{p}$ orbital which is used to form $\pi$ bonds.
It is the $\text{p}$ orbitals that are responsible for the conjugation and their combinations according to the LCAO model are the interesting part since the HOMO and LUMO of the system will be among the molecular orbitals formed from the conjugated $\text{p}$ orbitals.
For a start take ethene, the simplest $\pi$-system, being comprised of only 2 carbon atoms.
When you combine two atomic orbitals you get two molecular orbitals.
These result from combining the $\text{p}$ orbitals either in-phase or out-of-phase.
The in-phase combination is lower in energy than the original $\text{p}$ orbitals and the out-of-phase combination is higher in energy than the original $\text{p}$ orbitals.
The in-phase combination accounts for the bonding molecular orbital ($\pi$), whilst the out-of-phase combination accounts for the antibonding molecular orbital ($\pi^{*}$).

Now, what happens when you lengthen the conjugated system by combining two ethene fragments?
You get to butadiene.
Butadiene has two $\pi$ bonds and so four electrons in the $\pi$ system.
Which molecular orbitals are these electrons in?
Since each molecular orbital can hold two electrons, only the two molecular orbitals
lowest in energy are filled.
Let's have a closer look at these orbitals.
In $\Psi_1$, the lowest-energy bonding orbital, the electrons are spread out over all four carbon atoms (above and below the plane) in one continuous orbital.
There is bonding between all the atoms.
The other two electrons are in $\Psi_2$.
This orbital has bonding interactions between carbon atoms 1 and 2, and also between 3 and 4 but an antibonding interaction between carbons 2 and 3.
Overall, in both the occupied $\pi$ orbitals there are electrons between carbons 1 and 2 and between 3 and 4, but the antibonding interaction between carbons 2 and 3 in $\Psi_2$ partially cancels out the bonding interaction in $\Psi_1$.
This explains why all the bonds in butadiene are not the same and why the middle bond is more like a single bond while the end bonds are double bonds.
If we look closely at the coefficients on each atom in orbitals $\Psi_1$ and $\Psi_2$, it can be seen that the bonding interaction between the central carbon atoms in $\Psi_1$ is greater than the antibonding one in $\Psi_2$.
Thus butadiene does have some double bond character between carbons 2 and 3, which explains why there is the slight barrier to rotation about this bond.

You can construct the molecular orbitals of butadiene by combining the molecular orbitals of the two ethene fragments in-phase and out-of-phase.

This method of construction also shows why the HOMO-LUMO gap of butadiene is smaller than that of ethene.
The molecular orbital $\Psi_2$, which is the HOMO of butadiene, is the out-of-phase combination of two ethene $\pi$ orbitals, which are the HOMO of ethene.
Thus, the HOMO of butadiene is higher in energy than the HOMO of ethene.
Furthermore, the molecular orbital $\Psi_3$, which is the LUMO of butadiene, is the in-phase combination of two ethene $\pi^{*}$ orbitals, which are the LUMO of ethene.
Thus, the LUMO of butadiene is lower in energy than the LUMO of ethene.
It follows that the HOMO-LUMO energy gap is smaller in butadiene than in ethene and thus butadiene absorbs light with longer wavelenghts than ethene.

If you continue to lengthen the $\pi$ system by adding more ethene fragments you will see that the HOMO and LUMO are getting closer and closer together the longer the $\pi$ system becomes.

Looks like it took some work... so can we interpret jumping the gap from one particular HOMO to one particular LUMO configuration an example of promoting the dye to a particular singlet excitation state?

@user4717 Yes, promoting an electron from a filled to an empty molecular orbital will lead to an excited state. Another thing that the MO picture conveys very nicely is that such a promotion would lead to a change in the bonding situation in the molecule. For example, if you promote an electron from $\Psi_2$ to $\Psi_3$ the middle bond between carbons 2 and 3 is strengthened while the outer bonds between carbons 1 and 2 and carbons 3 and 4 are weakened. You can draw a lot of useful conclusions from MO pictures.

That makes a lot of sense.  So if we look here: http://www.files.chem.vt.edu/chem-ed/quantum/jablonsk.html, $S_1$, $S_2$, ... correspond to events where an electron is promoted from a HOMO to a LUMO (the exact jump depending on the particular electronic configuration), and the intermediate eigenstates correspond to different vibrational/etc. modes?

And intersystem crossing happens when we either flip the spin of an electron somewhere (going to a triplet state, $T_n$) or move it between orbitals?

@user4717 Exactly. If you take butadiene as an example you might say promoting from $\Psi_2$ to $\Psi_3$ (without spin-flip) would lead to $S_1$ while promoting from $\Psi_2$ to $\Psi_4$ would lead to $S_2$. And each electronic state also has a multitude of different vibronic states, e.g. $\Psi_4$ can perform symmetric and asymmetric stretching vibration and so on.

@user4717 Yes, intersystem crossing which leads to a triplet state means that you promote an electron from a filled orbital to an empty orbital accompanied by a spin-flip. But you cannot have two electrons with the same spin in the same molecular orbital. So, an intersystem crossing where you just flip the spin-down electron in, say, the completely filled $\Psi_2$ to another spin-up electron is not possible.

That makes sense given Pauli exclusion, but can we also call moving an electron around without a spin flip and intersystem crossing event?

@user4717 When you promote an electron without spin-flip, the overall spin of your system remains the same. So, if you start from a singlet state you end up in a singlet state. But intersystem crossing means that you change the overall spin of your system. So, without spin-flip no intersystem crossing.

Right, I thought so as well, but that would mean this diagram (http://www.files.chem.vt.edu/chem-ed/quantum/jablonsk.html) is wrong right?  That "IC" for intersystem crossing appears to show transitions of electrons between unoccupied orbitals without spin flips (where we have intersystem crossing events leading to jumps between singlet or between triplet states).

@user4717 The "IC" means "internal conversion" not "intersystem crosssing" (which is abbreviated as "ISC"). Internal conversion is the radiationless transition between energy states of the same spin state, e.g. by quenching vibrational modes through collisions between atoms/molecules etc.

@Philipp Hi, your excellent answer generated a doubt in my mind. Generally, as conjugation increases, a red shift is observed. But in the case of butadiene, π1 and π*4 orbitals are generated whose energy gap is larger than that between the π orbitals of ethylene: does this not correspond to a blue shift?
So you have a red shift for the HOMO-LUMO transition and a blue shift for the π1 -> π4* transition?

The effect is already "visible" on the comparison of ethene ($\ce{H2C=CH2}$) and 1,3-butadiene ($\ce{H2C=CH-CH=CH2}$).
Let's build butadiene by combination of the molecular orbitals (MOs) of of ethene.
Combination of the $\pi$ orbitals gives one new occupied MO without a nodal plane (this is lower in energy than the $\pi$ orbital of ethene) and another occupied one with one nodal plane. This is our HOMO (highest occupied molecular orbital) of butadiene.
The HOMO of butadiene has a higher energy than that of ethene.
Now let's have look at the antibonding $\pi^*$ orbitals of ethene and their combination. Again, we obtain two new MOs, one higher in energy than the $\pi^*$ of ethene and another lying lower. This is our new LUMO (lowest unoccupied molecular orbital.
The LUMO of butadiene has a lower energy than that of ethene.
As a result, the HOMO-LUMO gap is lower, and the absorption shifted to higher wavelength.

While you certainly show that the energy gap closes with increasing size of the $\pi$ system, I'm not sure it appropriately addresses the why part of the original question.

@tschoppi Let's wait and see :) In the worst case, you'll have to give an in-depth lecture on HMO theory ;)

Let's have an extreme example of 9 conjugated double bonds. That makes your $\pi$ HOMO orbital. The LUMO, $\pi^\star$, has one node, the middle double bond being antibonding, so it will consist of two times 4 conjugated $\pi$ bonds. To say, the difference in energy between HOMO and LUMO is quite small, therefore the excitation energy is low.

The simplest physical picture is surely that of a particle in a box. (Of course its more complicated otherwise we would not need to do accurate quantum calculations but this model retains the basic ideas)
If we consider the conjugated $\pi$ orbitals to define the box length then as the box gets bigger the energy levels become closer to one another. On filling with electrons, two for each orbital, we then find that the energy gaps between HOMO and LUMO get smaller as the number of double bonds increase. This means that excitation  of an electron from HOMO to LUMO (i.e. absorbing a photon, which gives rise to the colour of the molecule) takes less energy as the number of double bonds increases. Thus ethene absorbs in the uv, whereas carotene, well is sort of carrot coloured, and absorbs in the visible. Polyacetylene (hundreds of conjugated double bonds) formed as micrometre thick films is brown and absorbs all across the visible spectrum and into the infra-red. In the bulk it looks metallic black as the gap between HOMO and LUMO has almost disappeared.
A similar argument applied to aromatic molecules of increasing ring size, say benzene to porphyrins and chlorophylls.


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Remove empty space from structures with R and S groups
We receive structures from a server within our company. Those structures are transferred as MOL files. Some of these structures contain R and/or S groups. Some of the creators of these structures create a really large empty space between the actual structure and the groups. We have currently ~30.000 structures in our database. Around 2.000 of these structures are flawed.
I have attached how such a structure might look at the bottom, as the image is quite large.
What I am trying to do now is to remove that empty space.
The process is currently receiving the MOL files wrapped within an XML file that does have additional information to store along with the structure. That process is written in Java. Up to now we do not have any library within the project to process and manipulate structures, as we have just parsed the MOL as String from the XML file and written it into a CLOB field in the database.
I know that libraries like the CDK exist, but right now I do not know if it can help me with that task.
I have pasted the actual mol file here.
If we open it with a viewer it looks like this:

Are you sure that the R and S are in the mol file and not part of the xml envelope?
Stereochemical R and S descriptors would not make any sense for a molecule like anthracene!
Is it conceivable that the letters come from Risk and Safety statements?
It would be helpful if you could upload the original mol file somewhere.

EDIT 1
The file that you kindly uploaded is supposed to be a V2000 mol file, but it is not standards-compliant.
(The full specifications can be obtained from Accelrys upon registration.)
The first flaw is already in the header! The header is supposed to have a length of three lines.
The following CTAB should start with the counts line (in line 4 - note that we're not using zero-based indexing here).
With this flawed format, OpenBabel, a standard tool for the conversion of chemical data formats, casts an error.

EDIT 2
The rest of the file, including the stereochemical descriptors, seemed to be fine.
Once I had inserted the missing blank line in the header, a PNG could be generated using
obabel -imol 498268e9.txt -opng -O stuff.png

OpenBabel comes with bindings for Python, Java, C++ and whatnot. You might want to talk to your IT about some refactoring ;)

You are right about anthracene. I had made up the sample, as I was not sure if I was allowed to past the real ones. I have added a concrete example in the question.

So the R and S groups may just be removed? I ask as I do not find them in your image any more :)

By default, they are omitted and I haven't searched for a way to blend them in again ;)

I'll have a look into it. Although it is helpful to get to know open babel and the formatting issue, I still need to know how to get rid of the whitespace, without loosing the R and S groups ...

There's a misunderstanding. The information is still in the mol file. It just isn't shown somewhere in the image when a well-formed mol file is processed.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


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Rate of reaction increases exponentially with concentration
In our chemistry practical we dropped 0.07g of Magnesium into various concentrations of an acid (sulfuric or hydrochloric - only one was used, not sure which), and measured the rate of reaction by the rate of the H2 produced (used a gas syringe). We then collected data at a suitable interval of the culmulative amount of H2 produced.
We drew the graph (with multiple concentrations - 0.25, 05, 0.75, 1, 2). Then we measured the steepest point (fastest rate) and took the gradient.
Plotting rate against concentrations (0.25, 05, 0.75, 1, 2), we came to the conclusion that rate varies exponentially with concentration.
But the rate law of a typical reaction is:
r = k * [A] * [B]

, which would mean doubling [B] would double the rate law, not have an exponential effect on the reaction. Why did we come to the conclusion that concentration has an exponential effect?
In our case the rate law, I would think, would be elementary, but we are not dealing with concentrations of both Mg and HCl, just HCl, as the Mg was solid - therefore no concentration of Mg in solution.
My question is: In a reaction between two liquids, is the rate of reaction linearly dependent on the concentration?
AND: What is the rate law of our reaction? - Why did we come to the conclusion we did?
Thanks for the help!! :)

You need to be careful when stating what kind of a curve fits a collection of data points. For sufficiently small datasets, it is possible to interpolate any curve with good accuracy. Would you be able to provide the raw data and graphs you made so we can see for ourselves?

What is the rate-limiting step? The faster Mg evolves hydrogen, the less surface area is exposed for reaction but the faster the convection. Mg heats by reaction and cools by liquid flow. Mg surface area changes over time, residual dimensions vs. pitting. Confounded variables cannot be disentangled by varying only one at a time (Theory of Experimentation).  The world is dirty and complicated.  Computer speech was not solved not by modeling anatomy.

@NicolauSakerNeto, OK good point. Our teacher also implied that this relationship was normal.

Rate equation is experimentally determined. So, you can't directly write $r=k[A][B]$ (where r is rate, k is rate constant). Probably you might have performed first order reaction. For which $r=k[A]=\frac{2.303}{t}log\frac{[R]}{[R_0]}=\frac{2.303}{t}log{e^{kt}}$.
The above equation suggests that rate is proportional to exponential function.

Fair point. I was assuming too much it seems. Is there a less mathematical way of describing it? Like why is it proportional to exponential function? Thanks

Where does the 2.303 come from??? I'm used to principles of chemistry, but not the mathematics behind it.. Thanks! :)

In my statement I had assumed that we were dealing we an elementary reaction, and thus could write the rate law from the chemical equation.

@SwedishArchitect: $\frac{1}{t}In\frac{[R]_0}{[R]}=\frac{2.303}{t}log\frac{[R]_0}{[R]}$. It may be noted that concentration of pure solid remains constant and can become the cause for not being in the rate law expression as like the reactant taken in excess in pseudo first order reactions.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


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Does the emission spectrum for a dye depend on the frequency of light used to excite the dye?
Take a look at the emission spectrum for an arbitrary fluorescent dye: http://en.wikipedia.org/wiki/File:Fluorescein_spectra.jpg
Does the emission spectrum depend on exciting the dye at its peak absorbance wavelength $\lambda_{max}$?  Consider that there is going to be a spread of eigenstates around a particular excited state (http://www.olympusmicro.com/primer/java/jablonski/jabintro/jablonskijavafigure1.jpg), corresponding to particular molecular vibration or rotational modes.
The thought is that, if we excite a dye at a wavelength slightly red-shifted with respect to $\lambda_{max}$, or slightly blue-shifted with respect to $\lambda_{max}$, we'd start out at a different eigenstate clustered around the same excited state (say $S_1$).  I imagine there's a probability of decaying from each of these eigenstates in $S_1$, which gives us the spread for the emission spectrum, so if we bias the initial eigenstate, we'd achieve a different emission spectrum (presumably "statistically" red-shifted with respect to the excitation frequency).  Or is the fluorophore taking a rapid random walk through all accessible eigenstates belonging to a particular excited state?
I suppose we could also ask a similar question with respect to the photoelectric effect in terms of the energy provided to the excited electron (conditioned on having a threshold frequency to span a band gap).

Does the emission spectrum depend on exciting the dye at its peak absorbance wavelength $\lambda_{max}$?

No, it is independent of that. The excitation $S_0 \rightarrow S_1$ is vertical, bond length are not changed in the excitation process. Compare the mass of the nuclei with that of the electrons and you know why. You'll just end up in a different vibronic level of the electronically excited state.
However, before the emission takes place, this initial Franck_condon states relaxes to the vibrational ground state of $S_1$ with an equilibrated geometry and emission occurs from that (Kasha's rule).

Thank you very much for your answer, and for pointing out Kasha's rule.  I wasn't sure what you meant by "...bond length are not changed in the excitation process...", I don't think I was suggesting this?

From Kasha's rule I can extrapolate that Stokes shifts are more likely than Anti-Stokes shifts.  And can I also understand that the spread of the emission spectrum is due to IC to different vibrational eigenstates in the ground state?

In organic molecular photochemistry in solution, I've personally never observed any Anti-Stokes shifts for the fluorescence. But i've seen strong and solvent-dependent Stokes shifts.

The literature and references for Kasha's rule is exactly what I was looking for.

The Jablonski diagram usually shown (as in your web page reference) is that for a molecules in solution and, as has been explained in other, answers the vibrational deactivation is generally fast so that it cannot compete with fluorescence so generally fluorescence occurs from the lowest excited vibrational level of the excited state. In Azulene, a naphthalene isomer with a 5 and 7 member rings, fluorescence comes mainly from the S$_2$ excited state to the ground state as internal conversion S$_2$-S$_1$ is slow compared to fluorescence. This occurs because of the very large S$_2$ to S$_1$ energy gap.  Some benzanthracenes also show emission from S$_2$ . The vibrational transitions in fluorescence spectra in solution are broadened due to collisions and solvent perturbation effects.
In the vapour phase when molecules are at low pressure and there are no collisions during the excited state lifetime, fluorescence arises from the level excited and to various vibrational levels in the ground state. In this case the emission spectrum does depend on the level excited, and the spectrum is now a series of sharp lines, rather than broad vibrational bands observed in solution.


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Can dust conduct electricity?
Just out of curiosity, can dust conduct electricity? The source of dust varies, is there any form of dust that can conduct electricity?

The answer to this question hinges on two key questions:

What kind of dust are we dealing with?
How is the dust arranged (i.e. what is your experimental setup)?

Which Dust?
Firstly, I'm going to assume you meant domestic dust (since there are more kinds of dust than you'd imagine).
I quote Wikipedia:

Dust in homes, offices, and other human environments contains small amounts of plant pollen, human and animal hairs, textile fibers, paper fibers, minerals from outdoor soil, human skin cells, burnt meteorite particles, and many other materials which may be found in the local environment.

Arrangement
The arrangement of the dust particles is key. Are we considering an aerosol, or are we carefully laying a line of dust on the table and connecting it with two electrodes?
Aerosol - We can safely assume that the conductance of air has a far higher impact on the measurement than the conductance of the particles in the aerosol (the dust). So here, dust does not conduct electricity. Note: I ignore the "conducting" via static electricity processes here, but they may play a role. I do not consider that "classical" conductance.
Line of Dust - Organic material is usually insulating, as is our skin (if it wasn't for the moisture that is needed for, you know, living), paper, textiles, hair and so on. So the main constituents of domestic dust does not conduct electricity. So also here, we will not observe conductance.
There is one exception to the above, where conductance is a possibility: Metal dust. This is kind of cheating, but consider the waste of some metalworking process, in which metal dust (or powder) is generated. This dust, being composed of metals and metal oxides, would likely conduct electricity (when laying it out in a line, still not as an aerosol).

An example I like involves space stations and vehicles in orbit. To write in space, an option would be to use pencils. However, when dragging a graphite tip across a surface, a small amount of graphite dust is released. Normally this fine dust settles quickly, but in space it keeps floating for a while. Though it may seem innocuous, the dust can eventually deposit on electronics, forming dangerous short-circuits as the graphite dust is electrically conductive. This is why pens must be used in space.

Yes, dust can carry static charge. That's why people are cautious with dust on the inside of computers as they can short-circuit the electronics or inside.
Interestingly, dust particles in those huge grain silos; if given an ignition source - usually a spark - can produce such a large explosion that it blows off the rough of the silo.
While it wouldn't conduct full current flow per se; the suspended particles are able to carry electric charge and differences between the charges of the particles allows dust to have conductivity.

Isn't the main problem with dust inside computers that it clogs fanblades (restricting their motion) and heatsinks (reducing their effective surface)?

@KlausWarzecha That's one reason. I was reading up on other forms, that if dust goes inside, i.e. once you open it, the static electricity discharges creating a large enough charge difference to short circuit some electronic components. I believe that this is exacerbated with more expensive electronic components.


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Molecular Oxygen in Suzuki Homocoupling
I'm curious as to the role of molecular oxygen in Suzuki homocoupling reactions, as discussed here.
In a standard Suzuki cross-coupling reaction, molecular oxygen is not required. However, this paper claims that the homocoupling reaction will not proceed in an Argon environment. Is there a (relatively) straight forward justification for this that I am missing?
This is the reaction mechanism from the paper:

They are not using a base either here, are they? Typically, Suzuki-Miyaura couplings (with arylhalides as the reaction partners) are performed in the presence of base to "activate" the arylboronic acid.

I don't remember a direct palladation of arylboronic acids or their derivatives in any of the couplings. Typically, $\ce{Pd(0)}$ inserts into the $\ce{C-Hal}$ bond of arylhalides and is thus oxidized to $\ce{Pd(II)}$. Ouite often, $\ce{Hal}$ is then displaced with $\ce{OR}$ by the action of a base. Think in $\ce{KOt-Bu}$ often used in the Kumada coupling.
In the homo coupling that you described here, there is - for obvious reasons - no $\ce{Ar-Hal}$ for the oxiative addition step. Without $\ce{O2}$ for the oxidative activation of $\ce{Pd(Ph3P)2}$, the mixture will thus remain inactive.


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Flavor transference through the air and plastic bags
I am not sure if this is a chemistry question; however, I was wondering:
In my desk I have two plastic, zip-lock bags.  One contains cinnamon tea, the other contains nuts.
After about a week of these bags being next to each other, the flavor of the cinnamon somehow leached through both bags and now the nuts taste like cinnamon!
Is there a chemical explanation for this or isn't this a question of chemistry?

Cinnamon tea is most likely flavoured with cinnamalaldehyde (1) and eugenol (2), the main components in cinnamon oil. The latter compound also gives cloves the characteristic aroma.

To my knowledge, ziplock bags are made from polyethylene. They are not meant to be aroma-tight, like vacuum-sealed coffee packages.

No problem! Nicely flavoured question :)

Cinnamaldehyde dissolves in and diffuses through most polymers within sensory threshold. Use glass containers with lined metal caps.  Isn't ground up tree bark amazing?

Try taking the nuts out and putting them in a bowl or cup for a little. See if you just smelling the Cinnamon from the bag is making you "taste" it.
Just an idea.

I can try that but I am pretty certain it actually tastes like cinnamon because it doesn't taste very good.  I like cinnamon but it doesn't go well with these nuts.


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How to properly relate quantum mechanical energy to the classical world
The Equation of Interest
The energy equation derived from the "Particle in a 1-D box" problem is as follows:
$E=\dfrac{n^2h^2}{8mL^2}$
where $n$ is the principle quantum number, $h$ is Planck's constant, $m$ is the mass of the particle (e.g. electron), and $L$ is the length of region of space (i.e. box).
The Goal
I would like to illustrate the harmonious nature of this energy equation derived from the quantum mechanical nature of the system and classical mechanics.  In theory we should be able to define a mass and region of space for a clear comparison.
The Example: Electron vs. Car
I want to compute the lowest possible energy of an electron using the equation above.  We define $m=9.11\times10^{-31} \textrm{kg}$ and the region of space as a reasonable value for the diameter of an atom, $L=0.2 \textrm{nm}$.
Plugging this into our equation along with Planck's constant ($h=6.626\times10^{-34} \textrm{ m}^2 \textrm{kg s}^{-1}$) we obtain
$E_{\textrm{electron}} = \dfrac{(6.626\times10^{-34} \textrm{ m}^2 \textrm{kg s}^{-1})^2}{8 *9.11\times10^{-31} \textrm{ kg} * (2\times 10^{-10} \textrm{m})^2} = 1.51 \times 10^{-18} \textrm{ J}$.
Now let us imagine a car with $m=2000 \textrm{ kg}$ and a region of space as $L=1 \textrm{ m}$.  The resulting energy is
$E_{\textrm{car}}=2.74\times10^{-71} \textrm{ J}$.
Now in both cases, the energy is so small that we might as well say its zero (from a classical standpoint anyway).  Of course $E_{\textrm{car}} << E_{\textrm{elecron}}$.
*So we cannot stop here as quantum theory and classical mechanics are virtually indistinguishable (with respect to raw energies) at this point correct?
We can, however, convert these energies (which are purely kinetic energies) into something which is clearly distinguishable such as velocity by using the equation
$E = \dfrac{1}{2}mv^2$
Alternatively,
$v = \sqrt{\dfrac{2*E}{m}}$
where $m$ is the mass of the object and $v$ is the velocity.  If we plug in our previously determined energies we get the following results...
$v_{\textrm{electron}} = 1.8\times 10^6 \textrm{ m s}^{-1}$
and
$v_{\textrm{car}} = 1.7\times 10^{-37} \textrm{ m s}^{-1}$
Here we see that the velocity of the car is essentially zero (or technically speaking, so slow that it wouldn't ever be observed as moving in billions of lifetimes).  This jives with classical theory.  However, we see that the velocity of the electron is staggering, even though the energy of the electron is 'essentially zero'
The Conclusions
So, in order to properly relate the energy equation derived from the quantum mechanical system of a particle in a 1-D box, one cannot simply compare the raw energies of the particle (i.e. electron) to that of an ordinary object such as a car because both energies result in a number that is essentially zero which defeats the point.  As such, one must relate the energy to something appropriate such as the velocity of the object to properly illustrate how the energy equation still relates to everyday life.
The Question
Are my conclusions correct or can the raw energetics simply be used to make a proper comparison between quantum/classical theory without resorting to recasing the results into a physical property such as velocity?
#not-homework

There is a great series of popular science books by George Gamow about a character called Mr Tompkins who experiences the world as if quantum effects are large enough to perceive (the idea being: what would the world be like if Planck's constant were very much larger). They are well worth reading as an explanation of quantum effects.

*So we cannot stop here as quantum theory and classical mechanics are virtually indistinguishable (with respect to raw energies) at this point correct?

This is wrong. Just because you can't immediately, instinctively comprehend the difference between the $10^{-18}\  \mathrm{J}$ and $10^{-71}\  \mathrm{J}$ and therefore think they are both "indistinguishably small" doesn't mean it's true. The difference is massive. Our brains just haven't evolved to deal with numbers whose logarithms are very far from 0, because for almost all of our existence as a species, it never had to. Perhaps this site may give some very qualitative though fun context on how brains perceive the magnitude of a number.
In order to cope with this limitation of ours, we have developed some strategies. One can be understood as a form of chunking, where we simply modify numbers in a way so they become closer to something we understand. You probably won't be able to immediately give an example of an object that weighs 200,000 ounces or $3\times 10^{-27}$ solar masses, though you may find it quite a bit easier if I say 6 tonnes instead. This is because even if we inherently know the concept of an ounce or a solar mass, we can't really comprehend the magnitudes of 200,000 or $3\times 10^{-27}$ (the latter of course being much worse), and in order to get our heads around it more easily, we instead invent a new unit to measure weights in which we can express the same information using numbers closer to 1. The whole idea of scientific notation for numbers and SI prefixes for units can be understood as a consequence of this.
Another way is the one you use; instead of reframing the units and numbers, reframe the situation. You may feel there is no difference between $10^{-18}\  \mathrm{J}$ and $10^{-71}\  \mathrm{J}$, but the 53 orders of magnitude in difference is about the same as the number of atoms in a human body and the number of atoms in the entire observable Universe! Notice how I simply took the ratio into a slightly more familiar context (you know the Earth is just a small part of the observable Universe, and a human being is puny in comparison to the Earth).
So all of your calculations are correct, but if Homo sapiens were smarter, they would be entirely unnecessary!

Yes the difference between the two numbers is very large, but at the same time, the difference between 1.0E-1230 and 1.0E-9999 is massive.  At the end of the day, both numbers are zero.  This is the point I was trying to make.

But they aren't both just zero! One is much more of a zero than the other! Your brain is trying to contextualize those numbers, and it does so improperly because it has never had to deal with such magnitudes.

But we are relating these values to the classical world which is the point.  The energy equation derived from quantum mechanics should still apply to the 'real world' by producing something measurable but how can one measure something so small?  This is why I recast the concept into terms of velocity where there is a noticeable difference.  Can E-18 J be measured in reality or would that show up as 0?

@LordStryker That amount of energy is very much measurable, and actually on a molecular level an energy of 1eV is quite massive, and that is E-19J.  You surely don't measure it with your hand, but energies~E-20J can cause molecules to behave fundamentally differently and change the course of reactions.  From the outcome of reactions the effects of QM is quite evident.

1E-18 J can be measured quite easily; it's on the order of 10 eV. 1E-71 J most likely cannot be measured currently in the Universe according to the way we understand physics. For comparison, the photon with the largest wavelength that fits in the current observable Universe (half of a wavelength equalling the diameter of the observable Universe) has an energy of about 1E-52 J. I still think your confusion is entirely independent of a specific physical theory, though.

Huh, I guess I never thought of J in terms of another unit such as eV.  10 eV is definitely very measurable.

That just goes to show how reliant we are on number chunking!

You are treating the car as an elemental particle without inner structure, where the only degree of freedom is the overall translation.  Within this model, it is true that QM seems to produce very little effect for cars.
However, cars that have only overall movement but no other structure is not at all useful.  You would not call something a "car" if it does not have internal combustion.  You need materials with different properties to construct different mechanical and electronic components of the car to make it useful, safe and comfortable.   None of these are possible without quantum mechanical effects.
Even though combustion and ignition seem quite violent and energetic reactions, on a molecular level it is quite rare to achieve energies E-18J.  In fact, using Bolzmann distribution, one find that such energy level actually correspond to extremely high temperature of almost 60000K, that is 10 times the temperature of the center of the sun.   Of course, we rarely actually deal with electron in boxes, and in chemical reactions one is dealing with the CHANGE of energy, not the energies themselves, so normally one come across energy figures magnitudes smaller than that, but nevertheless quantum mechanical effect is enormous on a chemical level.
In fact, without quantum mechanical effect, there will be no stable atoms, no stable molecules, no reaction and no chemistry.   It will just be a bunch of chaotic movement of particles that no one can understand, and there will be no computers to help either.  In fact, there would be no life as well.
You may be baffled at the thought of this, because rarely does such small number seem to matter in life.  In life, when we think about energy, we think about events that only happen a small number of times, like monthly electricity bills or a high MPG car, or say the electricity supply of a billion household.  But the number of molecules is much greater, usually ~10^23, and each carrying a piece of quantum mechanical effect.  Put together, a small energy on a microscopic level can be magnified to reach quite amazing results.

A small correction - the temperature at the centre of the Sun is actually around 15 MK. It's the surface that is around 6000 K.

without resorting to recasing the results into a physical property such as velocity?

The energy is a perfectly physical property. In fact, for many quantum mechanical systems, the energy is the easiest physical property to measure.

can the raw energetics simply be used to make a proper comparison between quantum/classical theory

Yes, the scales of the energies are sufficient to make a comparison between quantum and classical systems. As you and Nicolau Saker Neto suggest in teh comments to his answer, convert the energy of the electron into eV: $1.5\times 10^{-18}\mathrm{J} \times \frac{1\mathrm{eV}}{1.6\times 10^{-19}\mathrm{J}} \approx 10 \mathrm{eV}$. The ground state of the Hydrogen atom is $-13.6 \mathrm{eV}$, comparable to the electron in the $2.0\mathrm{nm}$ box. So the energy you calculated for the electron is similar to that of most of chemistry, which is dominated by quantum mechanics. So that's a reasonably-sized quantum mechanical energy. But it's not noticeable compared to the joules of macroscopically-sized objects moving at macroscopic speeds.
The $1.7\times 10^{-37}\mathrm{J} \approx 1.0 \times 10^{-18}\mathrm{eV}$, on the other hand, is negligible compared to both macroscopic physics and to ordinary chemistry.


text:
Can I damage a biotinylated antibody in solution by vortexing or mixing too vigorously?
I am running an immunoassay using biotinylated antibodies and have noticed a decreasing trend from duplicate assay runs (completed within minutes of each other) from the same vial.
In between each run:

I switch to a new low protein binding pipette tip.
I vortex the biotinylated antibody in solution in a low protein binding vial.
Then I prewet the pipette tip and deliver the reagent.

Note that degradation seems unlikely.
Any ideas on what may cause this?
Could it be vortexing or shearing during pipette tip mixing?

What kind of assay is this? i.e. ELISA, etc.?

Its a sandwich ELISA assay.

Was there any foam? 2) Is there any reason to think that solutes will settle, or is there condensation on the top of the tube, or has the reagent been frozen? If not, there might not be any reason to even mix the reagent.

I never worked in the field but applied mechanical stress (French Press, sonication) while extracting chlorophyll a from cyanobacteria.
Anyway, a cursory search furnished some sources like this, where vigorous vortexing of antibody samples is advised against. Instead, gentle or pulse vortexing or simple spinning is suggested.
But please take this with a grain of salt since I'm not really familiar with the standard protocols here.

It is good practice to avoid vortex mixing any protein solutions, especially vigorously.  Vigorous vortex mixing can produce foaming and proteins can denature (unfold and lose activity) when they are present at an air/liquid interface (as in a foam).  This is because the hydrophobic regions of the protein will want to be in the air region and the hydrophilic regions will want to be in solution, and so unfolding may occur.
In practice, antibodies are usually quite stable as proteins go and you can often vortex them without much negative effect to your experiment. That said, if you routinely vortex proteins in the lab and come across a less stable one you will suffer degradation.  Since you will not always know in advance which proteins will be OK to vortex, it is best practice to mix gently using a pipetman (again avoid introducing air that creates a foam) or inversion, or other method. If you must vortex, then be gentle with short pulses, avoiding foaming.
I have purchased some antibodies where the datasheet specifically says not to vortex mix.

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


text:
What are the products for the redox reaction HCl + SnCl₂ + KMnO₄?
In the redox reaction, where $\ce{HCl}$ is the excess reactant do these reactions produce following half reactions?
\begin{aligned}
(1)&&\ce{SnCl2 &-> Sn^{4+} + 2e-}\\
(2)&&\ce{8H+ + KMnO4 + 5e- &-> Mn^{2+} + 4H2O}
\end{aligned}
Where does the $\ce{HCl}$ come into play here, or do you consider it at the end when you have a final equation?
This is a redox reaction where $\ce{KMnO4}$ was added to $\ce{HCl}$ and $\ce{SnCl2}$ (hydrochloric acid was simply used as an excess reagent)

Welcome to Chemistry.SE. I edited your post to make the chemical equations pretty using MathJax. More information here:http://chemistry.stackexchange.com/help/notation

@BenNorris Dear Ben, I believe it is $H_2O$ not $H_{20}$. :)

@Jun-GooKwak - fixed

$$\ce{KMnO4 + SnCl2 + HCl -> KCl + MnCl2 + SnCl4 + H2O}$$
Oxidation states are for reactants:
\begin{aligned}\ce{
K  &= +1\\
Mn &= +7\\
O  &= -2 \\
Sn &= +2\\
Cl &= -1\\
H  &= +1\\
}\end{aligned}
Oxidation states for products:
\begin{aligned}\ce{
K  &= +1\\
Cl &= -1\\
Mn &= +2\\
Sn &= +4\\
H  &= +1\\
O  &= -2\\
}\end{aligned}
This is a redox reaction carried out in acidic solution. $\ce{SnCl2}$ and $\ce{KMnO4}$ are titrated with hydrochloric acid. What is being oxidized and what is being reduced? Potassium is being reduced, magnesium and tin are oxidized.
Let's write out our half reactions:
\begin{aligned}\ce{
HCl + SnCl2 &-> SnCl\\
HCl + KMnO4 &-> MnCl2 + KCl + H2O
}\end{aligned}
If we look at the oxidation states of our reactants, and products, $\ce{H}$ has +1, $\ce{Cl}$ -1, and for the products $\ce{Sn}$ +2, $\ce{Cl}$ -1 for the first reaction. The first reaction gains two electrons and thus is reduced.
For the 2nd reaction, $\ce{H}$ again has +1 and $\ce{Cl}$ -1. Potassium has +1 and the permanganate ion has -2. Specifically, $\ce{Mn}$ has a theoretical oxidation state of +7 and oxygen -2. In the products side, magnesium has an oxidation state of +2 and $\ce{Cl}$ -1, $\ce{H}$ is +1, $\ce{O}$ is 2-, and $\ce{K}$ becomes +1. Chlorine as an element has an oxidation state of 0.
For redox reactions, we have to make sure our reactants and products are first balanced for our elements other than H and O in acidic solution. It is clear that the chlorines are not balanced, therefore:
\begin{aligned}\ce{
2HCl + SnCl2 &-> SnCl4 \\
3HCl + KMnO4 &-> MnCl2 + KCl + H2O
}\end{aligned}
We balance oxygen atoms with adding molecules of water:
\begin{aligned}\ce{
2HCl + SnCl2 &-> SnCl4\\
3HCl + KMnO4 &-> MnCl2 + KCl + 4H2O
}\end{aligned}
Now we add H+ to balance H:
\begin{aligned}\ce{
2HCl + SnCl2 &-> SnCl4 + 2H+\\
3HCl + KMnO4 + 5H+ &-> MnCl2 + KCl + 4H2O
}\end{aligned}
We now add electrons to balance charge:
\begin{aligned}\ce{
2HCl + SnCl2 + 2e- &-> SnCl4 + 2H+\\
3HCl + KMnO4 + 5H+ &-> MnCl2 + KCl + 4H2O + 5e-
}\end{aligned}
We multiply our half-reactions by integers so that the electrons are the same number:
\begin{aligned}\ce{
10HCl + 5SnCl2 + 10e- &-> 5SnCl4 + 10H+\\
6 HCl + 2KMnO4 + 10H+ &-> 2MnCl2 + 2KCl + 8H2O + 10e-
}\end{aligned}
Now we add 'em up and cancel:
\begin{aligned}\ce{
16HCl + 5SnCl2 + 2KMnO4 &-> 5SnCl4 + 2MnCl2 + 2KCl + 8H2O
}\end{aligned}

I think Sn loses 2 electrons, while Mn gains 5 electrons

Ah... this question is long. I almost got it. I think I wrote one of the half-reactions wrong, let me look it up.

The final equation should be 16H+ + 2KMnO4 + 5SnCl2 ==> 5SnCl4 + 2MnCl2 + 8H2O + 2KCl

@Jonie Sn loses 2. Mn gains 4.

no Mn gains 5, it goes from +7 to +2

oxidation number of K is +1

Are you sure that the equation you provided is correct? I added up everything and it checks out. Ah ok, thank you. I thought potassium was an alkaline earth. Thanks, I'll rework that.

No problem, and thank you for taking the time out to help me, i really appreciate it! I just dont understand why Cl2 forms in your products

Got it. Few more edits. I used the wrong half reactions.

what happened to the 10H+?

btw since Sn is being oxidized, the electrons should be in the products, and since the Mn is being reduced, the electrons should be in the reactants, but this is just a technical error

In hindsight that's correct, the reduction phase of a cathode half-reaction has the electrons in the reactants side. I think I can leave that for you to figure out and I believe this question is resolved. Thanks.

but what happens to the 10 H+

It's canceled out on both sides just like the electrons.

My pleasure. :)

When using $\ce{MnO4-}$ as an oxidant, pH is crucial.
In acidic medium: $\ce{Mn(VII) -> Mn(II)}$
$$\ce{8 H3O+ + MnO4^{-} + 5 e- -> Mn^{2+} + 12 H2O}$$
Under neutral conditions: $\ce{Mn(VII) -> Mn(IV)}$
$$\ce{2 H2O + MnO4^{−} + 3 e− -> MnO2 + 4 OH^{−}}$$
In strongly alkaline medium: $\ce{Mn(VII) -> Mn(VI)}$
$$\ce{MnO4^{−} + e− -> MnO4^{2−}}$$
As far as an excess of $\ce{HCl}$ is concerned, one might take into account that $\ce{Sn(II)}$ and $\ce{Sn(IV)}$ possibly exist as their anionic chloro complexes
$\ce{[Sn(Cl)4]^{2-}}$ and $\ce{[Sn(Cl)6]^{2-}}$, respectively.


text:
Why does nitric acid not turn yellow everytime?
I have dilutions of different samples, which are made with 0.1 M $\ce{HNO3}$. Only I left them on the lab table for a few days and one turned yellow.
So my question is why would one turn yellow and not the other ones? They all have the same concentration of sample. The matrix in all the samples are the same. The difference is the concentration of $\ce{Cl-}$ in the samples. The one with the lowest concentration of $\ce{Cl-}$ turned yellow.
Any ideas why?

Pure nitric acid is colourless, the brown colour is caused by dissolved nitrogen dioxide, $\ce{NO2}$.
Nitrogen dioxide is formed by the slow decomposition of nitric acid, according to $$\ce{4 HNO3 -> 4 NO2 + 2 H2O + O2}$$
Nitrogen dioxide is a good oxidant and it does so - supposed that it finds a suitable partner.
This would be chloride. So, the more chloride in solution the more redox chemistry furnishing colourless products.

I don't understand the answer.  First, the potentials don't work out; dilute $\ce{HNO3}$ cannot oxidize Cl$^-$ to Cl$_2$.  Second, the OP says  "the one with the lowest concentration of  Cl$^-$   turned yellow".

Nitric acid takes yellowish brown colour due to the presence of dissolved nitrogen dioxide.
Nitric acid slowly decomposes even at room temperature, specially in the presence of sunlight. $$\ce{4HNO3 -> 4NO2 + 2H2O + O2}$$ Liberated nitrogen dioxide dissolves in the acid and gives a yellow colour.
Brown vapours are seen in the bottle and the nitric acid turns yellowish in colour.


text:
What is the inert pair effect?
I was reading about the p-block elements and found that the inert pair effect is mentioned everywhere in this topic. However, the book does not explain it very well. So, what is the inert pair effect? Please give a full explanation (and an example would be great!).

The inert pair effect is an example of a relativistic effect. The simplified explanation for it is given here.

The inert pair effect describes the preference of late p-block elements (elements of the 3rd to 6th main group, starting from the 4th period but getting really important for elements from the 6th period onward) to form ions whose oxidation state is 2 less than the group valency.
So much for the phenomenological part. But what's the reason for this preference? The 1s electrons of heavier elements have such high momenta that they move at speeds close to the speed of light which means relativistic corrections become important. This leads to an increase of the electron mass. Since it's known from the quantum mechanical calculations of the hydrogen atom that the electron mass is inversely proportional to the orbital radius, this results in a contraction of the 1s orbital. Now, this contraction of the 1s orbital leads to a decreased degree of shielding for the outer s electrons (the reason for this is a decreased "core repulsion" whose origin is explained in this answer of mine, see the part titled "Why do states with the same $n$ but lower $\ell$ values have lower energy eigenvalues?") which in turn leads to a cascade of contractions of those outer s orbitals. The result of this relativistic contraction of the s orbitals is that the valence s electrons behave less like valence electrons and more like core electrons, i.e. they are less likely to take part in chemical reactions and they are harder to remove via ionization, because the s orbitals' decreased size lessens the orbital overlap with potential reaction partners' orbitals and leads to a lower energy. So, while lighter p-block elements (like $\ce{Al}$) usually "give away" their s  and p electrons when they form chemical compounds, heavier p-block elements (like $\ce{Tl}$) tend to "give away" their p electrons but keep their s electrons. That's the reason why for example $\ce{Al(III)}$ is preferred over $\ce{Al(I)}$ but $\ce{Tl(I)}$ is preferred over $\ce{Tl(III)}$.

Please don't refer to relativistic mass. Mass is invariant, the energies associated with it are not.

@gsurfer04 Could you elaborate on what you mean by your second sentence, because the part "the energies associated with it are not" needs more explanation. If you are talking about rest mass, which indeed would be a constant, then your argument is correct. I know that the whole argument that mass increases with speed is not the most correct way of expressing what happens as the term mass used here is not identical with the rest mass but I didn't want to dive into that discussion in my answer. Thanks for raising awareness of this issue.

For reference: http://physics.stackexchange.com/q/1686/7768

@Philipp I am not an expert to judge the correctness as gsurfero4 did but your answer is absolutely amazing. Such good explanation is nowhere available!

@Philipp But in wave mechanics electron is not supposed to be object so why are we assuming electron to behave as a planet and spin?

Inert pair effect is mostly shown by the 15-17th group elements.
That is, the oxidation state reduces by 2 for elements below ($\ce{As}$, $\ce{Sb}$), which is more stable than the other oxidation states.
The reason for this is the inertness of the inner $s$ electrons due to poor shielding.

However, the s and p orbitals are thought to be the most effective in terms of shielding, whereas the d and f orbitals have poor shielding. Why is it the other way around in this situation?

As we move down the group, the intervening d and f electrons shield the ns2 electrons poorly and hence the ns2 electrons are pushed inside towards the nucleus This results in the reluctance of ns electrons to participate in bond formation and this phenomenon is called Inert pair effect.


text:
What chemical reaction can cause this blue-ish deposit?
Recently I noticed that some of our glasses and cutlery had developed a blue-ish deposit as you can see for the two sets of spoons below. For each set the spoon on the left has the blue-ish deposit whereas the spoon on the right is the original.

I traced the cause of this blue-ish deposit to a specific effect. It only occurs when the spoon or glass has been used for a medicine called Questran-A, which is a powder that needs to be dissolved in water and then ingested. And more specifically, it only happens when these glasses/spoons are subsequently washed in our dishwasher.
This leads me to believe that some reaction occurs between the (left-overs) of the medicine and the dishwasher tablets. My question is: does anyone know what reaction between the chemicals in the medicine and those in the dishwasher tablets could cause this blue-ish deposit?
Obviously this will be impossible to say without the ingredients of both, so here is a list of the contents of the dishwasher tablets and the medicine:
Dishwasher tablets (Albert Heijn):

pentasodium triphosphate
sodium carbonate and sodium carbonate peroxide
taed
sodium silicate
PEG-90
alcohols, C16-18, ethoxylated mind 30 EO
acrylic copolymer
PEG-4
maleic acid/acrylic acid copolymer,sodium salt
alcohols, C12-18 ethoxylated propoxylated
zinc sulfate
beta-alanine, N-(2-carboxyethyl)-, N-coco alkyl derivs, disodium salt
mn-complex
benzotriazole
subtilisin
tetrasodium etidronate
alpha-amylase
benzisothiazolinone

Questran-A:

4 mg colestyramine
30 mg Aspartame(E 951) equal to 17 mg Phenylalanine
Citric acid(E 330)
kelcoloïd
Silicon dioxide(E551)
Orange flavoring
Xantham gum (E 415).

Let's speculate a bit ;)

Colestyramine is a twodimensional polymer network, build from styrene and and 1,4-divinylbenzene.
The polymer has a lot of benzyltrimethylammonium groups. That means it can act as an anion exchanger.
It is used to strongly bind bile acids
Neither the polymer itself, nor the adduct with bile acid are supposed to degrade in the bowel. The loaded polymer is supposed to be excreted.

The dishwasher tablets, on the other hand, contain the sodium salt of a maleic acid/acrylic acid copolymer.
That means we have two polymers, one with a lot of $\ce{Ph-CH2N^+(CH3)3}$ groups and another one with lots of $\ce{-COO-}$ groups.
Sounds like a mass wedding of two chain gangs :D
Edit
It is conceivable that this deposits a thin film on the spoons and the glassware.
In order to explain the bluish colour, we have two options:

The polymer aggregate absorbs in the 580 nm range - we observe the complementary colour. Due to lack of conjugation, I don't see how this could happen here.
The colour results from some dispersion effect related to opalescence, as suggested in Michiel's comment.

Nice discussion! Small follow-up: could the mass wedding be responsible for the blue-ish deposit? Maybe just on the basis of layer thickness, or otherwise chemically?


text:
Why are excited Slater determinants used to describe electron correlation?
Usually the first step in describing the electronic wave function of a molecule or atom is to describe it with a single Slater determinant. I get that this is an independent particle approximation, and that in the context of Hartree-Fock, each electron moves in the average field of the other electrons. That's fine.
But when we correlate the electrons, this is done by generating excitations out of the Slater determinant, and then taking weighted combinations of them. For example, configuration interaction mixes the ground state Slater determinant with singly, doubly, triply, etc excited determinants. A fancier way of doing this is coupled cluster, where cluster operators generate excitations out of the Slater determinant.
Hermann Kummel says that we can think about electron correlation this way:

"The first thing one may imagine happening is that two particles mutually interact, thereby lifting themselves out of the Fermi sea, so that after the interaction both are in unoccupied orbitals"

Kümmel, Hermann. "Origins of the coupled cluster method." Theoretica chimica acta 80.2-3 (1991): 81-89.
Why can we (or should we) think about electron correlation that way?

In what way? In terms of linear combinations of Slater determinants or in terms of 'interacting electrons being in unoccupied orbitals'?

@LordStryker Both, really. The standard mathematical way to treat dynamic electron correlation is to "mix in" excited Slater determinants, which is equivalent to saying that when two electron interact, they "excite" each other out of their ground state occupied orbitals and into previously unoccupied virtual orbitals. In other words, an excitation from occupied to unoccupied.

@jjgoings, for a simple physical reasoning watch this video by professor Jack Simons. You could also read almost identical material in his book "Quantum Mechanics in Chemistry" (Chapter 8) freely available here.

I guess, it is obvious why additional determinants are used to describe non-dynamical correlation (sort of, by definition of non-dynamical correlation), so I interpret the question as it is about dynamical correlation (i.e. electron avoidance).

Don't want to spam, but this seemed like the easiest way to contact you. I don't know if you had seen it, but there is a new site on the network specifically about computational chemistry/materials: Matter Modeling SE. One of your papers came up in an answer I wrote there and I realized you might be interested in the site.

As I mentioned in comments, professor Jack Simons offers a relatively simple physical interpretation of the idea of mixing in excited detrminant for treating dynamical correlation in his book "Quantum Mechanics in Chemistry" (Chapter 8) freely available here, as well as in this video which is also authored by him.

An important mathematical finding is that a linear combination of a reference determinant and a doubly-exicted one can be expressed as linear combination of two other determinants. Namely,
\begin{multline}
 c_1 | \dotsc \ \phi_a \alpha \ \phi_a \beta \ \dotsc |
 -
 c_2 | \dotsc \ \phi_r \alpha \ \phi_r \beta \ \dotsc |
 = \\
 \frac{c_1}{2}
 \Big(
 | \dotsc \ (\phi_a - c \phi_r) \alpha \ (\phi_a + c \phi_r) \beta  \ \dotsc |
 -
 | \dotsc \ (\phi_a - c \phi_r) \beta \ (\phi_a + c \phi_r) \alpha \ \dotsc |
 \Big) \, ,
\end{multline}
where $c = \sqrt{c_2/c_1}$.1
Here, the determinants to the left differ by a doubly occupied spatial orbital $\phi_a$ being replaced by a doubly occupied spatial orbital $\phi_r$, while the determinants to the right describe the singlet $(\phi_a - c \phi_r)^1 (\phi_a + c \phi_r)^1$ state. So, a state created by adding the doubly-excited $| \dotsc \ \phi_r \alpha \ \phi_r \beta \ \dotsc |$ determinant to the reference $| \dotsc \ \phi_a \alpha \ \phi_a \beta \ \dotsc |$ one is equivalent to a state in which one electron occupies $\phi_a - c \phi_r$ spatial orbital (being in any spin state) while another electron occupies $\phi_a + c \phi_r$ spatial orbital (also being in any spin state). And this is the way electrons "avoid" each other: by occupying these different spatial orbitals.

For example, $\pi^2 \rightarrow \pi^{*2}$ configuration mixing in alkenes or $\mathrm{2s^2} \rightarrow \mathrm{2p^2}$ configuration mixing in alkaline earth atoms produce left-right polarized and top-bottom polarized spatial orbital pairs shown below.

Here one electron stays closer to the left carbon atom by occupying $\pi^2 + c \pi^{*2}$ orbital, while another avoids it staying closer to the right carbon atom by occupying $\pi^2 - c \pi^{*2}$ orbital.

In this case one electron stays closer to the top of an atom by occupying $\mathrm{2s} + c \mathrm{2p}$ orbital, while another avoids it staying closer to the bottom of the atom by occupying $\mathrm{2s} - c \mathrm{2p}$ orbital.

1) Here is the proof for $2 \times 2$ determinants, where for brevity $\phi_i = \phi_i \alpha$ and $\phi_i^* = \phi_i \beta$.
\begin{align}
 c_1 | \phi_a \ \phi_a^* | - c_2 | \phi_r \ \phi_r^* |
 &=
 \frac{c_1}{2} \Big( 2 | \phi_a \ \phi_a^* | - 2 c^2 | \phi_r \ \phi_r^* | \Big) \\
 &=
 \frac{c_1}{2} \Big( 2 (\color{red}{\phi_a \phi_a^*} - \color{green}{\phi_a^* \phi_a}) - 2 (\color{blue}{c^2 \phi_r \phi_r^*} - \color{purple}{c^2 \phi_r^* \phi_r}) \Big) \\
 &=
 \frac{c_1}{2} \Big( 2 (\color{red}{\phi_a \phi_a^*} - \color{blue}{c^2 \phi_r \phi_r^*}) - 2 (\color{green}{\phi_a^* \phi_a} - \color{purple}{c^2 \phi_r^* \phi_r}) \Big) \\
 &=
 \frac{c_1}{2}
 \Big(
 (\color{red}{\phi_a \phi_a^*} - \color{blue}{c^2 \phi_r \phi_r^*})
 -
 (\color{green}{\phi_a^* \phi_a} - \color{purple}{c^2 \phi_r^* \phi_r}) \\
 &\phantom{=\frac{c_1}{2}}-
 (\color{green}{\phi_a^* \phi_a} - \color{purple}{c^2 \phi_r^* \phi_r})
 +
 (\color{red}{\phi_a \phi_a^*} - \color{blue}{c^2 \phi_r \phi_r^*})
 \Big) \\
 &=
 \frac{c_1}{2}
 \Big(
 (\color{red}{\phi_a \phi_a^*} + c \phi_a \phi_r^* - c \phi_r \phi_a^* - \color{blue}{c^2 \phi_r \phi_r^*})
 -
 (\color{green}{\phi_a^* \phi_a} - c \phi_a^* \phi_r + c \phi_r^* \phi_a - \color{purple}{c^2 \phi_r^* \phi_r}) \\
 &\phantom{=\frac{c_1}{2}}-
 (\color{green}{\phi_a^* \phi_a} + c \phi_a^* \phi_r - c \phi_r^* \phi_a - \color{purple}{c^2 \phi_r^* \phi_r})
 +
 (\color{red}{\phi_a \phi_a^*} - c \phi_a \phi_r^* + c \phi_r \phi_a^* - \color{blue}{c^2 \phi_r \phi_r^*})
 \Big) \\
 &=
 \frac{c_1}{2}
 \Big(
 (\phi_a - c \phi_r) (\phi_a^* + c \phi_r^*)
 -
 (\phi_a^* + c \phi_r^*) (\phi_a - c \phi_r) \\
 &\phantom{=\frac{c_1}{2}}-
 (\phi_a^* - c \phi_r^*) (\phi_a + c \phi_r)
 +
 (\phi_a + c \phi_r) (\phi_a^* - c \phi_r^* - c^2 \phi_r \phi_r^*)
 \Big) \\
 &=
 \frac{c_1}{2}
 \Big(
 | (\phi_a - c \phi_r) \ (\phi_a^* + c \phi_r^*) |
 -
 | (\phi_a^* - c \phi_r^*) \ (\phi_a + c \phi_r) |
 \Big) \, .
\end{align}

It perhaps isn’t a very satisfying answer, but mathematically the reason is that this is our only option!
The atomic spin-orbitals form a complete basis set: we are guaranteed to be able to represent the true wavefunction as some linear combination of them. Under Hartree-Fock, as you point out, this is a single Slater determinant. If we want to do better than this, the extra chemistry that we want to add is electron correlation. The extra maths we need to add to cope with this is more basis elements… and the way to add these is to mix in more Slater determinants. Which is why I say this is our only option: if we start with the HF ground state, and want to mix in more basis functions still, then by necessity we will have to mix in some HF excited states.
(There is one extra point that might help in understanding the Kümmel description, which is that if we could get a lower energy for the ground state by mixing in a single-electron excitation, then the HF procedure would already have done this. So we need to consider two-electron excitations, which as Kümmel says can be interpreted as interactions between these two electrons that raise them above the HF ground state.)

Unfortunately I have to agree with you -- it seems to be one of those places that we have to rely on math instead of intuition. The HF Slater determinant forms a basis for an N-electron wavefunction, just like our single electron basis functions form a basis for an independent particle wavefunction.

Regarding your point about single electron excitations: they do show up in many correlated methods (CCSD and CISD, for example)! While you are correct in saying that single excitations don't mix with the ground state directly (e.g. Brillouin's theorem), they do mix with higher excitations. According to Thouless, single excitations create any single Slater determinant out of any other single Slater determinant. Basically it optimizes your reference in response to including dynamic electron correlation. It's a big reason CCSD is pretty insensitive to reference (which makes it a robust method).

Of course you are correct that my statement in the final paragraph only applies to the ground state, and I’ve edited to make this clear. Apologies if the level of explanation was rather too basic for you!

The concept of 'mixing' these determinants (i.e. ground and some excited state) is fine with me mathematically.  However, the physical interpretation of this mathematical approach is very dissatisfying to me.

@LordStryker, there is some simple physical interpretation. Please, watch this video.

Actually there is a reason for this.
The exact mathematical way to expand a function of N variables $\Psi(x1, x2...)$ - in terms of a basis - is a multi-determinant expansion, given the requirement below.
The determinant appears "magically" when the function is required to be anti-symmetrical wrt the exchange of coordinates of any two electrons.
$$\Psi(1,2) = -\Psi(2,1)$$
See Szabo and Ostlund, formulas 2.70-2.73

Analogy: we can expand any well behaved, monoelectronic orbital (a single variable function) as a set of infinite basis functions, and we can expand any multi-electronic system as an infinite number of determinants.

Downvoters are invited to say why, and get an explanation. There is even a reference to the exact equations in a book that answer the question.

The only value of this answer is a reference to a book you need to own and which isn't even cited properly. This is not an answer, this is a comment.

@Martin-マーチン Its a bit of a rude comment. The explanation is literally the second paragraph. Most readers will know the book I lazily cited. I would expect that a sharp eye would value it more. I may include more details, but I think it is a good start. You dont, no problem.

I don't think my comment was rude, but I'll pass your complaint on to the other moderators to check. I still want to clarify though. I know the book, I have read it, parts of it I understood, but since then I've passed it on. I do not recall the equation you referred to. You could have taken my explanation as a hint to fill it in, but you have chosen a different path. What appears to you as a complete answer doesn't to me. This is how the voting works on the site, and now that my view is locked in, I cannot (and also would not) change it

I have to agree with Martin. This seems to answer the question "why does a determinant provide an appropriate electronic wavefunction?" rather than answer the original posted Q.

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


text:
Finding the average rate of consumption
The question is:

If $2.0 \cdot 10^{-4}$ moles of dye in $\pu{50 mL}$ of solution is consumed in 188 seconds, what is the average rate of consumption of dye in $\pu{mol L-1 s-1}$?

I am not sure if I have done this correctly. I use the rate of reaction formula which is $\text{Rate} = -\frac{\mathrm{d[A]}}{\mathrm dt}$ There is a negative sign because the dye is being consumed and it is disappearing.
Given:

[dye] = $2.0 \cdot 10^{-4}$ moles in $\pu{50 mL}$

volume of solution = $\pu{50 mL}$ = $\pu{0.05 L}$

$t = \pu{188 s}$

Since we have $2.0 \cdot 10^{-4}$ moles of dye per $\pu{50 mL}$ of solution, that means in $\pu{1 L}$ of solution, we get $2*(2.0 \cdot 10^{-4})$ moles of dye = $4.0 \cdot 10^{-4}$ moles
\begin{align}
\text{Rate of consumption/disappearance} &= \frac{-4.0 \cdot 10^{-4} \, \pu{mol}}{\pu{188 s}} \\
 &= -2.1 \cdot 10^{-6} \pu{mol L-1 s-1}
\end{align}
Is my answer correct? Please point out my mistakes if I did it incorrectly.

Since we have $2.0 \times 10^{-4}\text{ mol}$ per $50\text{ mL}$ of solution, to get the amount of substance per liter, you multiply by:
$$ \frac{1000\text{ mL}}{50\text{ mL}}=20$$
So we get $4.0 \times 10^{-3}\text{ mol L}^{-1}$ as Klaus says.
You're being asked to find a rate of dye consumption. This should be a positive number, if the dye is being consumed - which it is. If dye was being consumed at a negative rate, more dye would appear over time!
Imagine we begin with $c = 1.0\text{ mol L}^{-1}$. We end with $c = 0.5\text{ mol L}^{-1}$. Thus we might say that $\Delta c = -0.5\text{ mol L}^{-1}$. We see that the change is negative, signifying that dye has been consumed. If this occured over the course of 180 seconds, we might write:
$$\text{average rate of consumption} = -\frac{\Delta c}{\Delta t} = -\frac{-0.5\text{ mol L}^{-1}}{180\text{ s}} = 2.77\times 10^{-3}\text{ mol L}^{-1}\text{ s}^{-1}$$
So even though the change in dye concentration is negative, the rate of dye consumption should be a positive number.

Thank you for providing me a very clear explanation and pointing out the rate of consumption part! And I later realized that I messed up the conversion between mL to L.

Since we have 2.0*10^-4 moles of dye per 50 mL of solution, that means in 1 L of solution, we get 2*(2.0*10^-4 moles) of dye = 4.0*10^-4 moles

Unless I'm too tired, the concentration is
$ 4.0 \times 10^{-3}\ \mathrm{mol} \cdot \mathrm{l}^{-1}$

Thank you for pointing this out! I guess I was too tired so I messed up the conversion.

@Jesse No problem! Honestly, my first mental arithmetic guess was also off by one order of magnitude :D


text:
In synthesizing PbS nanocrystals will there be any problems if I use lead IV acetate instead of lead II acetate?
I am just about to start my senior project synthesis of PbS quantum dots and we ordered the wrong lead acetate. The method calls for lead (II) acetate and we got lead (IV) acetate. If we use the lead (IV) acetate will our synthesis be contaminated and not work or can we still use that chemical with two extra acetates attached?

Perhaps it's possible to turn the lead(IV) into lead(II) quantitatively with a mild reducing agent before you actually start the quantum dot syntheses?

You might run into problems. $\ce{Pb(OAc)4}$ is a pretty strong oxidant!
What will be your sulfide source? Are you sure that it isn't oxidized these conditions?
Edit 1
At least N,N'-diarylthioureas are efficiently desulfurized by lead tetraacetate and furnish ureas (and sulfur) via a carbodiimide intermediate, see Bull.Chem. Soc. Jpn., 1984, 57, 315-316.
Edit 2
If you don't mind a personal advice: I wouldn't want to add complexity to the protocol just to maneuver around problems (and create others) I wouldn't have if I were using the right starting materials in the first. (The German term for that is verschlimmbessern, which translates to something like disimprove).

Sulfur is coming from a sulfurea/thiourea solution. The other reagents used are SDS CTAB as stabilizing agents and EDTA as a chelating agent to disperse Pb. Could adding a base such as aqua ammonia stabilize the excess acetate?

In the present procedure aqua ammonia is already used in the preparation of the reaction mixture of water EDTA, SDS, and CTAB before the lead and sulfur precursors are added. the aqua ammonia is used to change the pH to 10 because EDTAs ligand formation is pH dependent and needs to be at a high pH around 10 for the fully deprotonated EDTA4− form to be prevalent. Does this make a difference now?

You would still have the wrong oxidation state, isn't it? I know that Pb(II) does form complexes with EDTA, what about Pb(IV)? I have no idea.


text:
In Graham's law, is the rate of effusion only dependent on molar mass?
According to Graham's Law, the effusion rate of a gas is inversely proportional to the square root of the molar mass of the gas.
However, consider this situation: We have s balloon with 1 mole of neon gas and another with 2 moles of neon gas. The gases are in thermal equilibrium and occupy the same volume. In this scenario, would the rate of effusion be the same?
The molar mass of the neon in both balloons is the same, so according to Graham's Law the rates would be the same. But intuitively, thinking about 2 moles versus 1, the pressure is doubled if volume and temperature are kept constant. And if the pressure is doubled, more collisions per second occur, so I would think that despite Graham's Law, the rate of effusion is also affected by the number of moles.

Molar mass is just one of the factors affecting effusion. Graham's law states that everything else remaining same, the rate of effusion of a gas is inversely proportional to $\sqrt M$. In other words, Graham's law describes the relation of the rate of effusion with only one of the factors involved in determining the overall rate.
For the general rate of effusion of a gas at pressure $P$, and the hole area $A$ at a temperature $T$, using kinetic molecular theory, it turns out to be: (derivation on Physics.SE)
$$r=\frac{PA}{\sqrt{2\pi MRT}}$$
Now, in your question, the balloon with two moles will have a greater pressure (more moles in the same volume) and therefore it will have a greater rate of effusion for the same hole area and temperature and molecular mass.


text:
Why does helium seem to exhibit an ionization energy affected by the shielding of electrons?
According to Wikipedia, the shielding effect only happen in atoms which have more than 1 electron shells. The core electrons repel the electron in the valence shell. However, I have heard some lectures mention the shielding effect on helium atom when they explained why the first ionization energy of $\ce{He}$ is not exactly 2 times more than $\ce{H}$.
Why is this? According to the definition of shielding effect, there shouldn't be one in $\ce{He}$ atom.

In $\ce{He}$ you have 2 electrons instead of 1 for $\ce{He^{+}}$ (which would have two times the ionization potential of $\ce{H}$). Loosely speaking, the electrons repel each other, thus raising the energy of the electron state compared to that in $\ce{He^{+}}$. Thus, the ionization energy is smaller than 2 times that of $\ce{H}$. I'm not sure whether one should call that shielding but it has a similar effect - only that the effect is weaker than with "traditional" shielding by core electron shells.

I'm pretty sure it still counts as shielding. Even the simplified Slater rules take same-subshell electron shielding into effect. More elaborate schemes like Clementi-Raimondi shielding actually incorporate shielding of inner electrons by outer shells, too!

@NicolauSakerNeto I was a bit unsure whether this is in general referred to as shielding. Since in my courses in general chemistry the term shielding was introduced only in the context of core electron shielding I thought there might be a different term for same-subshell shielding in the general chemistry community.

The premise is problematic as there is no 'shielding' effects in an isolated Helium atom.
However, Phillipp was correct in explaining why the ionization energy of Helium is not exactly twice that of Hydrogen.  This is because the energies of each electron are higher due to their correlated nature (i.e. they repel each other into higher, non-ground-state-like energies).  Therefore, it takes LESS energy to rip one of those electrons out of the system.

It may help to think about the reverse process : the capture of a free electron by a $\ce{He+}$ ion to form $\ce{He}$. When this free electron is still far, it is intuitive that the attractive force between it and the two protons is indeed electrically shielded by the electron that is already circling the nucleus : it is attracted by the $+2$ charge and repelled by the $-1$ charge. (We will ignore magnetic effects, which are typically much smaller)
If there was no shielding, then the second electron would feel the entire $Z=2$ charge and settle at a radius of $r=\frac{1}{2}$ of the Bohr radius, just like the first electron. Its ionization potential would be $\displaystyle\frac{Z}{r}=4$ times that of hydrogen. From this perspective, the difference that shielding makes isn't between "almost twice" and "twice", but between "almost twice" and "four times".
How each of the calculation methods (e.g. Slater) translates this physical shielding effect into factors is another matter. The clearest is perhaps Mills* model, where every electron completely shields one proton. The (electrical part of) the first ionization energy for $\ce{He}$-like atoms is thus simply $\displaystyle\frac{Z-1}{r}$, with $r$ determined by a force balance equation $(0.567$ for $\ce{He)}$. The second ionization energy is $\displaystyle\frac{Z-0}{r}$, with $r=\frac{1}{Z}$ as per the classical Bohr model $(0.500$ for $\ce{He+)}$. This formula extends to subsequent electrons, but determining $r$ becomes a bit more cumbersome.

*Randell L. Mills, The Grand Unified Theory of Classical Physics, 2018 edition (ISBN: 978-0-9635171-5-9) Chapter 7 "Two-electron atoms", eqs. (7.35) and (7.45). The general formula for the radius $r$ of a $\ce{He}$-like atom is $\frac{1-\frac{1}{Z}\sqrt{s(s+1)}}{Z-1}$, with $s=\frac{1}{2}$ the electron spin. For $\ce{He}$ this works out to $1-\frac{1}{4}\sqrt{3}$, or $0.567$ times the Bohr radius.


text:
Is the scientific explanation behind Smartkleen legit?
I'm trying to have a conversation/debate about the SmartKleen laundry detergent alternative, but lack the knowledge to understand if their explanation is science or pseudoscience.
Far Infrared emission at cold water temp, smaller water molecule clusters, magnets that increase levels of dissolved oxygen, etc., can someone offer a solid critique?

Take several wiki articles, shake the sentences randomly, and, voila, Smartkleen science! Not about to waste time on woo-woo.

It suffices to say that I was originally interested in answering this, but reading that page made my head hurt.

@chipbuster Agreed! To me, it's marketing technobabble that you would use to sell penny stocks for a perpetuum motion company that also offers remote healing and time travel :D

These guys are good. They even tell you to ignore claims that their product doesn't work!
Just check http://www.smartklean.com/html/wikipedia.html out and the subsequent wikipedia page (http://en.wikipedia.org/wiki/Laundry_ball) if you want an answer to your question.

I think the idea of "smaller water clusters" penetrating the cloth more efficiently is actually quite smart as the same idea is utilised in steam cleaners. But the rest of the article is a bit too psychotic. How can they say that colours are preserved while the Ball constantly generates H$_2$O$_2$ (in every second)? And why do they recommend the Stain Remover for stains if the ball was such a miracle...

Do a wash without adding anything.  Given lightly soiled fabric, there is enough detergent residue to get things "sparkling chean."  Absence of perfume will give a nice linen scent.  Water soluble dirt does not need detergent to disappear.  HOWEVER, oily stuff will stay put and accumulate. Things likes starch and protein generally adhere and accumulate.
A laundry ball is nonsense no matter what its claimed mechanism.  A nasty accumulation of crud requires weeks or months to accumulate.  The laundry ball "wears out."


text:
Calculation of the specific rate constant (k)

Suppose we found that the reaction of Red#3 with bleach is 1st order in hypochlorite. The observed rate constant ($k_\mathrm{obs}$) for one of the trials was measured to be $\pu{0.00400 s^{-1}}$. If the concentration of hypochlorite was fixed at $\pu{0.134 M}$, calculate the specific rate constant ($k$) for that trial.

I am very confused in which equation I should use to solve this problem. Am I to use this equation?
$$\mathrm{Rate} = k[A]^m[B]^n$$
If that is the case, for trial #1:
$$\mathrm{Rate} = \pu{0.00400 s^{-1}} \times \pu{0.134 M} = \pu{5.36e-4 mol L^{-1} s^{-1}}$$
But then what should I do next? This does not sound right. Please point out my mistakes.

use the correct equation for first order kinetics. Yours does not fit.

We assume 1st order kinetics in hypochlorite and 1st order kinetics in the color. Putting this in the equation we get the following:
$$\frac{\text{d}[\text{Red}]}{\text{d}t} = -k\, [\text{Red}]^1 \, [\text{bleach}]^1 $$
Because you keep the concentration of bleach the same throughout the trial, you can write
$$k_\text{obs} = k\times [\text{bleach}]$$
Now, we can continue onwards as follows:
$$ k_\text{obs} = k \times [\text{bleach}] \Leftrightarrow k = \frac{k_\text{obs}}{[\text{bleach}]} = 0.0299~\mathrm{s^{-1}\; M^{-1}}$$

Thank you very much for the clear explanation! :) I get it now.


text:
Formation of 3-chloromethylbenzene VS 3-chloromethylbenzene VS 4-chloromethylbenzene
For methylbenzene (with the methyl group at position 1), is it easier for chlorine to substitute the hydrogen at position 2, 3 or 4?
I was thinking somewhere along the lines of different charge densities around the benzene ring, and that the charge density closest to the already existing methyl group has been diminished due to the methyl group itself. However, I am unable to explain it properly enough.

I just corrected a little typo in the product names, but i left them wrong for didactic purposes ;) See below for an explanation :)

We are talking about the electrophilic aromatic substitution of methylbenzene (A) with chlorine.

The possible products are 2-methylchlorobenzene (2), 3-methylchlorobenzene (3), and 4-methylchlorobenzene (4)
Chloromethylbenzene (B) is something different, also known as benzylchloride. Chloromethyl is the name for the  $\ce{-CH2Cl}$ substituent :)
Please have a look at this site for an animated explanation of the mechanism and the selectivities.

In order to emphasize the differences, you might want to consider using the systematic IUPAC names 1-chloro-2-methylbenzene (2), 1-chloro-3-methylbenzene (3), 1-chloro-4-methylbenzene (4), and (chloromethyl)benzene (B).

It very much depends on the reagents and reaction type you are using to halogenate the aromatic ring, however, the most common method taught during organic chemistry is electrophilic hologenation, which in this case the methyl will be para directing.
http://en.wikipedia.org/wiki/Electrophilic_halogenation#Scope


text:
How to calculate ionic strength of zwitterionic buffers?
Using the formula for the molar ionic strength from Wikipedia — Ionic strength
$$I = \frac 1 2\sum_{i = 1}^nc_iz_i^2,$$
I have calculated that $\pu{0.1 M}$ of HEPES $(\ce{C8H18N2O4S})$ would have a strength of $\pu{1.7 M}.$ Is this correct?

The Ionic Strength is calculated as:
$$I = \frac{1}{2}\cdot\sum{c_iz_i^2},$$
where where $c_i$ is the molar concentration of ion i ($M$, $\pu{mol/L}$), $z_i$ is the charge number of that ion, and the sum is taken over all ions in the solution.
This is your HEPES molecule ($\ce{C8H18N2O4S}$):

Only one proton at $\ce{-SO3H}$ group can dissociate, as far as I see. So in your solution you will have two different types of ions with charge equal to 1 (assuming that molecule will dissociates 100%, and than:
$$I = 0.5\cdot(0.1M\cdot1^2+0.1M\cdot(-1)^2) = 0.1M$$
So your calculations are wrong.
But also be aware that this molecule is Zwitterion!
READ THIS Do zwitterions contribute to the ionic strength of a solution? (Stellwagen et. al., Anal Biochem. 2008, 373(2): 407–409)
The resume of the articles:
"Hence, electrostatic effects due to zwitterions are important, even if the zwitterions do
not contribute directly to ionic strength."
My calculations are primitive. They are also far from the truth, but they show how the principle works for you.

I think calculation of the ionic strength may be complicated due to the zwitterionic character of HEPES. According to this source, zwitterions have no net contribution to ionic strength, so that HEPES produces only a very small (though likely still non-zero) effect on ionic strength.


text:
Does it matter which end of a chain one sets to be 1 when drawing the structure of organic compounds?
For example, if I were to draw the structure of 1,2,4-trimethylcyclohexane, then would one or both of these be wrong?

Or are both right? I'm wondering if it matters where I start to count from 1 in these cases, if there is some convention to it or if it doesn't matter much.

Both of those are the same molecule (flip the one on the left over its horizontal axis, and they are identical).  It does matter where you start counting from, because the numbers chosen for the functional groups should be as low as possible:

Numbering of the chain. This is done by first numbering the chain in both directions (left to right and right to left), and then choosing the numbering which follows these rules, in order of precedence:

Has the lowest-numbered locant (or locants) for the suffix functional group. Locants are the numbers on the carbons to which the substituent is directly attached.
Has the lowest-numbered locants for multiple bonds (The locant of a multiple bond is the number of the adjacent carbon with a lower number).
Has the lowest-numbered locants for double bonds.
Has the lowest-numbered locants for prefixes.

 From IUPAC nomenclature
So, given the choice between 1,3,4-trimethylcyclohexane and 1,2,4-trimethylcyclohexane, you would choose the latter (for both molecules).

I don't quite understand. I don't know what a locant is.. Are you saying I should count from the top, so that the first diagram is 1, 3, 4 and the second one is 1, 2, 4?

@Threethumb They are both 1,2,4-.  Flip the diagram on the left over its horizontal axis.  They are the same molecule (since all of the carbons of the hexane are "created equal" -- there are no double bonds there).  The locants are the numbers of the carbons.

Ah right, so technically $1, 3, 4-$ and $1, 2 4-$ is the same thing, but $1, 2, 4$ is the preferred way of writing it because it has the lowest numbers?

@Threethumb Exactly.  Everything should always be as low as possible.


text:
Make tin nitrate?
How can tin nitrate be synthesized? Mixing nitric acid and tin makes tin oxide (I did this in chemistry class) so how can tin nitrate be created?

A quick look into some search engine results for tin nitrate yielded this source, which rather interestingly even comes with a video of the reaction. That probably isn't among the main routes to generate tin nitrate, though.
It seems the direct reaction of tin and nitric acid actually can yield tin nitrate, but only in cold dilute nitric acid, and even then only as a metastable solution. This article goes deeper into the synthesis of tin nitrate.

It seems the first link actually has a great deal of short videos for experiments. That's an unexpectedly good find!

It appears that a clean preparation of stannous (or stannic nitrate), other than by the possible direct action of gaseous NO2 on SnO (or SnO2?), as suggested generally by Wikipedia below, is not readily found. To quote:

Conversion to nitrates
NO2 is used to generate anhydrous metal nitrates from the oxides:[10]
$\ce{MO + 3 NO2 → M(NO3)2 + NO}$

Or with, for example, stannous oxide possibly creating stannous nitrate:
$\ce{SnO + 3 NO2 → Sn(NO3)2 + NO}$
Research finds the following source relating to a detection sensor employing nano-SnO, to quote:

We attributed this to the larger depletion layer change induced by the stronger adsorption of the reactive species at the surface of the material coupled with faster charge transfer kinetics [7,12]. Besides, first-principles density functional theory (DFT) calculations showed that adsorbed NO2 species have relatively high mobility over the Sn-O surface [36], especially at high temperatures and small area surfaces like in a single-element device, allowing two NO2 molecules to react in the material surface generating NO and NO3, in which the latter is much more effective to remove electrons from the semiconductor rather than the NO2 [36], thereby achieving overall a 1 ppm limit of detection - well below the toxic limits. Moreover, even with NO3 having a stronger bonding energy than NO2, it helps to stabilize the baseline once there is no evidence of NO3 desorbing and leaving oxygen species on the semiconductor surface.

So, it may be possible employing mildly heated nano-SnO with NO2 exposure to create stannous nitrate.
An example of a not clean path, per a source (which extracts its content from old, and at times very dated, chemistry journals), to quote:

According to C. H. H. Walker, tin dissolves in nitric acid forming stannous and stannic nitrates, the relative proportion of these two salts produced depending on the temperature and the strength of the acid; moreover, the yellowish white precipitate which separates when the somewhat concentrated acid is employed is said to be a hydrated, ill-defined, stannic nitrate.

Here is a suggested possible path being based on the known REDOX reaction of elemental tin with a cupric salt, which, in the current case would be cupric nitrate forming stannous nitrate:
$\ce{Sn(s) + 2 Cu(II) (aq) -> Sn(II) (aq) + 2 Cu(I)}$
Source: Page 29 at this reference.
Unclear if the formed Cuprous nitrate is even soluble here, where its precipitation would be ideal leaving just stannous nitrate. If the Cu(I) is actually soluble, as long as the solution remains acidic, pump in air to recycle the cuprous to cupric. In the presence of metal Tin, the REDOX will stay active forming more stannous nitrate! Note: this path requires copper nitrate which can be easily prepared from available reagents. Namely, create an aqueous stoichiometric mix of KNO3 and CuSO4, and freeze-out the K2SO4 hydrate leaving highly soluble aqueous Cu(NO3)2. Just discovered a reference noting that the very route (KNO3+CuSO4 acting on Tin) is cited as route to a basic stannous nitrate at this source.


text:
Why use glacial acetic acid in bromination of anisole?
In my organic chemistry lab (next week, the experiment has not been conducted yet) we are being asked to react anisole with bromine to get 2-bromoanisole (as well as 2,4-dibromoanisole and all the others).  In our lab instructions we are told to add anisole, bromine and glacial acetic acid to our mixture.  No mention of other compounds (such as $\ce{FeBr3}$) was made.  I was under the impression that bromine is not reactive enough to brominate an aromatic without $\ce{FeBr3}$.  What purpose does the acid serve, and how does it make this reaction work without the iron compound?

[...] bromine is not reactive enough to brominate an aromatic without FeBr3

That's true for benzene, but the methoxy-group in anisole isn't just o/p-directing.
It is activating as well. Actually, it's that much activating that anisole will react with bromine even without the presence of a Lewis acid.
So, why glacial acetic acid? It's a polar solvent and thus nice to stabilize the Wheland intermediates. Moreover, it's easily removed in aqueous workup.
Edit 1
Does the acetic acid react with the (mostly undissociated) acetic acid? At least a polarization of bromine ($\ce{Br^{\delta+}-Br^{\delta-}...HOAc}$) is conceivable and I remember that $\ce{BrOAc}$ has been postulated as the electrophilic species.
Edit 2
I remembered some brominations on dimethoxy-substituted hydroxymethylnaphthalines I did back in the days. The reaction was performed at 60 °C in $\ce{CHCl3}$ in the presence of about 3 equivalents of anhydrous sodium acetate. Excess of bromine was reduced by addition of aqueous sodium sulfite. The yield was in the 85% range and I don't find any notes on dibrominated products.
Edit3

And with excess Bromine you would (theoretically) react all of the anisole?

Pretty much, with 4-bromoanisole as the main product.

So you don't actually get any intermediate compound like you would with $FeBr_3$?  And with excess Bromine you would (theoretically) react all of the anisole?

No not polybrominated products - just if the reaction would go to completion

Two things: ‘It is activating as too’ (1st paragraph); ‘Does the acetic acid react with the […] acetic acid?’ (edit 1). You might want to clear up what you wrote there. Otherwise +1.


text:
How to determine the vapor pressure of a glycerine and propylene glycol mixture?
I am looking to calculate the minimum heat needed to create vapor from a specific mixture. I have a mixture of 60 % $\ce{C3H8O3}$ (glycerine) and 40 % $\ce{C3H8O2} $ (propylene glycol, propane-1,2-diol). Total volume of the mixture is 3 ml. Here are some facts that I know
 C3H8O2
 Density = 1.04 g/cm³
 Boiling point = 370.8 °F (188.2 °C)
 Molar mass = 76.09 g/mol
 Specific Heat = 2.5 kJ/(kg·K) or 0.60 Btu/(lb·°F)

 C3H8O3
 Density = 1.26 g/cm³
 Boiling point = 554 °F (290 °C)
 Molar mass = 92.09382 g/mol
 Viscosity = 1.412 Pa·s
 Specific Heat = 2.43 kJ/(kg·K) or 0.576 Btu/(lb·°F)

At what heat will the mixture begin to create vapor? I am looking for an answer, although I am not looking for someone to come in and just leave an answer. I would rather understand how you came about the answer, i.e. equations, additional info, so that I can adjust the percentages and volume on my own. The percentages of the mixture will vary. The volume of the mixture will vary depending on the amount of heat needed to create vapor, i.e. decrease volume if not enough heat is present.

For this, you will probably need the heat of vaporization and not just the specific heat.

Your best bet for this will be to use the Clausius-Clapeyron relation. This is an exponential function that describes the vapor pressure of a substance as a function of temperature, and uses the enthalpy of vaporization (latent heat) and a constant as parameters.
The constant is dependent on the particular substance, but there is a way to eliminate it from the equation if you can get one data point: the vapor pressure at a specific temperature. If you can do that, then you can use the two-point form to predict vapor pressure at another temperature:
$\ln\frac{p_2}{p_1} = \frac{-\Delta H_{vap}}{R}(\frac{1}{T_2}-\frac{1}{T_1})$
This version only has one parameter - the enthalpy of vaporization ($\Delta H_{vap}$). $p_1$ and $T_1$ are the vapor pressure and temperature at point 1 (an arbitrary point), and $p_2$ and $T_2$ are the vapor pressure and temperature at a different arbitrary point.
For a pure substance, if you know the enthalpy of vaporization and the boiling point, you an calculate the vapor pressure at any other temperature. That is because the boiling point is the temperature at which the vapor pressure equals one atmosphere. In other words, you would set:
$p_1 = 1 \space \mathrm{atm}$
$T_1 = T_b$,
Then solve for $p_2$. The equation would be:
$p_2 = p_1\exp([\frac{-\Delta H_{vap}}{R}](\frac{1}{T_2}-\frac{1}{T_1})) \tag{1}$
When you use this equation, make sure you first convert the temperatures to Kelvin. This only works with an absolute temperature scale. Pressure can be in whatever units you want, as long as both pressures have the same units. In other words, convert 1 atm to whatever units you want to use for vapor pressure.
The data you listed didn't have the enthalpy of vaporization for either component, but they are available from NIST:
$\Delta H_{vap} = 91.7 \space \mathrm{\frac{kJ}{mol}}$ - glycerol
$\Delta H_{vap} = 71.2 \space \mathrm{\frac{kJ}{mol}}$ - ppg
Incidentally, NIST lists different values for the boiling point than you have. I would double check those before plugging numbers in.
Now, you don't have a pure substance; you have a mixture. Depending on how non-ideal it is (which basically means how much the molecules of one component will affect the other components) you might have to account for that if you want to get very accurate results. On the other hand, if you just want an estimate and are planning to experimentally adjust the power output from there, you can get a good starting point by assuming that the mixture and the vapor are both ideal. Under those conditions, the total vapor pressure of the mixture will be equal to the sum of the partial pressures...
$p_t = p_{gly} + p_{ppg}$
... and you can use the Clausius Clapeyron equation from above for each component individually to find the respective vapor pressures. Then you can add them together if you want the total vapor pressure. However, you may be more interested in the vapor pressure of each component individually, since (from what I have heard) glycerol is more responsible for the "cloud" that forms, while propylene glycol is the better organic solvent. It would be helpful to know how much vapor of each component a given mixture would produce at a given temperature.
That's how you can predict vapor pressure from temperature. To relate this to the amount of heat you need, you will need to do another step. Temperature is a function of heat supplied and the specific heat of the substance:
$q = mC\Delta T \tag{2}$
where $q$ is the amount of heat transferred, $m$ is mass, and $C$ is the specific heat. You know the specific heat of the mixture, and the mass is an independent variable since you will control the composition and amount of the mixture. If you assume that volumes and heat capacities are additive, that mixing is fast, and that heat transfer is fast (at least locally), you can estimate the temperature (or rather, the maximum temperature) by connecting this to the power supplied by your device. Power in watts is the same as Joules/second. Therefore, you can multiply the power output by the time to get a total (maximum) amount of heat transfer (in Joules):
$q = P*t \tag{3}$
We can combine equations $(3)$ and $(2)$ and solve for the operating temperature:
$T=\frac{Pt}{mC}+T_i \tag{4}$
where $T_i$ is the starting temperature of the mixture (before you apply power).
Now we can substitute this back into $(1)$ to come up with an estimate of the vapor pressure of each component as a function of the mass of the liquid, the power applied, and the length of time that it is applied for:
$p_{vap} = (1 \space \mathrm{atm}) \exp([\frac{-\Delta H_{vap}}{R}](\frac{1}{\frac{Pt}{mC}+T_i}-\frac{1}{T_b})) \tag{5}$
The real vapor pressure will likely be lower, but I think this would get you to a good starting point.
A couple of things to remember:

$m$ refers to the mass of each component that is in close proximity to the heating element.
$Delta H_{vap}$ is in kJ/mol - make sure that the $R$ you use has the right units.
$T$ must be in Kelvin (both of them)
The units of $P$, $m$, and $C$ have to cancel out to give K.

If you want to get a (rough) upper estimate of the mass that will be vaporized, you could assume an ideal gas and use
$m=\frac{pVM}{RT}$
where $V$ is the volume of the vaporizer chamber, and $M$ is the molar mass of the component.

Vapor is efficiently evolved when the sum of the vapor pressures is one atmosphere (boiling) given each mole fraction in homogeneous solution, rimmed by non-idealities.  You then must input the latent enthalpies of vaporization to move mass.
That is the juice for a vapor inhaler.  The sum of the enthalpies of heating and vaporization will define the battery you need.  However, do you want vapor or micronized (respirable) fog?  Pedagogy aside, it is more efficient to use an ultrasonic nebulizer in which the mist exits at ambient temperature.  Phase changes are expensive to fuel.  Unless all of the fluid parcel is vaporized, you will get one theoretical plate of fractionation.
If you run hot in air, you might pyrolyze glycerin into glycidol and acrolein, and propylene glycol into propylene oxide and allyl alcohol (then oxidize to the aldehyde).  Nasty for chronic inhalation.  You might thermally degrade inhalable substances (yeah, like "flavoring") with high boiling points.