text:
Why are the boiling points of aldehydes less than those of ketones?
Both aldehydes and ketones have a carbonyl group, but since the carbonyl group of ketone is between alkyl groups, wouldn't it be more difficult for it to form hydrogen bonds than aldehydes? Why then, do ketones have a higher boiling point than aldehydes?

I'm not sure invoking hydrogen bonding is helpful, as simple ketones/aldehydes have very little enol character and therefore their physical properties should be determined by the keto forms.

Have a look at the comments to the answer of Rauru Ferro. He has posted a link that shows data which imply that it is not generally true that ketones have a higher boiling point than aldehydes - in fact they seem to be very similar and which one is higher changes within the homologous series. Maybe the explanations given in the comments already give you the explanation you seek. If they still don't answer your question you could specify the points you want to know more about.

To follow up on Neto's answer more explicitly: Due to the nature of the sp2 hybridized carbon, aldehydes have very little hydrogen bonding character in the non-enol form.

It isn't that good a generalization: always look at the data first.
Here is a table of most of the aldehydes and ketones with 6 or fewer carbons (the labels are used in the chart later):

Now plot this on a chart:

Branches is the number of branches in the carbon chain.
Note that while for 3 and 4 carbons the ketones do have higher boiling points, it is not clear this is true for 5 carbon compounds and certainly not true for 6-carbon ones.
So, I'd argue the pattern of boiling points is complicated and there is no simple pattern that needs to be explained.

Amongst aldehydes and ketones, ketones have higher boiling point. This is due to the presence of two electron donating alkyl groups around the $\ce{C=O}$ group which makes them more polar.
For example: the boiling point of $\ce{CH3-CHO}$ is 322 K and dipole moment is 2.52 D.
Boiling point of $\ce{CH3-CO-CH3}$ is 329 K and dipole moment is 2.88D
Dipole moment of $\ce{CH3-CO-CH3}$ is greater than of $\ce{CH3-CHO}$. This because there are two electron donating $\ce{CH3}$ groups around $\ce{C=O}$ bond whereas there is only one $\ce{CH3}$ group around $\ce{C=O}$ in $\ce{CH3CHO}$.
As dipole moment is greater so it is more polar and hence has higher boiling point.

Just a minor point: the comparison between acetaldehyde and acetone is not entirely fair, since acetone is also a heavier molecule, so its comparatively greater London dispersion forces will also come into play (albeit with a lesser degree of significance than difference in dipole moments). I think a better comparison would be between acetone and propanal, since that would be comparing molecules of the same size. Still, as the document Rauru Ferro linked to notes, the b.p. trend isn't entirely consistent.

Considering that the carbon atom is positively polarised, why don't the electron donating groups reduce this polarization rather than increasing the polarization?

How would electron donating groups increase polarity, this makes no sense.

For ketones and aldehydes of similar molecular mass, ketones have higher boiling point due to the fact that its carbonyl group is more polarized than in aldehydes. So, interactions between molecules of ketones is stronger than between molecules of aldehydes, and that gives a higher boiling point.

Could you explain why the carbonyl group is more polarized in ketones than it is in aldehydes? Or could you provide a reference for this?

As I remember from my Organic chemistry basement, the difference between ketones and aldehydes is based in the electronic distribution between carbon and oxygen. This distribution is more accused in ketones than in aldehydes, because in aldehydes, the hydrogen bounded the carbon of the carbonyl group give electronic density to the carbon (because hydrogen is less electronegative than carbon).

Hmm, but shouldn't there also be hyperconjugation between the additional alkyl group and the carbonyl group's $\pi$ orbital which would also give electron density to the carbonyl carbon? Is this effect that much weaker than the small difference in electronegativities between carbon and hydrogen that it leads to a significantly higher polarization of the carbonyl group?

I've found this document: https://www.google.es/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&ved=0CDkQFjAB&url=http%3A%2F%2Fwww.uwec.edu%2Flewisd%2Fchem323%2Faldehydes_ketones_nitriles.pdf&ei=SsaMUrSYJKmq0QWElYGQBQ&usg=AFQjCNEqR93PmuE-W4OJoI27RRjHF44LGQ&sig2=1Gm7loe2qnIyz7VnK0_8ig, where in pages 411-412 describes two canonical forms neutral and polarized, where the polarized form have a minor contribution. May be this could explain the little differences between boiling points (page 419).

Your linked document seems to suggest that the assumption this question is based upon is wrong since there are cases where the boiling point of the aldehyde is (slightly) higher than that of the corresponding ketone. The fact that the boiling points are quite similar might mean that the effects I described in my previous comment are nearly equal in strength.


text:
Why does NaBH4 reduce double bonds conjugated to carbonyl groups, while LiAlH4 does not?
I have been going through reduction of aldehydes using $\ce{LiAlH4}$ and $\ce{NaBH4}$. If there is a double bond conjugated with the carbonyl group, $\ce{LiAlH4}$ doesn't reduce it, leading to an allylic alcohol. However, using $\ce{NaBH4}$, some of the fully reduced alcohol will also be formed. Why is this so?

Better look at this: https://chemistry.stackexchange.com/questions/87138/why-does-lialh4-reduce-the-double-bond-of-cinnamaldehyde-whereas-nabh4-does-not

Neither $\ce{LiAlH_4}$ nor $\ce{NaBH4}$ are able to reduce an isolated $\ce{C=C}$ bond. But if you have an enal (a conjugated aldehyde) it can react (as an electrophile) either at the β-carbon or at the carbonyle group's carbon. According to the HSAB Principle the β-carbon is a “soft” center and would react preferably with “soft” nucleophiles while the carbonyle carbon is a “hard” center and prefers to react with “hard” nucleophiles.
Now, $\ce{NaBH4}$ is a rather soft nucleophile and thus it reacts with an enal at the β-carbon. After this reaction the $\ce{C=C}$ bond is gone. But what is left behind is still a very reactive simple aldehyde that gets reduced to the alcohol by $\ce{NaBH4}$ in a second step.
$\ce{LiAlH4}$ on the other hand is a rather hard nucleophile and thus reacts with an enal at the carbonyl carbon. After this reaction there is just an isolated double bond left that can't be reduced by $\ce{LiAlH4}$ in a second step.

Cinnamaldehyde is reduced to Hydrocinnamyl alcohol by $\ce{LiAlH4}$. Both the double bond and carbonyl group are reduced

For cinnamaldehyde, both reductants first effect direct reduction of the carbonyl group; however, LiAlH4 is capable of further reducing the double bond. This has been recently explained here.

Isn't HSAB a lie? https://onlinelibrary.wiley.com/doi/full/10.1002/anie.201007100

Reduction by LAH: What would be the product in case the double bond is conjugated to both carbonyl group and benzene ring?

@orthocresol Can I conclude that NaBH4 can reduce the conjugated π-bonds?

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text:
An alternative basis set for analytical computation of two electron integrals
It is well known that the usage of the Gaussian basis set, in contrast to Slater basis set, leads relative simple semi-analytical expressions for the two electron repulsion integral $(ab|cd)$. Could you propose any other basis sets allowing simple computations of two-electron integrals? This is quite general questions meaning that these functions should not fit atomic orbitals well (the problem is formulated for wider class of potentials rather than for the Coulomb one only).

Plane wave, wavelet, although i am not sure if the integrals are simple

One of the main objectives is the orthogonality of the basis functions. So you can choose any reasonable orthogonal basis.

But, which of them gives analytical solution of the two-electron integral? For example, plane waves are ok. Independently of number of cites, they lead to Fourier transform of the rational function. Gaussian functions and plane wave, what else?

Orthogonality of basis set (function) usually does not bring problem in computation. The issue arises when MO vs VB

A few things come to mind: 1) my understanding is that (ab|cd) in a gaussian basis is exactly solved for arbitrary "angular momenta" using recurrence relations such as the Obara-Saika method. The O-S method also has ways of generating the requisite derivatives with respect to coefficients, without which it is difficult to minimize the energy or do anything practical.
2) On the question of alternative basis sets to the Gaussian, one simply needs a complete basis set (of which there are many), helps to be orthogonal (almost as many), has to describe the boundary conditions of vanishing at infinity or being periodic with unit cell (fewer), hopefully is cusp-like near the nuclei (very few), and integration has to be tractable without resorting to quadrature, since there are four indices (each with 3 dimensions) but also millions of these integrals to compute -- and that leaves you with the Gaussian.

Another good point for effective computations within the gaussian basis functions is that you can use fitting procedures $(ab|cd)\approx(ab|p)(q|cd)$ and transform four centres into three.

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Specific gas constant for two gases
I have a mixture with 1 part of gas A, and 7 parts gas B. This ratio is in terms of mass.
Gas   Molar Mass (kg/mol)
A     0.17034
B     0.034

I'm trying to find the specific gas constant of the mixture.
$$
R_{\text{sp}} = \frac{R}{M}
$$
I thought it might be
$$
M = 0.17034\frac{1}{8} + 0.034\frac{7}{8}
$$
but I can't work out if that makes sense given the units, or whether I should be using atomic mass or something.

The molar mass $(M)$ of a mixture with known mass fractions $(w_{i})$ is calculated from:
$$
\frac{1}{M} = \sum_{i}\frac{w_{i}}{M_{i}}
$$
So in this case
$$
\frac{1}{M} = \frac{1/8}{0.17034}+\frac{7/8}{0.034} = 26.469\ldots\\
M = \frac{1}{26.469\ldots} = 0.037779872\ \mathrm{kg}/\mathrm{mol}
$$


text:
Coffee and tap water reaction - results in gray tint
I use filtered water to make my coffee.
If I mix a few drops of said coffee with our tap water, it turns a grayish color; it's much darker than the coffee would turn the water by itself (I compared by doing a side by side comparison and adding an equal amount of coffee to filtered water, resulting in a much lighter brown (not gray) color that is closer to what I'd expect from coffee). It's a very pronounced effect (i.e. easily visible when you rinse out a coffee mug that has just a few drops of coffee at the bottom). Because of the distinct color change and the fact that it's significantly darker than it should be from simple mixing, it's obviously a chemical reaction.
What common tap water contaminants would react with coffee in this manner? Iron perhaps?

Tap water doesn't contain a lot of iron (see http://www.auroville.info/ACUR/documents/laboratory/chemical_analysis_of_water.pdf) however that can vary depending on where you live - wouldn't it be more likely that calcium or magnesium reacted?

@LievenB That's why I'm asking. ;) I had assumed maybe iron or manganese (manganese in particular can apparently cause the grayish appearance) but both are supposed to react with air. This is reacting with coffee. It may be basic since coffee is acidic, and it doesn't seem to produce any gases. I left a glass of filtered water next to the tap water for several hours and the color of the tap water did change a bit (it got slightly yellow) but iron/manganese are supposed to turn the water red/gray respectively (and also cause deposits at the bottom, which I don't see).

this might be strange to think of, but could the coffee solution contain a pH indicating compound? (tea does) (that would clarify the colour change) I would suggest that you check the pH value of the different solutions that you made?

@LievenB Interesting thought. I don't have any pure acids/bases handy to test it on, but I'll definitely try that once I find some (most likely vinegar/bleach) to test it. If that's true, and if it works like tea, the darker color may indicate alkalinity.

yes - vinegar is a good idea to test it. Normally a coffee solution would be slightly acidic - if you want to use milder bases then I suggest you use soap (what is reasonably alkali)

Soap would be an option, but are all soaps alkaline, or just some? Watered down bleach might be more consistent.

you're right - let me know what the results are after you added the acid/base

@JustinMrkva Did you perform the experiments? Did you get some results?

The tap water might be slightly hard.  Hard water has a high content of minerals such as calcium or magnesium which makes the water more alkaline. Alkaline water will draw tannin out of the coffee much quicker than pure water.
Filtering the water with something like activated carbon filter then an ion exchange filter will lower the mineral content, thus lowering the pH. With a lower pH the dissolution of tannin will be slower.
Slowing the dissolution of tannin relative to more soluble portions like caffeine is part of the reason for steeping tea and coffee for only a limited amount of time.  More steep time, or more alkalinity in the steeping solution,  releases more tannin.  More tannin makes the coffee more anti-nutritional, more bitter, and it darkens the color.

I doubt that filtering with activated carbon will lower the mineral content.


text:
Split carbonate ions?
Can you split carbonate ions into their constituent elements directly in the carbonate form? (so not first producing $\ce{CO2}$ and then decomposing the carbon dioxide)

(edit:)
I know that when running electricity through an aqueous solution containing carbonate ions and other element ions, the carbonate ions will act as individual groups and will not split (only after chemical reactions (for example the thermolysis of metal carbonates will produce carbon dioxide)) So this was just out of interest - how can the carbonate ion itself be split into its constituent elements if it can't be split with electricity (or at high voltage, could it then be split?)?

This seems to be a homework question.
Can you please add what you have tried yet to solve your problem?

I'm not sure if it's homework, but questions of the type “why is X stable?” are not very suitable for our site… more interesting are questions of the type “I would have thought X decomposes into Y and Z, with this equation, because of S and T. Why doesn't in happen in these conditions?”

@F'x It isn't a homework question, it was just out of interest. I've edited my question, (done some more effort)

I tend to agree with @F'x.  The present question boils down to "once $CO_2$ is produced in the decomposition of carbonate, how can the carbon dioxide be further decomposed into its constituent elements."

I understand what you mean - but can the CO32- ion immediately (without producing CO2) be split into its constituent elements?

Plasma torch will disintegrate in just fine, only it's not a typical chemistry modus operandi.


text:
Internal vs external steam distillation
To extract essential oils from plant matter (amongst other purposes), steam distillation is often used. I've seen a lot of different glassware setups for this. I've sometimes seen them broken into two categories:

Internal steam distillation
External steam distillation

Here's my (admittedly shaky) understanding of each:
Internal
With internal steam distillation, you have one boiling flask containing plant matter and liquid water. You simply boil the mixture and send the gas to to the condenser.
External
With external steam distillation, you have two flasks. One contains water only. You heat this one and route the steam into the second flask. The second flask contains a mixture of water and plant matter. The steam from the first flask combines with the liquid mixture in the second flask. Optionally, the second flask also has a heating mantle. The output of the second flask then goes to the condenser.
Differences?
What is the effect of using one or the other setup? External is substantially more complex, so I can't imagine it would be used if it didn't offer some advantages in some circumstances. When would you use external, and why?

The main purpose is to distill the component (essential oil) with quite high boiling point.
You have to think of distillation of immiscible liquids, http://www.chemguide.co.uk/physical/phaseeqia/immiscible.html
where you can take advantage of the equillibria in liquid and gas phase separately.
In short - two very beneficial things happen:

boiling point of the mixture is lower than boiling point of water (and substantially lower than BP of the oil)
the composition of vapor is independent of composition of liquid, i.e. you can distill out even the smallest fraction of oil (c.f. Raoult's law for miscible liquids)

On the other hand, the partial pressure of the oil is small, let's say 5% that of the water, therefore you need to evaporate and condense 20x more of water than of oil. To achieve this, you can either put all the needed water in the flask and do the "internal" distillation, or add just a small amount of hot water and use steam from external source in "external" setup.
In general, it is simpler to just boil pure water (or even have a direct source of steam) and use it for distillation, than to take care of the whole process in one flask.
As such, I would see the "external" setup simpler and more robust, even though it needs little bit more glassware.

I've never understood this too well myself. I assume that internal steam distillation is worse because you rely on dissolving the compounds you want in the boiling water (often quite low solubilities) before they're allowed to exert their vapour pressure (most of the material being under a sheet of water)? I imagine the vapour pressure of a substance in a dilute solution is smaller than that of the pure substance at a given temperature.

Meanwhile external steam distillation theoretically has the undissolved substances directly exposed to the atmosphere in the flask, so they have maximum evaporation while the steam flow ensures matter transport outwards of the flask containing the desired substances without using a pump. Is that it?

"The main purpose is to distill the component (essential oil) with quite high boiling point."--This is the description of steam distillation in general. But my question is about two different sub-types of steam distillation: internal and external. Both are steam distillation, and both are said to allow distillation of substances with high boiling points. But what are the advantages and disadvantages of each?

Summarized illustrated here: http://www.erowid.org/archive/rhodium/chemistry/equipment/distillation4dummies.html#class4
The partial pressure of water is much higher, than of the compound distilled, therefore you need LOT of water to go through. You have to supply it, somehow. To supply directly steam looks as simple solution... Once you are happy with explanation, I will reiterate it into the answer itself.

Suppose you have 100 tonnes of harvested spearmint from a farm.  It is degrading in air and biologically as you look at it.  Batch internal steam distillation will be a procedural nightmare.  You will burn the biological where it touches the heated inside of the flask.  With external steam distillation, you continuously feed in biological and live steam, capture the mixed distillate (carvone), then discard the exhausted outflow.  You now have excellent process control for the temperature and volume of steam admitted, and efficient continuous throughput.
If it were peppermint, the menthol product is a solid at room temp.  That is a whole different ballgame re clogging.  A production line requires trained skilled people.  If it requires intelligent people, you only have a pilot plant.


text:
Unified theory of chemical bonds
My question is related to the edge between Subatomic Physics and Chemistry, but I decided that here is the best place to discuss it.
As we all know, Theoretical Physics is a well developed field of knowledge, able to describe the very tiny forms of interactions. However, after some researches for some kind of formula to perfectly describe the interactions on the atomic level (not subatomic), more specifically, a way to calculate the strength of chemical bounds, I keep comming across frustrating highschool explanations, such as

ionic > covalent > H-bond > dipole-dipole > van der Waals

The most precise data that I was able to find comes at page number 20 and 21 of this pdf, that apparently shows some experimental results (I'm guessing).
So here is my question: is there really a unified formula? What if I want to simulate chemical interactions (proton-by-proton, electron-by-electron) on a computer, is it that complex to do, even assuming that I want a simple simulation of about 20 atoms?
Edit: When I reffer to calculating chemical bonds strength, I also include predicting many other chemical properties, as electron affinity or reactivity of the elements, which are all related.
Note: I can accept as answer any pdf that clarifies my question, because I wasn't able to find. Maybe I'm looking it with the wrong key-words.

Welcome to Chemistry.SE! While I don't have the time to give you a full-blown answer at the moment I at least can try point you in the right direction. I think the formula you are looking for is the Schrödinger Equation. It describes the behaviour of particles on an atomic level (in the non-relativistic limit). But it is not directly solvable for systems consisting of more than two particles. For those cases, a lot of methods have been developed to find an approximate solution, e.g. density functional theory (DFT), Hartree-Fock, etc.

The whole field of theoretical chemistry is devoted to developing better and better approximations to the many-electron wavefunctions that model the bonding between atoms in molecules and the interactions between molecules.

$H | \psi \rangle = E | \psi \rangle$

The short answer: no. We do not know how to solve the Schrodinger equation analytically, even for systems as simple as the He atom.
The much longer answer:
The Schrodinger equation is the basis for almost all theoretical chemistry: chemistry that attempts to predict properties of compounds without an actual experiment. The problem is that the Schrodinger equation is unsolvable for any compound with more than two electrons, and so we must make some approximations to solve it. This leads to methods such as Hartree-Fock and Density Functional Theory.
While these approximations are usually pretty good, they are still only approximations, and sometimes stop working correctly. There is an added problem that the Schrodinger equation is only an approximation, and the proper theory to use to describe light-matter interaction is Quantum Electrodynamics, which is incredibly difficult to solve for a single photon reflecting off of a mirror.
Long story short: there may or may not be a "unified theory" for describing exactly how all bonds behave in every situation, but there are currently methods that are good enough to predict a large portion of the behavior that we observe.

I have a question. The Schrödinger equation has no analytical solution for most systems, but we use classical computations to achieve a numerical approximation for the answer. However, solving a quantum problem with classical operations creates an exponential slowdown. If quantum computers ever become reality, would molecular modelling suddenly become almost trivial? If so, that will completely reshape the profession of chemistry as we know it; why attempt a synthesis in the lab with expensive reagents, when you can just calculate the result in silico!

As far as I know, it is still an open question. arXiv:1007.2648v1 "... there are broad classes of Hamiltonians for which finding the ground state energy (and therefore also a thermal state) is known to be qma-hard, that is, most likely hard even on a quantum computer (see Sec. 2.3) (70,4, 66, 98, 111). Nevertheless, the scaling of the ground- and thermal-state energy
problems for chemical systems on a quantum computer is an open question. It is possible that algorithms can be found that are not efficient for all qma-hard Hamiltonians, but nevertheless succeed for chemical problems."

I don't know about quantum modeling, but some of the other quantum algorithms (Grover search and Shor factorization) are non-deterministic, so you need a classical computer to check them afterward. Don't know if that would apply to this. @user26143 do you know if quantum computer simulations would have to be checked clasically?

@chipbuster, I know very little about quantum computing. As far as I know, I haven't awared any necessity of classical check for quantum chemical computing...

@Nicolau Theory will never be a sufficient substitute for the experiment - however, it will (or is) a very powerful tool for designing chemical reactions. The problem at hand is unfortunately the human brain: We are usually blinded by our own ideas, so that we are not capable of searching further away from them. The calculations tend to only be the tool to put some bone to the idea, but it might never be able to actually go a different way that that, that the user intended. Correct me if I am wrong or too negative...

@Martin I agree that theory will never substitute experiment but I thought I should highlight that many experiments RELY on theory for interpretation.  Therefore, without theory, many experiments could not exist in any meaningful way.

If you want to have a more comprehensive overview about your question, I would suggest you to read this book The Nobel Prize: The First 100 Years",
In this link The Nobel Prize in Physics 1901-2000 you'll find one of the chapters for free with a reasonable explanation about why we treat our problems using different hierarchies or different approaches (bottom-up or top-bottom).

formula to perfectly describe the interactions  Calculating the bond angle of water, vacuum vs. condensed phase for three small atoms, has a huge literature offering no definitive answer.  Something as simple as the pendulum equation (ideal - in vacuo, no friction or other dissipation), is not a closed form,
http://upload.wikimedia.org/math/6/4/c/64cd3db46ef18d401109423e81cdf6ce.png
http://users.df.uba.ar/sgil/physics_paper_doc/papers_phys/mechan/Pendulo1.pdf
http://users.df.uba.ar/sgil/physics_paper_doc/papers_phys/mechan/Pendulo2.pdf
http://fy.chalmers.se/~f7xiz/TIF080/Period.pdf
http://www.phys.uconn.edu/~hamilton/phys258/N/pend.pdf
http://www.scielo.org.mx/pdf/rmfe/v54n1/v54n1a10.pdf
HyperChem Lite for a very modest price has a lovely MM+ (semi-empirical) calculation engine.  I do blocks of 32500 iterations, update screen every 500 iterations, with 1e-020 limit until things look good enough, the method chokes, or the structure is ridiculous (pi bonds in high symmetry environments can go odd places).
http://www.hyper.com/Products/HyperChemLite/FeaturesLite/tabid/375/Default.aspx
ab initio Hartree Fock/6-31G(d) is better if you have two weeks in a cluster.

HF/6-31G(d) produces sometimes worse results than semi empiric methods or molecular modelling. It will not take two weeks in cluster. A molecule about 20 atoms should not take more than 20 h on a serial core i7 calculation. I am currently doing geo-opt dft with 100 atoms on 4 cores (2.8GHz) in less than a day.

How about CCSD(T)/aug-cc-pVQZ on 8 atoms using 6 cores in just a few hours? :)  Hartree-Fock just shouldn't be used... ever... This isn't the 1970s.  Correlation energy can be captured with cheap, more rigorous methods on today's computers.  Oh, and 6-31G(d)?  Great for lab exercises, preliminary starting points, etc.  I will try to axe any paper that uses such a small basis for production work without it being extensively calibrated and/or for good reason.

I can send you HyperChem mm+ and HF/6-31G(d) *.hin files for http://www.mazepath.com/uncleal/schwartz.png http://www.mazepath.com/uncleal/schwart3.png. When things are slow, certainly Christmas, grind it out to observe real world results. If you can unambiguously assign a sense of chirality to the central five atoms of the point group T molecule, I'll send you a very large lollipop or a bottle of Lagavulin 16. It is a Wreave bet that the bottle remains mine - against anybody who tries. "8^>)


text:
Are there any molecules with the same spectrum?
Are there any two distinct types of atom or molecule that have identical emission or absorption spectra?

R/S chiral molecules

In addition to the R/S chiral molecules mentioned by @user26143. There are also molecules with a distinct atomic composition that have experimentally indistinguishable spectra, even though theoretically there will be minor differences. One example, according to Rabitz et al. 2009 PRL (freely available for viewing) are flavin mononucleotide and riboflavine and in fact many other flavins. Due to there very similar structure you need unconventional techniques to be able to distinguish them.


text:
Isoelectric pH of colloids?
How did we derive the relation between the isoelectric pH and the dissociation constants for the amino and carboxyl groups on zwitterionic amino acids.
$$pI=\frac {pK_{a1}+pK_{a2}}{2}$$
where $$\ce{NH3^+-R2(COO^{-})<=>NH3^+-R2(COOH)-K_{a1}}$$
$$\ce{NH3^+-R2(COO^{-})<=>NH2-R2(COO^{-})-K_{a2}}$$
For more complicated cases (charged side groups) and their isoelectric pH, here is one site which just posts the equations without any derivation. Besides this, there is no reference I could find for this derivation.

Consider the equilibria in dilute aqueous solution:
$$\ce{H2A+ \rightleftharpoons HA + H+} \ \ \ \ \ \ \ K_{a_1}=\frac{[HA][H^+]}{[H_2A^+]}$$
$$\ce{HA \rightleftharpoons A^- + H+} \ \ \ \ \ \ \ K_{a_2}=\frac{[A^-][H^+]}{[HA]}$$
At the isoelectric point:
$$[H_2A^+]=[A^-]$$
Using the expressions for $K_{a_1}$ and $K_{a_2}$, we find:
$$\frac{{[HA]}[H^+]}{K_{a_1}}=\frac{K_{a_2}[HA]}{[H^+]}$$
$$[H^+]^2 = K_{a_1}K_{a_2}$$
$$pH_{isoelectric}=\frac{pK_{a_1}+pK_{a_2}}{2}$$
See also this section on Wikipedia.


text:
electrolysis - selection of what is going to be on anode and cathode
I have a question about electrolysis. How do we know what will be on anode and cathode if we have for example NaCl diluted in water. How do we know if it's gonna be Na+ or H+??

It depends on its oxidation/reduction potentials. For example, in the electrolysis of a NaCl solution (in water), the reduction potential of water to H2 is -0.83 V while the reduction potential for Na+ to Na is -2.71 V. As the potential for water is greater, water will be reduced instead Na+.
Another example, a solution of CuSO4 in water, the potential reduction for Cu2+ to Cu is +0.34 V, greater than the -0.83 V for water, so the Cu2+ will be reduced.


text:
pH range outside conventional 0-14
Is a pH value outside 0 - 14 possible?
I asked my teacher who said: yes, it is, but very difficult to achieve.
Then on the internet, I found multiple answers, one saying it is but because of a fault in the pH glass meter we can't be sure, the other said that it is not possible because of the intrinsic value of water to ionize.
As you can see, I'm getting mixed answers from different sources and I'm really confused now. Could somebody here give a concise answer whether or not it is possible? A deep argument will be valuable.

Let me add a bit more to the answers already given. As has been said, $\mathrm{pH}$ is nothing but a measure of the activity of protons ($\ce{H+}$) in a solvent - $\displaystyle\mathrm{pH} = -\log_{10} \ce{\ a_{H_{\text{solvated}}^{+}}}$. In dilute solutions, a solute's activity is approximately equal to its concentration, and so you can get away with saying $\mathrm{pH} = -\log_{10} \ce{[H_{solvated}^{+}]}$.
The "normal" aqueous $\mathrm{pH}$ scale that goes from 0-14 delimits the region where both $\ce{[H^{+}]}$ and $\ce{[OH^{-}]}$ are lower than $\pu{1 mol L^{-1}}$, which is about the upper limit where concentrations and activities of solutes in solution are approximately the same and they can be used interchangeably without introducing too much error.
(The reason this is true is because a solute in a very dilute solution behaves as if it surrounded by an infinite amount of solvent; there are so few solute species that they interact very little with each other on average. This view breaks down at high concentrations because now the solute species aren't far enough apart on average, and so the intermolecular interactions aren't quite the same.)
Not only is it rarer to talk about pH for very highly concentrated solutions, because activity has a much more subtle definition than concentration, but also because you have to produce quite concentrated solutions to get unusual values of pH. For example, very approximately, to get a litre of an aqueous solution of $\mathrm{pH} = -1$ using hydrogen perchlorate ($\ce{HClO4}$, a very strong monoprotic acid) you would need to dissolve around $\pu{1 kg}$ of $\ce{HClO4}$ in a few hundred $\pu{mL}$ of water. A litre of $\mathrm{pH} = 15$ aqueous solution using potassium hydroxide ($\ce{KOH}$) would also require several hundred grams of the hydroxide in about as much water. It's so much acid/base in a small amount of solvent that many acids/bases can't even dissolve well enough to reach the required concentrations.
There is also another way to come across unusual values of pH. We almost always talk about acids in bases in water, but they also exist in other media. Notice that in the definition of pH, no direct reference is made to water. It just happens that, for water, we have:
$$\ce{H2O (l) <=> H^+ (aq) + OH^{-} (aq)} \\ K^{\pu{25^\circ C}}_{\text{autodissociation}}=k^{\pu{25^\circ C}}_\mathrm{w}=a_{\ce{H^+(aq)}} \times a_{\ce{OH^- (aq)} } \simeq \ce{[H^{+}]} \times \ce{[OH^{-}]}
 = 1.01\times 10^{-14}$$
$$\mathrm{pH}+\mathrm{pOH}=-\log\ k_\mathrm{w} \simeq 14 \ (25^\circ C)$$
However, for liquid ammonia, one has:
$$\ce{NH3{(l)} <=> H^+{(am)} + NH2^{-}{(am)}}
\\
K^{\pu{-50^\circ C}}_{\text{autodissociation}}=k^{\pu{-50^\circ C}}_\mathrm{am}=a_{\ce{H^+(am)}} \times a_{\ce{NH2^- (am)} } \simeq \ce{[H^{+}]} \times \ce{[NH2^{-}]} = 10^{-33}$$
$$\mathrm{pH}+\mathrm{p}\ce{NH2}=-\log\ k_{\mathrm{am}} \simeq 33 \ (-\pu{50^\circ C})$$
($\ce{(am)}$ stands for a substance solvated by ammonia). Hence, in liquid ammonia at $-50^\circ C$, the pH scale can easily go all the way from 0 to 33 (of course, it can go a little lower and higher still, but again now activities become important), and that neutral pH is actually 16.5.
For pure liquid hydrogen sulfate (some difficulties arise as $\ce{H2SO4}$ is a diprotic acid and tends to decompose itself when pure, but putting that aside and looking only at the first dissociation):
$$\ce{H2SO4{(l)} <=> H^+{(hs)} + HSO4^{-}{(hs)}}\\
K^{\pu{25^\circ C}}_{\text{autodissociation}}=k^{\pu{25^\circ C}}_\mathrm{hs}=a_{\ce{H^+(hs)}} \times a_{\ce{HSO4^- (hs)} } \simeq \ce{[H^{+}]} \times \ce{[HSO4-]} \simeq 10^{-3}$$
$$\mathrm{pH}+\mathrm{p}\ce{HSO4}=-\log\ k_{\mathrm{hs}} \simeq 3 \ (\pu{25^\circ C})$$
($\ce{(hs)}$ stands for a substance solvated by hydrogen sulfate). Thus, the pH in liquid hydrogen sulfate has a much smaller range than in water and ammonia, not going much further than interval from 0 to 3.
Notice also that not only does the range of $\mathrm{pH}$s change in each solvent, but there is no direct relationship between the values of pH between different solvents; formic acid in ammonia would behave as a strong acid and as such a $\pu{1 M}$ solution in ammonia would have a pH close to 0, but formic acid is actually a base in liquid hydrogen sulfate, and as such a $\pu{1 M}$ solution would have a pH above 1.5. An approximate comparison between the pH ranges and their relative positions in different solvents can be found in the figure below, from A Unified pH Scale for All Phases.

WOW! Never thought about pH in this way. This really enlightens me! I always thought pH is simply the acidity of something, but apparently acidity is determined by other factors and pH itself is not always good. I especially like the example you gave with pH and pNH2, never knew that was even possible!

@user209347 One of the best easy ways I've seen to define acidity is to calculate how willing a substance $A$ is to accept $H^+$. The less willing it is, the stronger an acid the substance $HA^+$ is. For example, $NH_3$ can somewhat easily accept $H^+$ forming $NH^+_4$, so $NH^+_4$ isn't a very strong acid. Meanwhile, $He$ is notoriously unreactive, so it would very weakly accept $H^+$ to form $HeH^+$. This in turn means that $HeH^+$ is a fantastically strong acid, around $10^{50}$ times stronger than sulphuric acid.

Using the same concept, a strong base can be determined by how much energy would be released when a substance absorbs $H^+$ (the more energy released, the more willing it is). A common example of a superbase is the methide anion, $CH^-_3$ (in other words, methane ($CH_4$) is an exceptionally weak acid). There are other ways to analyse acidity and basicity via the Hammett acidity function and others, though in this case it's more difficult because again activity has to be used instead of concentrations.

A small errata: $HeH^+$ is approximately $10^{60}$ times stronger than sulphuric acid. That's a trillion trillion trillion trillion trillion times stronger! More information can be found here.

In short, it is possible to have pH outside of this range.  However, in such cases, pH fails to be a useful or even meaningful measure. In concentrated solutions, one should use concentration directly.
From a practical point of view, pH helps to bring extremely small H+ concentrations to a more familiar range of numbers, but as the concentrations become larger this practice loses its value and becomes counterproductive instead.
From a theoretical point of view, pH as an acidity measure is based on the assumption that water does not have a varying concentration.  However in concentrated solutions, because of the lowering concentration of H2O itself you may find H+ concentration decrease as more acid is added.
Further complicating this, in water H+ is a simplified notation for H3O+ as H+ by itself actually ever exist in any solution, in concentrated solutions, it becomes ambiguous what H+ may refer.
For example, the notation "98%(w/w) sulfuric acid" would be much more practically useful than "a sulfuric acid with a pH of -2.25".  Not to mention there is no such thing as a "H+" concentration in that situation because it is not clear which one of the many species (H3O+, H3SO4+, HSO3+,...) would be referred to as H+, and anyway you choose that H+ concentration will most likely turn out to be even lower than a 50% sulfuric acid solution.

I imagine that all solvated protons contribute to the acidity of a mixture, rather than picking one form as representative for a pH calculation. Of course, figuring out the exact quantities of each species and their activity coefficients must be a demanding task.

@NicolauSakerNeto Yes, summation over all solvated proton species would make the most sense. And it would be demanding indeed! I imagine one particular difficulty is that theories for predicting activity coefficients only work well toward high dilutions, and even those corrected for higher concentration will usually only treat binary mixtures of a single salt(or acid) and water, and also fail for some binary systems.

From its definition:
pH = -log [H+]
If [H+] is greater than 1, the pH will be negative. For example, HCl at 40% minimum.
For a pH greater than 14 are necessary highly concentrated basic solutions.

So it is possible? Could you elaborate more?


text:
How to store chloroform?
I'm going to be aliquoting out ethanol-stabilized chloroform into smaller containers. It's presently in a large bottle sealed with a rubber(y) sheet so I can only draw it out with a needle.
My questions are:

How do I seal them?
If I'm aliquoting to minimize exposure of the full bottle to air, can I get rid of the rubber sheet?
Are there any restrictions on the sort of cap I can use on the aliquot bottle?

What sorts of impurities are you worried about? If you're worried about chloroform oxidation products vs dissolved bits of seal, you'd have to use different methods.

A septum sealed chloroform bottle is likely to have been packed under nitrogen to increase shelf life. Unless you are doing this under a dry nitrogen atmosphere, I'm not sure that decanting into smaller containers is likely to improve shelf life.
Chloroform needs to be stored in a cool, dry, well-ventilated place, out of direct light. Your choice of storage container is likely to be influenced by how long you intend to store it for, but an amber glass bottle with a ptfe-lined cap is probably best suited for longer storage. Rigid polyethylene and polypropylene do not have the best chemical resistance for long-term storage, but if they are air-tight containers will be probably be ok only for short term storage.
Be sure to read and understand the MSDS for chloroform fully before doing this. You will find storage recommendations and incompatibilities listed. DO NOT use containers or lids made from or containing aluminium, magnesium and other metals.

I was mainly planning on decanting to make it more convenient to work with while minimizing the risk of large spills. It won't be sitting around for long.


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How does molecular orbital overlap affect energy?
If there is more overlap of the same sign wave function, this apparently leads to lower energy. I don't understand why. My professor said that there would be fewer nodes, and fewer nodes means lower energy. However, why would overlap in general cause the energy to decrease?

Just out of curiosity, why don't you ask your professor? :)

This is a particularly common topic in molecular orbital theory to have trouble with. Hopefully your question attracts a good answer.

user26143, I did ask. After an hour and a half of muddling through these topics during office hours, things only got more confusing than they were to begin with...

How deep can one go in mathematics and/or quantum mechanics in the explanation?

Although the nature of orbital and wave functions can be puzzling, by the end of the day chemical bonds are hold together by the familiar electrostatic forces.  Although the electrons don't have a well-defined position, the forces felt by the nuclei is dictated by the spacial probability distribution of the electron.
To form a covalent bond, electron density has to accumulate between the two atoms.   The energy is lowered because the electron is close to both nuclei.   In a chemical bond such electron-nuclear attraction cancels out the repulsion between nuclei and holds the atoms together.
Orbital overlap changes electron distributions.   Orbital with different signs cancel each other in and lower the relative probability electron appear between the bonding atoms.  In the most extreme case of nodal plane, electron density is completely annihilated.  Because the total number of electron is unchanged during interaction, the lowering of electron density between nuclei is accompanied by an increase of density to the far side.
In other words, when you have orbital with canceling phases interacting, electron density will move away from the bonding area, resulting in higher energy.
On the other hand, a strengthening interaction increase the electron density between nuclei, thus lowering the energy and creating a pulling force between nuclei.

This helps so much, thank you!! I could understand this part separately, but it makes a lot more sense now when you put it all together!

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


text:
Stability of branched alkenes
I'm having difficulty understand the relative stabilities of the compounds

2,4-dimethyl-3-methylidenepentane < (3Z)-2,5-dimethylhex-3-ene < (3E)-2,5-dimethylhex-3-ene

which, for simplicity’s sake, can be described as

1,1-diisopropylethene < (Z)-1,2-diisopropylethene < (E)-1,2-diisopropylethene

The compounds are listed in order of increasing stability.
Also, what does the $\ce{C=C-C}$ bond angle have do to with stability?

The Z isomer is destabilized by comparison to the E isomer due to steric strain. This is a consequence of the isopropyl groups being on the same side of the double bond in the Z isomer, resulting in electron-electron repulsion between the groups. This also, at least partially, accounts for the increase in the bond angle that should be observed in the Z isomer: increasing that angle also increases the distance between the isopropyl groups, minimizing the mutual repulsion between them (admittedly, this is at the expense of the other angles, but the bulkiness of the isopropyls by comparison to hydrogens necessitates that distortion from idealized geometry). A similar effect occurs with the geminal isopropyl groups in the 1,1- isomer.
Additionally, the degree of substitution of an alkene has a dramatic effect on stability. The presence of alkyl groups on the carbons of an alkene is generally stabilizing (as can be empirically confirmed by examining thermodynamic data, i.e., heats of hydrogenation and heats of combustion). Evidently, calculations grounded in molecular orbital theory indicate that the mixing of $\sigma$ type molecular orbitals of alkanes with the $\pi$ orbitals of less-substituted alkenes is typically energy-lowering overall, and that the $\pi^{*}$ antibonding orbitals are sufficiently close in energy to the $\sigma$ orbitals of alkanes that hyperconjugative stabilization occurs (albeit relatively small in magnitude, since the energies are not very close). In the 1,1- isomer, only one carbon of the alkene can be stabilized by hyperconjugation, while both carbons should be able to benefit from it in the 1,2- isomers. (Speculatively, I would also think that the extent of orbital overlap should be poorer in the 1,1- isomer, since the geminal isopropyl groups would likely be staggered with respect to one another, which might prevent proper alignment of the alkyl group orbitals with those of the alkene $\pi$ system.)


text:
Molding/shaping sodium polyacrylate?
I've been researching sodium polyacrylate, which normally comes as those "aqua gems" for plants, or fake snow. Specifically, I've been hoping to find information on molding, casting, or otherwise giving shape to the material - nothing terribly complex, just things like simple stars for example (while hoping no one mistakes them for Jell-o jigglers).
So far, I've only been able to determine that sodium polyacrylate is not a thermoplastic, so the obvious solution won't work. I heard from one source that those toys that grow in water may be made of the same stuff combined with another polymer, but couldn't get any further leads on the subject.
So the question is, what solutions are available for giving sodium polyacrylate a defined shape, while keeping most of its absorption/growth properties?


text:
Can these changes be reversed by changing temperature?

An egg turning hard when boiled.
A spoon full of batter cooking on a hot griddle.

I said that they both cannot be reversed, because I think if you throw both the egg and batter into the freezer, it will freeze, but it will not be in its original form before you cooked both of them. Then again, I guess I could try this for real on my own but I don't know if that would be a good idea...

Both processes involved the denaturation of proteins and complex reactions so is impossible reverse them only with a change of temperature.

Boiling an Egg
Although in most case denaturation is not reversible there are some cases where it is.
However in the case of Egg white denaturation is irreversible (see here) and so the whole process is by the way irreversible.
Cooking batter
The same for the batter even if is unclear which flours are used, and the composition of the batter  you can be sure that after cooking batter on a hot griddle surely some irreversible reactions occur: denaturation, Maillard reaction, Amadori rearrangment.

When you boil an egg, you denature its proteins, changing their structure. It can't be reversed.
Once the butter is melted by heat, you can solidify back by freezing.

I am not sure for part 1. I guess reassociation (renaturation) is quite common both in proteins and nucleic acids.

Renaturation could be done on certain proteins (I'm not quite sure about egg proteins), under some conditions and if denaturation has not produce drastically changes.

The question refers to bAtter, not bUtter

In the case of 'batter', I think that after loosing a great part of the water of the mixture, the flour become 'toasted' and this involved chemical processes, not physical, so is mostly irreversible.


text:
What conditions to use to crystallise Equine Myoglobin?
What are the crystallisation conditions for Equine Myoglobin?
I've tried repeating some from the PDB but no success.

Does the 1958 Nature paper by Kendrew include any useful details? (Sorry, I'm behind a paywall right now).


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Pair of analogous periods on the periodic table
On the periodic table, period 2 and 3, 4 and 5, 6 and 7 and so on have similar blocks, identical length and groups. What are such pairs of analogous periods called?

This is an interesting question. I am not sure these pairs of periods have any special names, but I would love to learn otherwise.

I agree with @BenNorris, the periods as such are referred to usually by number, the 2nd period is known among theoretical chemists as "light elements", but groups of periods are of no importance, afaik.

In a bit of a web search, there does not seem to be any particular names for families of periods in the Periodic Table.  However, according to WebElements,

The f-block elements are assigned to Periods 6 (lanthanoids) and 7 (actinoids) since that is where they are located in the full, or extended, version of the periodic table.

However, these, according to the image below, only apply to the Lanthanides and Actinides, not all of periods 6 and 7.

Image source


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Where to find Atmospheric data for various cities
Hi im searching for atmospheric data on the following:
Air quality, Dust Particle Concentration, Pollution, Average Temperature etc

for various cities such as Frankfurt, London, Hong Kong, Bangkok...
Are there any websites that can give me this data (or similar atmospheric measurements) for each city?

I'm not sure if any single resource exists for these data, but NASA maintains a list of websites that host atmospheric and climate data.

this would be more on topic on http://area51.stackexchange.com/proposals/36296/geoscience - no sugestion to close, but support Geoscience SE!

I'm voting to close this question as off-topic because this question is not about chemistry

Most cities used to have own pages about pollution data. But there are other sources. For example:
London: http://www.londonair.org.uk/LondonAir/Default.aspx
Bangkok: http://www.pcd.go.th/airquality/bangkok/
If you are looking for a single page, try this:
http://aqicn.org/city/all/


text:
Electrolysis Lab
How do I keep hydrogen or oxygen in a test tube when pulling them out of water? I have a Styrofoam cup with two metal thumbtacks pushed though the bottom, and about 100 ml of distilled water mixed with 6 grams of sodium hydroxide in the cup. The cup is sitting on a 9 volt battery which separates the hydrogen and oxygen in the cup through the thumbtacks.
I put a test tube over the thumbtack that is producing oxygen and one over the thumbtack producing hydrogen. I'm trying to capture the hydrogen and oxygen in the test tubes and light them on fire. My problem is that when I pull out the test tube, the little water left in the tube pulls out all the gas when it falls out. How do I keep that from happening?
Here's a video of what I'm trying to do.

Close the tubes with a rubber cap before pull them out of the water.

I assume you're always keeping the hydrogen tube facing downwards since hydrogen is much less dense than air. The oxygen tube has to face upwards outside of water since it is slightly denser than air, or it will spill out. A residual column of water inside the tube might cause your gasses to disperse because as the water leaves, atmospheric air enters suddenly and can push the pure gasses out. It should be fine if you wait for the tubes to be completely filled with gas, or if you slowly let air inside the tubes. You can also use your thumb to cap the tubes when necessary.

In a high school experiment we took plastic pipettes, cut the tubes off so it was just the bulb, and marked them at regular intervals. We set up the electrolysis with 2 wires, so oxygen was being made at one and hydrogen at the other. We filled the bulb with the electrolysis solution and placed it over one wire to collect 1 gas, then carefully moved it to the other wire. As the gas was collected, it displaced the liquid. We then carefully placed the bulb with the gas on exposed wires from a kitchen lighter and ignited it to blow it across the lab.

The cheapest and simplest modification to the procedure in the video is to use plastic test tubes with threaded caps. When a tube is full of gas, tighten the cap just short of making a seal. Then squeeze the tube to force out any water at the bottom and tighten the cap.
This will work best if you use a cap of the type having a plastic cone on the interior of the cap, rather than just a flat cap. One down side to this method is that there will be a slight negative pressure in the tube with respect to the ambient air, so when you open it a small amount of air will be sucked in. And remember that when you do remove the caps, hydrogen is much less dense than air and will want to rise, so keep the tube inverted until you do whatever you plan to do with it. Likewise, oxygen is slightly less dense than air, so keep it pointing upwards when uncapped.
Remember that you are working with a pretty strongly caustic solution here; wear gloves and goggles.


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Perfluorobutanesulfonic acid (PFBS) Safety and Toxicity
I was wondering if anyone had any experience with PFBS or PFOS (perflurooctane sulfonic acid)?
I am thinking about trying to purchase some of these fluorosurfactants in bulk in order to (hopefully) use as a solvent.  However, I have read that PFOS is now banned due to the bad effects on people's health and the environment.  However, they claim that PFBS is a little better (half like of ~1 month in humans compared to ~5.5 years).
So, I was wondering how terrible it would be to be dealing with the pure liquid on a daily basis, with the use of some gloves and a hood. The hood isn't great in terms of suction, but it's at least something. I should also add that the hood is of course open to the air, so there's moisture in the air.
I do deal with chlorosulfonic acid in the hood pretty regularly, and it isn't a huge problem, although I'm not aware of HCl getting trapped in your system for extended periods of time either.
I also hope to be contacting these liquids with chlorosulfonic acid, where I hope that no reactions will occur, but I am not so great at chemistry so I don't know if that is a fair assumption.
I also noticed that they state in the Wikipedia article for PFOS and PFBS that these substances are used in some consumer items.  So, if they are used in consumer items without degrading, how dilute do they have to be to not have strong acute or mild long term health effects?
I figure they would have to be pretty bad in order for 3M to ban PFOS, but I don't know if that was mainly in order to stop the mass use of them in consumer items.
The MSDS for PFBS states that the LD50 for mice is 430 mg/kg, and it shouldn't be contacted with any moisture in air at all  While the Chlorosulfonic acid MSDS states some similar, but more tame, statements about moisture, and it's LD50 in mice is 50mg/kg.
So, I guess my main question is, has anyone used this substance in it's pure form in a normal hood type of environment?

"has anyone used this substance in it's pure form in a normal hood type of environment?" - Probably. "has anyone used this substance in it's pure form in a normal hood type of environment without long term adverse health impacts?"  - I think this is the question you are really asking. Please note that LD50 data is really only relevant for when all of your risk control measures have failed, which may be likely if your hood "isn't great in terms of suction, but it's at least something".

Thanks for the help.  I should have been more clear about the first question; obviously I know someone has handled this stuff, but I was just seeing if anyone on this site is familiar with the real use of the material.  For example, the technicians at my work deal with HF pretty regularly and they can speak about the real dangers of it, or simply say "it's not too bad if you don't get it on you" because they have more experience.
I'm just trying to figure out if I'm going to be in danger by opening it or if I have to spill it on me, etc.


text:
Why are wave functions orthogonal?
It seems that all wave functions studied in physical chemistry are orthogonal (e.g. particle in a box, hydrogen atomic orbitals). Does this come about because we purposefully make them orthogonal, or are they derived that way naturally? Can there be useful wave functions that are not orthogonal?

They are eigenfunctions of a hermitian operator.

In general, orthogonal wavefunctions are much easier to treat. In some cases they appear naturally, but usually, the orthogonality is imposed as a constrain while constructing the wavefunction.
For example, if you construct electronic wavefunction in the atomic orbital basis, you try to construct the orthogonal basis. This guarantees that the AOs are linearly independent. (Implication, not equivalence). Would you fail to fulfill this, the solution might still be possible, but much more difficult.
If you manage to solve the eigenvalue - eigenvector problem, the solutions are by definition orthogonal to each other. This is the case for the examples you provided.

When AOs are linearly independent, does this mean that we are ignoring any interaction between the AOs? How does this relate to Huckel MO theory when neighboring AOs' interactions are taken into account?

Sorry, I messed some things together. The atomic orbitals on one given atom are orthogonal (have zero overlap). Orbitals on two different atoms usually do have non-zero overlap (can be zero due to symmetry).

Eigenvectors are not orthogonal "by definition". It's just that two eigenvectors of a Hermitian operator $A$, to different eigenvalues, are necessarily orthogonal (else e.g. decomposing in the middle of $A^2$ would lead to contradiction). They're however typically normalised by definition, thus orthonormal.

One should be more clear about if the topic is orthonormal BASISes or WAVEFUNCTIONs. The answer seems to talk about the first, the question  about the second.

I will try to add another point to the discussion which is statistics. Quantum mechanics is a non deterministic theory which can only make assumptions about expectation values and probabilities of obtaining a special value. In general your wave functions or vectors correspond to states while the projection of one wave function onto another (aka the scalar product of both functions where the explicit version depends on the norm of the vector space) gives you probability amplitudes and its square a probability. Now probabilities are between 0 and 1 - the same goes for its square at the extreme points 0 and 1. Orthogonality just means in that sense that for a chosen eigenbasis your probability of switching the basis vector by measurement is zero. The normalization occurs as the probability of staying in that eigenstate upon measuring the same operator again must be one (as the probability of reaching another basis vector is zero) but it cannot be greater than one - this is why you always use properly normalized wave functions. To be precise you normalize every time that you are working with a mapping between scalar products, projections and states - you could also see parallels between the time correlation function and a scalar product in statistical mechanics.


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When and how in anaerobic digestion is water dissociated?
Anaerobic digestion and methanisation of organic feedstocks is a complex series of processes that result in biogas. One way to look at this is a very top-down, stochiometric approach as in the Buswell equation
$$\ce{C_aH_bO_c + \left($a - \frac{b}{4} - \frac{c}{2}$\right)H2O -> \left($\frac{a}{2} + \frac{b}{8} - \frac{c}{4}$\right)CH4 + \left($\frac{a}{2} - \frac{b}{8} + \frac{c}{4}$\right)CO2}$$
There's also a Buswell-Boyle equation that takes nitrogen and sulfur into account, but right now I just want to understand one aspect.
During the process, at some point, water is dissociated. When does this happen (hydrolysis, acidification, ...), what role does the water play and what are sample reactions that illustrate this?

This is not a trivial question to answer, as many diagrams are often simplified and do not state the full chemical reactions that occur, such as

from Aerobic and Anaerobic Digestion and
Types of Decomposition education webpage from the Mountain Empire Community College.
However, according to this image (below) from the Ecocorp page - Metabolic Chain of the Anaerobic Process for the Utilization of Organic Wastes, water is a reactant in the acetogenesis
 stage and is removed as a product in the methanogenesis stage.
(Sample reactions are included in the image)

(Just in case the image becomes a dead link), the chemical processes involving water from the diagram are:
acetogenesis
$$\ce{CH3CH2COOH + 2H2O -> CH3COOH + CO2 + 3H2}$$
methanogensis
$$\ce{4H2 + CO2 -> CH4 + 2H2O}$$

$CH_3CH_2COOH$ is Propanoic acid? If you look at buswell boyle, water is only needed as an educt (for the whole methanization chain) for proteins and fats. So propanoic acid would only be created as a product of proteins and fats?

Oh, and the answer is great, I think! I'll wait a bit with the accept, see what the bounty brings, but this looks good!

scratch my first question, I found the answer on wikipedia.


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Why is Charcoal such an excellent adsorbent?
Charcoal (also activated charcoal) is known to adsorb a huge variety of substances including a variety of paints, dyes and many different kinds of ions. Moreover, the amount of adsorption at normal pressure and temperature is high. I know physisorption requires high capacity of Van-der-waal bonding with the adsorbate which is favoured by surface atoms with large, easily-dispersed electron clouds and chemisorption requires actual chemical bond formation which requires highly reactive surface atoms. Charged species get adsorbed to form a double layer when the surface is charged, either due to ionization or due to polarity.

But charcoal possesses none of the structural specialities required to
  be a great adsorbent. Neither it has charged centres (unlike colloidal
  proteins, or metal sols) nor it has a very high reactivity (Charcoal
  is almost the final product of combustion of wood, bone, sugar etc and
  is supposed to be immune to further oxidation even at considerable
  temperature). Why is it then an excellent adsorbent?

I see three reasons:

Activated carbon is commonly used to adsorb organics, that should bind well on coal.
For many purposes, activated carbon is treated with potassium or iodine to provide ions for charged centers.
Charring organic stuff is a cheap and easy way to create large surfaces for adsorption.

I am not asking "Why is charcoal used as an adsorbent so often?" but rather "What property of charcoal surface makes it a great adsorbent for most substances?".

i believe what he meant is that charcoal has an amazingly big surface, and once activated it gain the property you are looking for on its large surface. and also its used because quite cheap.

Activated charcoal is a great adsorbent because of it's huge surface area. While it doesn't bind very many ions/atoms/molecules per surface area (which is the characteristic of a 'good' adsorbent), due to very big surface area per unit of mass it can adsorb a lot of particles. Actually, process of 'activating' charcoal is designed to maximize the surface area to mass ratio.

Charcoal isn't a particularly good adsorbent even though it is chemically very similar to activated carbon.
Activated carbon is usually made by more specialised processes that guarantee the final product will have a very large surface area (often >1,000m2/g). Manufacture usually involves pyrolysation with hot gasses, but many forms are also further activated by partially oxidising the surface (using gas containing oxygen) or bay adding chemicals such as phosphoric acid or potassium hydroxide before pyrolysis. These activation reactions create some functional groups on the surface that can enhance specific types of adsorption on the surface.
But the primary feature that matters is the high surface area, which normal methods for making charcoal do not achieve.

Charcoal is a good absorbent because its quite selective and cheap. Its high surface area means nothing if it doesn't bond to the components that you want absorbed and not a lot else- witness a household sponge which has high surface area and mercury- the mercury just slides off.  Charcoal is used in the gold mining/processing industry. The gold(and other metals) is oxidised to form cyanide complexes. The charcoal mostly only absorbs the cyanide complexes not all the other rubbish and water. Charcoal wouldn't be much good for the water purification industry if it absorbed water. Charcoal absorbation processes are a multitudinous- its thousands of papers of primary metallurgy(gold).

can cyanides be replaced in gold mining?

It is not quire mining rather leaching which separately or together and yes there are processes to do without the cyanide.

Coal is more effective and efficient in adsorbing fumes due to several aspects:

We can increase the surface area so it can adsorb higher volume of fumes according to the low of mass and volume; we can use that to develop the efficiency of coal to adsorb a large amount of adsorbent matter
Low cost
Low transportation
Doesn't react with fumes


text:
The entropy change during micellization
I was reviewing the thermodynamics of micelle formation from the Wikipedia page and this book. I understood most of the arguments but micellization is supposed to be an entropy driven process with a positive entropy change in the process of micellization. But isn't this a bit counter-intuitive?

In the unassociated form, the surfactant molecules are free to roam
  about in the solution much like normal solutes. But in the form of a
  micelle, they get associated, grouped into an associated whole, which
  reduces the freedom of independent movement and should therefore
  decrease the entropy (randomness) of the system? The entropy change
  should be negative during micellization. Why is it positive?

I understood the mathematical arguments leading to the develpment of the approximate formula $\Delta G=RT\ln(\text{CMC})$.

Where are you quoting that part from? It doesn't seem to be quoted...

The micellization releases water molecules which initially solvate the hydrophilic head groups. The release of these waters greatly increases the entropy.
Reference: Dong et al., Chem. Rev. 2010, 110, 4978-5022.

This is due to release of solvent molecules attached with non polar tails of surfactant monomer by hydration during self organistaion.
The entropy gain by this process exceeds the entropy loss due to amphiphillic association or solvation making the overall entropy change positive.

In order to accommodate the hydrophobic tail of the free surfactant, the hydrogen-bonded structure of bulk water must be distorted. A nonpolar hydrocarbon cannot form hydrogen bonds with water and thus water molecules in the immediate vicinity of a hydrocarbon are constrained in the possible orientations that it can assume.
Therefore, water molecules around a hydrophobic chain are more ordered which is unfavourable as it results in a decrease in conformational entropy. When all the hydrocarbon chains are hidden inside micelles, the structure of bulk water is largely restored leading to an increase in the conformational entropy. Thus, micellization is an entropy-driven process that seeks to minimize the free energy of a system by minimizing the interface between hydrophobic molecules and water. The increase in the conformational entropy of water more than compensates for the negative aggregation entropy of the surfactants themselves (surfactants are becoming more ordered on micelle formation).

When micelles have been formed why they don't aggregate further? For example small solid particles will eventually aggregate in water and settle out. Are micelles kinetically stable?


text:
How will these hydrogen atoms interact?
Assume that we have a space with just two $\ce{H}$ atoms and their distance to each other is $d$. Let's say they don't have initial velocity. What is the force with which they will act on each other?
This question might have some mistakes in itself, but the main thing I'm trying to learn is how an atom of element $X$ interact with an atom of element $Y$.

Something else besides F = - dE / dr ?

@ssavec: Actually, I'm not familiar with that formula. Can you give a link where I can find somethings about it?

The force is determined by the gradient of the potential energy. See e.g. (http://en.wikipedia.org/wiki/Scalar_potential)
That means, the system tends to adopt the configuration with lowest possible energy.

If you just have two H atoms, nothing else then the only forces acting are gravity and interatomic forces, in particular, the van-der-waals force (note slide 5 which says the van der waals potential is applicable to monatomic gases) . The van der walls force varies and the inverse sixth power, so it will be zero until the hydrogen atoms are quite close, so weak ol' gravity will be doing most of the work until then.
So, you should set up your potential around the center of mass of the system with U including the gravitational and van-der-waals potentials. For all but the closest distances, be prepared for very slow dynamics.
If you don't know how to use potentials, then take the negative of the derivative of the potential for the van der waals and then add in the force of gravity, both acting in the same direction.

The van der waals energy is inverse sixth power. The force, as a derivative of energy, is inverse seventh power.

@user26143  my mistake, yes, I was looking at the potential energy function when I was writing this...the force is inverse seventh.

How one atom interact with another atom is an difficult question and requires quite some calculations, even for a system with only two hydrogen atoms.  Basically, this is done by solving the Schrödinger equation for the system.  The answer will also depend on the electronics state of the system, especially the spin state.
If the system is in a triplet electronic state, the two atom will experience a repulsive force that approach infinity as you pull the two atoms closer, and that force approaches zero as the distance approaches infinity.
If the system is in a singlet state, the force is zero at the equilibrium bond length of $H_2$ atom(~0.74Å).  As the distance gets smaller the atoms experience an increasing repulsive force which approaches infinity at 0.  As the two atoms are pulled apart from the equilibrium there is a decreasing attractive force that approaches 0 at infinity.


text:
Safety concerns for radical halogenations of methane in sunlight?
I wanted to know about the substitution reaction of alkanes particularly interested in the reaction of methane with fluorine and chlorine.

Please use caution and discuss this with someone who is knowledgeable in this area before undertaking this experiment on your own.

Let's have a look at some fundamental data and the chemistry involved.

Chlorine is an inflammable gas with a boiling point of -35 °C and a nice yellowish-green colour. It is heavier than air.

It is highly irritant and toxic. It is a strong oxidant. With water, it partly decomposes to hydrogen chloride and hypochlorous acid. Your mucous membranes are wet. Inhalation of chlorine results in severe corrosive injuries. Inhale a tad longer to face dyspnoea, coughing of blood and asphyxia. Did I mention the pulmonary oedema and fatality?

Methane is a flammable gas with a boiling point around -160 °C and a flash point around -180 °C. The lower explosion limit of methane in air is around 5%.

Open flames and sparks are not your friend here.

For the mixture of methane and chlorine, red light districts are safe.
The bond dissociation energy of chlorine is $243\, \mathrm{kJ \cdot mol^{-1}}$, you will need blue light ($\lambda$ < 490 nm) to initiate the self-propagating reaction:
$$\ce{Cl2 ->[h\nu] 2 Cl*}$$
Then the chain reaction starts:
$$\ce{Cl* +\  CH4 -> HCl + CH3*}$$
$$\ce{CH3* +\  Cl2 -> CH3Cl + Cl*}$$
All that goes on and on; termination steps are radical recombinations:
$$\ce{2 Cl*  -> Cl2}$$
$$\ce{2 CH3*  ->  C2H6}$$
$$\ce{ CH3* +Cl* ->  CH3Cl}$$
Apparently hydrogen chloride gas is is produced in the reaction, together with methyl chloride.

Methyl chloride is a flammable gas (boiling point around -24 °C, flash point around -20 °C, lower explosion limit in air around 8%).
GHS hazard statements file it with H-351: suspected of causing cancer.

If you perform the reaction with a large excess of chlorine, $\ce{Cl*}$ radicals will attack methyl chloride, generate dichloromethane via the chloromethyl radical $\ce{*CH2Cl}$, etc. Tetrachloromethane ($\ce{CCl4}$) would be the final product then.

That was the healthy and boring part!
When fluorine talks to chlorine, it goes like in the musical "Annie, Get Your Gun":
Anything you can do I can do better!
Particularly when it comes to running havoc on your tissue!
As far as the photolysis of fluorine in the presence of methane is concerned: People have done that!
Typical conditions were:

Deposit methane in a fluorine matrix at 15 K
Co-deposit methane/argon and fluorine/argon samples at 12 K
Trap fluorine in a methane matrix at 15 K

Yes, these guys had no sense for adventure - but they weren't blown away either!


text:
What are the properties of various solutions at pH <2 that may cause injuries?
For example, why do some solutions not cause any injuries at pH 2 while HCl does?

Injury is not solely related to how strongly acidic a solution is, as there are several mechanisms by which organic matter is damaged from exposure to acids, not just protonation. It is also highly dependent on the type and length of exposure. Aqueous $\ce{HF}$ is likely considerably corrosive after skin contact even at pH 2, mostly because of $\ce{F^-}$ which is a rather reactive ion with a strong tendency to abstract $\ce{Ca^{2+}}$ (in fact, the fact that  $\ce{HF}$ is a weak acid makes it even more dangerous because the undissociated $\ce{HF}$ is much less polar and the molecules can cross the skin barrier more easily, allowing direct access to muscle and bone which can be severely damaged). Aqueous $\ce{HCN}$ also forms weak acids, but again $\ce{CN^-}$ ions are extremely toxic as they block cell respiration. The risk is compounded by the fact that $\ce{HCN}$ is a very volatile substance and can be released as a gas from concentrated or acidic solutions.
With regards to $\ce{HCl}$, you need to be a bit more specific. What kind of injury? $\ce{HCl}$ itself is a relatively safe acid in dilute solution, even though it's a strong acid, as it does not have strongly oxidising/reducing/dehydrating/toxic properties and is only volatile at rather high concentrations. Remember that $\ce{HCl}$ is even used as an integral part of our digestive system, being the species that allows stomach acids to reach pH 1-2.


text:
Shortcut for calculating pH of mixtures
My teacher gave me a shortcut to calculate the resultant pH if 2 or 3 solutions are mixed, which are relevant only if the difference between the pH of individual solutions is 1.

If there are two solutions, the pH of the mixture $\approx$ (the mean pH − 0.24).
If there are three solutions, the pH of the mixture $\approx$ (the mean pH + 0.56).

What is the logic behind this approximation?

Case 1: Mixing equal volumes of two solutions with a pH difference of 1.
Initially we have:
\begin{array}{|c|c|c|}
\hline
&[\ce{H+}]& \mathrm{pH} &n_\ce{H+}\\ \hline
a& x & \log x & xV\\
b&\frac{x}{10} & -\log x + 1 & \frac{xV}{10}\\
\hline
\end{array}
\begin{align}
\overline{\mathrm{pH}}=\frac{\mathrm{pH}_a+\mathrm{pH}_b}{2}=\frac{-\log x-\log x+1}{2}=-\log x+0.5
\end{align}
After complete mixing, we obtain:
$$
({n_{\ce{H+}}})_\mathrm{mix}=({n_{\ce{H+}}})_{a}+({n_{\ce{H+}}})_{b}=xV+\frac{xV}{10}=\frac{11xV}{10}\\
[\ce{H+}]_\mathrm{mix}=\frac{1.1xV}{2V}=\frac{11x}{20}\\
\mathrm{pH}_\mathrm{mix}=-\log \frac{11x}{20}=-\log x - \log\frac{11}{20}=\overline{\mathrm{pH}}-0.5+0.2596=\overline{\mathrm{pH}}-0.2404
$$
Case 2: Mixing equal volumes of three solutions with consecutive pH differences of 1.
Before mixing:
\begin{array}{|c|c|c|}
\hline
&[\ce{H+}]& \mathrm{pH} &n_\ce{H+}\\ \hline
a& x & \log x & xV\\
b&\frac{x}{10} & -\log x + 1 & \frac{xV}{10}\\
c&\frac{x}{100} & -\log x + 2 & \frac{xV}{100}\\
\hline
\end{array}
\begin{align}
\overline{\mathrm{pH}}=\frac{\mathrm{pH}_a+\mathrm{pH}_b+\mathrm{pH}_c}{3}=\frac{-\log x-\log x+1-\log x+2}{3}=-\log x+1
\end{align}
After complete mixing, we obtain:
$$
({n_{\ce{H+}}})_\mathrm{mix}=({n_{\ce{H+}}})_{a}+({n_{\ce{H+}}})_{b}+({n_{\ce{H+}}})_{c}=\frac{111xV}{100}\\
[\ce{H+}]_\mathrm{mix}=\frac{1.11xV}{3V}=\frac{111x}{300}\\
\mathrm{pH}_\mathrm{mix}=-\log \frac{111x}{300}=-\log x - \log\frac{111}{300}=\overline{\mathrm{pH}}-1+0.4318=\overline{\mathrm{pH}}-0.5682
$$
(It seems you got the sign wrong on your last equation.)
Caveats:
Technically these calculations only work well if we assume that the activity of the solvated proton is the same as its concentration, which is a poor approximation below $\mathrm{pH}$ of 0 and above $\mathrm{pH}$ of 14.
You also get a slight deviation if you mix solutions that have $\mathrm{pH}$ too close to neutral, because the auto-dissociation of water affects the equilibrium values of $[\ce{H+}]$ and $[\ce{OH-}]$.
For example, if we mix two equal amounts of liquid, one being pure water($\mathrm{pH} = 7$) and the other being aqueous $\ce{HCl}$ solution with $[\ce{HCl}] = \pu{10^{-6}M}$. Considering the auto-dissociation constant for water as $K_\mathrm{w} = 10^{-14}$, the $\mathrm{pH}$ for the last solution is actually approximately $5.9957$ because water auto-dissociation already affects it slightly.
Neglecting that, we mix the liquids creating a solution with $[\ce{HCl}] = \pu{5\times 10^{-7} M}$. Directly using $[\ce{HCl}] = \pu{5\times 10^{-7}M} = [\ce{H+}]$ yields a $\mathrm{pH}$ of $6.3010$.
Using the trick, we get the pH for $[\ce{HCl}] = \pu{5.5\times 10^{-7}M}$ as $6.2596$. However, the real value for $[\ce{H+}]$, taking into account the shifting of the water auto-dissociation equilibrium, is $[\ce{H+}] = \pu{5.19258\times 10^{-7}M}$, such that the real $\mathrm{pH}$ is $6.28461$. A small deviation surely, but it's present.


text:
New technology that has replaced the old radioisotope
What technology has the radioisotope Cobalt 60 replaced in medicine?
E.g.,PET preferred over CT scans

Please see my comment on your other question.  I gave you a bit of latitude with this one, as you have included a few more details, but attempting to answer it on your own will help people know how to direct their assistance.

After searching on Google (now there's an idea)...
If you are asking about what technology was used in medicine before Co-60 therapy was developed:
Co-60 therapy was developed in the early 1950's and was seen as an alternative to other forms of X-ray radiotherapy, or radiation therapy. It is a form of external beam radiation therapy. Radiation sources at the time (mostly derived from radium) required very high energy sources, and proved largely ineffective against deep-set cancers. Co-60 therapy was able to penetrate deeper at lower energies, and therefore was more effective against a range of previously untreatable cancer types. The first Co-60 machine was from the The University of Saskatchewan, and they have an excellent site devoted to the history of Co-60 treatment.
If you asking about what technology is now being used in place of Co-60 therapy:
Conventional X-ray sources were improved with the development the linear accelerator which could produce X-ray energies exceeding 8MV. Nowadays, commercial units can produce photons up to 25MV. Here is a useful link to compare linear accelerator and Co-60 radiation sources for medical applications. There are still many developing areas of use, with some experts claiming there is still very much a place for Co-60 therapy. Co-60 therapy can be applied with high precision with low side-effects and is just one of many approaches available, often used in tandem with other medical treatments.

Measuring photon energies in volts is so strange. I assume it's actually a measure of the field applied to accelerate electrons that smash into targets and produce x-rays, but why not just specify the resulting photon energies in eV?

I'm no expert, but I believe gamma ray beams are expressed as Volts, while electron beams are expressed in eV. A 1MV beam from a linear accelerator will produce photons of 1MeV, at maximum.


text:
What is the actual reason for geometrical isomerism?

My teacher explained it as restricted rotation between the double bonds. But I think there is more to it. After all, why is it considered to be a type of isomerism itself? Whether an alkene is cis or trans only depends on how you make the structure on paper. So what is the basic idea behind it?

What do you mean exactly by "it only depends on how you make the structure on paper"? There are no rotational nor mirror symmetries that can convert a cis isomer into a trans, so they are different compounds. Another way to look at it is that not all the parts of the molecule are in the same position relative to each other in the isomers, so there should naturally be different steric interactions. You can convert one into the other by rotation of the double bond, but that is highly restricted as you need to temporarily break the $\pi$ bond, resulting in a high bond rotation energy barrier.

I've tried to explain my point via paper 53. You see, I get your explanation but the problem arises when you draw it on paper. If you specify cis or trans, I will make the compound. But if you ask me that I make 2-butene and without seeing my structure you ask me is it cis or trans, then it would only depend on how I MAKE the compound. That's the problem I've experienced.

It is very hard to see which problem you have. If I ask you to draw a hat I cannot pretend you draw it in the colour I want.  If I want a red hat I must ask for it.  Else it cab be blu or whatever. It as an analogy  E and Z are different ergo the E Z nomenclature to distinguish them.  Moreover the title does not reflect your problem thus the nice answers cannot solve it . If your teacher ask you to draw but-2-ene is up to you.  Or you draw both as differently from the hat analogy their are just two, it is manageable.

@Alchimista The actual intention behind this question was the fact that sometimes my teacher "obviously" referred to some geometric isomer and I would be lost and be looking for properties of the compound to make myself understand why it was so obvious to him.

Geometric isomerism occurs when two structures with the same connectivity are not interconvertible.
Cis-Trans isomerism is common and easy to recognize kind of geometric isomerism. The carbon-carbon double truly has limited rotation. The double bond in the alkene functional group consists of a $\sigma$-bond located within the plane of the molecule and a $\pi$-bond located perpendicular to the plane of the molecule. In the image below (created by David Pilz for Wikipedia), you can see the $\pi$-bond in a perpendicular plane.

The $\pi$-bond is made from $p$-orbitals, so it cannot have the same symmetry as the $\sigma$-bond. You can see the formation of the $\pi$-bond from $p$-orbitals in this picture by Wikipedia user JoJan. In order for the $\pi$-bond to exist the $p$-orbitals must be aligned.

In order for the molecule to rotate from $cis$ to $trans$, the bond axis would need to rotate so that the $p$-orbitals are no longer aligned and can no longer form the $\pi$-bond.  This slide show from James Condon at Roane State(PDF) explains the same concept. Rotation around the double bond requires enough energy input to break the $\pi$-bond, at least $250 \ \text{kJ/mol}$, which is a lot for room temperature.
The clearest demonstration that geometric isomers are clearly different molecules is to look at some examples with drastically different properties. It is important to remember that the experimental observation that these compounds with very similar structures have very different properties predates the explanation that the double bond has limited rotation.
The two isomers of 1,2-dichloroethene have different physical properties due both the difference in polarity and the difference in shape:

$cis$-1,2-dichloroethane

Dipole moment = $1.9 \ \text{D}$
Boiling point = $60.2 \ ^\circ\text{C}$
Melting point = $-81.47 \ ^\circ\text{C}$
Density = $1.28 \ \text{g/cm}^3$

$trans$-1,2-dichloroethane

Dipole moment = $0.0 \ \text{D}$
Boiling point = $48.5 \ ^\circ\text{C}$
Melting point = $-49.44 \ ^\circ\text{C}$
Density = $1.26 \ \text{g/cm}^3$

However, my favorites are maleic acid and fumaric acid. Notice the extreme difference in water solubility as well as the difference in acidity. These molecules behave very different chemically.

Maleic Acid $cis$

Melting point = $135 \ ^\circ\text{C}$ (decomposes)
Density = $1.59 \ \text{g/cm}^3$
Solubility in water $788 \ \text{g/L}$
Acidity $\text{p}K_{a_1} = 1.9$, $\text{p}K_{a_2} = 6.07$

Fumaric Acid $trans$

Melting point = $287 \ ^\circ\text{C}$ (does not decompose)
Density = $1.635 \ \text{g/cm}^3$
Solubility in water $6.3 \ \text{g/L}$
Acidity $\text{p}K_{a_1} = 3.03$, $\text{p}K_{a_2} = 4.44$

Thanks for explaining geometrical isomers such a nice way @BenNorris but please look at the picture I've attached and the comment I posted below it. Although in real life you can differentiate between them but when you make 2 butene on paper and I ask you what it is, your answer would depend on how you've made your structure. That is the real problem.

Well, since cis and trans 2-butene are different compounds with different properties, you either need to be told which one to draw, be given enough information (for example some reactions are stereospecific and would generate one isomer and not the other based on the reactant), or it does not matter. If you are worried about when it matters and when it does not, ask your instructor about it.

@BenNorris In higher temperatures they should interconvert. So why they have different boiling point and not a common one? I mean suppose that from a given reaction we took only one isomer. Now we isolated this isomer and start heating it up. As we heat it up shouldn't interconversion be possible? So if we can find a temperature that can interconverts these two (reaching equilibrium) then we should expect a mixture with both of them. Then shouldn't this mixture have a common boiling point?

"Geometric isomerism occurs when two structures with the same connectivity are not interconvertible" Does this mean that a pair of enantiomers are also geometrical isomers?

Correct me if I'm wrong; for some reason, the boiling points of all pairs of cis-trans isomers lie below the temperature required for interconversion. If this isn't the correct answer to @Anton's excellent question, could you explain it, Ben?


text:
What does "Anal. Calcd." stand for written out?
I'm wondering what the abbreviation "Anal. Calcd. stands for exactly. It is put in front of calculated (expected) values for elemental analyses. "Analytically calculated" makes no sense for me, but maybe it's more of a language problem for me in this particular case. According to a quick Google research, some are listing it in their lists of abbreviations as "Analytical Calculated". Using the adjective instead of an adverb doesn't make it any more correct to me.

They are two distinct terms, that just happen to frequently be found next to each other. Their definitions can be found in the Journal of Organic Chemistry Guidelines for Authors (see page 29).
Anal. - abbreviation for combustion elemental analysis.
Calcd - abbreviation for calculated.
It is not actually a single phrase, as you've suggested here. Combustion elemental analysis is what people often mean when referring to 'getting an analytical done'. Writing Anal. signifies the type of experimental data that is about to be listed, just as you might specify 1H NMR, IR or MS. A common alternative to combustion elemental analysis is high resolution MS, which is often written in experimental sections as HRMS (ESI) m/z calcd
The format for presenting elemental analysis is journal specific, and therefore varies. The format you're asking about is consistent with the ACS publication recommendations. RSC guidelines accept the the data in a slightly different format.


text:
How do I find the ratio in which the residue of a reaction combines with HCl?

A mixture weighing $4.08\ \mathrm g$ of $\ce{BaO}$ ($M(\ce{Ba})$ = $138\ \mathrm{g\ mol^{-1}}$) and an unknown carbonate $\ce{XCO3}$ was heated strongly. The residue weighed $3.64\ \mathrm g$. This was dissolved in $100\ \mathrm{ml}$ of $1\ \mathrm M$ $\ce{HCl}$. The excess acid required $16\ \mathrm{ml}$ of $2.5\ \mathrm M$ $\ce{NaOH}$ solution for complete neutralization. What is the molar mass of $\ce{X}?$

My approach:
$16\ \mathrm{ml}$ of $2.5\ \mathrm M$ $\ce{NaOH}$ means $0.04\ \mathrm{mol}$ of $\ce{NaOH}$. Thus, the excess acid was $0.04\ \mathrm{mol}$ of $\ce{HCl}$. $100\ \mathrm{ml}$ of $1\ \mathrm M$ $\ce{HCl}$ means $0.1\ \mathrm{mol}$ of $\ce{HCl}$. Thus, $0.06\ \mathrm{mol}$ of $\ce{HCl}$ reacted with the residue. Now, how can I find the ratio in which the residue reacted with $\ce{HCl}?$ What is the composition of the residue?

In the comments I mentioned one way to solve the problem, but it turns out that was the roundabout way! There is a simpler solution, as follows.
From the neutralization reaction, it is possible to calculate that the mixture reacted with $0.06\ \mathrm{mol}$ of $\ce{H+}$. The carbonate and the oxide react in a proportion of 1:2 with the hydrogen ions, so we know that $n_{\ce{BaO}}+n_{\ce{XO}}=0.03\ \mathrm{mol}$. The mass balance from the thermal decomposition then suggests that $m_{\ce{XCO3}}-m_{\ce{XO}}=0.44\ \mathrm g$. These two equations are already enough.
Notice that the for the molar masses of the unknown oxide and carbonate, we have $M_{\ce{XCO3}}-M_{\ce{XO}}=44\ \mathrm{g/mol}$. If one converts the masses in the second equation into expressions in $M$ and $n$, the result is $n_{\ce{XCO3}}M_{\ce{XCO3}}-n_{\ce{XO}}M_{\ce{XO}}=0.44\ \mathrm g$, which can be turned into $n_{\ce{XCO3}}M_{\ce{XCO3}}-n_{\ce{XO}}\left(M_{\ce{XCO3}}-44\ \mathrm{g/mol}\right)=0.44\ \mathrm g$. It is easy to see that the thermal decomposition does not alter the number of moles of compound containing $\ce{X}$, so that $n_{\ce{XCO3}}=n_{\ce{XO}}$. Thus:
$$n_{\ce{XCO3}}M_{\ce{XCO3}}-n_{\ce{XCO3}}\left(M_{\ce{XCO3}}-44\ \mathrm{g/mol}\right)=0.44\ \mathrm g$$
$$n_{\ce{XCO3}}\left(M_{\ce{XCO3}}-M_{\ce{XCO3}}+44\ \mathrm{g/mol}\right)=0.44\ \mathrm g$$
$$n_{\ce{XCO3}}\times 44\ \mathrm{g/mol}=0.44\ \mathrm g\quad\Rightarrow\quad n_{\ce{XCO3}}=0.01\ \mathrm{mol}$$
Now the problem unravels easily. We get that  $n_{\ce{BaO}}=0.02\ \mathrm{mol}$, which implies $m_{\ce{BaO}}=3.08\ \mathrm g$ and $m_{\ce{XCO3}}=1.00\ \mathrm g$. Hence, $M_{\ce{XCO3}}=100\ \mathrm{g/mol}$, and from there $M_{\ce{X}}=40\ \mathrm{g/mol}$. A quick look on a periodic table yields calcium as the most likely candidate, and indeed its carbonate has the formula $\ce{CaCO3}$ and a molar mass of $100\ \mathrm{g/mol}$.


text:
Silver electrorefining
Ok, so let's say I have raw silver which is polluted with a bit of gold, copper and lead.
So through electrorefining I want to make pure silver out of it.
Let's say I have it all in an acidic solution. What happens when I turn on the Voltage?
How would I prevent the other elements from getting electrolyzed?
Because when I have the electrorefining of copper it gets clean even when polluted with silver, so why doesn't the copper get electrolyzed as well?

It depeds on the acid. If it is HCl then you will get Wohlwill Process only the silver will be not disolved. So baisiclly what you want.


text:
Strong acid equilibrium
So a teacher at school told me that strong acids do not create an equilibrium in general, while weak acids do (thats the difference, strong acids namely ionize completely). However, I lately stumbled upon a passage reporting that when experimenting with extreme pH, so extreme concentrations of acids, strange things start to occur and strong acids are not able to ionize completely, resulting in an equilibrium. I would explain this by the extreme concentration -> less water molecules in the vicinity of acid to react -> less chance to ionize.
But then I thought, are strong acids in fact not also creating an equilibrium, but because they react so strongly, they immediately release their proton. The conjugate bases are extremely weak, but they should be able to pick up a proton right? They simply become and acid, and release it immediately again. Please clarify this entire situation for me, thanks already.

To be more strict, a strong acid is defined as one that completely ionize in aquatic solutions. And as always (in chemistry), completely means almost completely.
In a strict sense the strong acid do "have an equilibrium" because any reaction you can write down and balance will have an equilibrium constant, calculable through Nernst equation. However, the equilibrium is inconsequential unless in extreme conditions. Any acid with $\mathrm{p}K_\mathrm{a} < 0$, or equivalently $K_\mathrm{a}>1$ is a strong acid.
Chemistry is a practical science and there is a tendency to use practical instead of strict terms.  So when your teacher say something like "fully", "completely", "do not have", etc. in a chemistry classroom, it implies approximations.
Also remember your teacher's discussions will be all focused on highly diluted solutions.  Even concentrations otherwise the situation becomes extremely difficult to treat and unfit for a textbook example.  Therefore in the context of your textbook, $\mathrm{p}K_\mathrm{a} < 0$ guarantees essentially complete dissociation.

Whether a substance is a strong or weak acid/base depends on the solvent it's in. For example, $\ce{HNO_3}$ is a strong acid in water. This is partly because $\ce{HNO_3}$ has a relatively high tendency to donate a proton (the conjugate base $\ce{NO^{-}_3}$ is quite stable and therefore has a low tendency to accept a proton back), and partly because water has a certain basic character, accepting protons to create $\ce{H3O+}$. However, the same $\ce{HNO_3}$ in pure liquid $\ce{H3CCOOH}$ (also known as glacial acetic acid) is actually a weak acid. This change in character is largely due to the fact that $\ce{H3CCOOH}$ is a much, much weaker base than water, forming the $\ce{H3CCOOH^+_2}$ ion with significantly more reluctance. Even though $\ce{HNO_3}$ can easily lose its proton, $\ce{H3CCOOH}$ is even moreso unwilling to accept it, and thus $\ce{HNO_3}$ becomes a weak acid. An amusing situation arises in pure liquid $\ce{H_2SO_4}$. Adding $\ce{HNO_3}$ there, we find that $\ce{HNO_3}$ actually behaves as a weak base. $\ce{H_2SO_4}$ itself also has a high tendency to donate protons, and of the two possible reaction products from protonation, $\ce{H_2NO^+_3}$ is somewhat more stable than $\ce{H_3SO^+_4}$.
Hence, acidity is dependent on the solvent, not only on the solute. A strong acid such as $\ce{HNO_3}$ in aqueous solution at high dilution is completely dissociated, as we expect of a strong acid. However, if more and more water is removed, then the solution starts to behave differently. Even though $\ce{HNO_3}$ has a strong tendency to protonate water, you can have a solution that contains more moles of $\ce{HNO_3}$ than water. Then it becomes obvious that not all $\ce{HNO_3}$ molecules are deprotonated. Practically all of the water left in the solution will be in the form of $\ce{H_3O^+}$, but there will be left over undissociated $\ce{HNO_3}$, so most of the excess stays undissociated, except for a small amount that undergoes self-protonation, generating a bit of $\ce{H_2NO^+_3}$.
All in all, in extremely concentrated solutions, even strong acids/bases will not show 100% dissociation, and saying that "strong acids completely ionize" is just a rule of thumb which shouldn't be taken too far.

nice examples, I understand that it is possible to have unprotonated strong acids in this case now.