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Composition of a typical printed circuit board (PCB)?
For the purposes of x-ray absorption, I'm trying to discover the average chemical composition of a typical PCB.
My research so far has indicated that the most common type of PCB is designated FR-4.  It is composed of fiberglass and epoxy resin.  The resin often contain Bromine to add fire resistance.  Let's ignore any printed copper for the mean time.  So we are looking at H, C, O, Si, Br, and probably some other bits and pieces.  But what are the percentage of these elements?

This article should be useful for you...

Material Recovery and Characterization of PCB from Electronic Waste
  Jessica Hanafi, Eric Jobiliong, Agustina Christiani, Dhamma C. Soenarta, Juwan Kurniawan, Januar Irawan, (doi:10.1016/j.sbspro.2012.09.1194)
Abstract
  Printed Circuit Boards are commonly exist in computers, laptops and mobile phones. Since it contains of a variety of metals and semiconductors, it is difficult to recover the materials in PCB. This paper aims to characterize the composition of these materials and a method to properly recycle and recover the materials in them. This paper utilized simple mechanical and chemical procedures to recover PCBs. Collected PCBs were disassembled, pulverized and separated by using density and magnetic separation method. By using the chemical recycling method, 98.82% purity of CuSO4 hydrate and Al2(SO4)3 hydrate were recovered from the PCBs. It is found that the milling methods, size of the sample and the density of the separating solution determine the effectiveness of the materials recovered.

Could you maybe give a short resume of the article's most important points concerning the question. Link-only answers might sometimes be helpful but really good answers add some additional text to make it as easy as possible to get the required information and a linked article often contains more information than is actually needed.


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Order of dipole moments
What is the order of dipole moments for $\ce{SO3, SiO2, P2O5,}$ and $\ce{Cl2O7}$? This question is from a book and the answer given is $\ce{Cl2O7 < SO3 < P2O5 < SiO2}$. Can someone explain why? When I drew the Lewis structures and applied VSEPR theory, it seemed that all the given compounds have zero dipole moment.

Can you double check that it says SiO2 has the largest dipole moment?  SiO2 should have no dipole moment, it doesn't say SO2 does it?  That would have a dipole moment, and...maybe the largest of those listed.

Make sure you distinguish between ions and molecules first. Also, don't be shy to look on the internet for the structure of these compounds. My hunch is that you had P2O5 wrong, but I could be mistaken.

@CHM Regardless, something seems off, as SiO2 certainly should not have the largest dipole moment.  And if they really did mean the SO3, and not sulfite ion, then that also should have no dipole moment.  So, the answer key still needs explaining if that is really what was provided as the correct answer.

The answer is really strange. $SO_3$ is plane triangle and has dipole moment zero, $P_2O_5$ is symmetric tetrahedron as dimer and has dipole moment zero, $SiO_2$ is insoluble solid with covalent grid and the only one that can have dipole from this molecules is $Cl_2O_7$, which is asymmetric in O atom as it has structure $ClO_3-O-ClO_3$. The polarisation of bond Element-Oxigen, however, grows in this order, as electronegativity of the element in question falls.

Something is odd here. Assuming $\ce{P2O5}$ to not be a dimer, then a valid Lewis structure drawn for it of the type $\ce{O2P}-O-\ce{PO2}$, which would be polar similarly to $\ce{Cl2O7}$. I expect the question is about the bond dipole moment of the X-O bond, in which case, the answer given in the solutions manual makes sense. @Tyrell, could you clarify whether the question asked for bond dipole moments or molecular dipole moments.

It's for bond dipole!

I wondered about that for a long time, but Ben Norris and BigGenius have the truth of it: the question is about the bond dipole moment, otherwise the answer would make sense. Here's why.
Molecular dipoles

Sulfur trioxide has planar trigonal geometry, and its molecular dipole is zero. Similarly, molecular $\ce{SiO2}$ has the same geometry as $\ce{CO2}$, i.e. linear, and thus its molecular dipole is zero.

$\ce{Cl2O7}$ is bent at its central oxygen atom:

and thus has a non-zero dipole moment. While $\ce{P2O5}$ generally occurs as its “dimer” $\ce{P4O10}$, it is possible to write a molecular structure for $\ce{P2O5}$ which then strongly resembles that of $\ce{Cl2O7}$, being bent at the central O atom. It thus also has non-zero molecular dipole.

Finally, ordering between $\ce{Cl2O7}$ and $\ce{P2O5}$: because the P is less electronegative than Cl, the partial charges borne by the P atoms will be larger than those of the Cl atoms, and thus I expect the molecular dipole of $\ce{P2O5}$ to be greater than that of $\ce{Cl2O7}$.

Bond dipoles
If the question is talking about the $\ce{X–O}$ (or $\ce{X=O}$) bond dipole in the series where X = S, Si, P or Cl, it becomes easier to understand. For one thing, because you're asked about $\ce{X–O}$ bonds where X spans a row of the periodic table, which is typical of this sort of problems. The answer would then be that bond dipole moment get bigger when the electronegativity difference of X and O becomes larger (the sharing of electrons gets more asymmetric). This leads to the answer you gave.

source: Stack Exchange

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What is the energy package released to the outside world?
Consider a chemical reaction where two different particles form another one
$$\ce{O + O_2 -> O_3}$$
I find it confusing how this can be an exothermic process. How to picture the release of $+\Delta H$ to the surroundings.
And am I right in viewing this energy value only as the net energy coming from an integration of the particle potentials as they get closer along the reaction coordinate?

Can you provide a source for the thermodynamic data? I'd have thought it was an endothermic reaction.

@CHM: I thought this is a exothermic one, but that doesn't really matter - the question goes for any exothermic reaction. (One could argue that the way I wrote it down doesn't even claim that this particular reaction is exothermic, but it's of course somewhat implied.)

Essentially yes - if you consider the two particles colliding into a potential well, the kinetic energy associated with each will (depending on the angular specifics) be apportioned into rotational, vibrational, and even occasionally electronic excitations. So, the two particles collide, and are then vibrating and/or rotating about each other with all the translational energy converted in the collision.
Non-reaction collisions with other ambient particles allow relaxation into lower-energy states, transmitting energy to those particles. After some collisions the new species formed is in equilibrium, transferring between low-energy rotational and vibrational states as it collides further, and the change in potential energy has been dissipated as rotational, vibrational, and translational energy in other particles.
(I guess I could really use some sort of animation for this.)

Mhm, my understanding of the energy package given to the outside, the so called system, hasn't really improved.

The "packet" of energy exchanged fall into one of three forms: heat, work, or electromagnetic radiation. The first two are kinetic energy phenomena.
Expanding on Aesin's answer about potential and kinetic energy.
Reactions that are exothermic convert potential energy into kinetic energy. The potential energy is the energy stored in the chemical bonds (electrons), and the kinetic energy is the molecular motion (rotation, translation, and vibration). Consider your "system" to be the electrons and the "surroundings" to be the nuclei of the atoms.
According to thermochemistry data from the NIST chemistry webbook for ozone and atomic oxygen, this reaction is exothermic (the heat of formation of molecular oxygen is by definition zero):
$$\Delta H ^0_{rxn}=\sum  \Delta H ^0_{f}(products)-\sum  \Delta H ^0_{f}(reactants)   $$
$$\Delta H ^0_{rxn}= \Delta H ^0_{f}(\text{O}_3 )- \Delta H ^0_{f}(\text{O}) $$
$$\Delta H ^0_{rxn}= 142.67~~\text{kJ/mol}-249.18~~\text{kJ/mol}=-106.51~~\text{kJ/mol} $$
In this case the potential energy of the electrons in ozone (total bond order of 3) is lower than the potential energy of a molecule of oxygen and an atom of oxygen (total bond order of 2). That extra potential energy is needed to keep O and O2 apart. When that energy is no longer needed, it has to become some other kind of energy (First Law of Thermodynamics: Conservation of Energy). In all cases of exothermic reactions, the extra potential energy becomes kinetic energy. If this kinetic energy increase occurs without a change in volume, then the temperature increases, and we say that "heat" has been exchanged. If the volume is allowed to change, then the increase in kinetic energy is converted to work.
Conversely, for an endothermic reaction, kinetic energy (work or heat) is converted into potential energy (less stable electronic arrangements).
Electromagnetic radiation is a little different. Instead of increasing the kinetic energy of the nuclei, an exothermic reaction can release a photon of light. In the case of the ozone reaction, the photon would have a wavelenth of 1.12 micrometers, in the near-IR.

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Software for chemical reaction simulation
I wonder, what software packs / environments are available for chemical reaction simulation. The goal is to learn or just rehearse some basic knowledge in chemistry. I suppose there should be some to buy but are there any open source projects dedicated to this?

What sort of reaction simulation do you mean? Permeakra's answer covers simulation of collections of molecules using electronic structure theory, but if you're thinking rather simpler things like 5 ml of HCl + 5 ml of NaOH -> X pH, something like Crocodile Chemistry would be more what you're looking for (but isn't open source).

Thank you for suggestion. The reason I am asking is to get overview of existing sw for mere humans (not for professional usage of course); I have never used any and wanted to rehearse in chemistry on another level (not just using books)

Oh, it turns out, in fact, that the packages that used to be called Crocodile Chemistry, now called Yenka Chemistry, seem to offer free home licenses for personal, non-commercial, non-academic use only, so you might want to take a look if relatively uncomplicated macroscopic modelling is what you're interested in.

Pmod, it would be helpful if you would be more specific in your question as to the topics and level that you need.  Thanks.

Please see http://meta.chemistry.stackexchange.com/questions/73/are-software-questions-on-topic

@Manishearth I don't think this is offtopic - I don't have any experience with reaction simulation. This is why I am asking - I can't be more specific. And I also don't agree with "It's difficult to tell what is being asked here" - what is asked is obvious. I supported this site from the beginning - if you consider that it is only for experts (I am actually software developer, but my favorite subj in school was chemistry) - it's sad, because I will not contribute then.

@pmod: Whoopsies, meant to close as "not constructive" or "off topic". Such questions, known network-wide as "shopping questions", are discouraged due to problems they can create. The reason for this being closed is nothing related to expert-ness (See this page, we allow easy questions given a few restrictions which are applicable everywhere anyway). While the page I linked to initially talks about expert-ness (my misconception back then), the answer below it is the actual policy.

If you mean finding saddle point, then you should consider Gamess and Quantum Espresso packages: free, stable, mature packages, useful for molecule simulations and estimating reaction barriers. They, however, have to be manually compiled and require some skill to setup, as they are meant to be used in POSIX environment (various UNIX clones), not windows, and does not have user friendly result viewer or job builder/scheduler. It is not a problem for experienced linux user, though...
I'm currently using quantum espresso and I found it useful for my cases.

What level of theory are they based on?

gamess allows Hartree-Fock, mp2 and dft calculations and several ultra-fine method in gaussian-type basis sets with optional pseudopotencial and relativistic corrections. Quantum espresso implements dft in plane-wave basis set with mandatory pseudisation and periodicity. Both have experimantal branch with GPGPU support

And it is free? Why do people pay so much for Guassian? Also be careful publishing with it. Guassian has a reputation for blackballing anyone who publishes with other software.

@Canageek: As far as I'm aware, under the Gaussian license you aren't allowed to publish timing data that originate from Gaussian-derived software that isn't in the form they provided it to you (e.g. if they gave you the source and you compile it), and you aren't allowed to give it to people who compete with Gaussian. That's it. There seems to be a lot of FUD people spread about this.

gamess is free for academic and personal usage. It requires registration, but nothing more. For commercial usage one have to by license. Quantum Espresso is GPLed. To my knowledge, Gaussian has not critical advantage in timing, but it implements more advanced features. For example, when I used gamess last time, it didn't have analytic second derivatives for energy in default dft code and calculation of vibration spectra with dft was painfully slow.

@Aesin Ah, someone from MSU that said their university didn't have a licence because a prof had published with a rival software package, so Gaussian wouldn't sell to them any more.

@Canageek Pretty sure, they do not care much with licensing problems from the very beginning.

@Canageek: MSU do claim to be "banned by Gaussian", but I suspect that's more likely to be because several people there contribute to GAMESS (and thus compete with Gaussian). I don't know if that means they can't get a site-wide license or whether no one on staff can license the software, though. (Also, this thread may be getting a bit long and off-topic for comments now.)

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What is the name of cyclopropane with a double bond?
I was drawing out all the isomers of $\ce{C5H8}$ and came upon two structures of cyclopropane attached with double bonds. The group is neither a methylene nor an ethyl... what is it? What are the names of these two molecules?

 Ethylidenecyclopropane
 Vinylcyclopropane (or Ethenylcyclopropane)

Note that these aren't the IUPAC names, care to add those as well? Also a bit about how the two functional groups have (common) names as ethylidene and vinyl respectively. Thanks :)

Ethylidenecyclopropane is the IUPAC name for that molecule. What else could you call it? See http://www.acdlabs.com/iupac/nomenclature/93/r93_271.htm. The vinyl substituent has been recently accepted by IUPAC (in the last 10 years), but I haven't found supporting proof on the web.

As a side note, IUPAC names are nice and all, but completely worthless if practicing chemists do not use them: 2-propanone, methanoic acid, and 1,2-didehydrobenzene are all significantly more frequently referred to by their common names of acetone, formic acid, and benzyne.

@Manish, as already noted, "ethylidene" is in fact valid IUPAC nomenclature.


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How to calculate entropy change for a gas reaction using $S^\circ$?
I'm calculating molar changes in thermodynamic properties due to reactions between gasses (assumed to be ideal gases).  I can calculate $\Delta H$ easily enough, because it's just $\sum_i \nu_i\Delta_f H^\circ_i$, with $\nu_i$ the stoichiometric coefficients. $\Delta G$ (at standard pressure) can be calculated from $\sum_i \nu_i\mu_i$, with $\mu_i = \Delta_f G^\circ_i + RT\log[i]$. Since $\Delta G = \Delta H - T\Delta S$, I can calculate $\Delta S$ as $\frac{1}{T}(\Delta H - \Delta G)$.
So far so good. But I'm inexperienced in working with tabulated quantities (I'm a more of a theoretical physicist than a chemist) and keep making mistakes with signs and units. So I figured that as a sanity check I would calculate $\Delta S$ directly, using tabulated values of $S^\circ$. But then I realised I don't know how to do this, and it seems that none of the textbooks on my desk explain it either.
Based mostly on intuition, it seems like it should be $\Delta S = \sum_i\nu_i(S^\circ_i + R\log[i])$, with the $R\log[i]$ term having something to do with the entropy of mixing. For the example reaction I chose, this gave something with the approximately correct magnitude but the wrong sign (-15.6 instead of 14.2, which is what I get by calculating it from the Gibbs energy and the enthalpy).
So my question is, what is the correct way to calculate the entropy change due to an ideal gas reaction if, for some reason, you only have access to the concentrations and the standard entropies of the reactants?

Just to make sure that I get your question correctly; the process in question is a chemical reaction that takes place at a temperature and pressure close enough to the reference temperature and pressure (those used for the tabulated values) that any changes to the tabulated values are neglible? (I assume this, as this is what you have done in the enthalpy calculation)
In that case, it is probably easier that you would expect - you can use (almost) the same approach as you used to calculate $\Delta H$ as both entropy and enthalpy are state functions. That is:
$\Delta_{rx} S = \sum_i \nu_i S^\circ_i$
where $S^\circ_i$ is the tabulated value for each of the components. This is described on Wikipedia.
If the temperature deviates sufficiently from the reference temperature used then you should adjust the entropy change accordingly
$\Delta S_{T_1 \rightarrow T_2} = \int_{T_1}^{T_2}\frac{C_V}{T}dT=C_V\ln(T_2/T_1)$
where the last equality holds if $C_V$ can be assumed to stay constant in the interval $T_1 \rightarrow T_2$. A similar adjustment can be made if the pressure, volume etc. changes during the course of the reaction. Again, as the entropy is a state function, it is enough to adjust the total change in entropy for the reaction.
Hope this helps! Please let me know if I am not being sufficiently clear, and I'll try to elaborate things a bit.

Hmm, but that's just the calculation for $\Delta_r S^\circ$, not $\Delta_r S$. I.e. it doesn't take account of the entropy of mixing. For a reaction at equilibrium, we should have $\Delta S = \Delta H/T$ (because $\Delta G =0$ at equilibrium) and that won't happen unless $\Delta S$ is concentration-dependent.

Thanks for the comment about the temperature change though. These are reactions in the atmosphere near the surface, so they're at 1 atm but at around 288K instead of 273K or 298K.

Ahh, sorry - in that case I think I misunderstood you a bit. The entropy change of mixing two ideal gases should be $\Delta_{mix}S = -nR(x_1\ln(x_1) + x_2\ln(x_2))$. This article on Wikipedia is a bit more thorough. In my experience, the entropy change associated with the reaction itself is (usually) quite a bit larger than the entropy of mixing.


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How to determine stable electron states in ionic and covalent bonds?
I'm working on a program that needs to determine if a bond between two or more elements will result in a stable state.  I understand at a high-level how to fill electron subshells using the Aufbau principle, but I also read that in some cases, electrons will jump from a lower energy shell or orbital to a higher one in order to maintain a stable state (Filling Electron Shells).
For example, if I want to determine if Hydrogen and Nitrogen will form a stable bond I would fill the shells for Nitrogen like so:

When adding Hydrogen, would the two electrons in the 2s shell jump to the 2p shell because with the one electron from Hydrogen the HN would then have a full 2p shell? Like this:

Or is this not a stable state? Would it require a large amount of energy to excite or promote those two 2s electrons to the 2p shell?
I'm trying to understand if there are rules or heuristics I can use to estimate if two ore more elements will bond (on there own w/o adding a large amount of energy to the system) using their valence electron configuration like this, or if there are too many exceptions, making it not a simple programming task to estimate this.

This question in general is entire point of quantum chemistry, so there is no simple answer on it except "setup run for package [PACKAGE_NAME] and try". In some simple cases , like one you stated, it is possible to note, that for second row of elements 2s-2p gap is hight, so N in HN will have $1s^22s^22p^2$ electronic formula.

As an addition to what @permeakra said, it might be an idea to edit your question to go into a little more detail on what you're trying to do (and constraints involved) so that someone can either suggest an appropriate approximate method, or maybe tell you that it's impossible. :)

@Aesin, I've edited my question a bit to explain what I'm trying to do.  Perhaps it's not possible to code some simple rules and I need to run a package such at StochKit?

@MattPalmerlee: It's still not really clear what sort of level you're aiming at, and I'll try to explain why: generally, two atoms in a vacuum will attract each other. Even helium atoms attract each other, but there's no significant electron transfer between them. If you want to work out whether two atoms can form a covalent bond or an ionic bond, that's an entirely different kettle of fish, and there are some rules based on electronegativity that sort of work for elements down to about calcium - beyond that is dangerous transition zone.

@Aesin thanks for that, my first version of my program will only go to Calcium (I should have mentioned that before) for simplicity's sake, to basically ignore the transition metals at this point.  What I'm imagining is that I could say: I have 3 molecules (let's say no noble gasses), the total electron count is 9 (HLiB or LiC), this is a stable state [bond] because the 2s orbital is 1/2 full and the 2p orbital is full.  I'm guessing by the comments and answers so far that it's just not that simple, I have to look at the energies and the geometries, etc...

Maybe a better way to explain what I was imagining was to just look at all molecules' valence electron counts (electrons that can participate in bonding right?), so for H, Li, and B you have 5 valence electrons, and that could point to the fact that HLiB wouldn't be able to bond in a stable fashion (maybe unless you added a Hydrogen).

For anyone who is curious, I've finished the first version of my game this question was for: Chem Fight instead of trying to estimate if two or more atoms will form a compound I use a list of common compounds, if a combination is on the list, it's a valid compound, otherwise an attack will be successful.

I'm trying to understand if there are rules or heuristics I can use to estimate if two ore more elements will bond

You should try to find good textbook/course about Molecular Orbital theory. As I recall, good university-level textbooks on general/inorganic chemistry dive into this aspect and make analysis for dimers of second row elements. This heuristics, however, are of limited usability: they can be used to explain differences in family of compounds with similar structure, but are virtually useless for predicting geometry in complicated cases. And yet, this is the best you can get without diving into quantum chemistry.
Here is quickly found link on the subject. http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch8/mo.html#valence

Thanks for that @permeakra, for my project I'm trying to keep it simple and asking the question: "Will it bond?" and only looking at a handful of elements max.  Sounds like there is no simple or even "approximate" way without getting into the geometry.


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Iron nomenclature differences between Iron(III) and Ferric
As I was doing my Ap Chem homework, sometimes the question would use Iron(III) and sometimes it would use Ferric.  What's the difference in usage?  Also, what's the difference between Iron(II) and Ferrous?

Hi Just, While this is an important question (and you've gotten some answers that nailed it right on the head), I think that's it's found quite readily by searching online.

There is no difference, in both the examples you cite.  The forms Iron(II) and Iron(III) are preferred today, versus the older forms ferric and ferrous, respectively, because they more explicitly give the oxidation states of Fe, +2 and +3 in this example.

The difference in usage is simply a choice.  The usage of the Roman numerals to indicate oxidation state is known as Stock nomenclature, devised by Alfred Stock in 1919.
The other method, using the -ic and -ous suffixes is derived from Latin.  The -ic suffix is used for the higher oxidation state of the metal, while the -ous suffix is used for the lower oxidation state of the metal.
However, the use of the Latin suffixes is no longer recommended for indication of oxidation state, but it may be used for metals exhibiting not more than two valences.
From IUPAC:

The following systems are in use but not recommended:
The system of indicating valence by means of the suffixes -ous and -ic added to the root of the name of the cation may be retained for elements exhibiting not more than two valences.


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Bisulfate vs Hydrogen sulfate
What's the difference between bisulfate and "something" hydrogen sulfate?
Has there historically been a preference for one over the other?

Nope, no rule, and again there is no difference, like your question about the names for iron oxidation states.  The "bi" form is the older form, and the "hydrogen" form is now preferred, again because it more explicitly describes the form of the compound.
EDIT: Some additional detail was requested.  When I say "preferred", I mean the name sanctioned by the IUPAC (International Union of Pure & Applied Chemistry), also known as the systematic name, so called because the rules to form the name are "systematic"; i.e. consistent and clear. Older names are called common names, which were in used before the IUPAC.  So "bisulfate" is a common anion name, while "hydrogen sulfate" is systematic.

Could you elaborate a bit on whether there was an era in which one or the other was preferred?  That way this question will have more value for future visitors.


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Good acid for cleaning marine toilets
Problem:
On my boat I have a marine toilet. It is operated with salt water.
The problem is, on the interior of the outflow tubing it accumulates some sort of hard grime. I am looking for a good way to dissolve this grime as it is impossible to replace the tubing.
From experience it is known that citric acid work - but not very well. Also it leaves a "sandy" residue that is difficult to flush.
Does anybody have an idea of what this "grime" consists of - and thus how best to dissolve it?
Materials:
On hand I have:

citric acid
56% phosphoric acid
23% HCl

Materials:

tubing: PVC
toilet itself: porcelain
pump: ABS or polypropylene with selected parts in acetal resin, with 316 passivated stainless steel fastenings, brass weights and neoprene seals and gaskets
valve: grey, harder plastic

I know this quite vague but am hoping somebody can help anyway.

To my knowledge, marine water contains great amount of $\ce{Ca}$ and $\ce{Mg}$ ions, that gives insoluble carbonates and phosphates. So, most probably, you have mix of carbonates, phosphates and, possibly but unlikely, sulfates of this metals and possibly some organic salts, like oxalates. Carbonates are readily dissolved by $\ce{HCl}$ wich in weak solutions will not dissolve copper, but will slowly dissolve steel. It is more interesting to use $\ce{Na2H2edta}$ . It dissolves $\ce{CaCO_3}$ easily and, to my knowledge, in hight concentration dissolves $\ce{Ca_3(PO_4)_2}$ as well, but not sure in the latter.

Thanks for the answer. You say "insoluble carbonates". That means insoluble in water, right?  What is the name of Na2EDTA if I want to buy it? Is it better than HCl or did you recommend it only because it affects metals less than HCl?

Ok, I found EDTA on Wikipedia. However I did not find Na2EDTA. I did find Na2H2EDTA, Na4EDTA and CaNa2EDTA. Did you mean any of those? How would I use them? If I understand correctly those are salts, right? I assume I would dissolve them in water and let is sit in the tube?

I meant the first one. One should made reasonably concentrated solution and submerge surface to be cleaned under it for several hours or even several days. EDTA derivatives under different names may occur in everyday chemistry, like some machine washing products, so may be you can find prepared solution on shelfs with products to threaten hard water for wash machines. Citric acid is way to go, according to this article http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1173122/ but I never saw it used this way, unlike EDTA derivatives, so I cannot say for sure.

A cheaper and easier to find solution than EDTA (permeakra's answer) is citric acid. Citric acid reacts with metal carbonates (bicarbonates, phosphates, chlorides, etc.) to produce the soluble citrates. The citrate ion keeps the metals in solution by chelation, the same process that EDTA (ethylene diamine tetra acetate) does.
Citric acid can be purchased in the grocery store or convenience store as one of the principle components for any sour powdered drink mix (like Tang).
You could also use any substance or cleaner designed to remove hard water deposits or lime scale, such as CLR, which contains lactic and gluconic acids as chelating agents.


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What chemicals could absorb/adsorb ambient atmospheric methane?
From an article in the New Scientist in 2010,

Since pre-industrial times, methane levels in the atmosphere have risen from around 715 parts per billion to nearly 1800 ppb, claims Folberth. "If methane could be taken back to pre-industrial levels it would bring about half a degree of cooling and would be equivalent to removing one-third of anthropogenic carbon dioxide," says Folberth. Furthermore, the cooling would take place within 10 years.

Klaus Lackner and others have developed technology such as the moisture-swing sorbent to capture ambient $\ce{CO2}$
Where might be the chemical options for a material that would absorb or adsorb ambient methane?

Some bacteria metabolize methane. This is sometimes used to treat the outgassings of landfill sites.

@mart yes, that's mentioned in the article I linked to

Some Bacteria can metabolize methane, methane can also reach with ozone.

There is none. Alkanes are very hard to activate. They require either free radicals, or coordinated unsaturated sites either in the molecule or on surface of solid. But in both cases, the atmosphere contains molecules that will react more readily.
However, the average life-time of methane in atmosphere is quite small, less than 10 years.

There are materials that absorb methane under ambient conditions. A quick Google Scholar search for "metal organic framework methane" shows some of the most recent and popular synthesized materials that can do this. Dr. Shengqian Ma at University of South Florida is a big figure in the field, and he currently collaborates with my computational research group. This is a figure of methane sorption for a material his group made at different temperatures:

Granted the highest uptakes are at high pressures (1 bar ~= 1 atm), and the graph is somewhat muddled at ~1 atm, but this material sorbs (much) more than its own volume at that pressure, and there are other examples.
It is a somewhat new class of materials (~20 years). One of the first to be synthesized is MOF-5, which wasn't that useful but served as a good prototype example for future syntheses. But there are a ton of them that have been synthesized to date. They are crystalline materials that are made by self-assembly of metal cations and organic ligands (usually). Some have even been made with light elements e.g. beryllium.
All of my graduate research thus far (1.5 years) has been studying the mechanism of gas sorption in MOFs. Many can physisorb methane, hydrogen, carbon dioxide, nitrogen, and other gases quantitatively. To be specific, several can do so with methane at ambient T/P. Many copper MOFs, for example, induce a pretty strong interaction (thus binding effect) with hydrogen gas, which is not easy to contrive.
See the wiki also.wiki


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How to compute the residence time for given flow rate in standard cubic meters?
I'm given a flow rate of a gas mixture of $3\ \mathrm{Nm^3/h}$ in standard cubic meters through a cylinder of length $50\ \mathrm{cm}$ and cross section $70\ \mathrm{cm^2}$, can I compute how long a particle will stay inside of the cylinder (which would equal the time the particle can react inside the cylinder)?
And what if the temperature changes during the reaction?
If the flow rate would be given in volume per time, I had an idea to compute this, but the use of standard cubic meters is baffling to me.

$Nm^3/h$ is volume per time, you just need to transform it the the units you need. the correction for temperature can be done with the ideal gas law (Or Redlich-Kwong, if you want to be really exact). Also note that you will get a mean retention time, not all particles will spend the same time in the reactor.

Isn't the the unit with the N in from something else than just a length?

@chm: The English Wiki has only standard cubic feet...

The N stands for "norm" and should be avoided (in fact, the German Wiki page lists the Nm³ as wrong abbreviation) as it can create confusion.(qed)
So your flow should be written as "3 m³/h under standard condtions".
If you can guess German, have a look at the German Wiki page. It is much better than the English one.
Standard conditions vary a bit between different norms
T = 0 °C or 15 °C or 20 °C
p = 1.01325 bar (sometimes 1 bar)
The gas is supposed to be dry.

k, I'm still not sure how this solves the problem. Like does this mean that for other T and p values, the flow rate is different? Does this deviation depend on the specific gas? I'm really interested in the residence time, the unit thing is just keeping me from it.

cbeleites understood your question better than me. Think of the amount (in moles or kg) of gas you have in a m³ under standard conditions, and adjust for the p and T you actually have. In most cases, you can wotk with the ideal gas law and then it's not specific to a certain gas.


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Polymer Slurry used in boring foundation holes?
I watched a video on the construction of Burj Khalifa, Dubai. The construction engineers answered that they're using a special type of protection known as Cathodic protection to protect from the corrosion due to local ground water. They told a compound named "Polymer Slurry" (What's that?). They're using viscous Polymer Slurry as their basic coat to dump the high density & low permeability concrete in order to create a "GOOD FOUNDATION". What is it made of? (Simply any polymer?) And what does it do? Does it actually repel water? Many organic compounds repel water... Then Why this?

"Polymer slurry" just means a slurry of polymer. It could be anything.

After reading the link provided in the question, and doing a little googling, of "polymer slurry" and "foundation", I found something useful.
Polymer slurry seems to be made of one or more hydrophilic superabsorbent polymers, such as polyethylene glycol, polyvinyl alcohol, or carboxymethylcellulose that produce a thick viscous slurry when mixed with water. It does not repel water so much as the opposite. It is so hydrophilic that it absorbs water before the water can reach the concrete or the steel. Then this slurry becomes thicker the more water it absorbs so that the rate of diffusion of water through the slurry decreases. Ultimately, this action creates a hydrogel, which may be quite strong, despite being mostly water, due to extensive hydrogen bond crosslinks.
Other applications of polymer slurry seem to be in the drilling industry to modulate the viscosity of drilling fluids.

The polymer slurry has the purpose of keeping the hole stable, while it is still being worked on.
While building a 50 m deep hole, I think it's common to need to change or replace the tool, when it broke, or the type of ground changed.
Changing the thing on the other end of 50 m steel rod means to pull out the rod, make the change, and let it down again.
But even without any change, the rod needs to be pulled out before filling with concrete, I think.
The slurry does not repell water on the surface like oil would, but it repels ground water from entering the bore hole by having a higher pressure.
The pressure comes from a higher density and same depth below ground.
One useful effect of the polymer could be that it does not soak into dry parts of the hole wall.
Over 50 m of hole wall, a touching rod will ceartainly loosen some sand, but id will not just drop to the end - depending on the slurry and the particle size, it will sink slowly, or even not at all.

From Burj Dubai - Preparing the Foundation:

So, large boreholes were dug 164 feet (50 m) deep and then filled with viscous polymer slurry to maintain the stability of the hole. The slurry is denser than water, but lighter than concrete. The concrete, when poured into the hole, displaces the slurry and hardens to form a foundation pile.


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Comparison of acids to dissolve phosphates
I am currently reading this German paper on the dissolution of phosphates by acids and chelants.
The authors do a "fast-test" of the dissolving power of different acids which goes as follows:

For each test, increasing quantities (10-50ml) of 0.1 n acids are
added to 100 mg of tricalcium phosphate and agitated for 1 hour. The
residue is filtrated and weighed.
For 0.1 n hydrochloric acid 20ml suffice for complete dissolution of
100mg tricalcium phosphate. Citric acid [...] shows >90% dissolution
with 40ml.

Not being a chemist I am having trouble interpreting this. What quantity is effectively being controlled by specifying 0.1 n acid? The paper is from the 60s. I thus suspect that n stands for normality but I am not sure.
What exactly is being measured?

The speed at which different acids dissolve the material? (Why then vary the volume of acid?)
The quantity of acid needed to dissolve a given of material? (Measured in what exactly? What does the volume effectively stand for? Moles?)
Something else?

Edit 1
J M's answer makes this much clearer. I still have some problems though:

Regarding normality: Does this ensure that the same concentration of protons is in each solution? I assume this is not the same as "constant PH" as some acids may be strong acids while others may be weak?

If even more citric acid was added, there would still remain 10% residue? Why did they not try with 50ml citric acid? With this test there are two variables that change: volume and residue. Does this not make it difficult to compare the results of different acids? I would understand a test like this: how much acid is needed to dissolve the material 100%. Then one could compare easily.

Regarding weak acids: Once one of the available protons from the citric acid has reacted, should not another previously non-dissociated acid molecule dissociate (to keep the dissociation proportion in equilibrium) and keep everything going to dissolve the material without any residue? Given infinite time, why do not 30ml citric acid also dissolve 10 mg of 100 mg of tricalcium phosphate?

I am still not quite sure however what the point of this test is. Should it not be reasonably easy to calculate this using the reaction equations and $\mathrm{p}K_\mathrm{a}$ of the acids? (See reaction equation below.)

Thought experiment: If not the acid quantity was the limiting factor but the amount of solution what would limit the reaction: i.e. infinite supply of acid molecules (high acid concentration) and infinite supply of phosphates but finite volume for the solution to take up the reaction product: Would the reaction be limited by the solubility of the reaction products? I assume, in this case a strong acid would be advantageous, right?

Reaction Equation
If I understand the papers I read correctly, one can dissolve 15 moles of $\ce{Ca_3(PO_4)_2}$ with 60 moles of $\ce{HCl}$ or 20 moles of citric acid (disregarding the fact that it is a weak acid?!):
$$\ce{ 15 Ca_3(PO_4)_2 + 60HCl -> 15Ca(H_2PO_4)_2 + 30 CaCl_2 }$$
$$\ce{ 15 Ca_3(PO_4)_2 + 20 CitH_3 -> 10 Ca(Cit)_2 + 25 Ca(H_2PO_4)_2}$$

N, or Normality is a way of standardizing the # of acid equivalencies a particular acid has across multiple acids.  Acids that have only 1 acidic hydrogen for example $\text{HCl}$, the Normality will equal Molarity.
Normality is defined by:
$$
Normality = \frac{c_i}{f_{eq}}
$$
Where $c_i$ is the molar concentration of the acid, and $f_{eq}$ is the equivalency factor.  In the cases of something like $\text{H}_2\text{SO}_4$, the equivalency factor is 0.5 because it is a diprotic (two proton) acid.
This test also is showing the effect of weak vs strong acids.  $\text{HCl}$ is a strong acid, so it dissociates essentially 100%, while citric is a weak acid.  A weak acid exists in equilibrium between the dissociated acid species, and the neutral species.  This equilibrium is defined by the acid-dissociation constant $K_a$, or it's logarithmic version $pK_a$.
For citric acid though, it is triprotic, meaning there are 3 acidic hydrogens, so it's $f_{eq}$ value is $1/3=0.3\overline{3}$
Your numbered questions:
1)  The speed is being measured to a certain point.  However, the chance is that all of these reactions are complete at this point.  The hour timepoint is to normalize everything to the same time frame.  But as indicated, the reactions are likely complete at this point.  What is really being measured is your second point.
2)  Absolutely, the quantity of acid needed to dissolve the material is being tested.  As you use more and more volume of acid, you are providing larger amounts of acid to the reaction.  This helps it dissolve more of the powder (as there is more acid now present), and also increase the rate of the reaction.
They standardize the reaction to normality because if they used molarity, 1 mole of citric acid has 3 acid equivalents, while $\text{HCl}$ has only 1 acid equivalent.

Thanks J M. This really helped me along the path of understanding. However, as some things became clearer, new questions came up. Would you mind taking a look at the addition to my question?

Sure I will do my best to take a look later this evening and add an answer to those other points.  One difficulty is that I do not read German, so I cannot read through the paper to digest more information.

Thanks. The paper is from an age where German was still acceptable as publishing language. - Things have changed quite a bit and it is obvious why...

@ArikRaffaelFunke Just wanted to say I will be posting a response tonight or tomorrow morning, just finished submitting my MS Thesis last night

There really is no rush at all. Take all the time you want. I am merely curious because as a physicist I never even got an introduction into the most basic concepts in chemistry. Thanks for allowing me to pick your brain. - BTW: Congratulations on submitting your MS thesis. That is quite an accomplishment!

Indeed, n (pretty weird, chemists would use the capital N) is used for normality.
In their fast test, the authors try to dissolve 100 mg (0.322 mmol) of $\ce{Ca3(PO4)2}$ with different acids and buffers at room temperature while shaking for 1 hr.
For comparison, they state that 20 mL of 0.1N HCl is sufficient to dissolve all the triphosphate.
In the case of the other solvents (solution according to STUBY, etc.), where solution was incomplete, they determine the weight of the dried residue (i.e. how much solid material was left) and try to correlate it with the volume of the solvent used (cf. Fig. 1). From the volume used and the concentration of the citric acid in the stock solution the absolute amount can be calculated.


text:
Why is methane's molecular formula conventionally "CH₄", while water is "H₂O" (among others)?
While revisiting some of my old notes about the Miller-Urey experiment, I stumbled across the "equation"...

Electricity + $\ce{CH4~/~ NH3~/~H2O~/~CO}$ = Amino Acids

This got me thinking.
Conventionally, why are molecules like $\ce{CH4}$ and $\ce{NH3}$'s molecular formula written differently (in H placement) than others like $\ce{H2O}$ and $\ce{HF}$? Is there a particular reason, or did it just happen to be?
NOTE: I realize the $\ce{H4C}$ or $\ce{H3N}$ make perfect sense, especially when drawing out molecules. It's just that these are rarely used in literature.

There was a meta discussion about whether this question should have be considered a duplicate of Why is arsenous acid denoted H3AsO3?.

The choice of whether to write binary hydrides with the hydrogen first or second depends on whether that hydrogen is considered acidic, with water marking the delineation point.
By convention, if a binary hydride is more acidic than water, then they are written in the form $\ce{H}_{\text{n}}\ce{X}$. If a binary hydride is less acidic than water, they are written as $\ce{YH}_{\text{n}}$. It so happens that this changeover neatly divides the periodic table. Binary hydrides from groups 16 and 17 (VIA and VIIA) are written with the hydrogen atoms first: $\ce{H2O,~H2S,~H2Se,~H2Te,~HF,~HCl,~HBr,~HI}$.
Binary hydrides from groups 1 (IA) through 15 (VIA) are all written with the hydrogen atoms last, regardless of whether the hydride is ionic $(\ce{LiH,~NaH,~KH,~CaH2})$ or molecular $(\ce{BH3,~CH4,~SiH4,~NH3,~PH3})$.
This notation is in keeping with that for more complex compounds. If the compound is an acid, the acidic hydrogen atoms are put at the from of the formula, while nonacidic hydrogen atoms are placed in the main part of the formula. For example, the hydrogen atoms in phosphoric acid $(\ce{H3PO4})$ are considered acidic and the hydrogen atoms in octane $(\ce{C8H18})$ are not. Acetic acid $(\ce{HC2H3O2})$ has one hydrogen atom that is considered acidic and three that are not, this the formula $\ce{HC2H3O2}$ is considered more helpful than $\ce{C2H4O2}$, although both are "correct" by different sets of rules.

Note, as an exception, organic acids are often written with acidic hydrogens last, as a convention of indicating carboxyl groups, e.g. $\ce{CH3COOH}$ instead of $\ce{HC2H3O2}$

@khaverim - true. Mind you, I most often write $\ce{HOAc}$.


text:
Pipetting strong acid with plastic material
Can I pipette concentrated sulfuric acid using a plastic pipette tip? Will either the plastic or the acid be affected?

This depends on type of plastic. You definitely can if plastic is perfluoropolyetylene (Teflon). You probably can if plastic is polyethylene or polypropylene, but both will degrade in long term (polypropylene will do it faster). You should not, if plastic is PET or other polyester, or polystyrene.
It is probably most reliable to find which plastic you dealing with and search for 'chemical resistance chart' for this plastic.

If you cannot figure out which type of plastic it is, assume you cannot pipette concentrated sulfuric acid.


text:
Products of HMTD decomposition
Hexamethylene triperoxide diamine, "Hex" for short, is an explosive organic compound.
I want to know the products of this reaction:
$$\ce{C_6H_{12}N_2O_6 -> ? + Energy}$$

Answer
This article provide two reactions for the thermal decomposition of HMTD, depending on reaction conditions.

(at or below 150°C, air or vacuum)

$$\ce{C_6H_{12}N_2O_6 -> 3CO_2 + N(CH_3)_3 + NH_3}$$

(160°C to 180°C, vacuum)

$$\ce{C_6H_{12}N_2O_6 -> 3CO + N(CH_3)_3 + 1/2 N_2 + 3/2 H_2O + 3/4 O_2}$$

References

Oxley, J.C., J.L. Smith, H. Chen, and Eugene Cioffi. “Decomposition of Multi-peroxidic Compounds: Part II. Hexamethylene Triperoxide Diamine (HMTD).” Thermochimica Acta 388, no. 1–2 (June 18, 2002): 215–225.


text:
standard reference for thermodynamic properties?
I'm a non-chemist attempting to put together a chemistry paper. Up to now I've got my thermodynamic data from various random pdf files found on the web, together with Wikipedia. Now that the paper is nearly finished I need to replace these with values from a single, respectable source. But I've realised I don't know how to find such a source. I have a few chemistry textbooks on my desk, which tend to have tables in the back, but they're not terribly comprehensive and none of them have all the data I need.
So is there a standard resource for thermodynamic tables? I particularly need the Gibbs energy and enthalpies of formation for gases found in the atmosphere, such as $\ce{O2}$, $\ce{CO2}$, $\ce{CH4}$ and $\ce{H2O(vapour)}$?  In an ideal world this would either be something I can access online or a book that's likely to be found in my institute's rather small library.

A reference that I've used quite a lot (as a student) is SI Chemical Data, which I guess is quite common. It should contain the data you ask for, among other things.
If you are looking for an online source, the NIST-JANAF Thermochemical Tables or NIST Chemistry WebBook should be good places to start. I think both should be respectable enough. The Chemistry WebBook might be a bit more comprehensive, but I suspect that much of the data is from the same source anyway, since NIST is behind both services.
Hope you can use one or the other!

Thanks for the info. Our library doesn't have SI Chemical data. The NIST-Janaf table looks good (the WebBook is too detailed, as it gives multiple figures from different sources for each quantity) - but I currently have no idea how the NIST online tables should be cited in the bibliography. If you can tell me that I'll accept this answer.

@Nat, see this.

There is a credits page which you might already have found that could prove useful.

What you'll want to look at is the Landolt-Börnstein series of handbooks, particularly Group IV (Physical Chemistry). Volume 19 deals with the "Thermodynamic Properties of Inorganic Materials", which will have the Gibbs energy and formation enthalpy included among the tabulated properties.

I too usually refer to the CRC Handbook. Both of my copies (86th ed, 2005-06 and 46th ed, 1965-66) have a section on thermodynamic properties of materials. The CRC Handbook works for the most of the general data I need to look up. There is some field-specific data I need that isn't in the CRC Handbook; I use the Polymer Handbook for that.
In terms of actually citing the data in your manuscript, I consider it to be best practice to cite an original research paper if possible. The CRC Handbook and Polymer Handbook cite every datum they contain (sometimes with more than one source). Ideally, you would look up that citation, verify both the datum and the method, and then cite that source.
I personally consider a citation to the reference book itself to be sub-optimal, but acceptable. NIST is a slightly different beast, as they often are the original source of the data they compile. They usually supply a preferred format for citations, so I'm a little surprised that it's not obviously placed on the NIST-JANAF page that @Kjetil Sonerud linked to. Google led me to a citation page that you might find useful, in addition to the other page Kjetil linked. EDIT: The page @J.M. linked to in his comment is exactly what I was thinking of. I can't figure out if that's the same source as the NIST-JANAF tables.

I think the CRC Handbook of Chemistry and Physics is a very good source to look for all sorts of physical or chemical data and it can be considered as a kind of standard reference. You can also find an Internet version of the book which would be a bit more "lightweight" than the printed book with its more than 2600 pages.


text:
Is there a liquid crystal or similar material that gets cloudy when an electric field is applied, but is otherwise clear?
Is there a way to make a liquid, such as a liquid crystal material, transition from cloudy or clear and vice-versa using only electric current?
By cloudy, I mean, similar to the effect that cholesterol or 5CB exhibits, thermally.
If so, where could this be obtained, and how quickly does it switch between states?
If not, how else might such an effect be achieved using electric current? I've read about the Kerr effect, but it didn't change opacity so much as refractive index, and it also appeared to use a great deal of electricity. I'm thinking something within the single volts.

The would be sort of the reverse of what happens in a liquid crystal display pixel, so, it's possible.

I expect that answer is "no, or at least, no any common". However, if one is interested in shutting light, he may use safe idea as in liquid crystal which can be both transparent by default and nontransparent by default.

I would expect this to exist. One major class of stimuli-responsive polymers is thermosensitive polymers. These have a sudden drop in solubility with an increase in temperature above a certain point. Depending on the particular polymer and the concentration, this transition can be accompanied by a clouding of the solution.
A review of stimuli-responsive polymers from Brent Sumerlin's group contains a section about electric field sensitive polymers. These are usually in the context of the field causing a mechanical response from the polymer. Sumerlin's review cites this paper from a Hungarian group discussing polymer gels that change shape under electric field. You would want the electric field to affect the solubility of the polymer. Sumerlin's review doesn't appear to directly mention anything exactly like that, but I would expect it to be possible.
I'm not as familiar with ionic liquids, but this paper uses an electric field to turn a hydrophobic ionic liquid hydrophilic. That's what happens with the thermosensitive polymers, and more what I was thinking of.

Why was this answer voted down? I think it has merit, but I need more time to study what's been outlined here. :)

Hmm... I'm willing to examine thermosensitive polymers for this application, provided the following:

Hmm... I'm willing to examine thermosensitive polymers for this application, but it definitely complicates matters. I'd need a cooling jacket, and individual heating elements per cell. Further, it'd have to be a low enough temperature to not melt plastic, but not so low that I'd need refrigeration. It'd be nice if this material could be sourced easily and at low cost, also. ;)

@CryptoQuick, the polymers I'm used to working with will undergo their transitions at temperatures below plastic melting (usually between room temperature and body temperature). Sourced easily and low cost are harder; most of these materials are still in the synthesize-it-yourself phase.

Colin, I don't suppose you'd have any updates? Perhaps something is on the shelf these days?

Electrochromic materials are produced for this purpose and are typically sold as “smart glass”.

This is may be a good road to go down, and I believe Colin McFaul is providing more detail in another answer. I'm just curious if this property can be observed from all sides, or only from certain directions?


text:
Does the chemical in an alkaline battery make battery leaks unavoidable?
Battery leaks was an issue in the 80s and 90s, and since quality and innovation is constantly improving, I thought battery leak might be slowly going away.  But I found that even nowadays, name brand alkaline batteries still leak after a while, even if originally concealed in a tight package of the metal shell.  Is it true that the chemical in an alkaline battery inherently expands in volume over time, and make the alkaline battery leak issue unavoidable?

I'm betting that the alkaline solution just slowly eats the casing.

Nope, it eats zinc electrode. Wikipedia article about alkaline batteries has good discussion on the subject

@permeakra The wikipedia article is not bad, but one key feature is missing: Where is the KOH? Is it mixed with the Zn? Is it mixed with the manganese (II) oxide? KOH reacts with amphoteric oxides, like manganese (I) oxide and zinc oxide, which form as the battery discharges. I'm not sure where the hydrogen gas mentioned comes from; hydrogen usually forms in acidic redox. I would guess that water forms, and expands and/or vaporizes.

$KOH$ is electrolyte here, it has to have contact with both $MnO_2$ and $Zn$ . $Zn$ slowly reacts with solutions basic enough, the reaction is $Zn + 2KOH + 2H_2O = K_2[Zn(OH)_4] + H_2$. The reaction is slow and may be slowed even further by slowing diffusion in electrolyte solution, but not stopped. When battery is discharged (but not self-discharged), the reaction is different and no hydrogen produced. When battery cannot be discharged externally (too low voltage to give some current) it still can self-discharge and hydrogen produced will tear case apart if any Zn remains in battery.

@permeakra That is a pretty good explanation. You should turn it into an answer.

Also liquids tend to leak- it tough to make a container  liquid tight especially if its cheap like a battery. One of the factors in the research to create dynamite is nitroglycerin is a liquid so it leaks.

The reason for leaks in alkaline batteries is that as batteries discharge — either through usage or gradual self-discharge — the chemistry of the cells changes and some hydrogen gas is generated. This out-gassing increases pressure in the battery. Eventually, the excess pressure either ruptures the insulating seals at the end of the battery, or the outer metal canister, or both. In addition, as the battery ages, its steel outer canister may gradually corrode (in alkaline batteries there is some potassium hydroxide, a very strong base) or rust, which can further contribute to containment failure.
(see http://en.wikipedia.org/wiki/Alkaline_battery#Leaks )
I also found this online :
Heat is the biggest culprit. Storing your flashlights under the seat of your car, or in direct sunlight will cause the batteries to leak. The reason they leak is because they are filled with a chemical paste of which the liquid portion is composed of a base called Potassium Hydroxide. When the batteries get hot, the paste inside expands, and forces chemicals out of the seals in the bottom and top of the battery.
[...]
Regardless of the brand you buy,[...] all batteries will leak eventually, so if you don't use the flashlight for extended periods of 3 months or more, remove the batteries, or change them out every couple months.
(from here)


text:
Interpreting a formula tattoo
I saw

on the Internet and I'm curious as to what compound is represented by it.

The molecule seen in the tattoo is a protein-based mammalian hormone called oxytocin, commonly called "the love hormone" because it is involved in several aspects of sexual reproduction.

Image Source

IUPAC Name ,and other highlights?


text:
How can I identify and neutralize the chemical leaching out of new nitrile gloves?
I have a few cases of brand-new, nitrile gloves of a type and vendor that used to be good.  But apparently, the vendor has moved manufacturing to a different country and outsourced customer service to Hell. I can't return the gloves.
The gloves all have a chemical smell, resembling a burnt transistor with a hint of burnt rubber. If worn for a while, they irritate the skin.
Since this model of glove used to be outstanding, and the cases were not cheap, I'm hoping there's some way to salvage the gloves.  I've tried repeated washings in mild detergent, or just airing them out for 4 weeks. No change.
What chemical(s) is/are this likely to be?  How can I tell?
Is there a way to neutralize this/these compound(s) without otherwise destroying the gloves?

I'd just toss them out, a few cases of nitrile gloves aren't expensive and the amount of work it might take if it is at all possible would likely be not that cheap.

@MadScientist, 5 cases, retail value over $500 US. (and that's by no means the highest price for gloves with the same specs)   We currently are using substitutes and will trash these if needed, but I hate waste. (Think of the children!)

It seems we're talking about different kinds of gloves, I thought they were the standard disposable kind for around 5-10 EUR per 100 pack.

@MadScientist, disposable Nitrile gloves typically have either 500 or 1000 per case.  These are thick, exam-grade gloves with long cuffs.

@Brock Adams: Would it be possible to analyse the vapors using IR-spectroscopy?

@Spießbürger: Yes, if I can find a place that will do so cheaply.  My guess is that the chemical(s) is/are either a component of nitrile glove manufacture, or a cleaning solvent (or mold release, etc.), or some reaction of those types of substances.

Most probably it ain't organic solvent: they go easily. The best way to get rid of remains of organic solvents and other vaporous-able chemicals is to use vacuum: put gloves under mild (0.1-50 torr) vacuum and heat to about 50-80 Celsium (more heat - more result, but many plastics degrades or melts at hight temperature) for several hours (up to several days in some cases). The problem is that chemical may be unvaporousable and have to be washed with proper solvent. You have to know you enemy then, meaning close knowledge of manufacture process or simply drop this option as too costly.

Pack a glove into a glass container, let it sit for a few hours or days, get the container to your institute's mass spectroscopy and gas chromatography departments, and have them have a go at the gas phase in the container. I think that might be your best bet at finding out what is leaking out of these gloves. An IR or UV/Vis of the gas phase might also yield results, as Spießbürger suggested above.
Actually, if you got an analytics department at your university/institute, why not go to them, they might be able to have someone do a B.Sc. or M.Sc. thesis on that. Even if you end up trashing the gloves, at least then it won't have been a complete waste.

Okay, I'll see if the local JC has that equipment and is game.  Nearest university is 2 hours away by car. (I'll call them too.)

Wasn't able to do this, but this is the best answer so far.  The gloves were scrapped; what a waste.


text:
What does the reaction of chromium with ozone form?
I know almost nothing beyond 10th grade chemistry. I'm trying to name a web application, and was wondering what the products are of a reaction between the element Chromium and the gas Ozone (ozonolysis of chromium?).

Chromium Oxide, Chromia, and Chrome Green would all be options, as names for various oxides of Chromium

Ozone is a strong oxidizing agent, even stronger than molecular oxygen. But it is still a donor of oxygen atoms. Thus, depending on conditions, you'll have different chromium oxides. The most stable and common chromium oxide is chromium (III) oxide $\ce{Cr2O3}$. It has numerous nicknames. Other oxides are much harder to get.
Substances named ozonides do exist and in two different flavours.
Inorganic ozonides are salts of big, inoxidizable cations, like $\ce{Cs}$ and have formulas like $\ce{KO3}$. I'm not sure which cations have ozonides that have been already produced, but to my knowledge the only reasonably easy-to-produce ozonides are $\ce{KO3}$, $\ce{RbO3}$ and $\ce{CsO3}$ by reaction of the corresponding hydroxides with ozone.
Another type of ozonides is organic ozonides that are quite weird heterocyclic compound, which are product of reaction between alkenes and ozone.
If you need a name based on word games, you may consider that chromium is a derivative of an ancient word for 'color' and ozone, as a strong oxidizer, has bleaching and antiseptic properties.

It might well form CrO3- Chromium (VI) oxide.

It might just as well form nothing. I don't see why the passivation of chromium would be broken by ozone.


text:
Which of the following possesses the highest energy electron?

Which of the following possesses the highest energy electron?

$\ce{Br-}$
$\ce{Ca^2+}$
$\ce{Cr+}$
$\ce{As}$

Don't bromine and arsenic have equally high energy electrons (i.e. in the $\ce{4p}$ subshell)? My book says that arsenic is the correct answer. But I don't see why $\ce{Br^{–}}$ is wrong.

My best guess is that your book wants you to argue that, since the bromine nucleus has a higher charge than arsenic, electrons are more strongly attracted to it, lowering their energy. And indeed this would be true - if we only considered interactions between the electrons and the nucleus, and ignored electron-electron interactions. Unfortunately, however, the situation in real life is more complicated than this, because electrons repel one another, “shielding” each other from being attracted to the nucleus (which, of course, is the reason why subshells have different energies in the first place). And the bromide ion has three more electrons than the arsenic atom, which actually turns out to be quite significant.
In fact Webelements gives the electron affinity of $\ce{Br}$ (i.e., the first ionisation energy of $\ce{Br^-}$) as 324.8 kJ mol$^{-1}$ while the first ionisation energy of As is 947.0 kJ mol$^{-1}$. So in fact it seems that, experimentally, the bromide ion has the highest energy electron (which therefore requires the lowest energy to remove it).
This is a more difficult question than it looks - the problem of how to deal with interactions between lots of electrons is still an active research field! A fairer question to do in your head would be to compare, say, $\ce{Br^{2+}}$ with As: in this case the number of electrons of each species is the same, so that the argument I put forward at the beginning of this response is actually correct here.

The energy of orbital changes via row in periodic table and falls to the end of row. Because of it in orbitals for particles from the end of row have lesser energy.
The concept, however, seems... questionable.

$\ce{Br-}$ = $\ce{[Kr]}$. $\ce{Kr}$ is a noble gas and has a full  octet, meaning that the energy is SUPER low because of how stable it is.

Welcome to Chemistry.se! In its current state, your answer is not addressing the original issue of the question. Could you expand it by [edit]ing in some more details, please. Also have a look at the [help] and take the [tour].


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How to properly dispose of sodium...or store it properly
While performing an inventory of my school's chemical storage closet today, my fellow teacher and I found some sodium tucked away in the back corner of our flammables cabinet.  One container appears to have leaked the oil the sodium was being stored in, and we aren't sure about the second container.  We don't know how long it's been there, but neither of us is going to use it, and, honestly, we don't want it.  How do we dispose of this?  Can someone with a little more chemistry know-how point me in the right direction?  Are we stuck with this?
Or...
In the event that the administration won't let us dispose of the sodium, can you tell me how to remedy the storage situation?  We don't feel that there is an adequate amount of oil in the container for proper storage of the remaining sodium, and if we have to keep it I want to make sure it is being stored correctly.

3g of sodium + water http://youtu.be/_WY7ayBipBM, keep in mind if wish to dispose it using open water. Small pieces (about 0.1-0.2 g) dies without such pornography, but it will be loooong.

It might be useful to know how much sodium we are talking about, a few grams or much more?

@permeakra ..."pornography"?

You can use what's left as lye if it has oxidized all the way

If you handle sodium, the first thing you need is a way to extinguish it if it catches on fire. You should have a bucket of sand ready to use, and additionally a class D fire extinguisher, if available. I'm assuming you know this, but don't let the sodium come into contact with water.
You always want the sodium stored in an inert liquid, usually mineral oil is used for that. Organic solvents like n-hexane are also used for short time periods. You should refill the mineral oil immediately.
The usual way to neutralize sodium is to put small pieces in an organic solvent like n-hexane and slowly add either isopropyl alcohol or ethanol to it. Isopropyl alcohol reacts slower, and is therefore a bit safer. When all the sodium is dissolved, you can add water to make sure there is no elementary sodium left.
How dangerous this is depends a lot on the scale, it's easy to do if you are only getting rid of a bit of sodium you haven't used in your reaction. I have no experience with neutralizing large amounts of sodium, this method should still work but it might take a long time if you are careful as you'll have to do it in small batches.
To be honest, the fact that you asked this question here on the internet indicates to me that you should at least let someone more experienced supervise the procedure.

I just wanted to comment I cannot agree more.  If neither you nor your teacher are sure how to handle the disposal, neither of you really should be doing it.

I would only add that if the school does not have a designated person to handle this, the best option is probably to call the local fire department. They will either be able to dispose of the sodium, or will put the OP in touch with someone who will.

If there is a nearby college or university with a chemistry department, someone at that department would likely be willing to help with disposal.

Yeah...I'm a biology teacher and have absolutely zero use for pure sodium.  It's only a few grams and is probably left over from a previous chemistry teacher who no longer works with us.  There is no way I'm undertaking a neutralization like this on my own--and quite honestly I don't have the time to really do this properly.  For all of sodium's interesting properties, I have no desire to blow up my classroom/science lab.  We have permission to dispose of it and Emory University is right around the corner.  We might get some help from them.  Thanks, guys!

$+1$ for the last paragraph.

Roesky published a wounderful method based on flowerpots and dry sand, and a shallow tray of water. I regard this as my excalibur for slaying even the worst and most disgusting alkali metal monsters.
What you do is to put some dry sand in a large flower pot. Add the sodium or potassium scrap. Next fill the pot up with dry sand. Now stand the flower pot in a shallow tray with a little water. The water slowly rises up inside the flower pot over several days. The sodium is quenched by the water, the sodium hydroxide tends to react with the sand to form solid silicate. If something happens then this method has far less flammable matter which can become involved in a fire. This is why I like it. With the IPA method the use of IPA greatly increases the size of the fire which you can have if something goes wrong.
I hate the disposal method based on isopropanol / ethanol with a deep passion. I have had to fight a fire resulting from when this method went wrong. One of my lab mates added some sodium scrap to a large bowl of IPA and which already contains a lot of sodium scrap. The whole lot went up like the olympic torch.
I was out of the lab at the time going to see my PhD supervisor, I wanted him to buy me a chemical. He was out of his office, as I turned around to go back to the lab I saw the yellow flames licking up to about waist height. I rushed back to the lab.
Knowing that the bowls in question were typically used for the disposal of sodium, I grabbed a dry powder extingisher. A young lady with a CO2 extingisher came at the fire at a path at 90 degrees from mine. She gave the fire a squirt of CO2. The fire meekly complied and died. But as soon as she stopped discharging CO2 it jumped back up and continued its reign of chaos.
I then set to work upon the vile monster with my ABC powder extingisher, I had pulled the safety pin (droped it) and I took aim at the vile dragon and I squaezed hard on the levers. to my horror the bowl started to slide backwards away from me as soon as my jet of white powder hit the fire. I stopped my discharge and quickly reconsidered what I was doing. I then looked at the still burning fire and then I squezed more in a more gentle way and I let out a gentle stream of powder into the fire. I laid down white powder into the fire and as the fire became smaller I stepped closer and closer until I was right at the edge of the bowl which was the size of a frying pan. As I covered the fire with powder it died out and I found myself standing next to bowl of white solid in a christmas card scene. The floor around me was covered in white powder.
Chubb sell a special metal fires extingisher under the name "Pyromet" this is a special dry powder system which applies powder in a very gentle way. It uses graphite which is delivered with a special lance. I was impressed when I saw a film of it being demonstrated on a pile of burning magnesium turnings.
My own cheap and cheerful version of the pyromet system is to fill a polyethene bag with dry sand. I keep a bucket of loose sand in my lab and a bucket filled with these sand bags. If you have a sodium fire then lift one of these sand bags onto the fire with tongs or some other method which keeps your hands away from the fire. The heat of the fire will melt the bag and the sand comes out onto the fire thus covering it with sand.
I think that this is much more easy than trying to deliver sand with a spade or a large spoon onto a fire. I have used this method to control burning sodium during what I regard as controlled sodium burns while disposing of the residues from using sodium sand to convert ethyl esters into TMS protected hydroxy ketones.
My method is to remove the sodium and sodium chloride from the reaction mixture by filtration through celite under nitrogen gas using schlenk methods to handle the mixture. After finishing the workup and using the vacuum pump to remove solvents from the celite pad. I would clear all flammables from the workplace and deal with the leftover sodium sand. I would use a large steel can with a hole in the bottom as my flower pot. I would put in some dry sand. I would then unstopper the schlenk filter and with great care tip the solid residue out into the steel can. A few times it ignited during this operation and then started to burn in the can. I would then normally add a bag of dry sand to the top of the steel can and then wait. This method quickly brought the burning sodium under control.
Looking back I think that with 20 years of experience and my sandbag method I would have managed the sodium fire I put out as a postgrad a lot faster and with greater ease. But it was the first fire I had to fight in my life as a chemist. I hold the view that when you have had to use a fire extingisher for real that you will become a better chemist. I also hold the view that practise fires do not count, when you fight a practise fire you have someone who is experienced in fire control with you. If things get out of hand then they will step in.
I was taught a few simple rules when fighting a fire

Always approuch a fire in a direction which allows you to escape, it is best to be between the fire and the door to the room. Rather than have the fire between you and the door.
If the fire gets worse despite your best efforts, then retreat away from it.
Before you do battle with the monster, give your extingisher a quick squirt at a direction away from the fire. If it fails to operate then check the safety pin is out and with dry powder then give it a mighty shake. Also if something unexpected comes out of the extingisher then retreat away from the fire. I have heard of horror stories of fire extingisher service workers who fail to add the chemcial additives to the water placed in a foam extingisher.
If you develop any breathing difficulties while doing battle with the fire then retreat away, stop doing your St-Geroge act and leave the dragon alone.
If you see gas cylinders threatened by the fire then back away, make sure you tell the fire fighters when they turn up about the cylinders.
If it is a fire which involves escaping gas then do not use an extingihser, doing so will set you up for a gas explosion. Instead if you can disconnect the gas supply by closing the valve on the cylinder or the main gas valve for the lab. Then if the fire is still burning then use an extingisher on it. If you can not put out the fire by cutting off the gas supply then leave the fire to a professional firefighter.
If in doubt then retreat away from it.
If in further doubt then see rule 7.

I later managed to see my PhD supervisor and he did buy me that chemical, and I was able to use the chemical to do a reaction which was the basis of a good paper. One of the best chemcials I have ever asked someone to buy for me.
Like the person who gave the other answer I strongly hold the view that if you have to ask a basic question on sodium metal disposal it suggests to me that you need to ask someone who is more trained / experienced than you to deal with the problem. One of the greatest books I have ever read on radiation safety has the simple crisp advice of "refuse any work which you are either not trained for or equipped for". While this advice was targetted at industrial radiographers, it also applies to unwanted sodium metal.
A final warning, while I have chosen to share some of my experience of sodium (and potassium) and fire control, when you choose to deal with these things you have be fit to do what you need (or want) to do. Before you work as a chemist you should learn how to put out fires and manage accidents / mishapes. If you do not know how to then you should stop work and seek expert advice / training. Also if you apply fire control methods wrongly or handle sodium wrongly you endanger yourself and others, so I would suggest you think long and hard before either choosing to be willing to fight fires or work with sodium metal. I accept no responsibility for what happens to your with either fires or sodium metal.

Your hatred about the isopropanol method is not a criticism of the method but of the way it was performed. It is a safe method as long as you don't try to dispose of large quantities in one go. Adding small quantities of the metal to the liquid is relatively safe.


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Why don't atoms with expanded valence shells have a formal charge?
I'm working through Chemistry - Principles and Practice, 3rd Edition and I have few complaints so I don't think it's the text. At one point, I was asked to draw the Lewis structure, with formal charge, of SO3. The central sulfur atom has six covalent bonds with the surrounding oxygen atoms. Based on the calculation for formal charge, you'd expect one of +3, but there is none. How am I to know when there is or isn't a formal charge, and why?

Formal charge is the difference between the number of valence electrons that a neutral unbonded atom has and the number of valence electrons assigned to that atom when after it forms bonds in a molecule. It looks like you are making your mistake in assigning the proper number of electrons to the sulfur atom.
The electrons assigned to an atom are:

All nonbonding (lone pair) electrons on that atom in the Lewis Structure.
One half of all bonding electrons (one electron for each single bond, two electrons for each double bond, etc.).

Consider water, $\ce{H2O}$. The oxygen atom has two bonds and two lone pairs. Oxygen originally had 6 valence electrons. We assign 6 electrons to it (4 nonbonding and 2 bonding). The formal charge is $6-6=0$.

Now consider the hydroxide ion, $\ce{OH-}$. The oxygen in hydroxide has 7 electron assigned to it: 6 nonbonding and 1 bonding. The formal charge is $6-7=-1$.

Now consider the nitrogen atom in the nitrate ion, $\ce{NO3-}$. Nitrogen originally has 5 electrons. In the nitrate ion we assign 4 electrons to that nitrogen (1 for each single bond and 2 for the double bond. There are no lone pairs). The formal charge is $5-4=+1$.

As a final example, let's look at the P in phosphate, $\ce{PO4^3-} $. Phosphorous original had 5 electrons, like nitrogen. We assign 5 electrons to phosphorous in phosphate (1 for each P-O bond and two for the P=O bond). The formal charge is $5-5-=0$
How many electrons are assigned to the sulfur atom in $\ce{SO3}$? From the question it sounds like you have the correct Lewis Structure. It looks like you are making your mistake in assigning the proper number of electrons based on the number of bonds.


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Can acidic conditions break disulfide bonds
I am denaturing a protein using organic solvent and acid (49:49:2% water:methanol:acetic acid), but I want to maintain the disulfide bonds.
My chemistry knowledge isn't good but disulfides are broken under reducing conditions and the reaction also uses free protons:

So I wanted to check if acids could break disulfide bonds without reducing conditions.

No, acidic conditions (i.e. low pH) shouldn't be enough by itself to break a disulfide bond. The main reaction that breaks a S–S bond is cleavage by reduction. This is in a way similar to peroxides, which are stable even at very low pH (see e.g. mixtures of sulfuric acid and hydrogen peroxide).

Short or long term exposure to reagents? If long term, any oxygen (dissolved or otherwise) presence? If yes, note this paper https://pubs.acs.org/doi/pdf/10.1021/acs.jpcc.5b10949 , describing a complex surface-based reaction of oxygen and water slowly acting on FeS2:
FeS2 + 7/2 O2 + H2O = Fe(2+) + 2 SO4(2-) + 2 H+


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What legal guidelines exist for completeness of a Materials Safety Data Sheet?
I know that the answer to this question will vary by country, but some aspects must be similar amongst those countries that have authorities overseeing chemical and industrial hygiene. The Wikipedia article on the topic is not so good. My understanding for the US is that a MSDS must be correct and complete to the best knowledge of the preparer of the sheet.
My question is thus:

Is there are legal requirement to the correctness and completeness (to the best of the preparer's knowledge) for a MSDS? If not, then why bother with MSDS at all?

This question was inspired by the commercial for this product from 3M. You can view the commercial by clicking on the videos link at the bottom of the site. Never mind how much nonsense their claim of "botanically pure ingredients" is, I was intrigued by the MSDS link at the top of the page. I wanted to know what the "botanically pure" substances were.
There are two versions of this product, with different scents. The MSDS for each is linked: Breeze and Lemongrass. Section 2: Ingredients lists the same four items for both products:

Water 65-80%
Cloth 20-30%
Sodium lauryl sulfate <3% (surfactant, not a naturally occurring substance)
Thyme Oil <0.1% (antibacterial, at least thymol is, and botanical in origin, but certainly not a pure substance)

No odorant is listed in the ingredients, but Section 3 clearly indicated they have different smells. 3M knows what odorant was added, and if legal requirements exist for completeness, then 3M is earning a finger-shaking and a slap on the wrist. So, again,

Is there are legal requirement to the correctness and completeness (to the best of the preparer's knowledge) for a MSDS? If not, then why bother with MSDS at all?

OSHA's FAQ says that chemicals that make up less than 1% (0.1% if it is a carcinogen) of a product do not need to be listed on the MSDS. So it looks like 3M could be in compliance if the oderants are present in trace amounts (in that case, I have no idea why they feel the need to include the thyme oil).
The only relevant thing OSHA seems to say about consumer products specifically is that employers are not required to provide an MSDS for consumer products used in the workplace, as long as the consumer product is used in the same manner as a consumer would.
The Consumer Product Safety Commission specifically disclaims jurisdiction over chemical products, and the Chemical Safety Board doesn't appear to have anything related to consumer chemical products.


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Is there a simple way to get the circular dichroism of a molecule from its structure?
Are there any heuristics to get the relative absorbtion of left and right circularly polarized light by a molecule from its molecular structure? Is it even possible to predict which polarization is selectively absorbed?

My understanding is that the general problem is an open research question. This is tangentially related to my own research, so I will check a couple of books tomorrow for relevant discussion.

Is there a simple way to get the circular dichroism of a molecule from
  it's structure?

Short Answer: Yes, for many molecules if you know the molecule's structure, then you can predict the shape of the optical rotary dispersion (ORD) or circular dichroism (CD) curve.  Conversely, and perhaps more importantly, if you know the shape of the ORD or CD curve, then you can predict the molecule's structure, in particular, the absolute configuration at chiral centers in the molecule. There are many empirical techniques, such as the "Octant" rule.  These methods are based on numerous experimental observations that have been codified into sets of rules for different classes of compounds.  These empirical rules can be used to determine the absolute configuration of cyclic ketones, conjugated dienes, alpha, beta-unsaturated ketones, twisted biphenyls as well as other systems.
Long Answer: When polarized light is passed through a chiral sample and the light is absorbed, there are two physical phenomenon that can be measured.  One is called the circular birefringence and it measures the difference in the refractive indices ($\ce{n_{R}, n_{L}}$) of the sample when a beam of right and left circularly polarized light is passed through the sample.  When this difference is plotted against wavelength an ORD curve results.  The other parameter is called circular dichroism and it measures the difference in the molar extinction coefficients ($\ce{\epsilon_{R}, \epsilon_{L}}$) when right and left circularly polarized light passes through the sample.  When this difference is plotted against wavelength a CD curve results.  The dependence of these two phenomenon on wavelength is termed the Cotton effect.
The following figure shows the uv-vis absorption (blue line), ORD curve (green line) and CD curve (red line) for the two enantiomers of camphor sulfonic acid.  Note how the ORD and CD curves for one enantiomer are mirror images of the same curves for the other enantiomer.  It is the carbonyl chromophore's $\ce{n->\pi^{\ast}}$ absorption that is responsible for the curves seen in this image.  In ORD spectra there are typically 2 maxima, one positive and one negative.  The positive maxima is a "peak", while the negative maxima is termed a "trough". If, when moving from longer to shorter wavelength, a peak is encountered first, then the ORD curve is said to be positive (+).  Conversely, if a trough is encountered first, then the ORD curve is negative (-).

image source
Chromophores can be divided into two groups, symmetric or inherently dissymmetric.  A carbonyl, nitro and olefin are examples of groups that, by themselves, are symmetric.  They may be perturbed by asymmetry on other atoms around them, but alone, by themselves, they possess no inherent asymmetry.  On the other hand, groups such as helicenes, twisted biphenyls and unsaturated ketones possess a twist or handedness; such molecules are inherently dissymmetric.  The intensity of an ORD or CD curve from a molecule containing an inherently dissymmetric chromophore is usually much greater than the peak intensity from a molecule containing a symmetric chromophore.
With that information as background, we can now apply these concepts to the examination of real molecules.  One of the most studied classes of molecules where these concepts have been applied are the substituted cyclohexanones.  In the early 60's  Djerassi, Moffitt, Woodward, Moscowitz and Klyne (some of the greats!) developed an empirical set of rules to predict the absolute configuration of chiral centers in substituted cyclohexanones.  The rule became known as the "Octant" rule and came about because of the work they were doing with complex natural products that often contained a substituted cyclohexanone ring.  They would run a reaction on a pure enantiomer of a cyclohexanone containing natural product and wanted a method to quickly determine if a chiral carbon in the molecule retained its absolute configuration or if the configuration was inverted in the reaction.  After comparing ORD curves from countless cyclohexanone containing natural products, they were able to compile their observations into the Octant rule.
As its name implies, the Octant rule divides the region around a cyclohexanone carbonyl into 8 regions or octants.  As the picture below (left part of figure) illustrates, the octants are generated by bisecting the carbonyl group with 3 planes.  Two of the planes contain the carbonyl group.  One divides the carbonyl into top and bottom regions and the other divides the carbonyl into right and left regions.  The third plane actually passes through the middle of the carbonyl and is perpendicular to the 2 other planes.  This third plane divides the molecule into front and back regions.  There are a few molecules that do have substituents in the front region, but most molecules do not, so we can usually simplify our analysis by only considering the 4 rear octants.  Each octant is also assigned a plus or minus symbol telling us whether we would expect a substituent in the octant to contribute in a positive or negative manner to the ORD or CD curves.

image source
The figure on the right shows how a cyclohexanone would be placed in the octant diagram and how its projection appears as a rectangle when sighted down the carbonyl group.
Let's examine the case of 3-methyl cyclohexanone.  First let’s draw the possible conformers of both S- and R-3-methylcyclohexanone

Now let’s take the 2 possible conformations for R-3-methylcyclohexanone (axial and equatorial methyl group) and project them onto the octant diagram.  We see that the equatorial conformation has a positive contribution to the Cotton effect, while the axial conformer has a negative contribution.  At equilibrium, the molecule exists predominantly in the conformation with the methyl group equatorial, therefore we expect the positive Cotton effect from this conformer to outweigh the negative contribution to the Cotton effect from the axial conformer.  Therefore, we expect a positive ORD and CD curve for R-3-methylcyclohexanone.  Similar analysis leads to the expectation of a negative ORD and CD curve for the S enantiomer.  Indeed, experimentally it is found that the ORD curve for R-3-methylcyclohexanone is positive and negative for S-3-methylcyclohexanone.  The full name for the R enantiomer is R-(+)-3-methylcyclohexanone

image source
More detail on the 3-cyclohexanone example, as well a nice, concise review of ORD can be found here.
Books have been written on this subject so there is much that I have left out.  For those who would like to explore further here are a few phrases that, in addition to the previous link, can help you get started:  Cotton effect, ORD biological applications, and magnetic circular dichroism

Predicting the circular dichroism of a molecule ab initio is possible, but pretty difficult. The only way I know of doing that is advanced quantum chemistry calculations, for example in the time-dependent density functional theory (TD-DFT) approach. You can find an example of its use on a small organic molecule in this publication, and a mention of this on the VCD page of Wikipedia.
Regarding heuristics, I don't think there are any… a fully characterization of the spectrum by quantum chemistry calculations seem unavoidable due to the complexity of the problem.


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Spectral analysis of periodic table of elements in time domain
I'm trying to get the spectral analysis data of the periodic table of elements in the time domain format or find a site that has the data in the time domain format.  All the sites
I know have the data in the frequency domain which is great but I need to do my own fft analysis starting from the time domain because the frequency domain does not include the phase angle data, which I also need to use for my calculations/animations.
Does anyone know where I can get this data or can someone send me the spectral analysis data for sodium and potassium in the time domain?
Thanks

What do you need the time domain for? Depending on what you need them for, you could just take the frequency domain data and do an inverse fourier transform.

@Dan That won't work since the phase angle information is not included when you use the frequency domain, that is the reason I need to start from the time domain

What do you need this for? Why do you need the phase angle?

@Dan sorry if the question wasn't clear but I need to run my own fft analysis starting from the time domain not the frequency domain because I also need the phase angle data to use in my calculations/animations. I'll update the question. If this is the wrong forum to ask this question can someone tell me the correct one.

The question was clear on that point. What I want to know is why you need to run your own FFT analysis and what you need that phase angle data for.

Also, what frequency range do you want the data in? If it's IR or higher, said tables probably don't exist. It's hard to get phase information at those frequencies, and the measurements are made directly in the frequency domain.

Its unclear what you want. Is it the lifetime of excited states of atoms, or IR spectra as a function of time as a species reacts or what? Absorption or emission spectra can only very rarely be used to get time information as the line shapes are distorted by process such as Doppler and collisional broadening and inverting is then pointless.


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Modern alternatives to Gaussian
Specifically, I'm curious if there are any programs which come close to Gaussian in breadth but also take advantage of things such as GPU processing.
There's a page on wikipedia that goes through several options but I'd like to get feedback from actual users.

Personally, GPU code is still very young. My lab won't even consider it as a viable resource until (much) more features are added and bugs fleshed out.  Off the top of my head?  No GPU code exists that comes close to mirroring Gaussian or other CPU alternative (its been a while since I've checked).  I leave this as a comment since I cannot address GPU programs directly but if you were interested in others I'd be happy to write something up.

Hi Radu, your question is very very broad. There are hundreds of computational chemistry codes out there, so you really need to specify what specific features (quantum chemistry methods, approximations, etc.) you're looking for. Then we can better help you…

Tempted to -1 for a question requesting an open-ended or list-based response, but I can't remember if we do that here.

Hi. You should look at this page for a VERY brief intro to some linear algebra GPU packages. If you're really into computational chemistry, you don't need to use Gaussian at all - every method available in Gaussian is first implemented by people writing their own code, then ported to Gaussian. You can develop your own GPU code in any language you want, and then port it to Gaussian. Don't expect to get paid, though.

And when I mean "ported to Gaussian", I mean written in FORTRAN and then added to the program. No personal experience with Gaussian, so that's from a couple of people working in the field who have.

"Specifically, I'm curious if there are any programs which come close to Gaussian in breadth"
Not really, as far as I can tell. Gaussian is a massive work by many, many people over many, many years, and the reason it's still one of the most popular packages, as far as I can tell, is that it does most things (notably excluding non-GTOs), even if it's not necessarily the fastest or the best at the things it does in every case.
"but also take advantage of things such as GPU processing."
Double no. As LordStryker says above, GPU programming is pretty new, and most of the resources I've seen for it are C-based, while a significant chunk of scientific development still takes place in Fortran. I suspect library-based encapsulation of GPU-running numerical routines may improve this over time (I mean, by introducing people to the idea of getting speedup by using the GPUs more easily), but I wouldn't hold my breath. It'll be interesting to see what hybrid chips with multiple types of core, or the Xeon Phis will be able to do, though.

Regarding the first part, I think QChem is the closest program to being a full-fledged Gaussian alternative/replacement. Their documentation is a thing of beauty as well.

I've always heard GAMESS was the canonical alternative to Gaussian.  I don't know much about either, though, aside from the fact that GAMESS doesn't use GPU processing either and is free as in beer.
I don't know how these programs work on a low level, but matrix-vector operations usually don't get much of a speed boost from using GPUs. If these sorts of packages need lots of matrix-vector or vector-vector operations, GPU processing would probably not speed things up very much.

Can you justify/explain your claims that linear algebra does not benefit from massive parallelism offered by GPUs?

At seminar given at my university, the speaker stated that Matrix-Matrix operations could have their run time reduced by a factor of ~10 over a single CPU, matrix-vector operations by a  factor of ~3, and essentially no speedup for vector-vector operations. It's not that BLAS level 2 operations wouldn't benefit from GPU, its that the benefit isn't as great as running it on several CPUs.

Ah, I was under the impression that such operations would benefit from massive parallelization.

@CHM: I've learned a bit more about GPU processing, and it seems that quantum chemistry programs can benefit quite a bit, but that that basis sets and algorithms used for multiple CPUs are totally unsuitable for GPUs. As a general rule, you usually have to build your program to use GPUs from the ground up.

There is Firefly (the package formerly known as PC GAMESS:).
Firefly also has CUDA capability. Here are some benchmarks.
Firefly is free and the project is active (latest release is from 2015.09.05).

Disclaimer: It's about two years since I've last used Firefly and I never used in on the GPUs.