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Hypervalency and the octet rule
I realize that the octet rule is more a suggestion than a rule, and that it applies mainly to non-transition metal compounds. Still, compounds that don't have an octet, like $\ce{BH3}$, tend to dimerize or disproportionate in order to achieve an octet.
In recent SE Chem posts discussing the bonding in phosphorous pentachloride and iodine heptafluoride, hypervalent bonding was used to describe the bonding in these molecules.

If we count the electrons around phosphorous in $\ce{PCl5}$ there are 6 in the three equatorial bonds (2x3) and 4 in the hypervalent axial bonds; a total of 10. However 2 of the electrons in the hypervalent bond are in a non-bonding orbital, leaving 8 bonding electrons, an octet.
If we count the electrons around iodine in $\ce{IF7}$ there are 4 in the two axial bonds (2x2) and 10 in the hypervalent equatorial bonds; a total of 14. However, 4 electrons are in a hypervalent non-bonding orbital and 2 more electrons can be viewed as being associated with a fluoride anion, leaving 8 bonding electrons, again, an octet.
Are these compounds considered to be exceptions to the octet "rule"? Do all compounds involving hypervalent bonding violate the octet "rule"? Or do these compounds comply with the octet "rule"?

Not sure if this would be of interest, but it seems related to your question - Identification of an iridium-containing compound with a formal oxidation state of IX Wang et al. 2014, Nature

Out there in the real world, university students and school pupils alike favour strict rules that are true as often as possible (or at least have clear, easily remembered exceptions), while their professors and teachers on the other hand try to tell them that these rules are merely to be considered ‘guidelines’ rather than actual rules (and attempt to welcome them aboard the Black Pearl).
And from this point on is where the question turns philosophical very quickly. One side will say

Ah, so the rule is still valid in this surrounding; I just need to remember to do A and B when applying it!

while the other side will reply

Ah, but this shows us that the so-called ‘rule’ can no longer be strictly applied because you need to consider data C and D that point towards the real truth.

The thing is, both sides are somehow correct while parts of their descriptions remain wrong. Maybe we can think of these as two mesomeric descriptions of the same principle?

As we all (should) know, all our concepts of describing bonding and electron location are mere approximations of the actual underlying truth. Hence, all the rules that were ever laid out — like a bond consists of two electrons between two atoms, oxidation states or the octet rule — are again just mere approximations. They cannot be correct nor can they be properly ‘valid’. Thus, what isn't properly true can't be fulfilled or invalidated.
I think that is the most theoretical answer to the question.

From a practical point of view, it mostly depends on how strict you postulated the octet rule, and how quickly you would consider it violated.
Traditionally, the octet rule wasn't valued highly for elements that weren't carbon, nitrogen, oxygen or fluorine. As such, anything that violated the octet rule in its strictest sense was considered a violation (and proof for its invalidity).
Nowadays, my observations imply that the octet rule is gaining support again, and some of the simpler cases that were hither considered violations are now something more like special cases or maybe just octet rule with additional mesomery. In the eyes of those supporting this side, these compounds comply with the octet rule.
And of course, there is the large grey space between the black and the white.

I for my part tend to favour rules like the octet rule very much — although I tend to use a not-as-standard defintion for it:

s- or p-block elements will always favour surroundings where the number of valence electrons is 8 or less. For the two elements of the first period, this reduces itself to 2 or less.
  Mesomeric structures that allow an all-octet are to be favoured over those that do not.

Under this definition, which relies heavily on mesomery for cases such as $\ce{IF7}$ or $\ce{PCl5}$ while it allows for a single, unambiguous description of $\ce{SO4^{2-}}$, these cases comply to the octet rule and actually further prove its validity. Others will surely disagree.

I do not understand your definition. What do you mean with with " all-octet are to be favoured"? How does that apply to the $\ce{IF7}$ case?

@Martin That sub-rule isn't really meant for the $\ce{IF7}$ case, although it can be applied there. It's meant for things like carbocations with adjacent oxygens where the positive charge can be mesomerically relocated to oxygen. For $\ce{IF7}$ it means I would have iodine as $\ce{I^{3+}}$ with four bonds to fluorides plus three fluoride anions; the whole lot of them mesomerrily swapping bonds. But considering I know and am familiar with 4-electron-3-centre bonds, I might just draw three of those plus one standard 'single' bond and mesomerically swap through those.

In general, the octet rule gets less useful as you move down the periodic table (one early teacher of mine cautioned against using it for elements other than $\ce{H}$, $\ce{C}$, $\ce{N}$, $\ce{O}$ and $\ce{F}$). You can "force compliance" with the octet rule for molecules like $\ce{IF7}$ in the manner you describe (semi-ionic character etc.) but you quickly get to the point where the rule loses its predictive power, because more subtle theories of bonding are required to justify the octet rule's conclusions.


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What is the relation between surface tension and initial molecular harmonic?
Initial harmonic is measured as Hartree/Bohr^2 ( like IHarmonic=n in Gaussian Software ). As I am from physics background I am used to look at features from their dimension point of view:

Actually surface tension is the only physical feature that I know is compatible with Force / Distance. I was wondering if there is any relation between the surface tension and initial harmonic.
Update: Another physical feature is spring constant which is not irrelevant also !

Aren't these the same units as the spring constant $k$ from a classical spring obeying Hooke's law $F=-kx$?

@CurtF. yes it is.  I was thinking that in Molecular Mechanics, atoms are assumed as oscillating particles on springs. The stronger the spring, the smaller the volume. If we look at the whole system as a bag of oscillating particles ( if we can ! ) , the surface tension can be proportional to this constant ( instead of springs avoiding outward movement, it can be assumed as surface tension doing the same job inward ) . Is there such a concept ?

Surface tension arises from the fewer neighbors that molecules at the surface have, relative to the number of neighbors for molecules in the bulk.
For example, consider a simple model of rigid identical spheres in a hexagonal closest packing.  In the bulk, there are 12 spheres in contact with a given sphere, which could be thought of as 6 spheres at vertices of a regular hexgon surrounding the given sphere, 3 spheres in a triangle below and 3 spheres in a tringle above.
At a surface, at least 3 of the neighboring spheres would be absent.
Considering an interaction energy between neighboring spheres, the loss of interaction energy (due to fewer than 12 neighbors per sphere) will be proportional to the surface area of the system of spheres.  The minimum energy of the system will be when surface area of the arrangement of spheres is minimized.
For the above reason, the units of surface tension are Energy/surface area, which the OP correctly points out is the same as force/distance.
From the above explanation, particularly that even for rigid spheres always in direct contact the Energy/surface area (force/distance) relationship arises, it can be understood that surface tension has no particular relationship with a force constant between molecules (or between atoms of a molecule).  Instead, surface tension is related to the energy difference between a molecule with maximum neighbors and a molecule with fewer neighbors.
Considering a simple harmonic oscillator, energy vs distance is a parabola.  There is no horizonal asymptote.  Regardless of what the force constant is, there is always infinite energy required to remove a neighbor. In reality, intermolecular force will approach zero at infinite distance between molecules. It is the difference between energy at equilibrium seperation and infinite separation that is relavent to surface tension, and force constant does not provide this information.
The terms "initial molecular harmonic" and "initial harmonic" are not standard chemistry terms and only have meaning in the context of a particular computer program.  "Initialharmonic" or "Iharmonic" refers to a force constant that can be for linear stretching, bending about a vertex that is an atom, or dihedral bending, etc.

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Can acid mine drainage form acid rain?
Can acid mine drainage form acid rain?
Since normal acid rain is formed via combustion of fossil fuels, the sulphur dioxide and nitrogen dioxide are all in the gas phase, whereas in acid mines the sulphuric acid is in solution.
Here's the reaction for acid rock drainage if it helps:
$$\ce{FeS2 + 3.75 O2 + 0.5 H2O → Fe^3+ + 2 SO4^2- + H+}$$

I don't think acid rain formation from liquid acid mine drainage is very likely.  The reason is that in acid rock drainage (a.k.a. acid mine drainage or AMD), the acidity is caused by sulfuric acid.  Sulfuric acid, $\ce{H2SO4}$ is much less volatile than water.  It simply won't appreciably evaporate from bulk liquids at normal conditions on the Earth.
There is one condition where acid mine drainage could possibly contribute to the formation of acid rain: if there were lots of aerosols formed from acid mine drainage waters.  If there were an AMD-laden stream or creek that flowed over a waterfall or was otherwise agitated enough to form very small droplets and splashes, some sulfuric acid might travel to the atmosphere in the small aerosol droplets.  That could contribute to acid rain, although a more likely effect would be a much more local (tens to hundreds of feet) acidification of the environment.
However, even then, I suspect that sulfur from coal combustion would be the dominant source of airborne sulfuric acid in most regions.  In the United States, coal consumption is roughly one gigaton per year, leading to about 20 megatons of sulfur emissions to the atmosphere.

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Nitration of pyrrole with sulfuric and nitric acids
Would 2,5-dinitropyrrole be a product of nitration of pyrrole with $\ce{H2SO4/HNO3}$? It seems that 2,5-dinitropyrrole is not formed when using $\ce{Ac2O/HNO3}$ (ref 1). Is this just because of the mild conditions used, or does this reflect the pyrrole ring being more deactivated by the first nitration than a benzene ring would be?
1 Cooksey, A. R.; Morgan, K. J.; Morrey, D. P. Nitropyrroles—II : The nitration of pyrrole. Tetrahedron 1970, 26 (21), 5101–5111. DOI:10.1016/S0040-4020(01)93163-1

Pyrrole is more activated towards electrophilic aromatic substitution, due to the electron donation of nitrogen atom.

Yes, but is 2-nitropyrrole activated towards electrophilic substitution relative to nitrobenzene? I assume it would be but then unfunctionalized dinitropyrroles should be relatively easy to make (unless the harsh conditions cause other side reactions). Sorry if I was being unclear !

Would 2,5-dinitropyrrole be a product of nitration of pyrrole with $\ce{H2SO4/HNO3}$?

I doubt it and would rather expect a lot of tar to be formed here. Pyrrole tends to undergo acid-catalyzed polymerization!
For decades, acetyl nitrate, formed by addition of fuming nitric acid to acetic anhydride, is the reagent of choice.
Mononitration of pyrrole predominantly gives 2-nitropyrrole, accompanied by minor amounts of 3-nitropyrrole. (K.J. Morgan and D.P. Morrey, Tetrahedron, 1966, 22, 57–62. DOI).
In the case of 1-methylpyrrole, the yield of the 3-nitropyrole is somewhat larger (H.J. Anderson, Canad. J. Chem, 1957, 35, 23–29). In you want even more of the 3-nitro isomer, a more bulky alkyl substituent (t-Bu) or N-silylation with $\ce{((CH3)CH)2SiCl}$ is a viable approach.
Nitration of 2-nitropyrrole at −15 °C with nitric acid in acetic anhydride  furnishes a 4:1 mixture of 2,4- and 2,5-dinitropyrrole with a total yield of 61%. (R. Alan Jones, Gerritt P. Bean, The Chemistry of Pyrroles: Organic Chemistry: A Series of Monographs, Vol. 34, Academic Press, p. 124)

Safety concerns should be indicated here, particularly operating with concentrated nitric acid and any organic layers.


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What are isodiaphers?
We know that isodiaphers have same difference in neutrons and protons, but I encountered another definition:

Atoms which have same isotopic axis are called isodiaphers.

I don't know what it means by "isotopic axis". Please try to explain in simple words.

These are basic terms of radiochemistry:
Isotopes are nuclides with a common number of protons $Z$.
Isobars are nuclides with a common number of nucleons $A$.
Isotones are nuclides with a common number of neutrons $N$.
Isodiaphers are nuclides with a common neutron excess $N-Z=A-2Z$.
Thus, in the chart of the nuclides, isodiaphers are found in diagonals from bottom left to top right. For example, the mother and daughter nuclides of alpha decay are isodiaphers.

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Can a meso compund be rotated in ANY sort of way in order to show its achirality?
Are there any limits to rotating a compound in order to prove that its chirality or achirality? (assuming that you're rotating about a single bond)

Usually no, then you would enter the realm of conformations. Some conformation are notoriously unstable, but still you cannot really talk about stereoisomerism.

Limit is rotation energy - hence existence of atropoisomery.

Are you talking about bond rotation or rotation of the entire molecule?

I'm talking about bond rotation

I think what you may be asking is: could there could be a nominally "meso" compound that has restricted rotation about its central single bond, causing it to exist in the form of two non-interconvertible stereoisomers. Example- start with meso-tartaric acid, and make the di-tert butyl ether from the OH groups and di-tert-butyl ester from the COOH groups. Maybe these groups would be bulky enough to restrict rotation. You could mentally rotate around the central C-C bond until the molecule has a plane of symmetry, so it is still a nominally "meso" compound, but physically you could separate the two stereoisomers.

Rotating about a single bond would be a conformational change and not a stereochemical one. Meso compounds are not defined as having rotational symmetry, either way. For example, 1,2-diphenylethane-1,2-diol does not have rotational symmetry since one carbon is R and the other S.  However, it's still a meso compound because its mirror image can be rotated to show that it is identical to the original compound.

There is no need to be sarcastic. Your molecule would  be indeed unstable, but main problem is are you talking about saturated or unsaturated molecule - trans suggests unsaturation - in this case it wouldn't have freedom of rotation around C=C, but rest of molecule's name says saturated

You're right, I'm sorry for being rude.  I just used that molecule because it's a common example they give in the textbooks.  I know it dissociates, though, but I think that's a discussion for a different topic.  I'm talking about an unsaturated molecule, because trans-1,2-dichloro-1,2-ethane diol is (1R, 2S) or (1S, 2R) which are the same meso compound.  It's just a common naming convention.

Hmm, here it's with trans - http://chemwiki.ucdavis.edu/Wikitexts/Purdue/Purdue_CHM_26100/Chapter_04%3A_Stereochemistry/4.3_Diastereomers but I'm pretty sure it shouldn't be, as it's not unsaturated or cyclic

The properties of a molecule don't change upon rotation about a single bond and such rotations are very fast and frequent at room temperature, so you can consider the rotated form and decied the chirality of the molecule.


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Why is acetylsalicylic acid still more acid than benzoic acid?
In this question, it is explained why salicylic acid is a stronger acid than benzoic acid. So, why acetylsalicylic acid (aspirin), whose conjugate base cannot hydrogen bond since there is no hydrogen, is still more acid than benzoic acid?

aspirin: $\mathrm{p}K_\text{a}$ = 3.5
benzoic acid: $\mathrm{p}K_\text{a}$ = 4.2

Acetoxyl is a moderate activating group for the aromatic electrophilic substitution, so it should mean that the resonance-driven electron-donating effect prevails over the inductive electron-withdrawing effect, therefore it should raise the $\mathrm{p}K_\text{a}$ of the acid.
Is it different between electrophilic substitution reactivity and acidity? Does the inductive effect prevail in this case?

The acetyl moiety in acetylsalicylic acid has a $(-I)$-effect, i.e. compared to a proton attached to the aromatic ring (as in the case of benzoic acid), it withdraws electron density out of the aromatic ring.  This provides an additional stabilization of the conjugated base, that is equivalent with rendering acetylsalicylic acid a stronger Brønsted acid than benzoic acid.  Hence, the $pK_a$ of acetylsalicylic acid is lower than the one of benzoic acid.
Regarding the second paragraph, please notice $pK_a$ is a thermodynamic property -- where time does not play a role.  In contrast, the substitution reactivity you likely refer to (Hammett parameters) refers to reaction kinetics and rate constants.  Here, time does play a role.


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Is it possible to oxidize ether to ester?
Is it possible to synthesize an ester from an ether in a single step? For example if I had an ether $\ce{(R-CH2-OR')}$, can I oxidize it so it will be an ester $\ce{(R-CO-OR')}$.

Yes, it is possible.  One reagent to do so is acidic bromate salts.  This paper discusses a number of optimizations and characterizations of the bromate-powered ether oxidation reaction.
$$\ce{R-CH2-O-CH2-R + BrO3- ->[H+][H2O; {r.t.}] R-CO-O-CH2-R + Br^-}$$
Hydrolysis of the formed ester leads to an undesired byproduct: the free acid $\ce{RCOOH}$.  In addition to that byproduct, overoxidation of the substrate can also happen, leading to anhydrides or even $\ce{CO2}$.
There may well be plenty of other oxidants and catalysts than can achieve the same reaction, but I thought this one was notable because it works in water at room temperature.  Yield depends strongly on reaction conditions and the vagaries of the particular substrates you use, though.


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What is the hybridization of the carbonyl oxygen in a carboxylic acid?
My professor sent out a review guide. One of the questions on this guide is

The carboxylic acid carbonyl oxygen is:
  1. $\ce{sp^2}$
  2. $\ce{sp^3}$
  3. $\ce{H}$-bond donor
  4. rapidly equilibrating between $\ce{sp^3}$ and $\ce{sp^2}$
  5. 1 and 3
  6. 1 and 4

I chose (4) because of the resonance that can be drawn with a negative oxygen and positive carbon. However apparently the answer is (6). I don't understand how something can be equilibrating between two states and still be one of those states.
Could anyone explain?

There are almost no occasions where an oxygen atom is sp3 hybridised. In most of the cases, oxygen is sp hybridised (first order approximation) when its coordination number is one, and sp2 hybridised when its coordination number is two. In special occasion, coordination number is higher, it can adopt a different hybridisation. But the flaw of reasoning in the question is the flaw of the hybridisation concept your teacher is trying to teach you, which is at least incomplete and outdated. Sorry.

I don't understand how "6" can be correct.  How can the carbonyl oxygen be $\ce{sp^2}$ (1) and also be rapidly equilibrating between $\ce{sp^3}$  and $\ce{sp^2}$ (4)?  Sometimes in introductory classes the carbonyl oxygen is presented as being $\ce{sp^2}$ hybridized (even though as @Martin points out this is not really correct either).

I absolutely agree with @ron. I did not even pay much attention to that detail. I guess what is meant is (and in my opinion this is wrong) that within the resonance framework we could describe a doubly bonded oxygen (max sp2) and a singly bonded oxygen (approx sp3), where the former would be predominant. It's 3am here, and I cannot answer this question now, I hope someone else will do that, but if not, pingback to me will work and I'll try to make the time.

'Equilibrating' is a nonsensical concept regarding wavefunctions.

I already pointed it out in the comments, but I believe it is time to give it some more thought and explanation. Let's deal with some conceptual issues first.
Hybridisation is a model, that can be used to describe a bonding situation. It is never cause for a certain geometric arrangement, it is always the result of a bonding arrangement. Strictly speaking it is not even necessary to be invoked, but it can be quite handy. Quoting from Aditya's answer - I could not have said it better:

First of all, 'hybridisation' is a hypothetical concept, i.e. orbitals don't really mix to form new orbitals, but 'hybridisations' are very successful in explaining structures of molecules and their physical properties. So knowing that a compound has $\ce{sp^2}$ hybridisation will just help you arrange the constituent atoms around the central atom. Physically it has no other meaning.
Note: Only orbitals can be hybridised, never compounds.

Having understood that, it is easy to see, that atoms in a molecule could use many different hybridisations schemes. There are some concepts, based on quantum chemical calculations, which would find the most appropriate fit between a Lewis structure and the molecular orbital approach. See for example this fine question: What is natural bond orbital theory used for? There is also a more quantitative approach, which is called Valence Bond theory, but that is beyond the scope of the answer.
Since this is just a mathematical concept, an equilibration between $\ce{sp^2}$ and $\ce{sp^3}$ would not change a thing. We can decide which way we want to look at it. Ron consequently raised the question in the comments:

I don't understand how "6" can be correct. How can the carbonyl oxygen be $\ce{sp^2}$ (1) and also be rapidly equilibrating between $\ce{sp^3}$ and $\ce{sp^2}$ (4)?

As I pointed out, (4) should not even be an option. The premise of the question is clearly flawed. Please do not take hybridisation too seriously.
This is how I come to my next point. Are the oxygen's orbitals in carbon acids really $\ce{sp^2}$ (or even $\ce{sp^3}$) hybridised? Well.

It is often appropriate to describe oxygen's orbitals as $\ce{sp^2}$, when it is bonded to two other elements. Two $\ce{sp^2}$ lobes are needed for the bonds, the remaining $\ce{sp^2}$ orbital hosts one lone pair, the remaining $\ce{p}$ orbital hosts the other lone pair.
More often the case is that oxygen has only one bonding partner. In these cases it is more appropriate to describe the oxygen with $\ce{sp}$ orbitals due to symmetry reasons. Consider the local symmetry to have an infinite rotational axis $C_\infty$, hence there have to be two degenerate, perpendicular orbitals. In most of these cases you will also have (partial) π-bonding and/or hyperconjugation.
There are very few exceptions, like $\ce{H3O+}$, where it is appropriate to describe the oxygen with $\ce{sp^3}$ orbitals.

In all of this it should be considered that hybridisation can be fractional. See the last note here and more consequently "What is Bent's rule?" and "Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?".
An important point to remember is, that when there is a π-bond, the hybridisation can be $\ce{sp^2}$ at maximum.
I believe that answer 6 is really about a wrong interpretation of resonance stabilisation of the molecule. (Please see Why is there a need for resonance? for more information)
Carbon acids can be described in the resonance scheme below. Now it appears, that the oxygen employs $\ce{sp^2}$ orbitals in the first structure (A) and $\ce{sp^3}$ in the others (B, C). It is important to recognise, that the bonding situation in this molecule is a superposition of all (possible) resonance structures. Theses structures are not in equilibrium.

In conclusion there are several issues with this question and you might want to address this with your teacher.

What about the notion of this molecule being dissolved in water? If the carbonyl oxygen started out as $\ce{^16O}$, and the hydroxyl oxygen as $\ce{^17O}$, then the hydrogen would flip from on oxygen atom to the other. In that sense there is an equilibrium.

@MaxW yes, that is correct and to go even further, this is true for any solution and you don't even have to mark the oxygens. Of course the electronic structure of the molecule will adapt in these situations. But interpreting the question like this is a stretch I think.

A great answer. I'll need to understand why does the oxygen appear to be in the sp3 hybridization state in (B) and (C) though. It has only one linkage to a C atom. It has 7 electrons, due to the -1 charge.. I'll try to wrap my head around it.


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Determination of structure with HNMR
I have been trying to solve this NMR question for about four hours and I feel like I'm always really close but never manage to find the perfect combination.
Anyway here's the info in the question
Formula is $\ce{C11H14O2}$
1.triplet at 2.4 ppm (2H)
2.singlet at 3.3 ppm (3H)
3.triplet at 3.5 ppm (2H)
4.doublet at 7.4 ppm (2H)
5.doublet at 7.5 ppm (2H)
6.singlet at 2.0 ppm integrates for 3H
I have tried a number of structures and the one I feel i got very close to was this one, each number on the side corresponds to the signal that I think it is. The X is just the area where I know i am missing a bond, I just don't know how to seal it up.

Edit: After the advice given by you all, I came up with this structure which seems to be pretty sound

Do the doublet and triplet come with coupling constants, or signals that can be picked to be translated in such?

unfortunately the information I have written is the only info available in the question. I just assumed that the triplet at 3.5 and the singlet at 3.3 were very close in proximity due to the similar peak values

I just wanted to make that sure. If it is not given in the question itself, then it is usually also not necessary for the problem to be solved. That assumption you make is very reasonable. We have a couple of experts on the network, I am pretty sure, this question won't be without answer long.

I do hope so, the real problem I am having with this question is managing to satisfy the atom count, I just can't find a combo of proper peak strucutre and atom that works....

Are you using table with NMR values - you really should.

We were too focused on the problem that we forgot to tell: Welcome to Chemistry.SE! :-D

The acetophenone looks good to me, I had the same idea.

Calculate your DBEs (double bond equivalents):

\[\mathrm{DBE} = \frac{2c-h+2}{2} = \frac{2\cdot 11 - 14 +2}{2} = 5\]
Note that a benzene has 4 DBEs!

Shifts at 7.5 ands 7.4 are definitely not alkine protons. Aromatic protons are more likely, aren't they?
Largest couplings between benzene protons are typically ortho couplings. How large are they in your case?
Two sets of doublets at 7.5 and 7.4 ppm might indicate a 1,4-disubstituted benzene. You might want to play with some increment rules here.
You have two triplets at 3.5 and 2.4 ppm with each 2 H. These might be two neighbouring methylene groups in $\ce{C_{aryl}-CH2-CH2-O-R}$.
R in (5) might be $\ce{CH3}$.

Thank you very much, I was way too focusd on getting a linear structure that I totally forgot about trying an aromatic structure.

Your DBE's are the same as what I call Index of hydrogen deficiency (IHD), right?


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Electrochemistry - Redox & Electroplating
I've stumbled across an electrochemistry problem that I need help on:

$$\ce{Fe~|~Fe^{+2}~||~Cl^{-},Cl2~|~Pt}$$

I understood that iron is at the anode and is being oxidized, however I do not know what was going on with chlorine, it looks like it is being oxidized while it is supposed to be reduced.
Maybe this is a non-spontaneous reaction? Can this be possible, can chlorine be used as an oxidation reaction? And if so how can we calculate the voltage?

Finely divided iron powder spontaneously combusts in chlorine gas, so, assuming aqueous solutions, iron is oxidized at the anode and dissolved chlorine gas is reduced to chloride ions at the cathode. The cell notation is simply incomplete.

Here the iron electrode is made by dipping an iron plate or wire in a $\ce{FeCl2}$ or $\ce{FeSO4}$ solution. It is an anode, where $\ce{Fe}$ is oxidized according to $$\ce{Fe -> Fe^{2+} + 2 e-}$$ The chlorine electrode is a cathode made by dipping a platinum electrode in a solution where some chlorine gas $\ce{Cl2}$ is arriving from under the platinum plate, so that a few bubbles remain adsorbed on its surface. This $\ce{Cl2}$ gets reduced to $\ce{Cl-}$ by absorbing the electrons coming from the iron anode according to the equation : $$\ce{Cl2 + 2 e- -> 2 Cl-}$$ The global effect of these reactions is equivalent to the following equation that describes the general change of composition in the cell : $$\ce{Fe + Cl2 -> Fe^{2+} + 2 Cl-}$$

I am very surprised this one went 6 years without even a comment! Anyway, (+1) for closing the books on it.


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How does composition of air affect vapor saturation limit?
Suppose we hold temperature, pressure, and gas composition (say, a standard atmosphere) constant.  Then we know that if we allow the water content of the air to vary it can range between 0 and some saturation limit S g/L.
How does that saturation limit vary with the composition of the air?  For example, does it depend on the molecular weight and proportions of the other constituents of the air?
Assuming the water (or whatever substance is in question) does not chemically react with any other constituents at that temperature and pressure are there other constituent chemical relationships that can affect the saturation limit?

To a first order approximation, no. But you seem to realize that if you start to consider the solubility of gas in the droplet of water at saturation, then there can be some effect on the vapor pressure. Try to define the question mathematically with a few equilibrium equations that include solubility constants of the gases in the mixture, and include a formula for the colligative effects, and you might get some interest in starting to develop an answer

@repurposer Are you saying there are solubility effects in gases?  If so I'm not familiar with those.  To clarify what I think you might be suggesting elsewhere in your comment: I'm interested in a "static" answer.  I.e., I'm not interested in whether, at and beyond saturation, condensates can dissolve other gases and thereby reduce air pressure and revaporize or anything like that.  Rather, the question is more generally something like, "Can air saturated with water vapor hold more or less alcohol vapor?"  Or, "Can an atmosphere of nitrogen hold more or less water than one of helium?"

then, I think the answer is no, the partial pressures will be independent of one another and add to the total pressure by Dalton's Law. However, if you are looking at low temperatures or high pressures compared to atmosphere, or particularly polar, polarizable, or large molecules then dispersion forces will affect the overall equation of state. But I would not expect substantial differences to say that any gas would 'hold' much more of another gas, unless you are, say making detailed thermodynamic measurements, say of Virial coefficients for some mixture, which almost nobody does

@repurposer So in layman's terms, in a gas state (below any applicable chemical reaction temperatures) molecules don't care about the size or mass or composition of the other molecules they're bumping into. They only care about the temp ("avg speed") and pressure ("crowdedness"), and those two parameters determine each chemical's vapor saturation limit?  But -- and this is worth an answer -- you suggest there is a point at which the chemical's size does come into play.  And other chemical properties would change the answer -- polarization (not sure what that is outside of light)? Ionization?

I think I won't write a full answer right at this moment. Polar molecules have an electric dipole, polarizable molecules form instantaneous dipole (see dispersion). It's good that you bring up ions, which behave more like an electric monopole. I should back up a little and say, I think the layman's terms as you put it will get you in trouble, see this link, and you can see there really is a sizable vapor composition difference once you have condensed some liquid in a system

Correct URL appears to be http://vle-calc.com/phase_diagram.html ... and this does look like an easy question with a number of complications.  I look forward to answers!

feetwet - Shouldn't you change your question to use "gas" throughout instead of "air"?

@MaxW – Maybe, but I intentionally used "air" to denote an ambient mixture of various gasses.

@feetwet - "Air" typically means the earth's atmosphere. A "gas" can be anything from $\ce{SiCl4}$ to HCl to He. // So in general questions in a chemistry book would assume that Dalton's Law holds. // If you want an answer that covers all sorts of "unusual" gases (gases not in our atmosphere) then there will be deviations.

I agree with "repurposer"'s comment that the saturation limit would be independent of the composition of the air.
It would just be a function of the partial vapour pressure of water.
$K_{p}=p_{H_2O}$, where $K_p$ is the equilibrium constant in pressure terms.
Consider this case similar to that of aqueous tension - it depends only on the temperature of the system, and not the other (non-reacting) components.
Whenever we want to find the pressure of the dry gas, i.e. pressure of the mixture of gases $-$ the pressure due to the water vapour.
Hope this helps!

Since dessicants reduce vapor pressure etc. any gaseous component that changes the chemical potential of the water drop will have an effect on the liquid vapor equilibrium. that is the rationale of cloud seeding to induce rain [I am not saying it is easy to do or it works]


text:
Co49Ni21Ga30 - molar mass and density
I need to calculate the molar mass and density of the shape memory alloy $\ce{Co49Ni21Ga30}$
(I suppose the numbers indicate the mass percentage).
To get the molar mass, is it correct to use the formula below?
Mol_mass_compound = Mol_mass_Co*0.49 + Mol_mass_Ni*0.21 + Mol_mass_Ga*0.30

Regarding the density, how do you suggest to proceed? Thanks!

Welcome to Chemistry.SE. Your equation would be correct if a sum formula would be a table of weight percentages. But please note that one of the most important concepts in chemistry is the mole. The molecular weight is the sum of products of atom weights (of an element) and the number of atoms (of that element) in a molecular formula.


text:
How come diluting a cell decreases the voltage?
I just want to know why, if we dilute a cell with water, it decreases the voltage. Assume it is a galvanic cell.

Your assertion the voltage in a cell is decreased by diluting the cell  is not necessarily true. According to the Nernst Equation:
$$E_{cell}=E^0-\frac{RT}{nF}\ln(Q)$$
Where $Q$ is the reaction quotient relating $\frac{[\mathrm{metal~ion~formed~by~oxidation}]}{[\mathrm{metal~ion~reduced}]}$.
The value of $Q$ can be increased by: increasing the concentration of the metal ion formed by oxidation, or by decreasing the concentration of the metal ion that is reduced. By increasing the value of $Q$, the magnitude of the term $-\frac{RT}{nF}\ln(Q)$ is increased. This means that a larger number is subtracted from the $E^0$ of the cell, and that the cell voltage decreases.
The value of $Q$ can be decreased by: decreasing the concentration of the metal ion formed by oxidation, or by increasing the concentration of the metal ion that is reduced. By decreasing the value of $Q$ to a fraction, the value of the term $-\frac{RT}{nF}\ln(Q)$ is made positive. This means that this term is added to the $E^0$ of the cell, and that the cell voltage increases.
This is consistent with the Le Châtelier's Theorem. Essentially, increasing the relative concentration of your reactants drives the reaction forward (higher $\Delta V$), and increasing the relative concentration of your products drives the reaction backwards (lower $\Delta V$).


text:
What is the precise definition of maximum work in the context of electrochemistry and thermodynamics?
I'd like clarification of the meaning of $W_\text{max}$ in the following equations (and thus in the combined result in the third equation):
\begin{align}
W_\text{max} &= \Delta G\\
W_\text{max} &= -nF\varepsilon\\
\Delta G &= -nF\varepsilon\\
\end{align}
In the first equation, we somehow come to the conclusion that the maximum useful work available from a process is that process' change in Gibbs Free Energy. I ask now what work that means - does it mean work done on the surroundings by release of energy from the system? Does it mean the maximum work done on the system via a chemical process (this makes more sense because it doesn't make sense that an increase in Gibbs Free Energy of a system would do work on anything but the system)?
Secondly, in the electrochemical definition of $W_\text{max}$, we have the point of view of the system; that is, positive work is done on the system and negative by the system. Therefore, we add the negative sign because when cell potential is positive the work done on the system is negative (it does work on the surroundings).
The only logical conclusion I can come to is that the two $W_\text{max}$ definitions must be both with respect to the system's point of view, which means that $W_\text{max} = \Delta G$ refers to the maximum amount of work done on a system during a chemical process (the intuitive interpretation that I posed). However, it still makes me uncomfortable to define "useful work" as work done on the system. How can we harness work from putting energy into something?

"Maximum Additional Work that can be extracted at constant pressure and temperature" to be precise, which may include electrical work, etc.  It means the work besides the expansion work.
The negative sign signifies the sign of the charge of the electron, since electrical work is proportional to the charge.

Therefore, we add the negative sign because when cell potential is
  positive the work done on the system is negative (it does work on the
  surroundings).

That's correct. This is merely a convention, though. We could define positive and negative work the other way around as long as we also flipped the sign in all the equations that use work: $\Delta G = w$ becomes $\Delta G = -w$; $\Delta U = q + w$ becomes $\Delta U = q - w$; etc.

How can we harness work from putting energy into something?

You can't, you harness the work when you take the energy out; you "store" the work when you put energy in. If, for example, you connect a battery to a DC motor and then have the motor raise a weight, you're doing ("harnassing") work by the cell on the surroundings, and a convention among chemists is to consider this negative work.


text:
Does the natural abundance of radioactive elements change?
Given that the half-lives differ between isotopes of a particular radioactive element, wouldn't the natural abundance change (fairly rapidly)?
For example, if we isolated a bunch of thorium, wouldn't the relative amount of the various isotopes of thorium in the sample differ over time, with the more slowly decaying isotope eventually making up the largest proportion?

Yes, you are right. The abundances of (radioactive and stable) nuclides change due to radioactive decay.
Regarding this, most environmental radionuclides can be divided into four categories:
Cosmogenic radionuclides are constantly produced by cosmic irradiation of the atmosphere. Therefore, their average concentrations in the atmosphere stay roughly constant even if their half-lives are short.
Primordial radionuclides have lifetimes comparable to the age of the earth. These nuclides have been formed at the time of the formation of the earth, the solar system, or even before. Their abundance is slowly decreasing due to radioactive decay. Therefore, the radiation exposure was higher some billion years ago when life evolved on earth than today. Furthermore, due to different half-lives, the relative abundance of these nuclides (e.g. the natural $\ce{^{235}U/^{238}U}$ isotopic ratio) also changes.
Radionuclides that are part of the natural decay chains of $\ce{^{238}U}$, $\ce{^{235}U}$ and $\ce{^{232}Th}$ are constantly reproduced by the radioactive decay of their mother nuclides. Therefore, their abundance is only slowly decreasing with the long half-life of $\ce{^{238}U}$, $\ce{^{235}U}$, or $\ce{^{232}Th}$ even if their own half-lives are short (this situation is called secular equilibrium).
Anthropogenic radionuclides are produced by modern technology. Since the discovery of nuclear fission, the activity concentrations of various anthropogenic radionuclides in the environment were increasing until the limited nuclear-test-ban treaty banned atmospheric tests of nuclear weapons. After that, the activity concentrations of most anthropogenic radionuclides decreased again.

If you isolate and chemically purify a radioactive sample, you remove any other mother or daughter nuclides and thus you disturb any decay chain or radioactive equilibrium. Thereby you start a radioactive clock.
For example, a solid rock sample may contain $\ce{^{226}Ra}$ $\left(t_{1/2}=1600\ \mathrm{a}\right)$ and its daughter nuclide $\ce{^{222}Rn}$ $\left(t_{1/2}=3.83\ \mathrm{d}\right)$ in a secular equilibrium. If you dissolve the sample and release the $\ce{^{222}Rn}$ (radon is a noble gas), the released $\ce{^{222}Rn}$ decays with its half-life of 3.83 days. If you purify the $\ce{^{226}Ra}$ and put it in a closed container, initially the container does not contain any $\ce{^{222}Rn}$, but new $\ce{^{222}Rn}$ is produced by radioactive decay of $\ce{^{226}Ra}$.
A similar thing occurs when natural minerals solidify. Radionuclides become trapped and decay. As they decay, decay products accumulate in the closed mineral.
By measuring the amount of mother and daughter nuclide, it is possible to calculate how long this sample has existed. This is the basis for nuclear dating.

If you isolate a sample that contains multiple radionuclides of the same chemical element, you can observe changing isotopic ratios – as you have expected in your question.
Your example, freshly purified thorium from a natural sample, mainly contains six different thorium nuclides:
$\ce{^{234}Th}$ $\left(t_{1/2}=24.10\ \mathrm{d}\right)$
$\ce{^{232}Th}$ $\left(t_{1/2}=1.405\cdot10^{10}\ \mathrm{a}\right)$
$\ce{^{231}Th}$ $\left(t_{1/2}=25.5\ \mathrm{h}\right)$
$\ce{^{230}Th}$ $\left(t_{1/2}=7.54\cdot10^{4}\ \mathrm{a}\right)$
$\ce{^{228}Th}$ $\left(t_{1/2}=1.91\ \mathrm{a}\right)$
$\ce{^{227}Th}$ $\left(t_{1/2}=18.7\ \mathrm{d}\right)$
The individual half-lives differ significantly; thus, the different thorium nuclides decay at different rates. The isotopic ratios change accordingly.
However, the calculation is slightly complicated since $\ce{^{234}Th}$ indirectly decays to $\ce{^{230}Th}$, $\ce{^{231}Th}$ indirectly decays to $\ce{^{227}Th}$, and $\ce{^{232}Th}$ indirectly decays to $\ce{^{228}Th}$.

Yes, you are correct; over time the proportion of the more stable isotopes (those with longer half-lives) increases. There can be alterations to this tendency if a less stable isotope happens to be continually replenished by the decay of a heavier element.
There is some very interesting physical proof of this. Consider uranium, whose dominant naturally-occurring isotopes are $\ce{^{238}U}$ and $\ce{^{235}U}$, with respective half-lives of $\mathrm{4.47\times 10^9\ yr}$ and $\mathrm{7.8\times 10^8\ yr}$. Currently, the uranium on Earth is composed of approximately 99.3% $\ce{^{238}U}$ and 0.7% $\ce{^{235}U}$. Given that $\ce{^{235}U}$ decays faster, we expect that in the distant past there was a higher proportion of $\ce{^{235}U}$ than we find now. Indeed that is the case - billions of years ago, the fraction of $\ce{^{235}U}$ in natural uranium was as high as 3%. It turns out that, due to its fissile nature, this concentration of $\ce{^{235}U}$ was high enough to sustain natural nuclear fission reactors. Nowadays, natural uranium contains too little of the fissile isotope to spontaneously form underground nuclear reactors, and so enrichment techniques must be used to obtain fissile uranium fuel, containing a higher proportion of $\ce{^{235}U}$.


text:
How to make a CP2K input file with Avogadro?
I am trying to learn to work with CP2K and now need to make its input file. CP2K has kind of adaptor for Avogadro2 but not Avogadro ( I don't know what is the relation between them !).  I also tried that adaptor plugin with Avogadro2 but it does not work for me ( tried on Mac OS X Yosemite and Windows 7).
I searched a lot but seems there is no way to make CP2K inputs with Avogadro . Is there any ?

If you look at the CP2K website, you'll see that the coordinates are only a small fraction of the input file. Most of it is text describing how a calculation should be run.
The key parts from a coordinates perspective look like this:
&CELL
  ABC 12.4138 12.4138 12.4138
&END CELL
&COORD
  O      12.235322       1.376642      10.869880
  H      12.415139       2.233125      11.257611
  H      11.922476       1.573799       9.986994
&END COORD

As it looks, these are the Cartesian coordinates of the atoms (in regular space) and the unit cell vectors. You can either save the file in .XYZ format or use an input generator like GAMESS or Gaussian to get these coordinates.
If you want a periodic boundary calculation, you just need to estimate the bounding box. You can do this manually, or you can go to Crystallography | Add Unit Cell and then tweak the volume. In Avogadro v1.1.x, you'll see the ABC lattice vectors needed for that CP2K block.
As it might seem, Avogadro2 will be v2.0 of Avogadro. It's still in development.

There is an experimentally CP2K input generator for Avogadro 1.1 in
https://github.com/brhr-iwao/libavogadro1cp2k
It is easy to install if you use Avogadro 1.1.1 for Windows, only to download cp2kinputextension.dll from the above site and put it in the C:\Program Files\Avogadro\lib\avogadro\1_1\extensions folder in your PC.
Now libavogadro1cp2k is still in making and its function may be unsatisfactory.
But you can edit its generated input file. It will be easier than making CP2K input file from scratch.

This is news to me - why not contact the developers about integrating into the full version?

It is an honor to be told that.  I would certainly like to do so.


text:
Why would sulfuric acid turn gray/black?
About 2 years ago I purchased 2 liters of 98% concentrated sulfuric acid. It's in a standard poly bottle with leakproof lids, and originally was clear. Recently I noticed the acids have turned a dark gray; almost black color. They have both never been opened.
I suspect there was something in the lid that was eroded by the acid, but is there any other reason sulfuric acid would darken over time?

The only reason I can think of that would cause the solutions would blacken over time is if the solutions reacted with organic substances. Sulfuric acid is an extremely strong oxidizing agent, and dehydrates many organic compounds to produce carbon in the form of graphite, and water. For example, the reaction of sulfuric acid with paper (cellulose) is:
$$(\ce{C6H10O5})_{n}\ce{(s)} \xrightarrow{\ce{H2SO4}} {6n~\ce{C(s)} + 5n~\ce{H2O(l)}}$$
and can be seen in the following photo:

If a considerable quantity of organic material was reacted in your solutions, the $[\ce{H+}]$ in your solutions would actually decrease, so I would recommend titrating one of your solutions to see if this has changed since you bought them.

I don't think it would be dehydrated completely - to graphite.

Sulfuric acid is a moderately strong oxidizing agent at best. Dehydrating ability is another story.


text:
What would happen if I mix salt and sugar in a cup of water?
If I mix water in a cup with sugar at some stage the solution will become saturated and no more sugar will dissolve into the water. If I now add salt to the solution will the salt dissolve into the water or is the solution already saturated?
I have not tried an experiment as I cannot figure out how to measure the dissolved component of the 2, intuitively I think that some salt will dissolve, but not as much as in a pure salt water mixture.

Unfortunately, sugar is very soluble in water and will make a thick syrup (i.e. karo syrup) before you reach saturation.


text:
Water in the ultrasonication bath
Why something jelly forms in the water if it is kept in in a ultrasonication bath for long periods?
 When water is exposed to ultrasonication for say a month, some jelly substances form in the ultrasonicator bath. is this something related to some bioorganism growth?

you should add more information.

Keep your lab equipment clean! Change the ultrasonication bath water regularly, it gets pretty mucky with dirt and impurities.

To see these jelly flakes in an ultrasound bath is a normal thing -- referring to ``it happens regardless if the water was tap-water or deionized water, a just a matter of time''.  Perhaps this occurs faster in the former, than in the later, especially if the bath is used to clean glassware.  But it equally happens with glassware judged as clean (by eye) for example if the bath is deployed to initiate / support a reaction inside a reaction flask.  Ultrasound-assisted generation of Grignard reagents is one example.
The ultrasound bath works by sonication, and the cavitation simply detaches the oily/fatty/grease surface film covering quickly all what was touched once by naked hand (e.g. sweat), and breaks this into even smaller pieces.  Over time -- when the ultrasound bath is switched off --, this thermodynamically unstable emulsion decomposes, splitting into a water soluble phase, and a hydrophobic phase that then tends to agglomerate.  (I did not check the presence of organisms living in or on them since discharging the bath is a chore regularly done, too.)
Side note:  For me, these residues appear to be more a liquid, then a solid -- hence I used the term emulsion.  If they were a solid, the thermodynamically unstable mixture would be an suspension.


text:
Infrared spectroscopy of substances on a OHP film
How can we record a IR spectrum of a substance coated on a Over Head Projector (OHP) sheet?
instrument has both transmission and ATR modes. IR spectrum is to be recorded for both the OHP sheet as well as the material that is coated on it

do you want the IR spectrum of the OHP material or maybe the material of ink (used to write on the OHP sheet) ?

Your are going to analyze both, the OHP foil alone, and the part of the OHP foil that is covered by the coating and calculate the difference of the later to get the data of your coating material.
If you do want to do the classical way (transmission IR), just cut-off a little of the sample (a hole puncher as in the office may do the job, too), attempt to grind it with KBr and press the pellet as normal.
If you want to stick to ATR, place the foil on the dish without prior sample preparation.  This is quicker, and to some degree allows a space-resolved analysis, too -- the window (center dish of the table) of a single reflection geometry instrument (like the one below, Nicolet iS10 with smart-ATR-IR module) is of about 3 mm in diameter).  The window material of the ATR module may be Germanium (as here), yet there are instruments with a window of Diamond, too.


text:
How can an aromatic ketone be reduced in the presence of a nitro group?
Is it possible to reduce an acyl group to an alkane, but still keep a nitro group intact without its reduction to an amino group? A Clemmensen reduction would reduce the nitro group, and it seems like other reducing agents will also reduce the nitro group to an amino group. Would some other mechanism be better suited for this reaction?
For instance, how can I selectively reduce 1-(3-nitrophenyl)propan-1-one to 1-nitro-3-propylbenzene?

You will most likely be successful if you convert the ketone to the corresponding tosylhydrazone and reduce that with $\ce{NaCNBH3}$ (DOI).
A previous study (Bull. Chem. Soc. Jpn., 1974, 47, 2323-2324) has shown that the tosylhydrazone of cyclohexanone can be reduced to cyclohexane with $\ce{NaBH4}$ in polar aprotic solvents.
$\ce{NaCNBH3}$ is an even milder hydride transfer agent, that has been widely used in the reduction of imines and iminium salts. A review by C.F.Lane, Sodium Cyanoborohydride - A Highly Selective Reducing Agent for Organic Functional Groups in Synthesis, 1975, 3, 135-146 gives some background information on the versatility of this reagent.
The nitro group is definitely safe with $\ce{NaCNBH3}$. However, considering the high toxicity and the price for this reagent, you might want to run a small scale reduction of your tosylhydrazone with the much cheaper $\ce{NaBH4}$ first.

A discussion about alternative reduction methods was here Clemmensen reduction vs. Wolff-Kishner reduction.  I've not seen nitro groups reduced by the acid-triethylsilane or tetramethyldisiloxane treatment. One used on plant that I've seen (and used)  is TFA - triethylsilane. The carbonyl is protonated and the first hydride delivery gives the alcohol. A second protonation and hydride addition gives the methylene. As a side note, any aromatics that are not electron deficient e.g. the solvent toluene, can undergo an electrophilic addition during the second reduction step when water is lost.


text:
Basis set in the computational software "Gaussian-03" for studying adsorption of a hydrogen atom on a carbon paste electrode dropcasted with Pt-CNT
What is the basis set in the computational software "Gaussian-03" that can be used to study the interaction of $\ce{H}$-atom with $\ce{Pt-CNT}$ dropcasted electrode? How can we study the energy differences that arises due to the interaction computationally?

Maybe you want to share how far you got so far with your literature search and considering different calculations methods and models for your system. Why did you choose this or that calculation method and such detail... The question in this form would suggest you have very little idea about what you want to do.

I agree with @Greg. I also do think, that Gaussian is usually not the right program to do calculations for surface chemistry. Also, 03 is quite old now and might perform very weakly for large systems.

Actually this is a part of our groups work in regard to electrocatalysis. We just wanted to support our findings computationally by studying the interactions between the chemical moiety and the modified electrode

then you could at least a little bit information about your system here.
I go with Greg it.


text:
Is lead radioactive?
Is lead (Pb) radioactive? I have heard about it being radioactive but have also heard it referred to as the heaviest stable element. Radioactivity refers to release of alpha, beta, gamma rays.

The major isotopes of lead are not radioactive.

Because this is a popular question with an incorrect premise,  i have started a discussion regarding the validity of closing down this question in meta. Kindly please give your views regarding this question there.

The question ‘Why is lead radioactive?’ is ambiguous, but it is not necessarily based on a false idea. Any ordinary lead sample is indeed radioactive, which makes the question ‘why?’ valid. Maybe the question could be edited to clarify the problem.

@Loong In the spirit of the OP's message on the meta post you should go ahead and extend/ clarify the question. I believe it would be not conflicting with the author's intent - I believe the main driving force for the question was to get some clarification on the matter. A little context can never hurt, and the OP can always roll back, if she/he thinks the edit is inappropriate.

@Martin I don’t want to simply write a question that matches my answer. Therefore, I wrote my suggestions in meta.

Saying "The major isotopes of lead are not radioactive" is sort of circular reasoning. The major isotopes are the major isotopes precisely because they are not radioactive. Lead is the highest atomic numbered element that has stable isotopes.

Important naturally-occurring radioactive lead nuclides are:
$\ce{^{214}Pb}$ $\left(t_{1/2}=26.8\ \mathrm{min}\right)$ from $\ce{^{238}U}$
$\ce{^{210}Pb}$ $\left(t_{1/2}=22.3\ \mathrm{a}\right)$ from $\ce{^{238}U}$
$\ce{^{211}Pb}$ $\left(t_{1/2}=36.1\ \mathrm{min}\right)$ from $\ce{^{235}U}$
$\ce{^{212}Pb}$ $\left(t_{1/2}=10.64\ \mathrm{h}\right)$ from $\ce{^{232}Th}$
(Note that 'a' here is the unit symbol for 'year'.) On a geological time scale, their half-lives are very short. Nevertheless, these lead nuclides are members of the $\ce{^{238}U}$, $\ce{^{235}U}$, and $\ce{^{232}Th}$ decay series. Thus, they are constantly reproduced by decay of their respective mother nuclides. Therefore, all environmental samples (including soil, water, air, plants, and animals) naturally contain significant amounts of radioactive lead nuclides. However, in various materials (e.g. residues from mining or oil and gas production, fertilizers, building materials), the concentration of natural lead nuclides can be strongly enhanced by technology.
When lead is freshly refined (chemically purified), most of the mother nuclides are removed and the short-lived lead nuclides quickly decay. Nevertheless, the lead still contains significant amounts of $\ce{^{210}Pb}$ $\left(t_{1/2}=22.3\ \mathrm{a}\right)$. The presence of $\ce{^{210}Pb}$ is a source of background mainly through the lead X-rays and bremsstrahlung caused by high-energy beta radiation emitted by its decay product $\ce{^{210}Bi}$. This background is a real problem when lead is used as a shielding material for low-background detectors. Therefore, some shields are made from selected lead with certified low $\ce{^{210}Pb}$ content. Preferably, very old lead is used. Since $\ce{^{210}Pb}$ decays with a half-life of 22.3 years, samples of lead that are many decades old are relatively free of this activity. In one extreme example, 2000-year-old lead was salvaged from a sunken Roman ship.

"all environmental samples [...] naturally contain significant amounts of radioactive lead nuclides" seems to be an extreme exaggeration, if not outright incorrect. You're claiming there's a "significant" amount of radioactive lead in every coffee I ever drink?

@E.P. Yes, as long as you don't use distilled water for your coffee.

@Loong In that case, can you clarify what you mean by "significant"? You're obviously using some not-particularly-standard understanding of the term, I think.

@E.P. "significant" means that the measurement result exceeds the decision threshold [which is calculated so that the error probability of taking the wrong decision shall not exceed a specified probability (typically 5 %; i.e. the decision threshold corresponds to the quantile of the normal distribution of the background count rate for the probability of 95 %), see ISO 11929]

I agree that the low levels of radioactive lead should be described as "detectable" not "significant."

here is a link to an article about using lead from sunken Roman ships for shielding. http://www.nature.com/news/2010/100415/full/news.2010.186.html

One radioactive isotope of lead is $\ce{^{210}_{82}Pb}$, which has a half-life of 22.20 years. The only reaction that $\ce{^{210}_{82}Pb}$ undergoes is:
$$\ce{^{210}_{82}Pb -> ^{210}_{83}Bi +^{0}_{-1}e}$$
The $\ce{^{0}_{-1}e}$ released in the reaction is a high speed electron, otherwise known as a $\beta^-$ particle. The fact a $\beta^-$ particle is released by the $\ce{^{210}_{82}Pb}$ nucleus would suggest that it is an excess number of neutrons causing the instability.

As noted in the comments, the premise is wrong.
Actually: Lead is the heaviest stable element.
Why are heavier elements unstable?
The atomic nucleus consists of neutrons and protons.
Without going into detail, these are held together by short range nuclear forces.
The protons also repel each other due to their positive charge.
As we add more protons, the charge repulsion continues to build. Each pair of protons generates a repulsive force, so this grows in total as the square of the number of protons.
The short range attractive force does not grow quadratically in the same way, as 'distant' nucleons do not feel as much of an effect.
The overall effect is that above lead the repulsion is strong enough that the overall energy of a lower number and the radiated particle is less than that of the heavy nucleus.
The difference in total energy drives the radioactive decay.
As the elements get heavier this energy difference increases resulting in faster decays. This is complicated somewhat by the nuclear structure (the same concept as the electronic structure) in that completed nuclear shells create stability.

Before explaining why lead is radioactive, I would like to briefly explain what radioactivity is.
According to Wikipedia:

Radioactive decay, also known as nuclear decay or radioactivity, is
   the process by which a nucleus of an unstable atom loses energy by
   emitting ionizing radiation. A material that spontaneously emits this
   kind of radiation — which includes the emission of alpha particles,
   beta particles, gamma rays and conversion electrons — is considered
   radioactive.

That is, radioactive decay occurs because an unstable atom "prefers" to release extra energy to become more stable. If you consider lead, it has different isotopes. A large percentage (>98%) of lead ordinarily found in nature is of stable isotopes whereas the rest of its isotopes are unstable. It is the unstable isotopes of lead that are radioactive.
Another example is carbon.
Carbon-12 with six neutrons is a stable isotope of carbon which has no radioactivity and carbon-14 with eight neutrons is not stable and shows radioactivity.

"Which means": I cannot see how the preceding sentence could in any way be used to  infer any percentage at all. And if you mean the linked table, I think all isotopes there are either marked "observationally stable" (and their mole fractions add up to 1) or have a mole fraction that is marked "trace" at best, so 98.6% seems just an error on your part; if anything it would be 100%

@MarcvanLeeuwen thanks for showing the mistake I would correct that error.

All isotopes of lead are metastable and radioactive (although nobody, so far as I know, has invented a sufficiently sensitive detector to confirm this quantum mechanically certain uncertainty).
Alpha particle decay of lead results in stable isotopes of mercury. So the rather improbable answer to the question "Which is the heaviest stable element/nucleus?" is, of all things Thallium. That is not what I learned at school. The article Barely Radioactive Elements leads you, very gently and humbly, so it seemed to me, to Avogadro's Number of years.

Actually, taking into account isotopes which are theoretically unstable but experimentally extremely long-lived, every nucleus beyond $\ce{^92_40Zr}$ is radioactive.

Please don’t add any variations or permutations of ‘please Google X’ to your answers. Rather, summarize and quote your sources and — most importantly — cite them. This is a scientific site, after all, and I have yet to read a paper saying ‘as the reader may confirm by Googling/SciFinder searching …’ ;)

Touche, Jan! No offence intended. I  am  neither chemist nor physicist,nor (I trust) an unwelcome intruder - merely a medical specialist who tries to remain eclectic in my winter burrow in Darkest Australia. Oh, and I was not purporting  to produce a "paper". On a point of order, why is a "scientific site source" any  less of a source solely because (if I understand you correctly) it can be accessed by not solely the cognoscenti?


text:
Mesh size of the silica gel
What does the Mesh in the mesh size of silica gel that is used for various chromatography techniques mean?

Mesh is a unit representing the grain-size distribution of a material. So it represents either the opening of the mesh, or the number of grain per surface unit depending on what you are considering.
You can find here the conversion in grain size and surface unit related to the "Mesh".


text:
Enolate Ion formation
I was wondering why when acetaldehyde is treated with $\ce{OH-}$, the $\ce{OH-}$ will attack the methyl group instead of the carbonyl carbon, forming an enolate anion. I would have though that the $\ce{OH-}$ would have attacked the carbonyl carbon due to it having a  relative positive charge.
Additionally, my professor said during lectures that form x is favored over form Y. Is this because a negative oxygen is able to delocalize the negative charge (and be protonated more easily) more efficently than a $\ce{CH2-}$ group? If so,  is the x form so dramatically favored that it would not make sense to draw any mechanisms involving the y form?

The y form is very important in Aldol Reactions.

Enolate vs hydrate formation

Both the enolate and the hydrate form.
Aldehydes (and ketones) react with water in the presence of acid or base to form hydrates. However, for more aldehydes in aqueous solution (even dilute solution), the hydrate is only a minor component of the mixture (~1%).
The other 99% of the aldehyde molecules are perfectly able to form the enolate, but only a small number of them do because the difference in pKa between most alpha hydrogens on aldehydes and ketones (~21) and water (15.7).
So, in actuality most of the aldehyde molecules at any given time are neither enolates nor hydrates. Now, if a reaction should occur to consume enolate, then Le Châtelier will come out and nudge the system to make more enolate.

"Favored" resonance contibutors

The two resonance contributors $x$ and $y$ are not separate compounds. It is not possible for one structure or another to be "favored", though we often use that language because we can relate to comparing the structures of two distinct compounds, like the cis and trans isomers of an alkene. What we really mean when we say "favored" is that the resonance contributor is more important for describing the structure and behavior of the ion than the other. The true structure of the enolate ion is some linear combination of resonance contributors $x$ and $y$ (as well as some other minor things). Thus, the true structure might be like a weighted average over the contributors with $x$ having higher weight.
For example:
$$enolate = \frac{2}{3}x + \frac{1}{3}y$$

I'm sorry if I am misintepreting resonance, but if they are resonance forms (ie all identical molecules), wouldn't their reaction with another molecule, say acetaldehyde, form always one product, in this case 3-hydroxybutanal


text:
What is the mass of 1 atom of carbon?
I know that relative atomic mass of $\ce{^{12}C}$ is $12~\mathrm{u}$.
Therefore mass of $1~\mathrm{mol}~\ce{C} = 12~\mathrm{g}$
\begin{align}
 \text{mass of }6.022\cdot 10^{23} \text{ C atoms} &= 12~\mathrm{g}\\
 \text{mass of }1 \text{ C atom} &= \frac{12~\mathrm{g}}{6.022\cdot10^{23}}\\
&=\boxed{1.99\cdot10^{-23}~\mathrm{g}},\\
\end{align}
but
\begin{align}
 1~\mathrm{u} &= 1.66\cdot10^{-24}~\mathrm{g}\\
\implies \text{Mass of }1~\ce{^{12}C}\text{ atom} &= 1.66\cdot 12 \cdot 10^{-24}~\mathrm{g}\\
&= 19.92\cdot 10^{-24}~\mathrm{g}\\
&= 1.992\cdot 10^{-23}~\mathrm{g}.\\
\end{align}
Which is correct?
Also what is dimension formula of relative atomic mass, molar mass?
Why is unit of molar mass $\dfrac{\text{gram}}{\text{mole}}$ and not just $\text{gram}$ Since when has this $\text{mol}$ become a unit? It's just a number.

The number in the rectangle was off by 46 orders of magnitude!  (Sign error: should have been $10^{-23}$ not $10^{23}$ as it was originally.)  I'm edited it to be correct, presuming it was just a typo, but let me know if you have any confusion about why.

The mole became a SI unit in 1971.

Thank you both! @NicolauSakerNeto actually it was just a typo! the question was about which approach is correct? some of my friends were saying that second approach is dimensionally incorrect.

Sorry if my previous comment seemed condescending, I should have realized it was a typo but recently I commented a question involving an incorrect exponent sign and that led me astray.

Both approaches are correct.
Avogadro's number is $6.02214129\times 10^{23}$ and represents the number of carbon-12 atoms in 12 grams of unbound carbon-12 in the ground electronic state.
$12$grams$/6.02214129\times 10^{23} = 1.9926467\times 10^{-23}$grams
The unified atomic mass unit (u) is $1.660538921 \times 10^{-24}$ grams
$12 \times 1.660538921 \times 10^{-24}$ grams $ = 1.9926467\times 10^{-23}$grams

The Avogadro constant has a unit: $L = 6.022 141 79 30 \cdot 10^{23}~\color{red}{\mathrm{mol}^{−1}}$.

The difference between the Avogadro constant (dimensional) and Avogadro's number (dimensionless) is quite subtle, and often overlooked.


text:
Do all gases occupy same volume at equal temperature and pressure conditions?
Why does one mole of all gases occupy 22.4 L at standard temperature and pressure? According to my thinking, it may be due to equal diffusion of all particles when at a particular temperature and  pressure.
So going by this, one mole of all gases should occupy the same $x~\mathrm{L}$ at some other temperature-pressure conditions. Is this true?

Why do all gases occupy 22.4 L [per mol] at STP?

The question is based on a false premise. Only ideal gases are guaranteed to occupy 22.4 L/mol at STP.  There are many gases that are not ideal.

So going by this, all gases should occupy same $x$ L at some other temperature-pressure conditions. Is this true?

No, again this would be true only for ideal gases, not all gases.
At STP, many commonly encountered gases on Earth behave nearly ideally, but certainly not all.  For example, vapor of acetic acid and other simple carboxylic acids show strong deviations from ideality.
Wikipedia is a good place to read about the assumptions of the ideal gas law and the kinetic theory of gases.  The reason that ideal gases share a molar volume of 22.4 L at STP is not related to diffusion, but rather to the assumption of: (a) non-interacting gas particles that are (b) effectively zero-sized relative to the mean distance between them.
You can read about equations of state for non-ideal gases on Wikipedia as well.  Popular equations to describe non-ideal gases include the van der Waals equation, the Redlich-Kwong equation, and a whole slew of others listed at Wikipedia's real gas page.

Thank you that was very helpful! Actually i was taught .. quoting exact words " all gases occupy 22.4 L at STP"

I love this answer! Exactly helpful to a learner! +1 since I can't upvote more than once.

Although, technically no gas is ideal, which is why we can be so certain of their properties.

I agree with PyRulez. No gas is ideal. It would be good to change that so that it isn't misleading.

This "no gas is ideal" idea needs empirical support.  What's the deviation from ideality for say helium 4 at .00001 PA pressure and a temperature of 1000 K

Does the second assumption also imply that the mean distance between molecules for any ideal gas is the same? If yes, what is that distance?

IUPAC has changed the definition of STP from [273 K, 1 atm (101.3 kPa)] to [273 K, 1 bar (100.0 kPa)]. Hence their molar volume at STP has changed from 22.4 L/mol to 22.7 L/mol.

do you have a reference you can include for the change of STP?

Reference:  Ideal Gas wikipedia article mentioned above, and follow the references therein.

I didn't realize that, thanks for the info.  Also, why in the world would IUPAC do that?  It's hard to think of a reason other than the self-perpetualization.


text:
How do I know which is the "correct" way to write the reaction?
We need to calculate the heat produced by a set of $\mathrm{N_R}$ reactions in our reactor model.
\begin{align}
\mathrm{Q_r = \sum_{j=1}^{Nr} Rr_j*\Delta H_{R,j}}
\end{align}
$\mathrm{Rr_j}$ is the reaction rate of reaction $\mathrm{j}$, in $\mathrm{\frac{kmol}{kg_{cat} s}}$
$\mathrm{\Delta H_{R,j}}$, in  $\mathrm{\frac{kJ}{kmol}}$, is constant and calculated as:
\begin{align}
\mathrm{\Delta H_{R,j} = \sum_{i=1}^{Nc}\nu_{i,j}*\Delta h_{i,j}^f}
\end{align}
Now consider these:
\begin{align}
\mathrm{R1:}\ \ce{H2 + 1/2O2 \rightarrow H_2O}
\end{align}
\begin{align}
\mathrm{R2:}\ \ce{2H2 + O2 \rightarrow 2H_2O}
\end{align}
They are undoubtedly the same reaction, but the R2 gives a  $\mathrm{\Delta H_R}$ that is twice the $\mathrm{\Delta H_R}$ in R1. It is naturally so, but greatly affects my energy balance.
How do I know which is the "correct" way to write the reaction?
Does it mean that the reaction rate should be specific to R1 or R2?

Since they are the same reaction, except for the fact that R2 is two R1 reactions, wouldn't the enthalpy of the R2 be twice R1's, but in twice the time as an R1 reaction? I think this would mean that they both produce the same amount of energy in per unit time.

How do I know which is the "correct" way to write the reaction?

There is no single correct way.  Picking a different way to write the reaction affects both $\Delta H_{R_j}$ and $Rr_j$ in opposite ways that cancel out.

Does it mean that the reaction rate should be specific to R1 or R2?

Yes.  The unit of your reaction rate $Rr_j$ is $\frac{\textrm{kmol}}{\textrm{kg}_{cat}}$.  But the key question is kmol of what?  A good way to think about the answer is "kmol of things that have a stoichiometric coefficient of 1 in the reaction".
So if the reaction is R2, then the rate of the reaction will be the rate of disappearance of $\ce{O2}$.  The rate of formation of water will be twice the reaction rate $Rr_2$.  But if the reaction is R1, then the reaction rate is equal to the rate of formation of water (or the rate of depletion of hydrogen).
But you don't directly measure "reaction" rates, you can only measure the formation or disappearance of products or reactants.  So say you measured the formation rate of water.  If you decide the reaction is R2, then $Rr_2 = \frac{1}{2}R_{\ce{H2O}}$.  If you decide the reaction is R1, then $Rr_1 = R_{\ce{H2O}}$.  The factor of $\frac{1}{2}$ will cancel out the factor of two in $\Delta H_R$.
So in the end it shouldn't matter what choice you make.
For the exact same idea in a totally different chemical context, see this se.chem question on electrochemistry.

This confirms my hypothesis then! Now, since I was given the reaction rates, I just need to figure out for which one they were calculated...


text:
Are carbocation intermediates in equilibrium?
So for the reaction of $\ce{H_2O}$ with an alkyl ethene, in the presence of $\ce{D^+}$ ions, forms two possible carbocations:
$\ce{ R-CHD-CH2^+}$ and $\ce{R-CH^+-CH2D}$
The major product formed is the secondary carbocation as it is stabler than the primary carbocation.
So my questions are:

Are the two carbocations in an equilibrium or are they formed in
specific, fixed quantities?
If they are in equilibrium, i.e., the primary and secondary
carbocation, why and how would the secondary carbocation revert back
to being a less stable primary carbocation?
Also, is the hydride shift that converts the primary to a secondary
carbocation the only shift that can occur; even if after the shift,
a more stable carbocation can be formed (depending on the attached
alkyl group)?

Are the two carbocations in an equilibrium or are they formed in specific, fixed quantities?

$\ce{R-CH=CH2 + H^+ <=> R-CH2-CH2^+ + RCH^+-CH3 <=> alcohol~products}$
Yes, the two carbocations formed are in equilibrium. Initially the two carbocations will be formed in a ratio dependent upon the difference in activation energies leading to them (kinetic control).  However, since all of the steps in this reaction involve equilibria, after a while when equilibrium is established several things can happen

many carbocations will revert back to starting material which can then be reprotonated and reform the carbocations
or alcoholic product can be reprotonated, eliminate water and reform the carbocations
or the carbocations themselves can equilibrate directly by hydrogen shift

Once the equilibrium is established, the ratio of carbocations formed will be dependent upon the energy difference between the carbocations themselves (thermodynamic control).  Often, but not always, the relative stabilities of the transition states leading to the carbocations matches the relative stabilities of the carbocation intermediates themselves.

If they are in equilibrium, i.e., the primary and secondary carbocation, why and how would the secondary carbocation revert back
  to being a less stable primary carbocation?

why - that's what happens in equibria; even though phenol is much, much more stable than its enol counterpart, there will be an exceedingly small amount of the enol present at equilibrium

how - by any of the 3 routes mentioned in the bullets above

Also, is the hydride shift that converts the primary to a secondary carbocation the only shift that can occur; even if after the shift, a
  more stable carbocation can be formed (depending on the attached alkyl
  group)?

Other groups like alkyl or phenyl can shift too, but hydrogen shift is usually faster than the others.

Say the alkyl group was a third degree carbon, would a possible hydride shift also be included in the equilibria; or would the reaction already have moved on to the next step?

Not sure I fully understand your question.  If you're asking if a tertiary alkyl group can compete with a hydride shift, usually the hydride shift will be faster.  The shift(s) will occur before the intermediate carbocation moves on to product.

I mean like if there are two or more successive shifts possible, would all the possible carbocations exist in an equilibrium?

Yes, in relative amounts that correspond to their relative stabilities.

@ron So at equilibrium can we say that the concentrations of intermediates are non-zero?

@adosar Yes non-zero, of course the concentration of one might be much, much larger than the other.

Actually four possible intermediates are there, two from autolysis of water and two from added deutrium.

1) $\rm 1>2\gg3>4$ Since autolysis of water is very minute and a secondary carbocation is always stable than a primary carbocation since it has more hyperconjugations.
2) They would be in equilibrium since every reaction is reversible, though it's rate might be minute for a specific direction. There could also be one in a gazillion hydride shift case.
3) This depends on the migratory aptitude which is generalised as $\rm H>Ph>1^\circ>2^\circ>3^\circ$


text:
Effect of impurities on enthalpy of fusion of ice
When I freeze water with some impurities (1% NaCl by mass), the mixture seems to have a much lower enthalpy of fusion that pure water. The impurities cause the freezing point to depress to 0.6°C. I only have a simple calorimeter available but it looks like the enthalpy of fusion is dropping by 25%+ compared to pure water (tested by melting the ice into water and measuring the temperature drop).
Additionally, the ice does not have a nice crystal structure. It is "mushy" and easily deformed.
This effect is surprising to me since colligative theory tells us that most properties changes in a pretty small linear way at low concentrations (like the freezing point).
Am I observing a real phenomenon or is there some problem with my testing?
If its a real phenomenon, is there a way to calculate/estimate the effect?

are you mesuring change in freezing point, $\Delta T_f$ or change in enthalpy of fusion $\Delta H_f$ ?

@ADG Both. The change in freeze point matches theory nicely- $\Delta T_f = iK_bm = 0.6°C$. I back-calculated the enthalpy of fusion from $m_{water}c_p(T_f-T_i) = m_{ice}*h_f$ (where the final temp is 0C so the ice has no sensible energy change.)

ok the change in freezing point is matching but not the heat of fusion?

Thats the question. I don't know of a way to calculate the enthalpy of fusion of a salt/water mixture to compare the experimental value against. Is there a formula for this?

Maybe 1% isn't "low" concentration

Just to be clear, you're saying you measured the 1% brine solution to freeze at -0.6ºC, not +0.6ºC, right?

According to the article An Empirical Correlation between the Enthalpy of Solution of Aqueous Salts and Their Ability to Form Hydrates (Pandelov et al. 2010), your observation of a lower enthalpy is correct:

We have found a correlation between the enthalpy of solution of the salt and its ability to form a hydrate, namely, that the salt’s enthalpy of solution is lower than the standard enthalpy of fusion of ice (6 kJ/mol)

However, the drop in enthalpy you observed of 25%+ seems excessive, as can be seen in the graph below:

Latent heat of fusion comparison between water, $\ce{NaCl}$ solution and ethanol solution.
Image source: Colligative properties of water cited from the original article Formulation of the latent heat of fusion of ice in aqueous solution (Kumano et al. 2009)
KUmano et al. 2009's methodology centred around the use of

Differential scanning calorimetry analysis was performed and the measurements and calculated values compared with respect to the specific enthalpy of the ice/solution mixture.

As for the 'mushy ice' that you observed, it is likely to be a mixture of ice and brine, as per the diagram below:

Image source: Wikipedia

Thanks for the references, these are very helpful. Update for anyone else who might ever be reading this: I put the mushy stuff in a chest freezer for a week and it is a nice solid block. The issue must have been in the way I was making the ice.

I think your methodology is a bit wrong. When the frozen brine melts into the water, the water won't be 100% water anymore. So, the specific heat capacity of the water changes, making the change in temperature different. Maybe that's why you have a 25% drop when compared to water. You can follow the steps below only if you have a heat source that generates a known amount of heat(energy), like an immersion heater with known voltage and wattage.

Weigh an empty container

Clamp a filter funnel to a retort stand (or anything that can hold
it off the table surface)

Place the container below the filter funnel

Dry the ice and crush it

Place the crushed ice into the filter funnel

Set a timer for a specific amount of time, place the heater into the crushed ice and turn the power on, start the timer at the same time.

Once the time is up, switch off the power to the heater and remove the container

Now, weigh the container with the melted ice

Use the same set up and follow the same steps in the steps above, but this time, don't turn on the heater (but still leave it in the ice) (this is to find the amount of ice that melted due to the room temperature)

Find the mass of ice that melted due to the heater by subtracting the mass of the container and melted ice without the heater on from the mass of the container and melted ice with the heater on

Find the amount of heat generated from the heater by taking the power rating of the heater (in Watts) and multiply it by the time the ice melted for.

The main equation you'll use is ML = Pt, where M is the mass of ice that melted due to the heater, L is the specific latent heat of fusion, P is the power rating of the heater in Watts and t is the time that you let the ice melt for. From there I think you can figure out L.
All this is done with the assumption that the heat from the heater that is lost to the surroundings is 0 or negligible.
As for the "mushy" ice, you probably didn't let it freeze long enough.
IDK if any of this will help now though, cus it's been more than 5 years, and I'm pretty sure you're done with the experiment.LOL


text:
What mechanisms might make a substance travel different distances on the same TLC plate?
When doing TLCs to monitor reactions, it is considered good common practice to not only spot the reactants and the reaction mixture (likely after some mini-workup), but also to include a ‘cross-spot’: every other substance (mixture) spotted somewhere on the plate is spotted together on one spot, commonly labelled something like X.
When asking why during my bachelor's thesis, the PhD student supervising me said something that roughly translates to:

You can't be sure that the components don't run differently in the mixture than by themselves.

But in a perfect world, they should. Aside from the obvious case of having a polar reaction solvent that didn't entirely evaporate and skewes the elution mixture on parts of the TLC, are there other reasons why the components of a mixture should have different Rf values when spotted separately?

I think you have the key point there about the reaction solvent affecting the Rf of your analyte. Using a mini-workup routinely is a good habit to get into. In my experience, the cross spot is primarily useful to distinguish compounds that are very close in polarity.

are there other reasons why the components of a mixture should have different Rf values when spotted separately?

One might speculate that two polar reactants form a less polar "adduct". This should however result in an additional spot, rather than in altered $\mathrm{R_f}$ values for the individual compounds.
But this is speculation. Admittedly, I have never used the cross spot method. Neither have I ever observed that spots of the reactants had different $\mathrm{R_f}$ values in the reaction mixture.
On a side note: What always worked for me was:

Don't apply spots but bands and overlap them.
"Sharpen" the bands on the base line by developping the TLC in methanol or another polar solvent first.

Together, these two tricks allow to destinguish between two compounds with very similar  $\mathrm{R_f}$ values.


text:
Instrumental methods of analysis
I am taking the instrumental methods of analysis course. Our job is to find the unknown  mass of $\ce{Fe}$ which is added, by measuring the absorbance of different solutions with ortho-fenantroline.
Is there any way I can know if my calculated mass of $\ce{Fe}$ is correct or not?   Is there any way I can prove it, since I know the absorbances?

Am I right that you are planning to simply calculate the $\ce{Fe}$ concentration using the Lambert-Beer equation?
The obvious proof would be to prepare $\ce{Fe}$ solutions with known concentration(s) to compare the absorbances. Using these solutions, you can also make sure that the assumption of linearity over the concentration range is actually fulfilled.

We already prepared those solutions with known concentrations, and then found the concentration of the unknown Fe solution, and then of course the mass. I found the Fe mass, but I don't know if that is the exact one. That is known only by our teacher, and we will be testet if the mass of Fe given is not that of our calculations. Hopw i was clear ...

@NdrinaLimani Then, I think you've done all one can do. If repeated measurements give identical results, there's no reason not to trust your own  work.


text:
Does quantum yield of reaction factor in number of electrons needed?
Can someone tell me if quantum yield of a reaction factors in the number of electrons needed to create the particular species? If not, is there a metric that does?
For example, I have a 4 electron reduction which means that even if I have 100% efficiency, my quantum yield is
$$\frac{\text{# molecules reduced}}{\text{# photons absorbed}} = 0.25$$
Is there a better metric I can use so that 1.0 will be 100% efficiency?

I'm not aware of a definition of quantum yield that factors the electrons in. The normative reference was/is Silvia Braslavsky's Glossary of terms used in photochemistry (SRC).
Unless I misunderstood, you do have an overall four electron reduction, but this happens in a sequence of four excitation events.
From a formal point of view, the quantum yield(s) should be given for each of these processes. I reckon that this isn't possible, so an overall quantum yield of 0.25 is all you can get. In the end, I'd simply explain that, rather that looking for some "tricks" to come up with a more impressive number.

You are absolutely right that I shouldn't look for tricks to artificially inflate a value, but I asked about this correction for number of electrons so that it can be more clearly compared against a photochemical process that only takes 2 electrons instead of 4.  I really appreciate your response!


text:
Can organomercury compounds synthesize accidentally from household cleaning products?
I had broken a CFL partially on carpet last week. It was a spent bulb, and the outer glass shell broke but not the fluorescent tube - apparently the tube did have a small hole in it which found out after all my other steps when I saw it in the garbage.
When I cleaned it up first, I used a mild spray cleaner which contained methyl esters and various surfactants - I found this out now, not then, or I wouldn't hzve used it. Wiped it all up with a paper towel, then vacuumed in case I missed small pieces. I did ventilated at that point.
About two hours later, my cat made a mess on the rug. At that point, I sprayed down the carpet with resolve (which says it contains hydrogen peroxide). Wiped that all down with a towel, then revacuumed...and realized I hadn't cleaned out the dust cup. I think I then emptied that outside and cleaned out the HEPA filter.
My concern now is what I may have exposed myself to in that botched cleanup. Any ideas on what may have formed? Do I need to be concerned?

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website.

I realize that I cleaned it up entirely wrong, and have ventilated the house for a week. Should I be worried about the interactions between any of the mercury/hydrogen peroxide/methyl esters that may have been in my vacuum?

Unfortunately, you seem to have already broken some of the protocols of cleaning up a broken CFL according to the EPA's website.
According to this site:

Vacuuming is not recommended unless broken glass remains after all other cleanup steps have been taken.  Vacuuming could spread mercury-containing powder or mercury vapor.

Also according to data from a study of the effect of the breakage of CFLs performed by the Maine Department of Environmental Protection:

Mercury concentration in the study room air often exceeds the Maine Ambient Air Guideline (MAAG) of 300 nanograms per cubic meter (ng/m3) for some period of time, with short excursions over 25,000 ng/m3, sometimes over 50,000 ng/m3, and possibly over 100,000 ng/m3 from the breakage of a single compact fluorescent lamp.

and also:

Vacuuming a carpet where a lamp has broken and been visibly cleaned up, even weeks after the cleanup, can elevate the mercury readings over the MAAG in an un-vented room.

At the end of the report, it states that although mercury vapor is extremely toxic:

...it is unclear what the exact health risks are from exposure to low levels of elemental mercury...

It seems as though the best course of action you could take now would be to ventilate the room and if possible your house extremely well for several hours, and to avoid vacuuming any more.


text:
Solve this chemical or biological mystery
I hope this is the best place to pose this unique question.  Please forgive me if it's not.
I went into a closet to get a prescription medication that was filled about a year ago.
The label on the pill bottle has visible signs of something.   Identifying that something is the purpose of this question.
The "something" can be described as light brownish orange circles of discoloration.  They look a lot like mold, but I'm hoping that's not the case.  I tried to photograph them, but they don't show up well.
I took a look at a few other prescription bottles from that time period, and the labels on them also have this appearance.  Labels on newer bottles from the same pharmacy do not have this appearance.
On one bottle, there are also discolorations that are yellow, similar to what a yellow highlighter would produce (but not quite).
My concern is mold, but I'm hoping it's not.  I don't see other signs of visible mold in the area.
The closet is not near a bathroom.  The room is air conditioned and heated (depending on the season) and is kept at approximately 76 degrees Fahrenheit, and never outside the range of 61-89 degrees.  There is a MERV 16 filter on the HVAC system.
The pharmacy is very clean and is fairly new (probably less than 10 years old).  The entire pharmacy chain is probably less than 15-20 years old.  They compound medications within the confines of a special compounding enclosure; none of my medications were compounded.
I don't want to bias the answers, but I am hoping that the brownish orange circles can be explained by a chemical reaction.  I'm thinking that perhaps pharmacist hands are so covered with different chemicals that it could cause a reaction on the label over time.  I don't think the labels had the discolorations when I picked up the medications.
The areas on the labels that show the light brownish orange circles could possibly be areas where the pharmacist pushed down the label.  Areas in which people make contact with items have increased probability of growing mold; but given that a pharmacist was the one who touched these labels, there is also the increased possibility that their fingers were coating with small amounts of medications.
Given the above facts and descriptions, are there any likely chemical reactions that would cause the observed phenomena?
Please note that I'm not a chemist, so please write answers accordingly!  :-)

Is the discoloration coming from underneath the label or is it present only on the top of the label, e.g. can you scrape it off?

Welcome to chemistry.SE! "On one bottle, there are also discolorations that are yellow, similar to what a yellow highlighter would produce (but not quite)." I think I know what you're talking about, but my instances are usually caused by the tap water drying and leaving who-knows-what and some corroded steel (which seems not to be the case here). It would really pay if you add an image of what you're talking about.

@ron Great question.  I rubbed the discolorations with my finger, and they did not come off.  I could not tell if they were coming from underneath the label.  I noticed that on 2 bottles, each had a tiny slit or cracks in the label, just a few mm long.  Surrounding those slits/cracks was the discoloration.  Other areas on the same labels have the discolorations, but have no visible slits/cracks.

So perhaps it is adhesive seeping through weak spots on the bottom of the label to the top surface and reacting with the label material when contact occurs


text:
When I place a straw in carbontated drink and blow, more gas bubbles rise to the surface?
When I place a straw in carbontated drink and blow (making bubbling noises), more gas bubbles rise to the surface. The gas bubbles which rise to the surface are from the carbon dioxide gas dissolved in the drink, I'm not talking about the air bubbles I made with the straw?
Why does this happen?

Can you be a little more specific about your "experimental" set up? Maybe while blowing your breath into the drink, you also stir it up a little?

When you blow into a solution, your warm breath (~98.6°F) warms the solution up. As a solution warms up, less $\ce{CO2}$ is soluble.  Therefore, blow into a solution, warm it up, less $\ce{CO2}$ is soluble, $\ce{CO2}$ escapes (bubbles out) from the warmer solution.

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Hmm I never considered this simple possibility. I thought the blown air was acting as a nucleation site for removal of the supersaturated $\ce{CO2}$ in solution. Bubbles are good at nucleating other bubbles, which is why opening shaken cans of soda is not a good idea!

I don't believe this answer is fully correct. I repeated the experiment, this time using one of those dust-off air blower things, which blow out COLD air, and I got the same effect. In fact the more intense the spray, the more bubbles. So there's something more here.

I downvoted the answer because its incomplete. Increasing temperature of the solution can't be the only reason and when the temperature actually decreases it's not a reason at all. I think Nicolaw Saker Neto has the other half with bubbles nucleating other bubbles.

@ron But blowing a few air bubbles into a glass of Coke doesn't increase its temperature by anything even close to 10°C. Heat conduction between a gas bubble and a liquid is very slow and the bubble exists in the liquid for, what, about a second? I'd be surprised if blowing bubbles had any significant effect on temperature.

@DavidRicherby wrote "blowing a few air bubbles into a glass", the lungs contain around 6 liters of 98.6°F air.  Also, I just used 10° as an example, a 1° change could release significant $\ce{CO2}$

@JoshuaBenabou  What was the temperature of the air from the cold blower and what was the temperature of the drink?

@ron Nothing in the question talks about blowing a whole lungful of air through the drink.

@JoshuaBenabou  How much air did you blow into the drink through the straw?

@DavidRicherby Nor does it say a few bubbles.  Just thinking back to when I was a kid and we'd blow a lot of air in.

The drink was at room temperature and the temperature of the air is less than that.

I think this is a very under-appreciated answer and I do not understand Joshua's reaction at all. It might not be complete, but it is far away from wrong. I am sorry about that. (In this case, if this was my answer, I would remove the answer, just because I could.)