text:
What is/are the product when 2,2-dimethylhex-5-en-3-ol reacts with Lucas reagent?

What would be the product(s) if 2,2-dimethylhex-5-en-3-ol reacts with Lucas reagent?
Would a Methyl shift happen? I've heard that in Groove's process, only primary and secondary alkyl chlorides are formed.

Actually, in this case, a hydride shift from C4 seems more likely. Allylic carbocations tend to be more stable than even tertiary non-conjugated carbocations. Under SN1 circumstances you'd get a mixture of 5,5-dimethyl-3-chloro-1-hexene and 5,5-dimethyl-1-chloro-2-hexene. Under forcing conditions further addition might take place, mainly to the 1,3-dichloroalkane.

A shift of the methyl group is likely, because this would yield a more stable tertiary carbocation. In concentrated hydrochloric acid, the double bond can also be protonated, producing a secondary carbocation which is captured by $\ce{Cl-}$. If both happens, the product would be 2,4-dichloro-4,5-dimethylhexane.


text:
Hyperlithiated Carbon Species
Below is replicated Question 1 from the Final Qualifying Exam of the Australian Chemistry Olympiad, 2004B, here.
Question:
The theory of promotion-hybridization is quite successful at explaining why covalent molecules with a central atom from period 3 or higher (e.g. $\ce{SF6}$, $\ce{PF5}$, $\ce{I3}$) are stable even though they are hypervalent (i.e. they appear to "exceed the octet rule"). However, hybridization is inadequate in explaining the remarkable stability of molecules such as the so-called "hyperlithiated" carbon species such as $\ce{CLi6}$. Molecular orbital theory can be applied to $\ce{CLi6}$ (Reed and Weinhold, 1985) for an alternate explanation to when and why hypervalent molecules with central atom of period 2 can be stable.
Of all the different types of atomic orbitals (e.g. $s$, $p_x$, $d_{yz}$, etc) of a general central atom, six of these will participate in the forming of sigma bonds with the molecular orbitals of six ligands in an octahedral geometry. The valence $s$ orbital is one of them. Note that the ligands approach the central atom along the three coordinate axes.
(a) Which other five atomic orbitals can also do this?
Now, we shall consider the molecular orbitals (MO) of the 6 $\ce{Li}$ $2s$ orbitals (in an octahedral geometry). The lowest energy MO is given below in figure 1:

According to quantitative calculations, the MO in figure 1 is the only MO lower in energy than the uncombined lithium $2s$ orbital; all other MOs are slightly higher in energy than uncombined $\ce{Li}$ $2s$ and can thus be considered as antibonding. (Note that this is generally true for 2 to 8 Li atoms arranged symmetrically around the central atom).
The $s$ orbital will overlap with the MO given in figure 1. Exactly one of the other five MOs will have the right symmetry to overlap (and form a sigma bond) with each of the other five atomic orbitals of the central atom.
(b) Hence draw the other five MOs obtained by linear combinations of the 6 $\ce{Li}$ $2s$ orbitals. Indicate their relative energies on an energy diagram.
When the central atom is carbon, three of the six MOs (of the 6 $\ce{Li}$ $2s$ orbitals) will not combine with the corresponding atomic orbital of carbon. One of these is the one drawn in figure 1. Because carbon is much more electronegative than lithium, the $2s$ orbital of carbon is too low in energy to mix with the MO in figure 1.
(c) Which other two MOs will not combine when carbon is the central atom? Provide a rationale as to why this is so.
(d) Hence, draw the molecular orbital energy diagram of $\ce{CLi6}$. Keep in mind the relative energies of carbon AOs and the six Li MOs (as carbon is more electronegative than lithium).
(e) Using the energy diagram in part (d), explain the stability of $\ce{CLi6}$.
(f) Propose an alternative to the "octet rule" when predicting stability of hyperlithiated species of period 2 atoms. Using this rule, identify possible neutral (i.e., not cations or anions) hyperlithiated species of nitrogen and oxygen that can exist due to the same stability reason as that of $\ce{CLi6}$.
(g) Can $\ce{CH6}$ exist? If so, why so? If not, why not?
My response:
(a) These are the $p_x$, $p_y$, $p_z$, $d_{z^2}$, and $d_{x^2 - y^2}$ orbitals.
(b) Assuming analogy with octahedral complexes, Googling provides this, pg. 10 of which appears to answer the question. Given that $E_d$ < $E_s$ < $E_p$, the $d$ orbitals represented would be found at the bottom and $p$ at the top of the energy diagram.

Is an octahedral complex analogous to $\ce{CLi6}$?
What is a linear combination of atomic orbitals?

(c) By logical reasoning, these would be the two similar $d_{z^2}$, and $d_{x^2 - y^2}$ orbitals. I am uncertain regarding the rationale.

Is it because it is unfeasible for carbon's electrons to be in $3d$ orbitals?

(d) I really have no idea.

If three of the six possible MOs will not combine, does that mean that they are not present in the MO diagram?
Since carbon is more electronegative than lithium, it should contribute more to the bonding orbitals, while lithium should contribute more to antibonding orbitals.
How are the 10 electrons in $\ce{CLi6}$ distributed? In order to impart stability, there must be more electrons in bonding than in antibonding orbitals. Can we simply have three bonding MO orbitals with 2 electrons and three antibonding MO orbitals with 4 electrons between them?

(e) By logical reasoning, the larger number of electrons in bonding rather than in antibonding orbitals imparts stability to the molecule since bonding electrons are not entirely counteracted by antibonding electrons.
(f) I really have no idea.

By guesswork, I would propose the decuplet rule, in which hyperlithiated species from period 2 are stable with 10 valence electrons, producing the species $\ce{NLi5}$ and $\ce{OLi4}$.

(g) By logical reasoning, yes: it satisfies the conditions of (f).

My gut says no.

Given that my background in chemistry is limited to high school AP Chemistry, please curtail the material provided in your responses. Thanks for all your help in advance!

I have to say, the test you linked is fantastic. I would be very hard-pressed to complete it in the allotted time. I am also especially interested in the particular question you've asked, but I don't think I can answer it too well myself, so I hope it comes to others' attention.

If you're interested, other such tests are here, here, and here.

Nice question, sadly the solutions for Australia's FSE can't be found anywhere.

(a) Correct.
(b) Let me start with Linear Combination of Atomic Orbitals.
That's a quantum calculation method that assumes that all molecular orbitals can be derived from (basic) atomic orbitals by a method called linear combination. You might have heard of linear combination from vector geometry. It's basically just adding things up with different preceding factors. So 2x - 3y is a linear combination of x and y.
The thing about linearly combining atomic orbitals is that you must end up with the same amount of orbitals as you started out with (so you combine 6 atomic orbitals to make 6 molecular orbitals). There's also additional stuff you need to take care of (the squares of all the preceding factors of one AO need to sum up to give one), but that's only of marginal importance here.
An example of linear combination would be your average H–H sigma bond. You have two atomic orbitals (1s of each of the hydrogens) which you combine. I'll call the 1s orbital of the left hydrogen l and that of the right hydrogen r, for simplicity. The two simplest linear combinations are l + r and l - r. In the first one, you get positive interference, i.e. you add a positive phase to a positive phase – think of it as the positive half of a sine wave added onto another positive half of a sine wave: it'll get bigger. This type of interaction is favourable, and will thus give you a molecular orbital energy that is less than that of a separated atomic orbital.
But you also need to remember to create a second MO, the l - r. Here, you would be adding plus to minus, which gives zero. So you have a plane in between the two protons where the resulting wave function has zero value, is non-existant (a nodal plane). Because the nodal plane is between the atoms, the resulting molecular orbital is antibonding, or higher in energy. The basic rule is that the more nodal planes an orbital contains, the higher its energy level will be.
The question asks you to find the further five linear combinations that will lead to those group orbitals capable of overlapping withthe  central carbon's orbitals. So you basically know what to look for; and yes: the CLi6 is analogous enough to an octahedral complex, that you can just copy that image. If you didn't have either the picture or the hint, you would need to use group theory and reduction formulae, which I won't cover here (for those interested: the group orbitals will transform as a1g, eg and t1u).
I don't understand your reasoning for putting the d-type orbitals lower in energy (i.e. more stable) than s- or p-type ones. Usually, it's s < p < d. But the better way to answer the question would be using nodal plane counting: The orbital that will mix with the carbon'S s orbital has no nodal plane, and will therefore be the lowest (a1g, for those interested). The three that will mix with the carbon's p orbitals are second, because they contain one nodal plane each (t1u). Finally, those that will mix with d orbitals are highest, because they contain two nodal planes (either parallel or perpendicular to each other. Their energy level will be the same, though, says group theory) (you guessed it, those are eg).
(c) Yes, that can be regarded as the easiest answer. Carbon, being an element of the second period, only 'has' (term is incorrect, but you get the picture) 1s, 2s and 2p orbitals. 3s and 3p might be accessible by irradiation. 3d is not, as it is even higher in energy than 4s.
The same rationale applies to the higher order main group elements. Even though silicon, being an element of the third period, could be considered to access d orbitals to create anions like $\mathrm{SiF}_6^{2-}$, the d orbitals' energy is too far removed from that of the s and p orbitals, that it cannot be considered to take part in bonding.
Note that transition metals – the elements usually associated with octahedral surroundings and d-orbitals – usually use 4s and 3d (and sometimes 4p) orbitals. 3d is a lot closer to 4s and 4p than to 3p.
(d) Start off by thinking what the relative energy levels of the orbitals in question are. Carbon, as was pointed out, is more electronegative than lithium, so all of its orbitals will be lower in energy than the lithium orbitals (the difference is significant here).
That should give you a picture of carbon orbitals on one side (2s, rather far-ish down, three 2p's, still pretty far down) and the three sets of lithium group orbitals on the other side (the one that I called a1g being the lowest, but probably higher than carbon's 2p's; the three I called t1u next and the two eg last; those inside a group at the same height, of course).
Next, you need to find out which orbitals can recombine favourably. If you attempted to combine the orbital shown in figure 1 with a carbon 2p orbital, that intereference would be non-bonding or zero (you have bonding interactions on one side cancelled out by exactly the same amount of antibonding interactions on the opposite side). Note that they helped you by saying that the figure 1 orbital (and thus also carbon's 2s) and the two lithium orbitals that would bond with d-type orbitals do not take part in favourable interactions; so only the p-type ones remain, each of them creating a bonding-antibonding pair of molecular orbitals.
Moving from the edges to the centre of the diagram, you would move the carbon's 2s plus the three lithium group orbitals not taking part horizontally into the middle. Carbon's 2p and the other three lithium group orbitals are then mixed, lowering the 2p's energy while raising the lithium orbitals' energy.
You should end up with an energy diagram of the following type:
    C(2s) - 3 C(2p) -- 1st Li -- 3 2nd Li - 2 3rd Li.
It's not clear (nor can it be deducted a priori) whether the 2p orbitals will end up lower than the 2s (or the 2nd Lithium group higher than the third), but it is to be assumed that they shouldn't change their places too much.
Finally, fill in the electrons in the way you're used to. Two go into C(2s), six into the three C(2p), and two into the first Li orbital (the one shown in figure 1). To answer your questions:

non-combining MOs can or can not be present in the MO diagram. They should, if they are part of the valence orbitals. They should not if they are core orbitals (note we always omitted the seven 1s orbitals). Not combining however means, that they do not change height from where you put them initially.
Yes, that is correct, but that is more of a consequence we can think of after mixing has been done.
If we include the nonparticipating orbitals, the question is no longer a question. If they weren't present, though, you would do exactly what you proposed. You would fill in three electrons spin up into three antibonding MOs and one further electron spin down. The resulting system would be degenerated. It would not be stable, and would reduce symmetry to force the three orbitals apart, creating a discrete energy level for the single paired electron.

(e) Yeah, that was a pretty good guess into the wild, that would always be correct ;)
To use the molecular orbital diagram as a reasoning, though: Consider the state before bonding. The four carbon orbitals have four electrons between them (2 in 2s, 2 degenerated in 2p). The six lithium orbitals have six electrons between them (2 in the figure 1 orbital, 4 in the three orbitals that will take part in bonding; this state is also degenerated). After the bonding, not only have we moved four electrons out of relatively unstable orbitals into a lot more stable ones (those from the lithium orbitals into the carbon-centered ones), but we have also removed two degenerated states and turned them into clearly defined ones. Both are good for the system's stability.
(f) I think your guesswork is pretty much spot on. (The orbital diagrams would be totally different ones, though; as NLi5 would be a trigonal bipyramid (symmetry group D3h) and OLi4 a tetrahedron (symmetry group Td) as opposed to the carbon octahedron (Oh).
(g) I'm pretty sure this is a yes-but case. Yes, just looking at what we have for Li would suggest that H would do the same thing. But, hydrogen's atomic orbitals are lower than lithium's, as it is a lot more electronegative. That in turn means that there is significant mixing to be observed between C(2s) and the figure 1 type orbital. Therefore, the C(2s) orbital would again be stabilised, while the hydrogen group orbital would be destabilised – most likely so far that it reaches the energy level of the three higher group orbitals. While they were clearly empty in the lithium case, they would most likely be populated in the hydrogen case, destabilising the system.
Another thing to consider is the relative stability of CH4 versus CLi4. Again, the all equal phase group orbital of the lithiums would be too high to significantly mix with the C(2s) orbital in a tetrahedric case. The all equal phase group orbital of hydrogens does mix quite significantly, creating a much more stable CH4 molecule than CLi4. Thus, CH6 would eliminate hydrogen to create CH4, while CLi6 is stable enough (and CLi4 unstable enough) to not eliminate Li2

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text:
Amphoteric amino acid titration

Glycine $(\ce{NH2CH2COOH})$ is an amino acid with two $\mathrm{p}K_a$ values $(\mathrm{p}K_{a,1}=2.0, \  \ce{COOH};\  \mathrm{p}K_{a,2}=10.0, \ \ce{NH2})$.
(a) If $\pu{0.01 mol}$ of this amino acid is used to prepare one $\pu{1 L}$ solution. Calculate the $\mathrm{p}\ce{H}$ and the $\ce{NH3+CH2COOH}$ concentration in the solution.
(b) If $\pu{10 mL}$ of the above glycine amino acid solution is mixed with $\pu{90 mL}$ of basic buffer system. If the concentration of the $\ce{NH2CH2COO-}$ was identified to be $\pu{1.05 \cdot 10^{-4} M}$ indicate the $\mathrm{p}\ce{H}$ of the glycine solution.
(c) If $\pu{10 mL}$ of strong base $\ce{NaOH}$ ($\pu{0.01 M}$) was added to solution in (a) what is the final $\mathrm{p}\ce{H}$.

For part (a):
I used $[\ce{H+}]=\sqrt{10^{-10}\cdot0.01} =10^{-6}$.
Then I used $\mathrm{p}\ce{H} =14-\log(10^{-6})$ to get a $\mathrm{p}\ce{H}$ of $8$.
While my $\ce{NH3+CH2COOH}$ is $\pu{10^{-6} L}$.

For part (b):
I used the Henderson-Hasselbalch equation for a basic buffer $$\mathrm{p}\ce{H} =\mathrm{p}K_a     +\log\left(\frac{\ce{[NH2CH2COO- ]}}{\ce{[NH2CH2COOH]}}\right)$$ I got a $\mathrm{p}\ce{H}$ of $10.97$. But I'm   unsure of my answer.

For part (c):
I suspect that it is a basic buffer but I'm not sure which value to use. could anyone help me out with this.

note that pH is written with the "p" in lowercase because it means "p function" that is $-logx$

As in aqueous solution:
$$\ce{NH3+CH2COOH_{[b]}<=> NH3+CH2COO-_{[z]} + H+_{[x]}\\NH3+CH2COO-_{[z]} <=> NH2CH2COO-_{[a]} + H+_{[x]}}$$
So,
$$10^{-2}=\frac{[z][x]}{[b]}\\10^{-10}=\frac{[a][x]}{[z]}\\\implies 10^{-12}=[x]^2$$
Notice that a=b as the solution is always neutral; also z is a neutral compund
As,
$$[H^+]=10^{-6}\implies p^H=6$$
$$\implies[b]=100[z][x]=10^{2-2-6}=10^{-6}$$

You should have done $p^H=-log[H^+]$, your [b] is correct.

For part (b), yes, using Henderson-Hasslebalch equation:
$$p^H=10+\log\left(\frac{1.05\times10^{-4}}{10^{-6}}\right)\approx 10.97$$

In basic solution, the second equilibrium shifts well(as a result the first one too) to the right.As NaOH is a strong acid, we can neglect any other bases.In presence of NaOH,
$$\ce{NH3+CH2COOH_{[b]} + 2NaOH_{[k]}<=> 2NH2CH2COONa_{[a]} + 2H2O_{[y]}}$$
In millimoles:
$$\begin{array}{c|c|c|c|c}
&entity&b&k&a&y\\\hline
&initial&10&.1&0&0\\\hline
&final&9.95&0&.1&.1+110(original)\\\hline
\end{array}$$
Now you can similiarly apply agin the Henderson-Hasslebalch equation:
$$p_H=12+log(.1/9.95)\approx 10.01$$
Note that I am treating the compund as a monoprotic acid of $pK_a =10(-NH_3^+)+2(-COOH)=12$

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text:
Avoiding snow melting with salt
I've been quite stunned during the last Winter Olympics in Sochi by the following fact.
Temperatures were quite high, sometimes 10° Celsius, and the organizers had to do something to prevent the snow from melting. I heard on the TV the commentators saying that they were discharging salt on the tracks.
However where I live (France), we have occasionnally snow in the winter, and to keep the roads clear for the vehicles or the sidewalks clear for the pedestrians, city services also discharge salt !
How is it possible to have these two different effects with the (seemingly) same component ?

Salt kinda covers ice. In Olympics, salt was used to cover and shield ice from high temperature. In your area, France, they use salt to shield it from further low temperature resulting in formation of more ice... Btw, IMO, you should ask this on Physics.SE (I might be wrong at this point so I am not flagging as off-topic...)

@Awal I do believe it is the right place to ask this question - I do not have an answer, but I would also like to know.

There are different kind of salts and interactions with water. Salt is a really generic term. When using salt on highway the objective is to lower the melting point of water in order to keep it liquid instead of solid. In Belgium where we use CaCl2 and it only work above -4°C. Bellow, salty water freeze. Do you know what kind of salt has been used in Sochi?

@bela83 Wouldn't it also be possible for the snow to melt then refreeze over night forming a layer of ice which is then able to be used by the skiers? NY Times article, http://www.nytimes.com/2014/02/17/sports/olympics/emergency-shipment-of-salt-averts-olympic-crisis.html

@jlandercy Totally agree that salt is a generic term. I don't know though what kind of salt has been used in Sochi. Sports commentators are seldom keen on sciences ! The point on CaCl2 is interesting. I would really enjoy an answer explaining this phenomenon.

@Jun-GooKwak Thanks for the interesting reference.

This article and this article (posted by @Jun-Goo Kwak in the comments above) say that the salt was used to lower the melting point of the snow on the surface, which would cause it melt. It would then re-freeze (I am assuming either at night or due to being in contact with snow beneath it), forming an icy layer on the surface. I don't know anything about Alpine skiing, but it seems that an icy surface is better than a slushy surface.
The ice would also form an insulating layer on the surface, which might help keep more snow from melting. Ice is a better conductor of heat than snow is (snow contains a lot of air, which is a terrible thermal conductor), so it is unlikely that it would act as an effective conductive insulator. However, it would shield the sub-surface snow from convective heat transfer caused by the movement of air. This is how igloos and snow caves work.
There could also be an effect due to the phase transition of melting - when ice melts, it is an endothermic transition, which means it absorbs heat. Some of that heat comes from the air, but some of it will also come from the snow directly below the surface. The result could be that the snow melts more slowly than it would if it were not covered by a layer of ice. The reverse of this effect is the reason that spraying an orange crop with water right before a hard freeze can save the crop (pg. 3) - as the water freezes, it releases heat (freezing is exothermic) and some of that heat goes into the orange trees.
It is hard to guess which effect is most important (insulation vs. endothermic heat transfer), but I would say that both should be considered.
So, to answer your question, in both cases the idea is to melt snow and/or ice, but in the case of roads, we want the ice to melt immediately and flow away, so we use lots of salt. In the case of Alpine ski events, we want a little snow to melt and then re-freeze, so we use a little salt.


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How can I clean my lab coat?
Over the course of my studies, my lab coat has absorbed a non-negligible amount of different chemicals, some of them nasty organic polymers, some are more insoluble inorganic compounds like $\ce{Cr2O3}$.
Even after washing the lab coat some stains remain. Is there a way to get rid of those, or should I rather buy a new lab coat?

Messy in the lab are we? :)  Most I ever have to deal with is coffee spills.

@LordStryker ... the perks of being in the theory department. Well, on the downside, goodbye ether...

Well, I only use my lab coat for my chemistry olympiads duties. And it really would make a better impression on the students if I didn't walk around looking like a hobo in a lab coat ;)

I think that you have have already answered your question in your last comment: If you have to wear it as a supervisor, you want a new, fresh and clean coat. The multi-coloured undergrad rag with holes from sulfuric acid spills and dirty areas on the butt - students typically wipe their hands there - will not leave the best impression ;)

Why do you want a clean lab coat?  Having a coat covered in stains, and full of holes, indicates that you've been doing real chemistry.

I am disgusted at the thought of the coffee coming anywhere close to a lab coat and I sincerely hope that this happened in the lab coat’s natural environment, not the coffee’s.

@Jan I do not know what gave you the impression that the stains on the lab coat were coffee (which they assuredly were not). The "coffee spills are trouble" come from computational chemists.

If you don't have access to a laundering service, you may want to check out disposable lab coats.  You can reuse them several times, and they come with snap closures, pockets and cuffs.  Then if you spill something? Just throw it away and grab another one.

A lab coat will eventually get dirty, that's why it is a good idea to wear it. I had a lot of students complaining about the condition of their coat and they asked me if they could just wash it in the washing machine. My answer was always no.
As far as I remember - I am a theoretician now - a labcoat costs much less than a pair of trousers and a T-shirt. It should be renewed annually and disposed of according to the norms how you would dispose of an unknown chemical. You never know where the chemicals incorporated in your coat will end up, so dispose of them as safely as you can.
In some countries, it is required by the employer to provide equipment like that and taking care of either cleaning or disposing of it eventually. (In Germany that should be the case.) Wear it as a disposable layer of protection and please - for the safety of the environment and yourself - do never ever put it in a normal washing machine again.

As an additional note, if you wash it in the same washer like your normal clothes, how can you assure that it will not end up in your everyday clothes? I can imagine, that having chromium whatever in my underwear would not make me feel alright.

Everything soluble will get diluted into oblivion, and I believe in the Swiss sewage treating plants. ;)
And actually, I had to pay quite a bit for my lab coat. It was about as expensive as a pair of jeans...

@tschoppi For heaven's sake man, it's only 17 dollars! http://www.justlabcoats.com/white-lab-coats.aspx

It's a bad idea to not wash a lab coat. You wash your fingers, hair, glassware etc., but want to wear a coat that inevitably gets contaminated with something for a year? And i don't just mean obvious colourful spills. Have your head of lab organise the washing, by all means, but if that's not possible: Put it in a cotton bag, carry that to your home washing machine, put it in there yourself, just the coat, and let it run at 90°C. And then find yourself a better workplace.

@ Superbest Those are crap coats. A few drops of conc. nitric acid and you're  an olympic flame with a little less media attention.

@Karl You might want to write that up as a separate answer. I disagree, that you should ever wash your lab coat yourself. I think it is negligent to take it out of a lab in some other way than a trash bag.

Well, suit yourself. I'm pretty sure that won't contaminate my washing machine, because I'm a chemist and know how detergents work. If other people neglect my safety, I rather ensure it myself instead.

@Karl If you are implying that I might not know how detergents work, then I am very sorry to disappoint you. I am very glad to hear that you care about your safety. I still think it is not a good idea to take a lab coat home, as I already stated, I do not think it is a good idea to remove it from the lab in the first place. I believe that  is negligent. But this discussion should not happen in the comments section, if you would like to continue talking about it, I can be found in [chat].

Lab coats, like other forms of PPE, are supposed to provide a final barrier of protection, for when all other risk mitigation strategies have failed. They are not just an expensive form of a fancy cooking apron. If you are getting spills on your lab coat, you should take this as an opportunity to re-examine your laboratory techniques to try to establish why this is happening.
Contamination (including wiping hands) on lab coats should be treated at the time of incident, and contaminated lab coats should not be removed from the laboratory, unless in a contaminated laundry or waste container. Most workplaces will have a well-defined plan for laundering lab coats, and strict guidelines for how often this should be done. You can search for just how common this is.
Good lab coats can be expensive, but think of them a little like fire extinguishers. Most days in the lab, you shouldn't even need one. However, you want to make sure that when you DO need them, they will work properly and do the job they are designed for. Regular laundering of a non-contaminated lab coat will keep it in top condition, but once contaminated with hazardous chemicals that cannot be safely neutralized, they should then be replaced.
So, the answer to your question is: no, you should not attempt to launder your lab coat. Buy a new one.
And a final 2 cents worth: Dirty lab coats are not a badge of honour; they are a sign of a sloppy chemist.

You should never wash your own lab coat. That defeats the whole purpose of lab coats: separating yourself from the potentially hazardous chemicals you work with.
But your employer may have a program for laundering lab coats. My university has a program for safely and properly laundering lab coats. This way you can keep it reasonably clean until replacement.

Use or discard.  There is no "clean."  DuPont Teflon Advanced or an old can of perfluoroctanesulfonate Scotchguard for the new lab coat?  Classic Scotchguard,

An eaten lab coat decorated with holes, stains, and blackened smears of stopcock grease along the sides is a badge of honor.  You keep a pressed new one aside for when hominims in suits visit (JAFOs).  Formal occasions demand the Black Lab Coat.  They cannot be purchased.  As with lightsabers, you must make your own.  Given the potential toxicity of aniline black, one recommends starting with a new 100% cotton lab coat washed (remove sizing) and double rinsed.  Then, Procion MX and/or Cibacon F reactive blacks (and lots of non-iodized salt).
http://www.worlddyevariety.com/disperse-dyes/disperse-blue-354.html
If you are a polyester sort of fellow, Foron Brilliant Blue disperse dyes.

Polyester labcoat sounds to me like 'among the top ten worst ideas'.

A polyester lab coat is a warning to others, yes.  it melts onto you.


text:
Why does the water in a Powerade bottle tastes a little like Powerade after many refills?
I have a bottle of Powerade and after probably around 20+ refills of water, the new water I put in still tastes/smells a little like Powerade.
How does that happen?

You could experiment a bit. Put in fresh water. Leave it for bit. Pour the water into a glass. Remove the bottle from the room. Leave the water there for a bit. Drink the water. Also do this double blindly and at least 42 times. Record your findings. Hypothesis to be tested: water tastes normal. Venture theory: bottle stinks or something is going on in your brain that associates the sight of the bottle with the taste of Powerade. To test that, round up 27 people, half of whom never drank or saw Powerade. Apply for funding.

Hey,  you've discovered homeopathic medicine!
Ok, seriously, I agree w/ GlenTU that you may be associating the bottle taste with the taste of the fluid.  Or possibly you never rinsed the bottlecap (?) and it's got smelly res

Probably residue from the Powerade has dried around the nozzle, and you are tasting that when you drink water from the nozzle. Decant it to a clean glass/clean the nozzle the see what happens.

Typically water bottles used in sports or biking are manufactured from high-density polyethylene (HDPE), which is essentially like a longer-chain version of paraffin wax. In short, it's basically a solid nonpolar hydrocarbon (with the exception of the Zeigler-Natta catalysts, which are only present in trace amounts).
Meanwhile, a quick stroll through Fenaroli's Handbook of Flavor Ingredients shows that many of the flavor compounds typically found in foods are either somewhat nonpolar, have nonpolar sections, or have (mostly unsubstituted) aromatic rings present. For example, ethyl valerate, one of the primary constituents of apple aroma, has a butyl chain, is fairly nonpolar, and is nearly insoluble in water.
However, it may be soluble in HDPE, and so it may infuse the bottle with the flavor agents so that the taste can remain for a long time.
Of course, doing a double-blind test may be your best bet for checking to see if this is a real phenomenon, or your imagination. But I've personally noticed this as well, so it could very well be real.


text:
What concentration of sulfur hexafluoride is safe to inhale
I want to do a density demonstration, and I want to inhale helium and sulfur hexafluoride as a fun demonstration. I am open to using any source to purchase from, and money is not an issue (but preferably as inexpensive as possible), but what concentration should I buy (what concentration is safe to inhale), and from where?

$SF_6$ is available from science supply companies like Flinn.  It's quite expensive, even for a lecture bottle. (~$350 for a bottle, or 70 per balloon.)  You could also order a larger bottle from a company like Praxair.

As a safety precaution, this site recommends a few deep breaths of pure oxygen first to saturate your lungs.    That sounds like sound common sense to me.    https://www.stevespanglerscience.com/lab/experiments/heavy-gas-sulfur-hexafluoride-sf6/

I'm afraid this answer doesn't address where you can get $\ce{SF6}$, as I don't know any suppliers myself (it would probably help to specify where you are, though). I just want to straighten out the safety details and make sure you (and anyone else who reads this in the future) know what you're doing. All chemistry experiments can take a turn for the worse if done without adequate preparation.
Pure sulfur hexafluoride is a rather chemically inert gas (unlike the related but very toxic sulfur tetrafluoride $\ce{SF4}$ and disulfur decafluoride $\ce{S2F10}$), and it can be safely handled in regular situations. Decomposition of $\ce{SF6}$ usually requires aggressive conditions, such as exposure to molten sodium metal at 200°C. There are some situations where $\ce{SF6}$ reacts at low temperatures, but that requires irradiation with deep-UV photons or direct contact with solvated electrons from alkali metals dissolved in ammonia. I don't suppose you'll have to worry about such situations. It also seems that $\ce{SF6}$ can decompose into a small amount of the other aforementioned toxic sulfur fluorides if it is exposed to electrical arcs, but since it is an electrically insulating gas, the electric fields required for arcing are even higher than in air, so this also shouldn't be a problem.
All in all, both pure $\ce{SF6}$ and helium are very unlikely to cause chemical damage to the body, even indirectly. However, that doesn't imply they are completely safe. Of course, they can still act as an asphyxiant simply by displacement of oxygen in the lungs (make sure to read this!). Be aware that the brain is incapable of directly measuring the oxygen content in your blood, and instead measures oxygenation rather indirectly, by the amount of carbon dioxide. It is possible to faint and even die from hypoxia without any warning. Additionally, since $\ce{SF6}$ is relatively heavy, it requires more effort from your diaphragm muscles to expel from your lungs, making $\ce{SF6}$ inhalation slightly more dangerous than helium.
As a safety measure, do not take consecutive breaths in the non-oxygenated gas and do not spend more than 5 seconds without inhaling air. While your lungs are devoid of oxygen, not only are they unable to transport oxygen into your blood, the lungs actually work in reverse and help expel oxygen from your blood. After performing the trick, take deep breaths in regular air for a minute or two before trying it again, to be sure all the previously inhaled $\ce{SF6}$ or helium has exited your lungs and your oxygen supply has returned to normal. Hyperventilating before or after inhaling $\ce{SF6}$/helium is likely a bad idea, as it further messes with the balance of oxygen and carbon dioxide in the blood. You should avoid any possibility of becoming light-headed and fainting, especially if you decide to do the trick multiple times. Also be mindful that the gasses can accumulate in poorly ventilated spaces, with $\ce{SF6}$ being a particular problem as it will tend to stay close to the ground and disperse with more difficulty.
If you're going through the trouble to get $\ce{SF6}$, you should do the floating aluminium boat display, too. It's pretty cool! Here's another video showing you can even pour the gas into the boat and make it sink.

I've seen people who inhale $\ce{SF6}$ invert themselves afterwards, either by lying "upside down" on an inclined surface, or by doing a handstand, etc., and then breath in and out deeply a few times.  The idea was that the inversion helped expel the $\ce{SF6}$ from the lungs.


text:
Thymolphthalein solubility
I already know that thymolphthalein is soluble in alcohols, very alkaline aqueous solutions, acetone, and propylene glycol methyl ether.  My questions are what other solvents dissolve thymolphthalein, and what additives increase solubility (for example, specific surfactants)?

One expects [poly(ethylene oxide)-co-(polypropylene oxide)-co-poly(ethylene oxide)] triblock surfactants will micellize it (e.g., Pluronics); possibly Tergitols.  It looks too bulky to fit inside even gamma-cyclodextrin.
Hang a suitable side chain off the phthalic anyhydride and the product will dissolve in pretty much anything.  What does pH mean outside water?


text:
What's unique about the most commonly used hetero-cyclic elements?
I have question about hetero-cycles: Why do we use (the most common hetero-cyclic elements) nitrogen, oxygen, sulfur as hetero compounds most of time?

A possible reason for this is that heterocycles with O, N or S are often easier to obtain and more stable than the corresponding compounds which contain other main group elements as heteroatoms. Phosphole, for example, is less stable than furan, pyrrole and thiophene because of its lower aromaticity. Borole, the structural analog with the lighter heteroatom B, has not yet been isolated in unsubstituted form. Ring strain and the stability of X-C and X-H bonds (X = heteroatom) also need to be considered. Bond lengths tend to increase with heavier heteroatoms. With decreasing electronegativity of X in relation to C, the X-C bond becomes less covalent and is more prone to be cleaved by hydrolysis, for example.


text:
Why does rearrangement not take place in Wurtz reaction?
The Wurtz reaction is $$\ce{R-X + Na ->[ether] R-R}$$ where $\ce{R}$ is alkyl. But if we take butylchloride with $\ce{Na}$, the product is n-octane and not 3,4-dimethylhexane. I am unable to understand the fact that rearrangement does not happen even when it's a free radical mechanism.

The Wurtz Reaction mechanism involves radical anions. Neither radicals nor carbanions rearrange the same way that carbocations do.
The first step of the mechanism is a single electron transfer from $\ce{Na}$ to $\ce{R-X}$:
$$\ce{Na + R-X -> Na+ +[R-X].^{-}}$$
The alkyl halide radical anion fragments to a carbon radical and a halide anion:
$$\ce{[R-X].^{-}->R. + X-}$$
The primary alkyl radical is the least stable radical, but is not as self-destructive as the primary carbocation. Most importantly, primary radicals decompose through disproportionation ...
$$\ce{R-CH2-CH2. +.CH2-CH2-R->R-CH=CH2 + CH3-CH2-R}$$
... or coupling:
$$\ce{R-CH2-CH2. + .CH2-CH2-R -> R-CH2-CH2-CH2-CH2-R}$$
The coupling pathway is a perfectly good explanation of the Wurtz reaction, but another pathway is possible. $\ce{Na}$ is a good reducing agent, and that primary radical is readily reduced to an anion:
$$\ce{R. + Na -> R:^{-} + Na+}$$
The carbanion then reacts with a molecule of alkyl halide in a SN2 reaction:
$$\ce{R:^{-} + X-R -> R-R + X-}$$
The formation of the anion is actually problematic, especially if $\ce{R-X}$ is secondary or tertiary, since E2 elimination will compete.
$$\ce{(CH3)3C:^{-} + (CH3)3C-X -> (CH3)3C-H + (CH3)C=CH2 + X-}$$

You are making it too complicated, but it's really simple: $$\ce{Bu-Cl + 2Na + Cl-Bu -> Bu-Bu + 2NaCl}$$ where $\ce{Bu-{}}$ is the n-butyl group and $\ce{Bu-Bu}$ is n-octane.
You can understand wurtz's reaction much easier this way.


text:
Would decomposing an object into molecules be an environmental problem?
In a thought experiment, a friend proposed to me a machine capable of decomposing any object (e.g., a brick) into its molecules. I'm not a chemist or physicist, but my initial reaction was that this was certainly an environmental issue if used in a larger scale.
Is that right? Or is a molecule so basic that there would be no health concern from such a (theoretical) decomposition.

EDIT @Aesin:
My thinking was:

objects can't disappear; the brick will be breathed in, get into
water etc.
if we start decomposing more advanced objects, we
get all kinds of harmful elements in the air (mercury etc.)
possible radiation from decomposition

As for the purpose: Let's assume it would be used for general waste destruction. Let's assume energy is not a concern (thanks for mentioning it!)

This question is too broad. Despites this fact, we can assume that if such system exists it will consumes huge amount of energy (1st problem) and will be a source of matter (2nd problem) which will be released in environment. It will be, off course, a matter of concern because it will change working point of a lot of reactions that naturally occurs around us (depending of ẃhat molecule is produced). Anyway, the matter fact is, each time human use and abuse of a technology, drawbacks and issues follows.

Bricks aren't made of molecules.

@Aesin mmm I think that for the most part yes, because they are made of clay: silica is a molecule have covalent bonds...

@FrankSeifert: It would help if you could clarify why you think it would be an environmental issue -- you haven't mentioned what you would be expecting it to be used at a large scale to do.

Molecular level
It depends on the molecules you obtain from the decomposition if you use Plasma gasification on mixed organic-inorganic materials you have syngas ($CO$, $H_2$ ... ) and vitrify matter. Watch this video maybe a little bit misleading but can give you an idea. If you use a complete combustion with organic materials you will have mainly water, carbon dioxide, sulfur dioxide etc.,  depending on what you are burning. Indeed if you breath sulfur dioxide or nitrogen dioxide is not good for your health, and acid rain are a plague for forests. Incomplete combustion can lead to more hazard for the environment because you can have more complex molecules, sometimes very stable.

objects can't disappear; the brick will be breathed in, get into water
etc.

Brick are composed mainly of silicon, not carbon, so you can't use combustion, at a very high temperature you can have silicon, a metal. Maybe some dangerous silane.

if we start decomposing more advanced objects, we get all kinds of
harmful elements in the air (mercury etc.)

Mercury is a liquid, not a gas so if you hypothetically discover a cheap technique to get it in the elemental form you will have a liquid.

possible radiation from decomposition

No, you will not have radioactive elements so you don't have radioactivity ( $\gamma$-ray) or other dangerous radiations.
Did you mean particulate?
I think that maybe what you mean if is a danger to powered materials to very tiny particles. Particulate is indeed  dangerous but in this case, you don't reach the molecular level. For your brick, this means that you have dust, something quite similar to what you have in your home.

Decomposing something into 'molecules' is immediately indicative of a non-chemistry background! To try to address your question a bit more directly, I can see a couple of ways of looking at the thought experiment:
1) It decomposes everything to simplest possible component stable molecular structures.
In the case of organic material, plastics etc, you will probably end up with a preponderance of various gases and water, as largely this disassembly is equivalent to complete combustion (CO₂, H₂O, NH₃) likely being the main resulting candidates. These would float out into the atmosphere unless captured.
In the case of inorganic material, e.g. a computer motherboard you will end up with many molecules of various metals, most of which would rapidly oxidise due to their very high surface area (being unimolecular), and various silicates and ceramic unit cells. All of these would be incredibly fine aerosols or dissolved in atmospheric water, similarly rapidly dispersing unless contained.
Results would likely include enhancing the greenhouse effect (on balance as some compounds will diminish it), inducing acid rain (sulfates, nitrates), causing algal blooms/biome damage.
If you had storage for different outputs then you would be producing several industrially useful compounds and if you partnered it with metal reclamation facilities to deal with all the metal oxides it might be a useful way of recycling, energy no object.
2) It decomposes and sorts everything by chemical element.
Assuming you have a nice little storage container for each element (reasonable given you are sorting into the simplest molecule of a pure element), you would end up with various cylinders of oxygen, hydrogen, nitrogen, lumps of metal, lumps/powders of inorganic materials like phosphorus, sulfur, silicon. This would be potentially useful, but incredibly energetically expensive even beyond the separation system outlined by Uncle Al, so would be considerable only in the case of having essentially limitless free energy (many nuclear fusion plants/orbital solar collectors/dyson sphere type energy availability I suspect).
Issues here would relate to various elements being highly reactive on contact with air (or rather the oxygen/water in it). Obvious examples include allotropes of phosphorus and the rare earth metals.
Other issues unaddressed include fractional distillation and cracking of small molecule gases into smallest molecule gases, how to prevent reaction inside the vessel i.e. actually separate things rather than just provide them with huge amounts of energy to react with each other etc


text:
Understanding Beta Decay
Reading Wikipedia on "Beta Decay" they list the equation of Carbon-14 decaying into N-14.
see equation https://en.wikipedia.org/wiki/Beta_decay
I presume the 14 refers to total protons and neutrons.  The 6 in C, I believe means six protons.  In N, I believe the 7 refers to protons. Where did the seventh proton come from the decay of  C-14 into N-14?  I understand neutrons can decay into protons, but the math doesn't work for me.

For all radioactive decay (or other nuclear reaction) of a nuclide into other nuclides, the atomic number $Z$ and mass number $A$ need to be conserved.
$$\ce{_{Z_1}^{A_1}X -> _{Z_2}^{A_2}Q + _{Z_3}^{A_3}R }$$
$$Z_1 = Z_2 + Z_3$$
$$A_1 = A_2 + A_3$$
Additionally, the charges must be conserved. If $\ce{_{Z_3}^{A_3}R }$ is an alpha particle $\ce{_2^2\alpha^{2+}}$ or $\ce{_2^2He^{2+}}$, then:
$$\ce{_{Z}^{A}X -> _{Z-2}^{A-2}Q^{2-} +_2^2He^{2+}}$$
Neutrons decay into protons and electrons (beta particles). Both particles are needed for the math to work. And in order for the math to work, you need to know that the atomic number $Z$ is $0$ for a neutron (no protons), and $1$ for a proton (1 proton, or the $\ce{H+}$ ion). The electron does not have an atomic number. In fact, it is not that $Z$ and charges must be conserved separately. The sum of atomic number and charge is conserved.
$$\ce{_0^1n -> _1^1p+ + _{0}^0e-}$$
For the decay of $\ce{^{14}_6C}$ into $\ce{_7^{14}N}$, we first have a neutron decay into a proton and an electron, which is captured by the newly formed $\ce{_7^{14}N^+}$. The nitrogen nuclide is initially a cation because carbon only had six electrons, and nitrogen needs seven.
$$\ce{_6^{14}C -> _7^{14}N+ +e- -> _7^{14}N}$$

If the emitted electron is "captured by the newly formed nitrogen", what does a Geiger counter detect?

Something is really wrong here. I mean wikipedia is talking about N and e- and not N+ and e-. To me your solution is logical because the charges are conserved, but everybody else appear to have the "wrong" equation... :D

Asked it again from physicists: http://physics.stackexchange.com/questions/228929/beta-decay-of-radiocarbon

By beta decay in the nucleus of the atom a neutron decays into a proton, an electron, and an electron antineutrino and a lot of energy. It will lose a little mass, since $E = mc^2$, so the decay energy is coming from the mass we lose. The electron (or beta particle) has a high energy and leaves the atom.
$$\ce{_0^1n -> _1^1p+ + _{0}^0e- + \overline{\nu_{e}}}$$
By the decay of radiocarbon $8(n) + 6(p) = 14$ the product will be a nitrogen $7(n) + 7(p) = 14$, which means we will have one less neutron and one more proton in the nucleus.
$$\ce{_6^{14}C -> _7^{14}N+ + _{0}^0e- + \overline{\nu_{e}}}$$
The total charge remains zero, since we will have one more electron (beta particle) and one more proton in the system.
According to wikipedia the decay energy of radiocarbon is $E_\mathrm{d} = 0.156476~\mathrm{MeV} = 2.50702189 \cdot 10^{-17}\rm~kJ$. For a mole atoms this means $E_\mathrm{d} \cdot N_\mathrm{A} = 1.509764 \cdot 10^7\rm~kJ/mol$. The first ionization energy of nitrogen is $1.402 \cdot 10^3\rm~kJ/mol$, so the nitrogen cation won't be able to recapture the beta particle, since it will carry too much energy. I am not certain about how big fraction of the decay energy the electron will carry, but certainly much more than the first ionization energy. Since the nitrogen has a relative high electronegativity (3.1), it is likely, that it will get an electron from the environment from an electron donor, for example hydrogen. Radiocarbon labelled biomolecules are likely to fall apart by nuclear transmutations, since the bond energies in these molecules are around $10^2-10^3\rm~kJ/mol$, so a single decay can break $10^4-10^5$ bonds. That's why beta radiations are dangerous.

Charge and energy are always conserved. Mass is a form of energy, and what we call 'mass number' - the sum of neutrons + protons - is a close approximation to mass (because electrons are so light, having just over a two-thousandth of the mass of a proton or a neutron).
The conversion of a neutron into a proton when an electron is released takes care of charge conservation, because the charge on the proton is exactly the opposite of the charge on the electron.
The energy needed to create the electron in beta decay comes from the reduced binding energy of the nucleons in N-14 compared with those in C-14.
Does that make sense of the maths for you?


text:
Atmospheric changes relating to gases
I have the following question:

Which of the following would not be used to measure changes in atmospheric gases?
i) Satellites
ii) Mass Spectrometer
iii) Digital Thermometer
iv) Fractional Distillation

I assumed it would be fractional distillation since you cannot really measure anything with it? You only separate the fractions of different materials such as crude oil.
However the answer was Digital Thermometer. I guess I can understand how this object would be useless at measuring atmospheric changes relating to gases however how would you detect atmospheric gases changes using fractional distillation?

Before answering such question, you must define properly what you means by 'change' and 'atmospheric gases'. Are talking about atmosphere composition or state variables? Could you please clarify your question.

The question is put at the back of a book with hundreds others. It does not gives me any more information.

Terrible answers. Not a "satellite" but remote sensors in orbit. MS is only as good as its sample. Fractional distillation does not quantitate parts-per-billion species: 1000 metric tonnes of air - with zero loses - to recover one gram of a ppb contaminant. Dry air has an average MW - 29. How many liters of atmosphere, 22.4 l/mole, are necessary to dry (no selective losses!) and cryogenically liquefy, then cryogenically spinning band distill (good luck with that, too)?

Assuming 'change' means modification of composition, the answer is Digital Thermometer because even if temperature is linked to atmosphere composition it does not provide you any information about it.

Satelite can embed LIDAR system which will provide you crude qualitative (spectrum) and quantitative (intensities) informations on atmosphere composition, but it requires complex mathematical processing and relies on weak hypothesis;
Mass spectrography is probably the most accurate way because it is highly qualitative (retention time) and quantitative (ionization current);
Distillation will allow you to qualitatively 'separate' (boiling point) and 'quantify' (mass flow rate) distinct components.

Obviously temperature measurement is not able to provide any informations of this kind. Further more, temperature gradient in atmosphere is not only dominated by its composition, there is a lot of other phenomena involved in it.


text:
Why does sodium sulfate have an unusual solubility-temperature curve?
For many salts there are solubility curves as a function of temperature that are smooth (don't have any kinks). Sodium sulfate, however, has a kink in the solubility-T curve at 30 degrees as shown below:

What happens when the temperature is about 30 degree?

"Sodium sulfate" is a mixture of hydrated phases in shifting equilibrium given temperature and concentration.

The phase diagram is also interesting here.  Notice the convergence of three lines at about 30C and 30 wt% NaSO4, right near where the kink in the solubility happens.

Do you still remember the source where you got this phase diagram?

No, sorry.  Wish I had cited it.

In case anyone needs it: the figure is from Angeli et al., 2010 [1], adapted from Hougen et al., 1954 [2].
[1]M. Angeli, R. Hébert, B. Menéndez, C. David, and J.-P. Bigas, ‘Influence of temperature and salt concentration on the salt weathering of a sedimentary stone with sodium sulphate’, Geological Society, London, Special Publications, vol. 333, no. 1, pp. 35–42, Jan. 2010.
[2]O. A. Hougen, K. W. Watson, and R. A. Ragatz, Chemical Process Principles. Part I. 2nd edn. Wiley, New York, 1954.


text:
Ideal gas law? implied volume?
Working on a practice exam. The question is

A volume of gas contains
  0.25 moles of $\ce{CO2}$
  0.30 moles of $\ce{N2}$
  0.15 moles of $\ce{O2}$
What is the partial pressure of $\ce{O2}$ if the total pressure was $690\ \mathrm{mmHg}$ and the temperature was $25.0\ \mathrm{^\circ C}$

I know the formula is $p_a=\dfrac{n_aRT}{V}$
So far I know that
$n_a = 0.25\ \mathrm{mol}$
$R  = 0.0821\ \mathrm{l\ atm\ K^{−1}\ mol^{−1}}$
$T = 298\ \mathrm K$
But I can't seem to figure out volume. I was reading somewhere that volume is directly proportional to The amount of moles, but maybe it has something to do with the total pressure. I'm a little lost.
The only thing I can think of would be getting volume via the Ideal Gas Law as well. Would that work?

Welcome to Chemsitry Exchange, I flagged your post as an homework even if it is not totally accurate. Here are my hints: (1) Keep in mind that pressure are intensive and amount of matter and volume are extensive. (2) Partial pressure is a convenience of Chemist which can be regarded to as the volume occupied by the substance if it has been taken alone. Good work.

If you read through this page, you should be able to determine the answer: http://en.wikipedia.org/wiki/Partial_pressure.

You don't need the Ideal Gas Law. You can use it (and I'll show you how). The question asks for partial pressure. You want Dalton's Law of Partial Pressures.

Dalton's Law

Dalton's Law states that the total pressure in a system is equal to the sum of the partial pressures of the components:
$$P_T=P_A + P_B + P_C + ...$$
At constant volume and temperature (which you have, even if you don't know the volume), the ideal gas law can be rearranged so that:
$$\frac{P}{n}=\frac{RT}{V}$$
$$\frac{P_T}{n_T}=\frac{P_A}{n_A}$$
$$P_A=\frac{n_A}{n_T}P_T$$

Ideal Gas Law

You can find the volume of the system because you have the total moles $n_T$ and the total pressure $P_T$:
$$V=\frac{n_T RT}{P_T}$$
You can then replace the number of moles with $n_A$ and solve for $P_A$;
$$P_A=\frac{n_A RT}{V}$$
This approach is basically equivalent to the previous except:

Using the ideal gas law you have to do two calculation (volume and partial pressure), while you only need to do one with Dalton's Law
Fewer math steps means less rounding, which means less error and higher precision.
If you use Dalton's Law, you don't have to worry about the units of temperature or pressure. Temperature is not in the equation, and the pressures are a ratio. You can use whatever pressure unit you want without converting.
You don't need to worry about $R$ and its units in the Dalton's Law approach.

Solve problems as ratios whenever you have the chance.

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


text:
Titration of sulfuric acid with sodium hydroxide

$55.0~\mathrm{mL}$ of $0.250~\mathrm{M}~\ce{NaOH}$ is used to titrate $35.0~\mathrm{ml}$ of $\ce{H2SO4}$.
  What is the molarity of $\ce{H2SO4}$?

I know that the equation for this reaction is:
$$\ce{H2SO4 + 2NaOH <=> Na2SO4 + 2H2O}$$
I know that I have to use $$i_\mathrm a\cdot M_\mathrm a\cdot V_\mathrm a = i_\mathrm b\cdot M_\mathrm b\cdot V_\mathrm b$$ formula where $i$ is the number of ions, $M_\mathrm a$ and $M_\mathrm b$ are the molarities (of the acid and base, respectively), and $V_\mathrm a$ and $V_\mathrm b$ are the volumes of the acid and base (in milliliters).
I starting doing this problem and got up to this point:
$$i\cdot M_\mathrm a\cdot 35~\mathrm{ml} = 2\cdot 0.250~\mathrm{M}\cdot 55~\mathrm{ml}$$
So my question is:
What should I put for the number of ions on the left hand side? Is it three or two?

You're almost right. $i_a$ is the number of $H^+$ ions and $i_b$ is the number of $OH^-$ ions.

That is one way of doing it.
In reality, this is a stoichiometric question where you are trying to find the molarity of an unknown polyprotic acid by titrating it with a strong base (woo-hoo!) of known molarity. I prefer using this method as it is easier to see the relationship between the units as well as use chain-ink conversion which most people are familiar with.
We know that
$$ \text{M} = \frac{\text{mol}}{\text{L}} $$
Therefore,
$$ 55 \ \text{mL} \ \ce{ NaOH} * \frac{250 \ \text{mmol} \ \ce{NaOH}}{1000 \ \text{mL} \ \ce{NaOH}} * \frac{1 \ \text{mmol} \ \ce{H2SO4}}{2 \ \text{mmol} \ \ce{NaOH}} * \frac{1}{ 35 \ \text{mL} \ \ce{H2SO4}} = $$
Which equals
$$ 0.19643 \ \frac{\text{mol}}{\text{L}} $$

In reality you would do this experiment probably with conductometry or for sulphate it is even better to do it via gravimetry via $$\ce{BaCl2 + H2SO4 -> BaSO4 v + 2 'HCl'}.$$ It should also be noted, that the second protolysis of suphuric acid does not behave like a strong acid. The equilibrium $$\ce{HSO4- (aq) + {}^{-}OH <=> H2O + SO4^{2-} (aq)}$$ is not fully on the right side. There is a significant amount of undissociated hydrogensulphate ions left (at pH=7).


text:
How can water catalyse a reaction between iodine and aluminium?
A few drops of water can initiate a reaction between iodine and aluminium.
https://www.youtube.com/watch?v=SKSU72-1ERc
How can this happen, since iodine is only slightly soluble in water?

A naked aluminum surface (freshly sanded) in air immediately grows about 4 nm of adherent aluminum oxide that passivates the surface.  Iodine slightly dissolves in water and disproportionates
$$\ce{I2_{(s)} + H2O_{(l)} -> HOI_{(aq)} + I^{-}_{(aq)} + H+_{(aq)}}$$
Iodine is strongly solubilized by forming triiodide.
$$\ce{I2 + I- -> [I3]-}$$
The acids chew through the alumina surface barrier.  When bare metal and triiodide meet, everything takes off, also igniting the bare metal in air.
http://www2.ucdsb.on.ca/tiss/stretton/database/inorganic_thermo.htm
$$\begin{alignat}{2}
&\ce{AlI3_{(s)}}\quad &&\Delta G_\text{f}^\circ = -300.8\ \mathrm{kJ/mol}\\
&\ce{Al2O3_{(s)}}\quad &&\Delta G_\text{f}^\circ = -1582.4\ \mathrm{kJ/mol}\\
\end{alignat}$$
Absent water the iodine must diffuse through the intact alumina barrier, react with naked aluminum, then swell and disrupt the surface passivation from underneath.

It is very difficult to have a solid react with another solid due to the low contact area and the inability of the molecules to obtain the activation energy. Even though there is little iodine dissolved in the water, the contact area is greatly increased.
Once the reaction starts, the heat of the reaction provides the activation energy and the gaseous iodine (which makes the contact area large).

On warming, the action of iodine on aluminum to a limited extent forms aluminum iodide:
$$\ce{Al + 3 I2 -> 2 AlI3}$$
But, in the presence of water, hydrolysis occurs:
$$\ce{2 AlI3 + 6 H2O -> 2 Al(OH)3 + 6 HI}$$
removing some water. Also, with the newly created $\ce{HI}$ (see the same Wikipedia link as above):
$$\ce{2 Al + 6 HI -> 2 AlI3 + 3 H2}$$
or, for all three reactions the net reaction is:
$$\ce{4Al + 6H2O + 3I2 -> 2Al(OH)3(s) + 3H2(g) + 2AlI3}$$
Upon adding 6 more $\ce{H2O}$'s to each side we get:
$$\ce{4Al + 12H2O + 3I2 -> 4Al(OH)3(s) + 3H2(g) + 6HI}$$
And adding 2 more $\ce{Al}$'s to each side gives:
$$\ce{6Al + 12H2O + 3I2 -> 4Al(OH)3(s) + 6 H2(g) + 2AlI3}$$
Note, the process can be repeated, but the quantities '$\ce{3I2}$' and '$\ce{2AlI3}$' will remain constant while the amount of $\ce{Al}$ and $\ce{H2O}$ increase. As such, the Iodine acts essentially as a catalyst (only a small amount is required to initiate the reaction regardless of the amount of water, and only that initial Iodine is consumed).
Note, energy arguments are generally uninformative on reaction mechanics and if one postulates that $\ce{HOI}$ (from $\ce{I2}$ plus water) attacks $\ce{Al2O3}$ layer, not likely as even $\ce{HOCl}$ but slowly reacts.


text:
Reaction between hydrogen peroxide and ozone
What is the reaction mechanism of the reaction:
$$\ce{H2O2 + O3 -> H2O + 2O2}$$
I have tried searching this on Google but am not getting good answers.

Based on this Google search("reaction of peroxide and ozone"), neither the reaction nor the mechanism is simple.

The compositional chemistry is complex and the electronic states of the intermediates and products are complex re singlet oxygen (of which there are several kinds), http://www.mosao2.org/Article%20-%20O3/images_O3/O3_102_production/O3_102_production_pdf.pdf

http://en.wikipedia.org/wiki/Singlet_oxygen#Orbital_states

This is basically an excerpt of the answer I have given to another quite similar question.
Ben is quite right, neither the reaction nor the mechanism is simple. It involves several different radical chains all resulting in the decomposition of ozone.[a]
It is most likely that the reaction will start with the formation of oxygen radicals in the gas phase ($\ce{O}$ atom transfer).$$\ce{O3 + O3 -> O3 + O2 + O}$$
In presence of Hydrogen donors like $\ce{H2O2}$, which "is expected to be an excellent $\ce{H}$ donor."[a] Those are forming the main chain carriers will be $\ce{HO, HO2}$.
 Finally stopping at some point by recombination of these species.
\begin{aligned}\ce{
O + H2O2 &-> OH + HO2\\
  HO + O3 &-> HO2 + O2\\
  HO2 + O3 &-> HO + 2O2\\
  2HO2 &->}\text{stable products}
\end{aligned}
In aqueous solution the decomposition is highly pH dependent, but may always work because of autoprotolysis. $$\ce{2H2O <=> H3+O + {}^{-}OH}$$
Especially important might also be protolysis of hydrogen peroxide. $$\ce{H2O2 + H2O <=> HO2- + H3+O}$$
Steahelin and Hoigné propose several hypotheses in which way ozone will be decomposed, but I will not go into absurd detail here.[b] The core assumptions involve $\ce{{}^{-}OH}$ as the initiator of a radical chain. Followed by radical (ionic) species like $\ce{.O2-, .O3-, HO2., .OH}$ resulting again in a chain reaction.
\begin{aligned}\ce{
  O3 + {}^{-}OH &-> .O2- + HO2.\\
  O3 + {}^{-}OH &-> HO2- + O2\\
  O3 + {}^{-}OH &-> .O3- + .OH\\
  HO2. &<=> H+ + .O2-\\
  O3 + HO2- &-> .OH + .O2- + O2\\
  O3 + .O2- &-> .O3- + O2\\
  .O3- + H2O &-> .OH + {}^{-}OH + O2
}\end{aligned}

Hi Martin ! In this reaction, hydrogen peroxide is the  reducing agent,right ?

@biogirl Yes and no. The oxidation state in ozone is $0$, but that is the same as in bimolecular oxygen. The oxidation state of oxygen in hydrogen peroxide is $-1$. So one oxygen is reduced when forming water OS$=-2$ and one is oxidised to $\ce{O2}$ OS$=0$. This kind of reaction is called disproportionation. So referring to itself it is both, referring to ozone it is none of it.

Ok. Makes sense.

What if you use Barium peroxide instead of Hydrogen peroxide?


text:
Different techniques of distillations
I know that fractional distillation is used to separate two liquids that are miscible in one another and have close boiling points.
However, I am really confused about vacuum and steam distillation - in which cases do we use these separation techniques? Can someone please give examples.

I saw this site which mentions steam distillation at the end and has a lot of other cool information.  -  https://www.chemguide.co.uk/physical/phaseeqiamenu.html

Vacuum distillation is employed when the boiling point of the substance is too high (400 Celsius for example). Under reduced pressure, the boiling point will be reduced to the range that is easier to handle.
Steam distillation take advantage of azeotropic effect, water steam is bubbled into your liquid (aromatic compound for example) and carry out the mixture of water vapor and your liquid vapor. This technique can also avoid handling the high boiling point distillation because the azeotropic boiling point is usually lower than pure liquid.

Now I understand why vacuum distillation is used but I am still confused about steam distillation...Could you please explain the azeotrophic effect (I know that azeotropes are mixtures that have no change in composition when distilled) which has a role to play in steam distillation.

I am actually little wrong on the steam distillation in the answer. It is not about azeotrope. Sorry for my confusion. I have a link to a good web page that explains it very well: http://www.chemguide.co.uk/physical/phaseeqia/immiscible.html

I'm confused again. I understand that steam distillation is used when the substance is immiscible in water. However, I am confused about the difference in the purpose of the two distillations. They both allow the substance to boil at a temperature below its normal boiling point. How will I know which procedure to use?

Steam distillation is mostly used in industry. Vacuum distillation is more preferred in lab.

But they both serve the same purpose of boiling a substance below its normal boiling point where it would otherwise decompose?

The steam distillation is much more troublesome to handle in lab, while the cost of steam distillation is lower in industry than reduced pressure.

@IanFang Could you edit in your comment please - that would improve your answer significantly.

http://www.umsl.edu/~orglab/documents/distillation/Image169.gif
   boiling point vs. pressure nomograph
The separation unit of a distillation (or countercurrent exchanges overall) is the theoretical plate (originally from petroleum distillation towers),
http://fch.upol.cz/skripta/fcc_and_zvem_english/rektif/rektif_teorie_clip_image002.jpg
http://www.vaxasoftware.com/alen/MCTH_EN.jpg
Each separation step must be at equilibrium.  Your basic lab rig is might be three theoretical plates at best, vapor vs. liquid in a hollow vertical column.  Efficient physical separation demands thin film countercurrent flow to maximize surface area/volume, hence packed columns then
http://www.solvent--recycling.com/spinning_band_packed_column.html
   spinning band distillation
wherein a tightly fitting teflon screw thread spinning like all get out jams fluid condensate downward against vapor flowing upward. Net output is a drop every now and again.
The natural evolution of phase exchange is gas chromatography then liquid chromatography (much denser medium - look up the exchange maths). HPLC using monodisperse spheres with surface exchange only can have theoretical plate heights of a few microns, giving tens of thousands of theoretical plates in a rather short column.
http://www.sigmaaldrich.com/content/dam/sigma-aldrich/docs/Supelco/Posters/1/T413105H.pdf.

This was a bit too complicated for me to understand... could you please explain to me when we should use steam distillation and when to use distillation under reduced pressure.


text:
What exactly happened when I added KOH to hydrogen peroxide?
I recently added a potassium hydroxide, $\ce{KOH}$, pellet to a small droplet of hydrogen peroxide (30 %), $\ce{H2O2}$. You could clearly see the potassium hydroxide dissolving in the droplet when instantaneously the solutions began to fizz strongly (boil and/or release oxygen gas, I'm not sure) and release a lot of heat.
I'm not sure what happened, but I've got a few ideas of what might have happened:

There is some kind of reaction between the $\ce{KOH}$ and $\ce{H2O2}$.
The $\ce{KOH}$ released a lot of energy when dissociating into the aqueous solution and then boiled the water, maybe reaching the temperature of  150.2 °C in which it also triggered the decomposition of $\ce{H2O2}$ into water and oxygen gas.
On Wikipedia I read:

The rate of decomposition increases with rising temperature, concentration and pH.

So when $\ce{KOH}$ is dissolved, the solution has a high pH and this might trigger the decomposition of hydrogen peroxide

I did notice after a pH test that the plastic lid on which I had condensed some of the evaporated water had a very high pH, so this would most likely contain some dissolved potassium hydroxide.
In case a reaction happened, what could that specific reaction be?

I have no time to answer, but I am pretty sure that $\ce{KOH}$ is capable of abstracting a proton from hydrogen peroxide, leaving protonated dioxygen ($\ce{HO2-}$), which should be a very strong acid.

@Martin While $\ce{KOH}$ can certainly deprotonate $\ce{H2O2}$, the resulting hydroperoxide anion $\ce{HO_2^{-}}$ isn't a strong acid. $\ce{H2O2}$ has a $pK_a$ of 11.75 in water according to Wikipedia, and the second ionization is probably several orders of magnitude harder.

@Nicolau You are very right, I confused it with $\ce{HO2+}$ - as I said I was in a hurry.

your cap is basic because of small aerosolized water droplets, not some gas phase product.

According to THE HOMOGENEOUS BASE-CATALYZED DECOMPOSITION OF HYDROGEN PEROXIDE J. Phys. Chem., 1961, vol. 65 , pages 304–306:
$\ce{H2O2 + OH- <=> HO2- + H2O}$
$\ce{H2O2 + HO2- ->}$ [6-membered ring with 2 "H"s bridging 2 "O-O"s, with another H bonded to one of the "O"s] $\ce{-> H2O + O2 + OH-}$

It is clear that the rate will go through a maximum
  when the peroxide is 50% ionized, that is
  $[\ce{H2O2}] = [\ce{HO2-}]$.

There is some kind of reaction between the $\ce{KOH}$ and $\ce{H_2O_2}$

There was definitely a reaction.

The $\ce{KOH}$ released a lot of energy when dissociating into the aqueous solution and then boiled the water, maybe reaching the temperature of 150.2 °C in which it also triggered the decomposition of $\ce{H_2O_2}$ into water and oxygen gas.

Water cannot reach a temperature above 100 °C unless it is pressurized.
Yes, $\ce{H_2O_2}$ can disassociate rather quickly under the right conditions. An enzyme called catalase, can break up $\ce{H_2O_2}$ at an astounding rate. If you add an amount you can barely see, like the size of a period at the end of a sentence, to 1 mL of 30% $\ce{H_2O_2}$ in a test tube, the oxygen production will push the liquid out of the tube.

Hydrogen peroxide is stable in acidic conditions. This fact is compounded by the fact that stabilizers (chelates) are commonly acids which do not chelate ions in their basic forms. Furthermore, potassium is a soft largeish ion which catalyzes the autodecomposition.
Bad idea, don't do it again. Hydrogen peroxide is nasty.


text:
Why lanthanides and actinides are shown separate from standard Periodic Table layout?
In the standard Periodic Table layout , all the elements up to 56 are in order i.e are in the same layout table. However, lanthanides and actinides are always shown separately from the layout like in this layout:

What is the reason behind this structure?
Is this standard layout or can I represent it like this too?

The Periodic Table arranges elements in blocks as each type of orbital fills with electrons - $s,p,d,f,g,h$.  Alkali metals and alkaline earths are $s$-block filling (but could be one $s$-block slot).  $p$-block six electrons to fill are trelides, tetralides, pnticides, chalcogenides, halides, inert gases (but could be one $p$-block slot).  Transition metal $d$-block is ten elements (but could be one $d$-block slot).  Filling the $f$-block are 14 elements, lanthanoids and actinoids.  That gets sloppy to print and the elements are (or at least were) overall obscure.  They get condensed.
Representing the Periodic Table has become an an art form.  The plain vanilla variety is terse and useful.
http://en.wikipedia.org/wiki/Alternative_periodic_tables

One of my science teachers told me that the only reason that they put away the lanthanides and actinides in a separate row was because it would mess up the order and like shape and structure and some rules and stuff... so does that mean my teacher told be BS?

Because if you put lanthanides and actinides in to the Periodic Table like transition metals, the table will be way too wide.

There is no scientific reason not to put the f-blocks between s- and d-blocks!
However, doing so will create a very wide periodic table: The s-, p- and d-block are 18 elements wide, the f-block would add another 14, so almost double the width. Therefore, it is customary to save space and put the f-block at the bottom separately. It is not mandatory, e.g. Wikipedia uses the fully horizontal layout on element pages.

So there is a reason - it looks better and fits better into most spaces.

The two rows of elements present at the bottom of the periodic table are called the 4f series or lanthanoids and 5f or actanoids. They are also called inner transition elements. With each series, the properties of component elements closely resembles each other so their chemistry becomes complicated than their corresponding lanthanoids. Actanoids have large number of oxidation states compared to lanthanoids. Actanoid are radioactive and are man-made in the laboratory. They are not completely studied. Hence they are placed separately in the periodic table.

Also the actinides and lanthanides have very similar chemical behaviour- I think nearly all have a possible +3 oxidation state and similar metallurgical behaviour. For a long time, lanthanides was added as mishmetal to steel alloys.

The Schrodinger equation allows you to essentially derive the "shape" of the entire periodic table. How you place the elements really doesn't mean anything, it's just a map on paper of the territory, which is really the (not so simple) fact of how electrons stack up onto increasingly larger numbers of protons concentrated in one spot.

To preserve symmetry shape lanthanides are placed below main table .actually vertical elements are similar and horizontal elements are dis similar .but lanthanides shows horizontal chemical similarities so placed in f block

Welcome to chem.SE. As I don't think you're referring to molecular symmetry, your answer seems a bit vague to me. It's better if you clarify your points and provide more examples.


text:
Atomic coordinates for DNA bases as they appear in a B-form or A-form duplex
Where can I find atomic coordinates for guanosine / cytosine / thymine / adenine (either the ribonucleotide or deoxyribonucleotide bases) where the atoms are configured in the manner we would expect in the context of an idealized B-form or A-form nucleic acid duplex?  I'd like to calculate the angle between various bonds in these bases, and to play around with a few other things, but I'm having trouble finding proper atomic coordinate sets.

I have no idea about this, but I found something you might want to check out: http://onlinelibrary.wiley.com/doi/10.1046/j.1432-1327.1999.00270.x/abstract If it has no supporting information you might find structures at http://www.ccdc.cam.ac.uk/

I played around with a program called GROMACS about 4 years ago and don't remember much, but I do remember doing a tutorial simulating a DNA double helix and part of it included making the DNA by giving it a sequence and running a script. It can probably make A or B form.
EDIT: This was a long time ago and I forgot. It wasn't GROMACS, it was Amber. Look here http://ambermd.org/tutorials/ Tutorial B1 looks like the same tutorial I did with the DNA.

In Protein Data Bank (RCBS), for example as molecule of the months:
http://www.rcsb.org/pdb/101/motm.do?momID=23&evtc=Suggest&evta=Moleculeof%20the%20Month&evtl=OtherOptions
In Nuclear Acid Database:
http://ndbserver.rutgers.edu/


text:
How can I protect my skin from KMnO4?
We have national elections coming up in two days, and like every time the election officials will mark people's thumbs with a dab of some purple solution that leaves a brown stain on the skin (and thumbnail) that takes a while to grow out. I'm pretty sure that's a potassium permanganate solution - that stuff behaves exactly the same.
Now I'm not into electoral fraud (there are other controls in any case) but I dislike having that stain on my fingernail for a month or more afterward. Previously I've had partial success by rubbing butter into my skin, but the stain still gets through. I've also tried making some plain old sugar or even glucose available, hoping that it would get oxidized first, but I can't conclusively say that that worked.
Any other tricks to avoid, or at least strongly mitigate, these stains? There's a window of about a minute or two where the purple solution sits on the cuticle while I'm unable to do anything about it.

If it is permanganate, the brown $\ce{MnO2}$ stain will decolorize and solubilize in acidic reductant - lemon juice, perhaps boosted with a little crushed and dissolved Vitamin C.   I'd go for crystal/gentian violet or Disperse Red 9 as a marker, or a reactive Cibacron or Procion dye.
Cthulhu for Fearless Leader!  Why vote for the lesser evil?

I dissolved some citric acid in water and immersed my thumb in it for a while. The stain seemed not to be as bad as in other years, although I did abrade some more off.

You need bvoth the reducing agent and the acid, hence citrus juice that contains both.


text:
Internal and external pressure during a gas expansion
There is a solid substance in a rigid cylinder with a piston placed on top such that the distance between the piston and the bottom of the cylinder is essentially zero, and then the substance decomposes and forms gas that expands and pushes the piston up against a constant external pressure to a maximum point. It then falls slowly as heat from the reaction is lost to the surroundings. Is it true that at the maximum point, the internal pressure is equal to the external pressure?

Welcome to Chemistry Stack Exchange! Please add what you have attempted towards solving the problem into the body of your question.  For more information, see the site's homework policy for how to ask homework questions. Thanks!

If the frictionless massless piston is not moving, then internal and external pressures must be equal.  If the piston has mass, the internal pressure must be greater when it is stationary, to support the weight.  If there is static friction, all bets are off.
Suppose the piston were a right cone rather than a right cylinder.  Would the excess surface area change the outcome, all other things being equal?


text:
Cyclopentadiene/etc. hydrophobicity after nitrogen atom substitution at a double bond?
If I take a compound like cyclopentadiene ($\ce{C5H6}$) and I substitute a nitrogen for a carbon at one of the $\ce{C=C}$ bonds (making a $\ce{C=N}$ bond), how does this typically change the hydrophobicity of the compound?
The problem is that I'm not really a chemist.  My thought here is that nitrogen is more electronegative than carbon, so this will polarize the bond, induce a dipole, and make the molecule more hydrophilic.  In general, it seems to me that mixing together atoms of differing electronegativity in alkane or alkene/etc. compounds should create dipoles that can interact with polar solvent molecules, improving hydrophilicity.
As an additional thought, how would the nitrogen substitution effect stacking interactions between the substituted ring and a regular molecule of cyclopentadiene?  My guess here would be that hydrophobic effects would cause a more favorable pi-stacking interaction between two unsubstituted copies of cyclopentadiene relative to cyclopentadiene and the aforementioned and described nitrogen substituted version of the molecule.
To clarify, by "stacking interactions" I mean the sort of $\ce{C=C}$ stacking interactions that lead to cycloaddition reactions, not that the two rings will lie on top of one-another (as in the case of something like a pair of benzene molecules).

The cyclic unsaturated imine (Schiff base) will be more hydrophilic for the nitrogen lone pair hydrogen bonding and being a Lewis base toward water.  You do know that CpH [4 + 2} Diels-Alder reaction reversibly thermally dimerizes, right?  The endo-dimer is an exo-flip then hydrogenation away from JP-10 cruise missile fuel.
The cyclic imine, ${2H}$-pyrrole is an isomer of pyrrole.  One expects it will be labile toward proton migration in base.
http://www.chemspider.com/Chemical-Structure.7330.html
http://www.chemspider.com/Chemical-Structure.5256848.html
Click the Properties tab, etc.

Thanks, didn't know about the jet fuel composition. :)  Do you know if the lone pair will effect stacking?

Neither the imine nor CpH is aromatic for the ring methylene.  There will be adverse steric interactions if two such molecules approach.

I thought one could perhaps have offset stacking interactions between the two CpH rings?  Where just a pair of double bonds stack on one-another?

As you say, there is a CpH / CpH cycloaddition reaction, so I thought this requires this sort of edge-wise stacking interaction?

@user5420 In the transition state of the Diels-Alder reaction, the 2 CpH molecules are stacked with their double bonds perpendicular to each other, as it is, for example,  shown in the image on http://orgo.curvedarrow.com/punbb/viewtopic.php?id=144.

@JannisAndreska Thank you for the graphic!  What about a 2+2 cycloaddition reaction?  Would we have a parallel stacking interaction?  Or am I off the mark with this suggestion?

For a 2+2 cycloaddition, parallel stacking should be favorable.

@JannisAndreska So, do you think C = C / C = C parallel stacking, as opposed to C = C / C = N parallel stacking, would be more energetically favorable?  This guess is based on the (here unprotonated) nitrogen having a lone pair and being more hydrophilic (as per Uncle Al's answer)?

In the transition state of a cycloaddition, parallel stacking of both can be possible. The stacking (perpendicular or parallel) also depends on the symmetry of the molecular orbitals which take part in the reaction.

@JannisAndreska But in the case of the C = N - R / C = C stack, there would perhaps be a penalty from displacing a water molecule if we're in aqueous solvent?

Yes, I think the compound with the C=N bond will be better hydrated and more soluble in water than the compound with C=C bond. Cyclopentadiene is actually poorly soluble in water (1.8 mg/L according to http://www.guidechem.com/reference/dic-4966.html).

@JannisAndreska Unfortunately, I don't have any experience to be able to determine whether this means that the cycloaddition reaction rate will be one- or three-/etc. orders of magnitude slower in aqueous solvent due to the increased hydrophilicity of the C = N bond relative to the C = C bond.

I am pretty sure that the reaction will proceed more slowly, if at all. For such a reaction, I would recommend using an organic solvent in which both reactants are readily soluble.


text:
Removing diamonds from a 14 karat gold ring with acid
I have a 14 karat gold ring with diamonds. I want to weaken the metal (to get the diamonds out without damaging them).
If I dip the ring in sulfuric acid, will I be able to dissolve the base metal in the alloy leaving the gold in place?

Visit a jeweler to open the mounting.  Gold is inert to any chemistry you can safely assemble.  Tapping a diamond parallel to [111] crystal orientation will spall off triangular pieces.

there is a solution aqua regia which can dissolve gold but i don't know whether it will dissolve diamond or might ... It's better to ask jeweller

Jewelers have tools to do this as @UncleAl suggests. Gold is fairly inert (even 14 carat), and the diamond is more reactive with many things.

Surely the gold alloy of the ring is far more mechanically weak than the diamond (given diamond is 10 on the Mohs scale) so you could just use needle-nose pliers to open out the mounting without risking the diamond?

It doesn't have a mounting that can be got at with pliars

If it is a gold alloy, I suspect you can bend the entire ring with a small g-clamp and the diamond may pop out of its fitting. There is also the possibility that the diamond is in someway soldered to the base of the fitting, in which event heating would release it (even if not, gold melts a good bit earlier than diamond)

Can the attachments be ground with a Dremel if you can't reach them with pliers? Dimond does not solder or braze , etc. Aqua regia is a bad idea for an armature.

I think that asking a jeweler is the best thing to do because gold is not something to throw away. However talking about the chemical concern, sulfuric acid can't oxidize gold, if the base metal (you should specify where and how it is) is iron theoretically you will be able to dissolve it with sulfuric acid:
$$\ce{Fe + H2SO4 ->FeSO4 + H2 ^}$$
Something similar can happen with copper and tin but I think that temperature of the acid is a determinant parameter for these reactions.
If the base metal is inside the alloy is quite unlikely that you can affect it because gold is not a porous material, so you can only interact with the superficial layer.


text:
How can one come to know how a catalyst affects a reaction when the reaction and the catalyst is known
How can I come to know how a catalyst affects a reaction when the reaction and the catalyst used is known.
I searched the internet for an answer but could not find one.

Welcome to Chem.SE! Please try to specify a little bit more about what catalysts you're talking, and if there's a specific reaction you had in mind...

Empirically, you dump the catalyst to be tested into the reaction and determine the rate with/without for various catalytic loadings. This will tell you whether it increases or decreases the rate of the reaction.
Assuming that you are vaguely asking about a specific reaction and catalyst, you aren't going to get a specific answer without specifically mentioning them in your question. (I think that was enough specificity for now!).
More generally, you will probably want to search for university level teaching resources on transition metal or biological catalysts wherein you might find your answer. Higher level reading would be to use something like Web of Knowledge or Reaxys to find publications on your catalyst/reaction of question.

There is no simple way to know it.
The mechanism in some of the cases of catalysis is known, therefore some predictions can be  made. Also there are many cases where the given catalytic reaction has a great practical significance, therefore it is thoughtlessly tested with different combinations of reagents, and the results are published.

At a basic level, a catalyst lowers the activation energy of a reaction, which makes it easier for the reaction to occur ("speeding it up"). The catalyst is not chemically consumed by the process. (Though corrosion, coking, and other side reactions can still deteriorate catalyst perofmance)
Further details depend on the specific reaction and catalyst.

Here is an example by which one can I come to understand how a catalyst affects a reaction given that the net reaction and the catalyst are known. Lets look at the summary net Haber–Weiss reaction in which iron ions under acidic conditions are said to be the catalytic:
.O2− + H2O2  ---H+,Fe(II),Fe(III)→ .OH + O2 + OH-   [R3]
The reaction steps are believed to proceed in the presence of iron ions (and more so also in the presence of Cu(II) forming a redox couple equilibrium, Cu(II) + Fe(II) = Cu(I) + Fe(III)):
Fe(III) + .O2- (+ H+ ) = Fe(II) + O2 (+ H+ )   [R1]
where Reaction R1 is the reverse of the oxidation of ferrous to ferric in the presence of acid and oxygen witnessed in a spontaneous electrochemical based corrosion of iron to rust which proceeds as follows:
Fe → Fe(II) + 2 e-
O2 + e- → .O2-
Fe(II) = Fe(III) + e-
where the net of the last two spontaneous electrochemical half cell reactions is:
O2 + Fe(II) = .O2- + Fe(III)
which is the reverse of Reaction R1, with the implication being that R1 is the slow backward (or reverse) reaction. Next is the well known Fenton reaction:
Fe(II) + H2O2  → Fe(III) + .OH + OH−  [R2]
where the net of R1 and R2 produces R3. The role of the iron ions catalyst in acidic conditions (and copper ions assisting in its recycling) is apparent as is also why R3 (per R1) is an observed slow reaction. The presence of H+ would remove OH- and be catalytic for R2. The addition of some NaCl, which I would argue serves as an electrolyte for the electrochemical aspects of the system, apparently accelerates the reaction rate, although researchers give other arguments.


text:
An example to illustrate the difference between an inversion operation and a twofold rotation
Reading in Atkin's Inorganic Chemistry book, I cam accross:

Although an inversion and a twofold rotation may sometimes achieve the
  same effect, that is not the case in general and the two operations
  must be distinguished.

He does not give an example of an occasion where such cases are not the same.
I will appreciate if someone can give an example of when inversion and twofold operation won't give the same result.

If the body is chiral, a ${C_2}$-axis maintains chirality.  An inversion (${S_2}$, ${i}$, the product of ${C_2}$ and ${σ}$h) inverts chirality.
https://ww2.chemistry.gatech.edu/~wilkinson/Class_notes/CHEM_3111_6170/Symmetry_and_group_theory.pdf
Frame 5
http://alpha.chem.umb.edu/chemistry/ch612/documents/MolecularGeometryandPointGroups_002.pdf
More technical
http://www.b-u.ac.in/sde_book/msc_phychem.pdf
The math

Of course, chiral objects do not have any Sn improper axes of symmetry.  Above describes operation of the transformation .


text:
Identification of lithium and strontium salts
I'm having trouble interpreting the results I got for my experiment.
As you should be well aware of, lithium and strontium both burn with crimson red when they are placed in a flame. Now, I can't decide which one it is.
Then I was told to do an experiment with the same salt where I had to add dilute hydrochloric acid and barium chloride. I had no visible change.
Does this last test help me with identifying between strontium and lithium? What element is it?

What are the salts you used? Generally in differentiating lithium and strontium you can have another reagent such as hydrochloric acid or barium chloride and create a solid precipitate.

@Jun-GooKwak Hi sir, We were given an unknown salt and have to work out the what cation and anion are available. We were only allowed to do two test which I have describe above in the question. Yes, I added hydrochloride acid and barium chloride and I had no change. What does that mean? Is it Lithium that is present or Strontium?

I really do not see, that $\ce{LiCl2}$ or $\ce{SrCl2}$ are insoluble. They should be separable with $\ce{(NH4)2CO3}$.

I'd concur with Martin's last comment: differentiating them by addition of carbonate seems logical, as the strontium salt should precipitate out.

If your solution contains only $\ce{Li+}$ or $\ce{Sr^2+}$, you can differentiate them by adding a sulfate solution (for example, $\ce{Na2SO4}$ or diluted sulfuric acid). If strontium is present, a white, fine-crystalline precipitate of $\ce{SrSO4}$ will form. In the case of lithium, the solution will remain clear. This is because $\ce{SrSO4}$ is much less soluble in water than $\ce{Li2SO4}$ ($\pu{0.0132 g}$ versus $\pu{34.8 g}$ in $\pu{100 g}$ of $\ce{H2O}$ at $\pu{20 ^\circ C}$ (reference)).
Adding diluted $\ce{HCl_{aq}}$ and $\ce{BaCl2}$ to a solution is commonly used as a test for $\ce{SO4^2-}$ (if it is present, $\ce{BaSO4}$ precipitates). Maybe this could help you to distinguish between $\ce{Li+}$ and $\ce{Sr^2+}$ given the different solubilities of their sulfates. If your unknown salt readily dissolves in water and $\ce{SO4^2-}$ is found to be present in the solution, then it is likely that you have lithium sulfate. Strontium sulfate as unknown salt would not readily dissolve and therefore not give a positive test.
However, if your solution can contain more ions besides $\ce{Li+}$ or $\ce{Sr^2+}$, I would suggest to try the separation with ammonium carbonate which is already mentioned in the comments to your question.

http://en.wikipedia.org/wiki/Solubility_table
http://www.chemicalbook.com/ChemicalProductProperty_EN_CB6723042.htm
http://www.csudh.edu/oliver/chemdata/data-ksp.htm
http://www.chemicalbook.com/ProductChemicalPropertiesCB1184992_EN.htm
Lithium chromate is listed as being exceptionally soluble in water.  Strontium chromate is listed as being insoluble.

The OP clearly states that there is no chromate involved.

Add a drop of potassium chromate solution on a white drop plate. Precipitate means strontium; no precipitate means lithium. If two horses differ in height by only a quarter hand, how do you tell the black horse from the white horse?


text:
Distinguish between symmetric and asymmetric molecules with IR
I get increasingly nervous as I get my organic lab course reports. For, I often get asked whether the peak in the IR comes from an asymmetric or symmetric stretch.
I've looked at about ten books and turned google upside down. But I still can't answer the question in every case, though our molecules are quite simple.
Can you help me out with a table or something equivalent?
The most urgent example is a $\ce{C-H}$ streching at $3006~ \mathrm{cm}^{-1}$ from p-tbutyl-(2-ethanol)-benzene (don't mind the nomenclature, please).

Is that peak for the hydrogens on the benzene ring, the t-butyl group, or the ethanol group?

But are you talking about symmetric and asymmetric molecules or symmetric and asymmetric vibration modes?9

Whether the stretch of a bond in a molecule is asymmetric or symmetric depends on the relative movement of the atoms connected by the bond. Lets take a $\ce{CH2}$ group as an example. If the two hydrogen atoms connected to the carbon move in the same direction during stretching, it s a symmetric stretch, if they are going in opposite directions, the stretch is asymmetric (see also here):

Left: symmetric stretch, right: asymmetric stretch
A bond vibration like stretching will only be IR-active (i.e. give a band in the IR spectrum) if it is accompanied by a change of dipole moment. In the case of the $\ce{C-H}$ bond, carbon is slightly negatively and hydrogen slightly positively polarized. A symmetric stretch increases and decreases the distance between these partial charges periodically, and therefore also changes the dipole moment, because the latter is dependent from the distance between charges (source). During an asymmetric stretch, the elongation of one $\ce{C-H}$ bond is compensated by the shortening of the other, so the overall change of dipole moment should be weaker.
In the case of 2-(4-tert-butylphenyl)ethanol (the correct name for your compond), the band at 3006 $\mathrm{cm}^{-1}$ can either be assigned to the $\ce{CH2}$ groups of the substituted ethanol or to $\ce{C-H}$ bonds in the benzene ring. In the first case, you can distinguish between symmetrical and asymmetrical stretch (symmetrical seems more likely to me), while in the latter case, a single hydrogen is attached to a carbon, and a distinction of the stretching, like in the case of two substituents at the same carbon, cannot be made.

first of all I have no practical experience with this field (yet), but we had covered this in theory (and also I don't have chemistry in English). I hope that I can help.
From what I recall, you have to first write the structural diagrams of the suspected molecules and compare them as it can be seen on the picture bellow
(I had izopropanol and propanol in my book, but I think that you will get this from this). Then you search in tables for appropriate absorption frequencies and write them down.
Then you check the experimental graph that you obtained in order to find out with what molecule you are dealing. So for example the corresponding graphs for these two substances are:

thanks, but this is not quite my problem. i know how to figure out if i got my compound from an ir. my problem is, that i have no idea how to know if a strech is smetric or not...

@Bulawa I cannot help you with that. The title mislead me. Cheer up! Somebody from the community is bound to know the answer :)