text:
How to calculate the theoretical yield of a Diels-Alder reaction?
A Diels-Alder reaction happens between 9-Anthracenemethanol (208.26 g/mol) and N-Methylmaleimide (111.10 g/mol).
I'm using 0.069 g 9-Anthracenemethanol (0.033 mol) and 0.11g N-Methylemaleimide (0.0009 mol, limiting reagent).
Since DA reactions have 100% atom economy, the product has the molar mass of both products combined, 319.36 g/mol.
Multiplying that by the moles of limiting reagent, we get 0.28 g product.
However, the mass of both starting materials combined (0.069g + 0.110g) together is not even 0.28g. What am I doing wrong?

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You are missing a  couple of zeros in the number of moles of your 9-anthracenemethanol.  I calculate 0.00033 mol of that reagent, which therefore becomes your limiting reagant, and I calculate a total yield of 0.105 g of product, with about 0.73 g of  N-Methylemaleimide left over.

Oh jeeze. Thank you so much! I was losing my mind thinking I suddenly forgot how to calculate theoretical yeild.

Writing lab reports late at night will do that to you.  Best regards.


text:
Order of electrophilic substitution
What will be the order of reactivity towards electrophilic substitution in case of the following compounds:
benzene, ethyl benzene, isopropyl benzene, tert-butyl benzene
The answer at the end of the book says that ethyl benzene will be most reactive (the book doesn't explain the cause though it's a MCQ) but as according to inductive effect I think the tert-butyl group will have more +I effect on the ring. Doesn't it?

In your series, all of the alkyl benzenes will have roughly the same +I inductive effect.  Where they differ is with regard to the resonance effect.  Hyperconjugated structures such as those drawn below for ethylbenzene are often invoked to explain these differences.

There are two hydrogens in ethylbenzene that are capable of donating electrons into the aromatic ring by hypercojugation.  In isopropyl benzene there is only one and there are none in *tert-*butylbenzene.  Hence ethylbenzene should be the most electron donating of the compounds in your series due to resonance involving hyperconjugation.
Realize that these effects are relatively small, for example the relative rates for the electrophilic nitration of various aromatic compounds is:
benzene=1, toluene=24, tert-butylbenzene=15.7 (reference, page 1060)
Benzene has no substituent (other than hydrogen), so neither resonance or inductive effects play a role, it would be the slowest.  The expected order in your series would therefore be:
ethylbenzene > isopropylbenzene > tert-butylbenzene > benzene

Wouldn't the steric hindrance be of more importance than minor difference in hyperconjugating effect? I'm not sure if steric hindrance is even the issue here at all, but it seems very reasonable. Less available reaction sides reduce reaction speed, however I cannot confirm it as it isn't dealt with in Clayden et al. or Vollhardt & Schore (i.e. the introductory texts I own). I cannot imagine the difference in reactivity is very important here at all.

Actually the final answer lies in experimental values. These questions come in Indian Competetive exams are very complex to answer on basis on theory. Sometimes back I came across a question where in a species 2 resonance effect was dominating over 16 hyper conjugative effects.

@Jori  Yes, steric hindrance plays a role, but more so in determining the ortho\para ratio.  Remember that a pi-intermediate is formed first and it should be less influenced by steric effects then the following sigma-intermediate (Wheland complex).

I think @Jori is on to something here. Looking at the referenced table, toluene actually nitrates preferentially at the ortho position (o:56%, m:3.5%, p:40%). The steric bulk of t-butyl slows down nitration at the ortho positions. Looking at the relative rates, it seems like the difference is about what you would expect from eliminating two potential sites of reactivity.

@jerepierre  Is the relative rate reduced because of 2 sites blocked by sterics or 2 less hydrogens available for hyperconjugation?  I come back to my earlier comment that the first-formed pi-complex shouldn't really be influenced by sterics.  We are talking about some relatively small differences.

@ron In the t-Bu case, can't the C-C bonds engage in hyperconjugation? As I understand it, hyperconjugation is just the overlap of a filled sigma orbital with an empty p-orbital, without limitations on the atoms involved.

@jerepierre  yes, but to a lesser extent vis-a-vis the relative stabilities of $\ce{H^{+}}$ and $\ce{CH3^{+}}$

@ron What exactly do you mean by pi-intermediate and sigma-intermediate? I have not yet seen these terms come across in my organic chemistry text.

@Jori The $\sigma$-complex is an intermediate you may know as the Wheland intermediate. In the arenium ion formation of a $\sigma$-bond between the arene and the electrophile has already taken place. Earlier along the reaction path, an interaction between the $\pi$-system and the electrophile is postulated. This earlier transition state is known as a $\pi$-complex.

Hmmm... The Macmillan group presentation slides on hyperconjugation you attached in a previous comment/answer the other time stated that C-C bonds are better hyperconjugative donors than C-H bonds. This seems to contradict what you have said about C-H bonds being better hyperconjugative donors.

Alkyl groups on benzene are ortho-para directing. The $\sigma$-conjugation effect of the alkyl group is not dependent on the type of carbon that is attached to is (methyl, primary, secondary or tertiary). Only the carbon connected directly to the ring can interact with it. More substituted alkyl groups will however hinder reaction at the ortho-site lowering the total reaction speed (e.g. for t-butyl benzene essentially no ortho attack can occur, it is too sterically hindered). Therefore the predicted order of reactivity from most reactive to least reactive is: ethylbenzene, isopropylbenzene, tert-butyl benzene, benzene.

But what about the para position. As you are saying that for the steric hinderance the ortho site can be blocked but the para site is not hindered.

I cannot imagine the para position to be extremely hindered by something that is happening on the other side of the molecule. But you loose two reactive sites which is bad for reaction speed.

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text:
What stopper will stop HCl fumes?
The local hardware stores actually sell chemicals in recycled beer bottles with plastic stoppers as lids. The bottle of hydrochloric acid I bought is clearly losing gas to diffusion through the plastic, or a seal that's not tight enough--it smells bad, and the beer bottle's metallic label is slowly corroding.
How can I best seal this bottle? Some ideas are:
Try to mold a stopper from silicone. Use a (synthetic) cork, whatever's used for homebrewing. Line the stopper with a few layers of thin polyethylene or polypropylene (from consumer materials, such as air-filled plastic bags for packing).
Or could I limit the diffusion by putting a plastic bag around it? This is tough to answer experimentally (without a lab) because resisting physical degradation isn't the same as resisting diffusion.

it smells bad, and the beer bottle's metallic label is slowly corroding.

That's, unfortunately, very common. At one of my former labs 'acid' cabinet  was corroded from within. For this reason it is recommended to keep such bottles under fume hood.
AFAIK, common stoppers are made from polyethylene or polypropylene. At one of my former labs concentrated hydrochloric acid was stored in stock plastic canister, similar to that used to store petrol (on the other hand, both nitric and sulfuric acids were stored in 20l glass bottles, even though the stoppers were made from same polymer).  I HIGHLY recommend to by (or order by mail) dedicated chemical bottles with caps, though. They are usually fairly cheap and pretty universal.

Okay, so no stopper will stop it? Right, then. I came up with an alternative solution that seems to work,  but time will tell: I put the stoppered bottle of acid next to an open wide-mouthed bottle of water and baking soda solution. These were both crammed together, within a couple of plastic bags. My idea is that any HCl that diffuses through the stopper will have ample chance to contact the water and be absorbed/neutralized before it makes its way to the air in the room.
I put my nose a foot from the bag and couldn't smell anything after 24 hours, so this homegrown hack seems to be working acceptably.

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Boiling aluminum in high-alkalinity water
This question ultimately relates to brewing beer, but more specifically to water treatment before brewing.
Though I can't get a water quality report, I suspect I have high levels of bicarbonate alkalinity in my tap water because boiling it causes a white precipitate (which I think is calcium carbonate) to form. I base this assumption on this formula, in the book Brewing:
$$\ce{Ca(HCO3)2 ->[heat] CaCO3 + CO2 + H2O}$$
I wanted to sanitize some aluminum foil so I boiled it in fresh tap water for about 15 minutes. I noticed a few things:

The precipitate formed much faster than it normally would without foil in it.
The foil began to blacken in certain parts.
I'm not positive, but the precipitate seems a bit finer in texture than it normally is.

I thought at first maybe the foil was just providing nucleation points for the carbonate to form around. But then I thought I remembered reading somewhere that aluminum competes with calcium in certain situations (I think it was regarding osteoporosis, maybe). The question I have, then, is whether or not it's feasible that aluminum could be affecting a different reaction than what I'm assuming normally occurs in this situation?

It appears your creating calcium aluminate by reacting aluminum with alkaline water. Since that is probably not wanted in the brew, you might use glass or anodized aluminum, which is more resistant to chemical attack, and a water softener, if the calcium carbonate isn't wanted, either.


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Comparing the stabilities of allylic, benzylic, and tertiary radicals

Which is the most stable free radical among the given species?
(1) $\ce{CH2=CH-CH2^.}$ (allyl radical)
(2) $\ce{C6H5-CH2^.}$ (benzyl radical)
(3) $\ce{(H3C)3C^.}$ (t-butyl radical)
(4) $\ce{C2H5^.}$ (ethyl radical)

The book claims the answer to be the allyl radical, option 1. But how can we definitely say that the allylic free radical is more stable, compared to the benzyl radical (option 2) which is also resonance-stabilised, or the tertiary radical (option 3) which has more hyperconjugative donation from α-hydrogen atoms?

I just want to say your book is correct. In simple words in this case benzyl free radical has resonance but the mesomeric effect is not transmitted to the free radical simply because it does not have a LONE PAIR of electrons. And besides Benzene has a -I effect thus further withdrawing the electrons and making the free radical unstable. (Coz obviously it needs electrons)

This is a tough question.  I think it might even be unfair to ask such a question on a test in non-advanced classes.  In advanced classes it could make an interesting topic of discussion, but I'm still not sure that the "real" answer is known.
What can be said is that due to resonance, both the allylic and benzylic radicals are more stable than the t-butyl or ethyl radicals which are not resonance stabilized.  You can see from the following diagram that you can draw two resonance structures for the allyl radical.

For the benzyl radical you can draw even more resonance structures.

Superficially, this might suggest that the benzylic radical is more stable than the allyl radical. This was the "answer" and reasoning provided here.
However, if you do some thermochemical calculations (as done here), you arrive at the opposite conclusion that the allyl radical is roughly 2 kcal/mol more stable than the benzyl radical. The number of resonance structures alone is not a perfect indicator of stability.
For me, that is a pretty small difference in energy, and I think it would be fair to answer that the allyl and benzyl radicals have comparable stabilities.

Anyways, but what about the 3rd option ? Its a tertiary free radical with lot of +I effecy from 3 methyl groups and also 9 hyper conjugations. Why is it not totally in the picture ?

Resonance makes a big difference in stability.

@ron as link you gave on your ans of pdf '.' As done here' its written below measurements results that, benzyl radical are 42 kJ ( 12k cal) more stable than comparably substituted NON BENZYLIC RADICALS. And you have said You arrive at the opposite conclusion that the allyl radical is roughly 2 kcal/m more stable than the benzyl radical. It was comparison between benzylic and non benzylic radical if iam not wrong. Please correct if I have wrongly interpreted.

@Akashv No, you have read it incorrectly. The Link notes that "the allylic radical is stabilized by 12 kcal/mol relative to the nonallylic radical" and that "benzylic radicals are about 10 kcal/mol more stable than comparably substituted nonbenzylic radicals".  Therefore, the allyl radical has approximately an extra 2 kcal/mol  of stabilization compared to the benzylic radical.

@ron , but still I didn't understand why is allylic radical equally or more stable than benzylic radical, is it because the resonance structure of benzylic radical would disrupt the aromatic ring structure?

@ron Hello, I had concluded the following order after reading your answer  ( the first being most stable ) : benzo , benzal, allyl, benzyl, t-butyl,sec-butyl? Please confirm is it correct order? Please.

Please confirm that I have concluded the right order after reading your answer ( the first being most stable) : Benzo, Benzal, Allyl, Benzyl, t-Butyl. Please confirm.

@Shiva The order is allyl ~ benzyl > t-butyl.  The question didn't address benzo or benzal (in any case, not sure what you mean by those terms).

I am extremely sorry. I am such a fool that I wrote benzo and benzal instead of tri-phenyl methyl and di-phenyl methyl respectively. Sorry. Now please confirm the order of radicals(the first being most stable): tri-phenyl methyl, 1,1-di-phenyl methyl, Allyl~benzyl,t-butyl. Please confirm.

@Shiva Confirmed, yes that order is correct.

@ron Hello, I am very sorry to disturb you again.But I will be very grateful if you reply like as you did before.Can you confirm the following order of carbocations (the first being most stable) : tri-phenyl methyl, 1,1-di-phenyl methyl, t-butyl, benzyl, allyl, 2°, 1° ,methyl, vinyl.

I had read this question https://chemistry.stackexchange.com/q/27086 , and the question got high votes but unfortunately the answers were incomplete,unsatisfactory and un-reliable. Please reply by confirming or atleast by giving correct order. Thanks very much.

Please notice that I am asking the order of carbo-cations and not free radicals. I had spent about more than hours to get a reliable answer but got nothing than confusions from many posts and internet. Please reply.

@Shiva Your order looks correct except for the allyl~benzyl>t-butyl part.  But remember that substituents can always cause exceptions.

@ron Thanks for the kind reply but please notice that I am talking about carbocations this time instead of free radical. I concluded that t-butyl is more stable than benzyl from this answer : https://chemistry.stackexchange.com/a/86692/121235  . Can you please confirm again :)

@Shiva  Your ordering is fine, just keep in mind that the difference between ally, benzyl and t-butyl is small and can easily change from solvent to gas phase or with substituents. But again, on a test I think your ordering would be accepted as correct.

Imagine your favorite activities are playing Video games (Most Favourite),
and other ones are reading chemistry, spending time in a chemical lab. (these two are activities are those activities in between you can't choose one that you want to do, or say you like them equally.)
You dislike equally: reading history, reading civics, reading economics.
Let's assume two cases:
Case 1.
You have a big house.
You have four rooms in your house containing these things:

Video games
History books
Civics books
Economics books

I don't think you are going to spend your so much time in rooms no. 2, 3 and 4. So The contribution of room 1 to your life is pretty high … you are going to live your life mostly living in room 1.
This the case of the benzylic radical. Room no. 1 is the canonical structure containing aromatic ring. In spite of having a big house you will not use it economically. Technically speaking, electrons have a big space to get delocalized but they are not delocalized efficiently.
Case No. 2
You have smaller house than in case 1, but you have two rooms containing these things:

Chemistry books
Chemical lab

So in this case you are going to spend your time in both of the rooms equally …
This is the case of the allylic radical.
All rooms are equivalent, i.e. similar (or say identical), thus the resonance structures bring more stability to the system.
You have a small house but you will use it economically.
Technically speaking, electrons are delocalized efficiently.
That why both (benzylic and allylic) radicals have s similar stability order.
Note: The number of resonance structures alone doesn’t determine stability.
Both quantity and quality are important.
In most case of chemistry, quantity wins.
Note: It's just an example for explaining the superposition principle in a simple manner, the analogy must not be taken too seriously or over-implicated.
Resonance is a static phenomenon, you can't say the electron is wandering from this carbon to that carbon (spending some time on that carbon or on this carbon). The only structure which exists is that of the resonance hybrid, which has a definite electronic distribution or precisely speaking has well defined  time independent wavefunction.

Using the same concept as to distinguish whether the t-butyl carbocation is more stable than the benzyl carbocation, I calculated the isodesmic reactions of the form in $\eqref{isodesmic}$ at the DF-B97D3/def2-TZVPP level of theory.
$$\ce{ R* + CH4 -> RH + *CH3 }\tag{1}\label{isodesmic}$$
I estimated the thermal correction at the standard conditions of $T=\pu{298.15 K}$ and $p=\pu{1 atm}$.
The results are within error compensation rate for the allyl and benzyl radicals. The level of theory might simply not be enough. If it is hard to distinguish the two with (more or less) elaborate calculations, then it clearly shows that this is a rather unfair question.
\begin{array}{llr}
  \ce{R*}  &  \ce{RH}  &  \Delta G / \pu{kJ mol-1}\\\hline
  \ce{*CH3}        &  \ce{CH4}         &     0.0  \\
  \ce{*CH2-CH3}    &  \ce{H3C-CH3}     &   -26.8  \\
  \ce{*CH2-CH=CH2} &  \ce{H3C-CH=CH2}  &   -75.7  \\
  \ce{*CH2-C6H5}   &  \ce{H3C-C6H5}    &   -68.1  \\
  \ce{*C(CH3)3}    &  \ce{HC(CH3)3}    &   -51.9  \\\hline
\end{array}
Here are the resulting optimised geometries (click to enhance).

(I won't attach geometries or absolute energies this time, because that would exceed the character limit.)

Does "level of theory" mean the accuracy of the theory? Like higher level theory meaning more accurate calculations?

@ApoorvPotnis Not necessarily. It is more difficult, see here, in principle it means higher level more computational effort. Ideally this would mean less errors, but that's not always the case, since different approximations have different sources of errors. There are examples where a seemingly high level of theory fails completely: Transition metal chemistry is a graveyard for MP2 :D

@Martin-マーチン Please confirm that I have concluded the right order after reading your answer ( the first being most stable) : Benzo, Benzal, Allyl, Benzyl, t-Butyl. Please confirm.

@Shiva I cannot confirm or debunk, as the answer already clearly states, this is not possible.

I am extremely sorry. I am such a fool that I wrote benzo and benzal instead of tri-phenyl methyl and di-phenyl methyl respectively. Sorry. Now please confirm the order of radicals(the first being most stable): tri-phenyl methyl, 1,1-di-phenyl methyl, Allyl, benzyl,t-butyl. Please confirm

@Shiva this answer already clearly states that with the data provided here this is not possible. I am also not intending to revisit this in a comment to this answer.

Indeed, this is a perplexing question. Stability compared to what? Are we considering their ease of formation from neutral molecule, i.e. the bond dissociation energy (BDE, or $DH^\circ$) of a carbon–hydrogen bond of propene, toluene and isobutane? [Ethane will be ignored.] Or are we comparing the three radicals from the same starting point, their constituent elements in the standard state, in which case we need to consider their heats of formation?
For the homolytic dissociation of a molecule $\ce{R–H}$ to two radicals $\ce{R^. + H^.}$, the sum of the heat of formation $\Delta_\mathrm f H^\circ$ of $\ce{RH}$ and the $\ce{R-H}$ bond dissociation energy (BDE) equals the sum of the heats of formation of the two radicals $\ce{R^.}$ and $\ce{H^.}$. This follows from Hess's law:

Gas-phase heats of formation are available from the NIST website, and the BDEs are available here. Comparison of $\ce{C-H}$ BDE's (in $\pu{kcal/mol}$) gives the order of stability as allyl ($+88.8$) < benzyl ($+89.7$) < t-butyl ($+96.5$).
If on the other hand the heat of formation of the three radicals is computed relative to the standard state, then the order is t-butyl ($+12.4$) < allyl ($+41.6$) < benzyl ($+49.7$). [The value of $\pu{+52 kcal/mol}$ for $\Delta_\mathrm f H^\circ(\ce{H^. (g)})$ is half the BDE of the hydrogen molecule.]

$$\begin{align}
 & & \Delta_\mathrm f H^\circ (\ce{R^.}) &= \Delta_\mathrm f H^\circ (\ce{RH}) + \text{BDE}(\ce{R-H}) - \Delta_\mathrm f H^\circ (\ce{H^.}) \\
\ce{R} &= \text{allyl:} & \Delta_\mathrm f H^\circ (\ce{R^.}) &= \pu{4.8 kcal/mol} + \pu{88.8 kcal/mol} - \pu{52 kcal/mol} \\
 & & &= \color{red}{\pu{+41.6 kcal/mol}} \\
\ce{R} &= \text{benzyl:} & \Delta_\mathrm f H^\circ (\ce{R^.}) &= \pu{12 kcal/mol} + \pu{89.7 kcal/mol} - \pu{52 kcal/mol} \\
 & & &= \color{red}{\pu{+49.7 kcal/mol}} \\
\ce{R} &= \textit{t}\text{-butyl:} & \Delta_\mathrm f H^\circ (\ce{R^.}) &= \pu{-32.1 kcal/mol} + \pu{96.5 kcal/mol} - \pu{52 kcal/mol} \\
 & & &= \color{red}{\pu{+12.4 kcal/mol}} \\
\end{align}$$

Stability, as defined by IUPAC Gold Book (http://goldbook.iupac.org/terms/view/S05900), should be measured in terms of the Gibbs free energies and not the enthalpy changes. The method which Martin used above is in agreement with the IUPAC definition while yours seems to be inconsistent. I am not trying to say that their method is flawed or incorrect. But perhaps, you may wish to explain how using absolute, quantitative measures of the heats of formation would serve as an acceptable and reasonable gauge of chemical stability?

The answer is benzylic radical.

It is the most stable because it resonates more compared to allylic and alkyl radicals.
Regardless of its radical class, be it primary, secondary or tertiary, benzenlic radicals will always be more stable than any other types of radical.

After all the evidence, which has already been presented here, this just restates a seemingly simple solution, which is likely to be wrong. And the choice of words is worse; it resonates more paints an absolutely wrong picture. The second point has been disproved on this very page. Apart from this, please support your claims with actual facts and evidence.

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Mechanism of Grignard addition to carbonyls
So from the general reaction we can see that the carbonyl is reduced while the Grignard is oxidized to an alkene.
So for this hypothetical reaction (does this really occur?) we are asked to provide a plausible mechanism. This is what I got, but the prof. wants something better - why transfer negative charge from oxygen to an $\text{sp}^3$ carbon?

Click on the image to enlarge, if you wish.
Here are his hints:

Remember, prior to Grignard reaction the $\ce{Mg}$ center swaps out an ether
  solvent molecule for a reactant oxygen.  This interaction will lead to
  the reaction but in this case, due to steric interference a hydride is
  delivered instead as the Grignard is oxidized to an alkene.  In this
  way the Grignard reagent is almost reacting in the way $\ce{NaBH4}$ would in
  the presence of this ketone. Try to get all of this action working
  together in a single, concerted step.

So ... I get the first part just fine. The $\ce{Mg}$ center has an octet; it's solvated by two ether molecules in solution usually. Okay and in this case the carbonyl oxygen helps solvate the $\ce{Mg}$ center. And then hydride transfer occurs ... but how exactly do we get the hydride to leave without creating a carboanion? And is it really that important that everything be concerted?
This is my second attempt ... if this is all concerted in real life then I must have hit the orbital alignment lottery! All the stars must have been aligned ... !
The only thing I see about my second attempt is that in the end the $\ce{Mg}$ center doesn't have a complete octet - perhaps I should have shown another ether molecule popping in to help $\ce{Mg}$ out with its octet fantasies.

are there two methyl groups missing in the third step and the final product?

@bon yes, sorry. Just imagine they are there; I don't think that makes a difference to my question.

...Gets me dubious. Do those images need to be that large?

@MARamezani I've fixed all the images with respect to chemistry. As far as the size of the images goes ... I don't know how to adjust them.

You can optimize what I did, and it wouldn't hurt to look at this: http://meta.stackoverflow.com/questions/253403/how-to-reduce-image-size-on-stack-overflow :)

This reaction can be viewed as an example of a pericyclic (concerted) Ene reaction.  The ene reaction is actually fairly common in organic chemistry.   The following diagram details the mechanism as a concerted reaction with a cyclic 6 membered transition state involving 6 electrons (an "aromatic" transition state).

A note on the Ene reaction...
The Ene reaction is really just a variation on the Diels-Alder theme.  The Diels-Alder reaction is a concerted [4 pi + 2 pi] reaction, whereas the Ene reaction is a [(2 pi + 2 sigma) + 2 pi] concerted reaction.

you know everything! Not even the text covers ene reactions.

I looked at your mech. and compared it to mine. Mine has 8 electrons moving around in the transition state ... so mine would likely be anti-aromatic, correct?

You can do this reaction with alkenes too; check out [https://books.google.co.uk/books?id=7gPnWf4UDLsC&pg=PA236#v=onepage&q&f=false Modern Methods of Organic Synthesis]

The simple answer is yes, but to be clear I need to make it a bit more complicated and maybe use some new words.  Transition states with 4n+2 electrons are aromatic when they proceed with no orbital symmetry sign inversion (suprafacially in a 1,5-hydrogen shift).  4n electron transition states are aromatic when they involve an orbital symmetry inversion (e.g. the antarafacial process in the 1,7-hydrogen shift).  The 4n+2 process becomes antiaromatic when there is an orbital symmetry inversion and the 4n process becomes antiaromatic when there is no orbital symmetry inversion.

See here for example.

@Ron How do we know if there's been an orbital symmetry inversion?

My comment above "The simple answer is" is inadequate.  It takes a chapter in a book to work through the major topics.  Perhaps it would be better if I said that reactions involving 4n or 4n+2 electrons can both proceed through an aromatic transition state, but that the trajectories followed will be different.  Positive overlap in the HOMO describing the bonds in the starting material that will break must be maintained in the bonds that will form.  To maintain this positive overlap can require very different motions (conrotatory-disrotatory, suprafacial-antarafacial, etc.). See the above link.

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text:
Which is more stable — a carbanion on a vinylic position or a carbanion on a benzylic position?
I am getting a dilemma over this. Should the negative charge on the benzylic position participate in resonance with the benzene ring and make it more stable or is it the electronegativity of the $\ce{sp^2}$ hybridized carbon of the vinylic group that will make the carbanion more stable on it ?

One of the more accurate methods to measure carbanion stabilities is negative ion photoelectron spectroscopy.  In this process, a carbanion is placed (generated) in the gas phase (in a high vacuum).  The gas-phase carbanion is then irradiated with light.  Each frequency of light provides a specific energy ($E=h\nu$).  The frequency of the light is varied (and the energy supplied changes as a consequence of the change in frequency) until enough energy is supplied to knocked out of one of the carbanion electrons (this is called photodetachment).
$$\ce{CH3^{-} + hν -> CH3• +~~ e^{-} +~ KE_{electron}}$$
The kinetic energy of the photodetached electron is then experimentally measured and the electron affinity ($EA$) of the carbanion is given by
$$EA = hν ~~– ~~KE_{electron}$$
The $EA$ is the energy required to remove one electron from the carbanion. The larger the electron affinity, the more tightly the electron was held.  Stable carbanions hold their electron more tightly than less stable carbanions. The electron affinity of the allyl anion was measured as 0.481 eV.  For the vinylic carbanion, the EA was measured as 0.667 eV.
The fact that the vinyl carbanion is more stable than the allyl carbanion suggests that hybridization effects outweigh resonance effects in terms of stabilizing these carbanions.  That sounds reasonable and is what I would have guessed.
reference for EA measurements

Thank You. I didnt understand the first part of your answer because I am in high school and prepearing for College Entrance Test. The later part was clear to me. Thank You.

What part \ words didn't you understand?

The photoelectron spectroscopy part.

Let's see if I can improve it.

Its beyond my scope I think. You may not take the trouble to get down to my level. As of now I will remember the facts.

I made a try, perhaps it will help.

It is crystal clear now.


text:
Separation of metal particles from fluid in ferrofluid
I purchased commercial Ferrofluid, and when suspended, the water and/or alcohol turns into a brown sludge.  In 95% alcohol, all the metal particles separated from the "tar" Why, and how can I prevent this?

Assuming the ferrofluid is the one described at product description, then it is "an extremely fine powder, coated with a soap-like material called a surfactant, suspended in a mineral oil liquid base." From that description, it might be possible to dilute or extend the ferrofluid with mineral (or even vegetable) oil, but one would not expect ferrofluid to be compatible with water or with alcohol. It might be possible to dilute it with other aliphatic solvents.
There are directions for making your own water-based ferrofluid at Synthesis of an Aqueous Ferrofluid.


text:
Deriving The Triple Point Temperature and Pressure as the following expression
Using the usual notations of $\Delta H_\text{fus}$, $\Delta H_\text{vap}$, $\Delta H_\text{sub}$, as well as ($T_\mathrm l,p_\mathrm l$) and ($T_\mathrm s,p_\mathrm s$), which corresponds to the temperature and vapor pressure of liquid and solid phase, I have to derive the triple point temperature and pressure.
My strategy is to use the Clausius-Clapeyron Relation:
$$\frac{\mathrm dp}{\mathrm dT}= \frac{\Delta S}{\Delta V}$$
but the rest is history...
Could someone help me with
$\Delta H_\text{fus}$ and ($T_\mathrm s,p_\mathrm s$)?
So that I know that I'm moving in the right direction?

On a phase diagram there are three 1-dimensional curves, the melting curve, the liquid vaporization curve, and the sublimation curve.
It is sufficient to consider just two of the three curve, and determine the point at which they intersect.
Given the limited information availible to you, (no information about how volume changes upon going from the solid to liquid phase), it is wise to choose the liquid vaporization curve and sublimation curve, rather than the melting curve, as $\Delta V$ can be approximated as the volume of gas, the volume of solid or liquid being approximately zero compare to the much larger volume of gas.
From the Clapeyron equation in the question, you need to change variables from  $\Delta S$ which you don't know to $\Delta H$ which you do know.
When two phases are in equilibrium, such as along any of the three curves discussed above, $\Delta G$ = 0.
$G = H-TS$
$0 = dG = dH-TdS -SdT$
Considering that $T$ is constant during a phase change:
$\Delta H = T \Delta S$
So substituting:
$\frac{dP}{dT}= \frac{ΔH}{TΔv}$
$\frac{dP}{dT}= \frac{ΔH_{vap}}{TΔv}$ and $\frac{dP}{dT}= \frac{ΔH_{sub}}{TΔv}$
Then, as mentioned above the approximation $\Delta V = V_{gas}$ can be made and the ideal gas law (or other equation of state) can be used to express $V$ as a function of $T$ and $P$.  Then (after intergrating) you have two equations with two unknowns, $T$ an $P$, the temperature and pressure at the triple point.


text:
What is the chemical composition of the grignard reagent?
What is the actual chemical composition of the grignard reagent and what is the method of preparation? Can it be prepared in college laboratories?

A Grignard reagent is an organomagnesium halide of the generic form $\ce{RMgX}$, where $\ce{R}$ is a hydrocarbon group and $\ce{X}$ is a halide, usually $\ce{Cl}$, $\ce{Br}$, or $\ce{I}$.
These reagents are produced by the reaction of magnesium metal with the corresponding organic halide in a non-protic (but coordinating) solvent, like diethyl ether.
For example, you would make phenylmagnesium bromide from bromobenzene and magnesium:
$$\ce{PhBr(C6H5Br) + Mg ->[][\ce{Et2O}]PhMgBr}$$
They are moisture and air sensitive, but not so much that you need to get crazy with inert atmosphere. These reagents are not stable outside of solution.
Grignard reagents can be made in most laboratories set up for organic chemistry. In fact, they are very common reactions to include in undergraduate organic chemistry lab courses because they are very well behaved.


text:
What is the chemical composition of biofuels?
What is the chemical composition of biodiesel and green diesel? How do they differ from each other?

Diesel fuel is a fraction from the destillation of crude oil and consists of alkanes, cycloalkanes and aromatic hydrocarbons. In Europe, the properties of this mixture are defined in DIN EN 590.
This includes a certain density range, a minimum cetane number (CN) of 51, a maximum content of 8 mass-% for PAHs, a maximum sulfur content of 50 mg/kg. After distillation at tempertures up to 360 °C, the residue may not be larger than 5 vol-%. Note that diesel fuel may contain up to 7 vol-% biodiesel.
Biodiesel mostly is FAME, fatty acid methyl ester, produced by transesterification of triglycerides from biomass (vegetable oil).
Green diesel, also known as renewable diesel, is a fuel derived from biomass that does not contain monoalkyl esters of fatty acids. This can be achieved by catalytic hydrodeoxygenation (hydrotreatment) of triglycerides or fatty acids and typically involves a cracking step too, leading to an alkane mixture.


text:
How do I determine if a process is reversible without numeric information?
How do I determine if a process is reversible without numeric information?
It would be nice to know how to determine whether a process is reversible in order to solve problems LIKE this one:
A weather balloon is filled with Helium gas and released from the ground. It goes up 18km and achieves a diameter of 15m. Determine if the following values are greater than zero, less than zero, or equal to zero: ΔV, ΔP, ΔT, ΔU, ΔH, S_sys, S_surr, S_tot.
Here is my attempt at solving this problem. I believe at least half of my error comes from not knowing how to determine if this process, or any other process merely described, is reversible.
Obviously, ΔV increases because the balloon expands as the diameter gets to be large compared to what it presumably was at the time of release, even though the diameter at that time isn't mentioned.
ΔV > 0
ΔP < 0 because pressure decreases as altitude increases.
ΔT < 0 because temperature increases as altitude increases.
ΔU < 0 because ΔT < 0
ΔH < 0 because ΔT < 0
It seems that I have assumed that the external properties of T and P apply to the inside of the balloon, so I guess the system is reversible.
Stot = 0
Ssys = -Ssurr
I don't really know where to go from here...
I want to argue that because the balloon actually does manage to overcome gravity, it's change in volume must be pretty substantial (making the w = -PdV be what causes the balloon to go up). Maybe it's more substantial than the change in temperature, giving a large positive term due to volume in the entropy state equation listed above.
Ssys > 0
However, if the system is being looked at as the balloon itself rather than the air inside the balloon, then when the balloon isn't inflated, it has more freedom of movement than after it's fully inflated and therefore more entropy, meaning that the final entropy change of the system would be negative.
If I am correct in guessing that this process is reversible AND in guessing that Ssys > 0, then Ssurr < 0.
I would rather not have all of my answers to physical chemistry problems be guesses. How can I solve this problem without guessing? How can I know if my guesses are correct or not? How can I better approach this problem? It's possible that the correct answer to this problem is due to something that I have completely overlooked. How do I not overlook considerations while doing these types of problems?

Any process is irreversible if it involves:

Friction
Heat transfer across a finite temperature difference
Expansion when there is a finite pressure difference
Mixing of difference substances under any real circumstances
Anything else where work is not being maximized

So air friction against the balloon rising at a non-zero velocity makes the process irreversible.
Heat transfer from the balloon when the inside temperature is different from the outside temperature makes the process irreversible.
For more information see: http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node34.html

This might help you understand why it is a reversible process
http://how-it-looks.blogspot.com/2010/10/reversible-processes.html
and check out this website in order to get an understanding of how hot air balloon works
http://socratic.org/questions/how-do-gas-laws-affect-hot-air-balloons
Charles Law
chemed (.) chem (.) purdue (.) edu/genchem/topicreview/bp/ch4/gaslaws3.html
When thinking of whether or not something is reversible, ask yourself can the system goes back to the way it was before without exerting external work on the system.


text:
Can someone induce cancer in another person
Is it possible to poison someone's food on a daily basis in order to give them cancer?
The thing here is that i'm worried someone at my workplace might be poisoning other people's water (each employee has his own packs of water bottles at their desks, also, this person seems to have some mental issues).
It is possible to put chemicals in a bottle with no water leaking via a syringe. So my fear is that this bad person (this person IS bad., believe me) may have poisoned, in a daily basis for 3 or 4 weeks, our water.
So the question is, could this person poison the bottles to make others get cancer? Or another highly dangerous disease?

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour]. What you are posting here is worrisome and you should consider other actions than asking this question here. But in short, the scenario you are describing does not seem impossible to me - depends however on the knowledge and skills of the person you are asking about.

[…] So the question is, could this person poison the bottles to make others get cancer? Or another highly dangerous disease?

Yes, in principle.
Information on known cancerogenic compounds can be found in CMR (Canceogenic, Mutagenic and toxic to Reproduction) lists published by national organizations responsible for workplace safety.
In Germany, this would be the IFA (Institut für Arbeitsschutz der Deutschen Gesetzlichen Unfallversicherung).
List with old nomenclature
List with new nomenclature
I have added these lists just for the sake of completeness!
If you suspect any of the activities that you described, contact the local authorities.


text:
Does resonance overrule hyper conjugation in terms of stability in alkenes?

Which of the following is more stable ?

$\ce{CH2=C(CH3)-CH=CH2}$
$\ce{CH3-CH=CH-CH3}$

I was going for answer 2 because it has 6 hyper conjugative effects. I found the answer given to be 1, which has 3 hyper conjugations but more resonance structures. Will the R-effect meet the deficit of 3 hyper conjugations and make 1 more stable than 2?

I do not understand in which context should the one be more stable than the other.

I dont get you, sorry can you elaborate ?

Is it more stable towards hydration, oxidation, bromination, substitution, elimination, heating, pressure, etc, pp.? The context is missing. It is impossible to compare the stability of two molecules without knowing how to relate them.

Overall stability (?) The question asks for stability only. And it is a verified question appeared in public exams in past years.

We can hydrogenate unsaturated compounds and measure the heat given off, this is called the heat of hydrogenation. As the following figure illustrates, if we hydrogenate the various butene isomers they all wind up producing the same molecule, butane.

image source
By comparing their heats of hydrogenation we can see which butene was lowest in energy to begin with, that is, which butene is the most stable. The heats of hydrogenation reported in the figure tell us that trans-2-butene is more stable than 1-butene by about 2.3 kcal/mol. This extra stability can be attributed to the additional hyperconjugative resonance structures involving hydrogen present in trans-2-butene (6) compared to 1-butene (2).
Next let’s consider 1,3-butadiene. Here when we hydrogenate it, butane is again produced and we might estimate that it will have a heat of hydrogenation equal to twice that reported for 1-butene (1-butene is the best model because its double bond has the same substitution pattern as found in 1,3-butadiene) or 60.6 kcal/mol of heat given off. When we actually run the experiment we find that only 57.1 kcal/mol of heat is given off, 3.5 kcal/mol less than we estimated. In other words 1,3-butadiene is about 3.5 kcal/mol more stable than isolated double bonds and this is due to the resonance interaction between the two double bonds in 1,3-butadiene.
The resonance stabilization in 1,3-butadiene (3.5 kcal/mol) is larger than the hyperconjugative stabilization in trans-2-butene (2.3 kcal/mol).
Your question actually asked about 2-methyl-1,3-butadiene. In this case we have the resonance stabilization found in 1,3-butadiene plus some hyperconjugative stabilization from the additional methyl group. It will be even more stable than 1,3-butadiene. Further, 2-methyl-1,3-butadiene is more stable than any butene, so answer #1 is correct. Resonance stabilization is larger than hyperconjugative stabilization in this series of compounds.


text:
How to Determine the Rate Law knowing time and concentrations
Consider the hypothetical reaction described by the following equation:
$$\ce{3 A + 2 B -> C}$$
Reagent B was added in excess and the following time concentration data was obtained.
\begin{array}{cr}
A\ [\mathrm{10^{-3}\ mol\cdot L^{-1}}] & \mathrm{time [s]} \\ \hline
5.0 & 10\\
2.9 & 50\\
1.9 & 80\\
1.1 & 120\\
0.7 & 150\\
\end{array}
(i) Derive the relevant integrated rate law and hence find the rate constant.
I am thinking that to solve this question you just sub the values into the integrate rate laws and see if it produces a linear relationship.
Is this correct? Also could you please suggest a faster method of solving this type of question.
Thank you.

You could plot the data to determine the order of the reaction.  I would use Microsoft Excel to do this.  Reproduce the data and then add columns that calculate the $\ln [A]$ and $\frac{1}{[A]}$.  Then plot $[A]$, $\ln [A]$ ,and $\frac{1}{[A]}$ vs. time (I would use separate plots).  The plot that gives the "best" linear fit (which you can assess using the $R^2$ value), corresponds to the appropriate integrated rate law.

If $[A]$ vs. time is linear, then the reaction is zero order.
If $\ln [A]$ vs. time is linear, then the reaction is first order.
If $\frac{1}{[A]}$ vs. time is linear, then the reaction is second
order.

Hope this helps.

In addition to Tami M.'s answer:
You might want to give Python, the IPython Notebook and matplotlib a try:

import numpy as np
import matplotlib.pyplot as plt

t = np.array([10, 50, 80, 120, 150])
a = np.array([5.0, 2.9, 1.9, 1.1, 0.7])

a_int = np.log(a)

plt.figure()
plt.subplot(121)
plt.plot(t, a, 'ro-')
plt.xlabel('t [s]')
plt.ylabel('A')

plt.subplot(122)
plt.plot(t, a_int, 'gs-')
plt.xlabel('t [s]')
plt.ylabel('ln(A)')

plt.show()

Note that the plots are not fine-tuned and that I omitted a further fitting ;-)

Least-squares regression theory can also provide estimates of the slope and intercept and their respective variance estimates. The computed slope and intercept terms are assumed to be unbiased and have a normal distribution for deviations from the true underlying value.
The regression model for the ordinary least-square case is:
Y = mX + b
with a design Matrix X:
| x1 1 |
| x2 2 |
| xn n |
where each point x is appropriately transformed depending on the presumed rate order.
The Variance-Covariance Matrix, which is central to assessing the precision of the rate order estimate, is created by taking the matrix inverse of [ matrix product of [transpose of Matrix X] times [Matrix X]]]. It is further scaled by an estimate of the variance of the regression model fit (computed from the differences between fitted and reported for each point which is further squared, summed for all points, and divided by (n-2). See Wikipedia discussion of statistical theory at https://en.wikipedia.org/wiki/Ordinary_least_squares.
Then, one could apply a hypothesis testing procedure (like the t-test) as to whether the slope is o or 1 or 2. However, this is not precisely correct, and one should examine other data or theory (like looking at the expected slow rate-determining step coefficients), and then proceed to selected just one value from (0, 1, 2) for testing.
However, in the case of getting a fractional value (as a regression model of the above data actually indicates), try a direct test for linearity (see, for example, a proposed idea of modifying the regression at http://essedunet.nsd.uib.no/cms/topics/regression/4/3.html ). This consists of introducing a quadratic term in time, and if the latter fails a t-test of being different from zero, the model is deemed to have passed a linearity test.


text:
Is the atom the smallest particle, which takes part in chemical reactions?
According to modern atomic theory, the atom is the smallest particle which can take part in a chemical reaction. But during the formation of hydronium ion, $\ce{H+}$ ion reacts with $\ce{H2O}$ to form $\ce{H3O+}$ ion. Here the $\ce{H+}$ ion, which is smaller than hydrogen atom, takes part in a chemical reaction. So is that postulate invalid?

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You could argue that electrons are the most important particles in (organic) chemistry, which are of course much smaller than the atoms they belong to.

There's no such thing as free proton in chemical reactions.

@Mithonon How many free protons reach a given square centimeter of the Earth's atmosphere every second? http://services.swpc.noaa.gov/images/ace-epam-p-3-day.gif What happens to these free protons? https://www.bu.edu/csp/uv/proton/proton_intro.html

@Mithoron consider proton tunneling reactions. Definitely a free proton.

Whoah, yes it was a generalisation on my side, but the example with protonation of water is bad.

Protons definitely participate in chemical reactions.  Free protons are generally not present in liquid water because a free proton is extremely reactive, but in the upper atmosphere or in other situations where the density of matter is low there can be free protons which participate in chemical reactions.
However, I completely agree with Jori that an electron, which is much smaller than a proton, takes part in chemical reactions.  Gain or loss of one tiny electron can completely change the properties of a large metalloenzyme.
How small is an electron vs. a proton? In terms of mass, the electron is 1836 times lighter, but what about in terms of radius. According to Hans Dehelt's Nobel Lecture:

Proton radius [is] $0.86 \times 10^{-15}$m...
Today everybody “knows” the electron is an
indivisible atomon, a Dirac point particle with
radius R = 0....
But is it?

And goes on to explain that it is still possible the electron could have a radius on the order of $10^{-20}$ m.
See also Limit on Electron Substructure
Also, photons, which have no mass, participate in chemical reactions.

Yes, my immediate thought was proton and electron transfer, but great point with photons!

It is questionable to say an electron is “much smaller than a proton”. True: the proton has an inner structure, the electron apparently doesn't – in that sense, its size is exactly zero. However, that inner structure is not relevant for chemistry; what's much more interesting as a characteristic size is the quantum-mechanical unsharpness. And that is much larger for the electron, precisely because it has a smaller mass! (And yet larger for optical photons.)

@leftaroundabout Aren't you really referring to the size of a region having a given electron probability density, rather than the size of an electron itself?

Neutrons induce many chemical transformations.

Well, even neutrinos may cause reactions


text:
Which factor should I take into priority to determine reactivity of of Halides?
Such as in the attached question I was taking into account the stability of the carbocations (by the number of resonating structures) formed after removal of Chlorine from and concluded to be the order I>II>III>IV>V the answer states the order to be I>III>IV>V>II
Can conclude what factors they are priority to. Help needed

Are you familiar with the concept of aromaticity? The Hückel rule can be applied to charged species too!

Yes I know Huckel rule. Kindly, enlighten me the procedure. In which way I should apply the rule.

If you need to consider the stability of the carbocations obtained after chloride leaves, we can, as Klaus commented already, use Hückels rule, which more or less states that fully $\ce{sp^2}$ (i.e. usually $\ce{sp^2}$, as long as there is atleast one full p-orbital) hyberdized  compounds with $4n+2$ electrons are particularly stable or aromatic. The same type of compounds that have $4n$ are particularly unstable, anti-aromatic. Of course a carbocation has an empty p-orbital, so compunds I and III will be aromatic and compound II anti-aromatic. Compound I is fully benzenoid and also less strained than compound III and will hence be more stable, so I > III. Then follows the allylic carbocation IV which is obviously more stable than its non-allylic counterpart V. And last but not least the anti-aromatic compound II.
In conclusion I > III > IV > V > II, as the answer indicated.

shouldn't it be 4n + 2

Yes, I was about to suggest the edit.


text:
compression set of tin catalyzed silicone better than platinum?
Platinum silicone is more dimensionaly stable, it shrinks less when it cures.
However, i just need silicone which has good compression set (small change by compression), and perhaps, once they are cured, platinum and tin catalyzed silicone become the same material - silicone elastomer. Perhaps that tin silicone, having shrunk, is more resistant to compression set?
Does tin silicone have better compression set? Are the highest quality o rings made from tin silicone?

Tin catalyzed silicone lasts for a couple of years and platinum catalyzed lasts for 10 years, after which they lose their elasticity, becoming brittle and splitting. So while they may have equal properties for the first year, After a while the Platinum silicone will perform better. Using less catalyst for longer cure times also increases durability, and vacuum treating the silicone prior to casting and heat treatng it afterwards increases it's compression set performance.


text:
Which is more stable? A phenyl carbanion or a disubstituted vinylic carbanion?
I came across a question which asks to work out the stability order for the following species-

b.
c. $\ce{R2C=CH-}$

Can the lone pair on phenyl carbanion participate in resonance?  If yes, then how? Also, how is that more stable than the vinylic carbanion?

No it doesn't participate in resonance.

Then how come it is more stable than the primary vinylic carbanion ?

Again... stable towards what exactly? I find these (I suppose exam) question rather weird asking about stability of compounds without a proper context.

@Jori, unfortunately these questions are a lot to come by, at least here. Hmm, I put them like this: The compound that is more stable produces lesser heat upon decomposition.

@MARamezani What do you mean exactly?

Also, why close this question? It is not homework. I think OP could have tried to add some more of his own thought, but I consider it on-topic. If we do not want to scare too many newcomers away I think we should not have a too harsh policy.

The lone pair on the benzene is in the plane of the ring and hence orthogonal to the conjugated $\pi$-system so that it cannot interact with it. It is $\ce{sp^2}$ hybridized just like the vinylic carbanion, which also cannot resonate. Furthermore, both compounds lack an electron withdrawing group, so there is no inductive effect either. Looking at the pKa values for both compounds, which usually gives a good indication about the stability of the anion, we see that there is indeed hardly any difference between them (pKa of benzene ~43 vs pKa of propene ~44 vs pKa of ethylene ~44).
In your question you only show us option (b) and (c). It would be nice if you could show the whole question including all possibilities (also as much as possible in text please, so the search engines can find it!)

I got the order partially right. So I did not the whole question.

Jori, there are basically two different types of protons in propene; those attached to an $\ce{sp^2}$ carbon and those attached to an $\ce{sp^3}$ carbon. Which type of proton in propene is responsible for the pKa you reported?

@ron I assumed it would be for the $\ce{sp^2}$ since I suspected that the $\ce{sp^2}$ hybridization was of more importance than the allylic conjugation. Comparison with the pKa of 3,3-Dimethyl-1-butene would give a more definite answer for sure, but I cannot find it anywhere. However comparing with the pKa of ethylene, which is also ~44, gives a good hint in this direction. I simply listed that table because it listed both types of compounds at the same place.

@Jori, I didnt umderstand the reason why it is not participating in resonance. Its SP2 hy ridized in benzene ring and also planar molucule. Which part of the rule does it exactly violate. Please, enlighten me a bit on this

In the deprotonated vinylic molecule, the lone pair is not in an sp2 hybridised orbital. As soon as the proton is detached (or even while it is detached) the hydrogen will not uphold the $120^\circ$ $\angle\ce{CCH}$, but adapts a linear geometry with $180^\circ$. The carbon itself is then sp hybridised and the lone pair is in a p orbital perpendicular to the $\pi$ bond.

@Martin That is interesting. I suppose that is part of more advanced molecular orbital theory, which is sadly badly treated by most undergraduate organic chemistry books. I think Clayden does a better job than most other books, but still maintains a relatively simple model of hybridization. To what extend does that change the answer to the question? You make it sound like it is that then the benzene anion, which cannot lock the lone pair away in a p-orbital I assume, would be more stable then, yet the pKa seems pretty constant at ~44.

The model of hybridisation is fine, but one must always keep in mind, that breaking bonds will result in a different geometry and therefore also hybridisation changes. Your point is valid though, both of the molecules should be about equally stable (read: they are both quite reactive towards electrophiles) and you did of course draw that conclusion from experimental measurements. From MO I would assume the ethylene derivate being a bit more stable due to hyperconjugation which cannot be present in the benzyl anion.

@Martin You commented that the vinyl carbanion will adopt a linear configuration.  That may be true for a vinyl cation, but is it also true for the vinyl carbanion? I would have thought that the carbanion would  be bent with a barrier to inversion (just like an imine).  The lone pair would be more stable in an sp2 orbital rather than a p orbital.  Hybridization is a way to answer the OP's question.  The C-H bond in benzene is $\ce{sp^2}$ while the C-H bond in ethylene is $\ce{sp^{2.2}}$, consequently the phenyl carbanion is more stable.  If this question is reopened I will write more.

I was wrong about the geometry, see J. Am. Chem. Soc., 1978, 100 (19), pp 6007–6012. The lone pair is still not sp2 though. I think hybridisation is a poor concept to explain stability in any case (as long as we are not talking about highly accurate VB theory calculations and even then it is a stretch). I am tired of talking about hybridisation though. (If you want this question to be reopened, you have to rework it yourself I fear.)

Can the lone pair on benzene carbanion participate in resonance?

As Jori has pointed out, the carbanion lone pair is orthogonal to the pi system of the aromatic ring.  Therefore it cannot participate in resonance.  Here is a drawing that illustrates this point.

how is it more stable than vinylic carbanion?

The $\mathrm{p}K_\mathrm{a}$ approach discussed in @Jori's answer becomes problematic when we are trying to look at a small difference between two large $\mathrm{p}K_\mathrm{a}$s.  A $\mathrm{p}K_\mathrm{a}$ of $43$ means $1$ ionized molecule exists in every $10^{43}$ molecules.  Certainly difficult to measure experimentally and so a significant error is attached to these estimates. In fact, using the data from Evan's $\mathrm{p}K_\mathrm{a}$ table we find a $\mathrm{p}K_\mathrm{a}$ of $43$ for the benzene $\ce{C-H}$ proton and a $\mathrm{p}K_\mathrm{a}$ of $50$ for an ethylenic proton, suggesting that the phenyl carbanion might be more stable than the vinyl carbanion.
An alternate way to examine this question involves comparing the hybridizations of the phenyl and ethylenic $\ce{C-H}$ bonds.  Since bonds with more $\mathrm{s}$-character will stabilize electrons better than bonds with less $\mathrm{s}$-character (because the $\mathrm{s}$-orbital is lower energy than a $\mathrm{p}$-orbital), whichever bond has more $\mathrm{s}$-character should produce a more stable carbanion.
The $\ce{C-H}$ bond in benzene is exactly $\mathrm{sp^2}$ hybridized as required by the symmetry of the benzene molecule.  Experimentally, we know that the $\ce{H-C-H}$ angle in ethylene is $117^\circ$.  Using this fact along with Coulson's theorem (reference1, reference2, reference3) we can determine that the $\ce{C-H}$ bond in ethylene is $\mathrm{sp}^{2.2}$  hybridized.
The $\mathrm{sp^2}$ hybridized orbital in benzene has more s-character than the $\mathrm{sp}^{2.2}$ hybridized orbital in ethylene.  This would lead us to suspect that the phenyl carbanion is more stable than the vinyl carbanion.
Of course, once we remove these protons we expect the resulting carbanions to undergo further rehybridization in order to further stabilize their lone pair (carbanion) electrons.  It seems like a reasonable first approximation that they would both rehybridize similar amounts and the "relaxed" phenyl carbanion would remain more stable than the "relaxed" vinyl carbanion.  But in any case, the transition state leading to the phenyl carbanion will be more stable than the transition state leading to the vinyl carbanion because of the hybridization \ $\mathrm{s}$-character reasons presented above.
Edit: OP's comment on hybridization
Molecules are not just $\mathrm{sp, sp^{2}}$ or $\mathrm{sp^3}$ hybridized.  While the $\ce{C-H}$ bonds in methane are exactly $\mathrm{sp^3}$ hybridized (as required by symmetry), the $\ce{C-H}$ bonds in $\ce{CH3F}$ are not (see here for a discussion).  Bonds can have any hybridization index (the superscript to the right of "$\mathrm{p}$") between zero and infinity.  Hybridization indices are not limited to integer values, nature can mix $\mathrm{s}$ and $\mathrm{p}$ orbitals in whatever ratio is required to create the most stable bonds and molecules.

Cant we say that the vinylic one is more unstable due to electron releasing effect of R groups?

@Abcd Yes, if there are alkyl groups located near the carbanionic center then they could destabilize either carbanion inductively as you suggest.  However, I think this effect would be secondary to the hybridization effect discussed above,


text:
Does hydrogen really have seven isotopes?
I know that there are 3 isotopes of Hydrogen. But according to this website it is showing 7 http://periodictable.com/Properties/A/KnownIsotopes.htmlWebsite

As you can see by your link, heavier isotopes have decay time measured in $10^{-13}$ fractions of nanosecond. They are of theoretical interest for nuclear physics, but from chemists point of view can be ignored completely.

Look at the data there. The half-life of 3H is 12 years. The half-life of 4H is 1.39×10-13 nanoseconds. Does that count as existence for you?

Yes, there are seven known isotopes of hydrogen, though only two ($\ce{^1H}$ and $\ce{^2H}$) are stable with respect to nuclear decay, and only three ($\ce{^1H}$, $\ce{^2H}$ and $\ce{^3H}$) exist/can be made in enough quantities to be relevant outside of nuclear physics. All other hydrogen isotopes have extremely small half-lives. The next most stable hydrogen isotope after $\ce{^3H}$ ($t_{1/2}=12.3\ yr$) seems to be $\ce{^5H}$ ($t_{1/2} \sim 10^{-21}\ s$). Given that even the fastest chemical reactions happen in timespans of the order of $10^{-15}\ s$, we can safely discard any influence of hydrogen isotopes heavier than $\ce{^3H}$ in chemistry.
Out of the currently 115 elements which have been confirmed to exist, there are approximately 254 isotopes which are considered to be indefinitely stable, or have half-lives so enormous (over trillions of times greater than the age of the Universe, 13.8 billion years) that no value could yet be determined. Theoretically, only 90 of those 254 isotopes are expected to be truly indefinitely stable (proton decay aside), but there are great difficulties in setting up experiments capable of measuring such rare decay events.
On top of the previous 254, another 34 isotopes have half-lives which have actually been measured in laboratory, but are still large enough (>50 million years) so that a small fraction of the isotope could conceivably have survived since the creation of the Earth 4.6 billion years ago, and may be of interest in geological timescales. There are also isotopes with smaller half-lives, but which are continuously replenished by the decay of heavier atoms or by the action of cosmic rays, so they can still be found on Earth today. In all, about 339 different isotopes can be found naturally on our planet.
However, as many as 3100-3300 different isotopes of the first 118 elements are claimed to have been detected in laboratories, most of them with very small half-lives (seconds or less). Nobody knows for sure how many elements and how many isotopes can exist, though the number is finite.

Would a combination of other singly-positively-charged particles with an electron (positronium, muonium, bottomonium, etc.) be considered an isotope of hydrogen? Their chemistry should mimic that of hydrogen.

@iad22agp Good question. According to IUPAC, isotope is by definition a term applicable strictly to differing combinations of protons and neutrons. Anything not made solely of protons, neutrons and electrons is not an atom, but rather an exotic atom, of which there are several types depending on what gets replaced.

Helium, for one has isotopes from no neutrons up to eight.  The former limit may have been observed experimentally and is an intermediate in proton-proton fusion.


text:
Salt analysis: Position of lead ion
Why is $\ce{Pb^2+}$ present both in Group I and II of the basic radicals?
Is it related to the solubity product of its salts?

Yes, the Ksp for $\ce{PbCl2}$ is ~$10^{-5}$ while for $\ce{PbS}$ is ~$10^{-28}$
So if the concentration of $\ce{Pb^2+}$ is too low to detect by dilute $\ce{HCl}$ precipitation, it is still possible it could be detected as the sulfide precipitate.


text:
4-Bromo, 2,4-di-methylhexane and 3-Bromo, 3,5-di-methylhexane are the same thing?
Came across the following question. The answer only says that option 3 is incorrect which is I agree to but dont you think option 4 is also incorrect. As per IUPAC rules isnt it the Halide that should get lowest possible number (in this case number 3 instead of 4) or is it both are same ? I am confused please help.

Both names, ‘4-bromo-2,4-dimethylhexane’ and ‘3-bromo-3,5-dimethylhexane’ are unambiguous and describe the same compound. However, only ‘4-bromo-2,4-dimethylhexane’ is in accordance with IUPAC recommendations.
First of all, the seniority order of functional classes, which is used to choose the functional class (here: …hexane), does not apply to numbering of locants within the same compound class.
Concerning numbering of locants within the same compound class, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(…)

Furthermore,

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with
other locant sets, each cited in order of increasing value, has the lowest term at the first
point of difference; (…)

Therefore, the example is correctly named as 4-bromo-2,4-dimethylhexane rather than 3-bromo-3,5-dimethylhexane since the locant set ‘2,4,4’ is lower than ‘3,3,5’.

According to the IUPAC organic chemistry nomenclature, alkyl groups take precedence over halide groups so (4) is correct.
Similarly, C=C double bonds also take precedence over halides so (3) is incorrect and should be 2-bromoprop-1-ene.
(2) is also incorrect as there are only five carbons in the chain not six so the correct name is 4-oxopentanoic acid.
A table of functional group priorities can be found here.

But if we follow that then 2-Bromo- 3-methyl butane never exists. But it pretty much does exist.

2-bromo-3-methyl butane exists but it is not the correct IUPAC name for that molecule. the correct name is 3-bromo-2-methyl butane

The additional comment is not correct. According to rule P-14.4 (g) in the current IUPAC Blue Book (2013), the systematic name is 2-bromo-3-methylbutane (not 3-bromo-2-methylbutane).

Nice catch on the oxopentanoic acid!

Acc. to IUPAC we have to give lowest locant no. to side chain which in this case is 2-Bromo-2,4-dimethyl.
Another reason of this ans. being correct is that we take halides as a normal side chain and it is not given priority over methyl, we just only see lowest locant in this case.


text:
Why do the two methylene protons in ethylbenzene have a chemical shift at 2-2.9?
The data sheet does not have any chemical shifts in this range for compounds containing only carbon and hydrogen

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Protons attached to a carbon that is connected to a benzene ring typically resonate in the range 2.2-2.9 ppm (see the Ar-CH range in this table).
Here is the proton nmr spectrum for ethylbenzene.  The methylene protons absorb around 2.7 ppm, right in the upper end of the expected range.

image source

Ok I believed Ar was for argon not Aromatic

Organic chemists use a number of abbreviations that look like element symbols. For example, $\ce{Ar}$ is aryl and $\ce{Ac}$ is acetyl, even though those symbols also stand for argon and actinium. We feel like we can get away with this because there are currently no known compounds with carbon-argon and carbon-actinium bonds.


text:
Molecular Assembly Program
I'm not totally sure which Stack Exchange community this would go in, but Chemistry seemed best.
I've been thinking for a long time now, that it would be cool to have a program where you could design your own atomic models, from the sub-atomic scale and up, then test them in a real situation. You could create anything from simple molecular structures to...anything.
I've looked around online, but there aren't any programs that you can build your own atomic structure. Anyone know if there are any out there like this?
EDIT: I've tried a few other programs, including Avogadro, this, and another one by PHET.

"...your own atomic models..."; do you mean "hypothetical molecules"?

What do you mean by 'hypothetical'?

I mean molecules that don't exist, but since we love logo, we make them up. :)

@MARamezani I mean a program where you could, maybe with a drag & drop interface, create your own models, molecular structures, etc., then put them into a real situation and see how they act. This could maybe be used to create new medicines, simple viruses that could go in and kill a disease, or even new species. Does that help?

William, you might want to have a look at this wikipedia article on docking.

This article in BMC Bioinformatics might be helpful too, this gives some more general information on computer-assisted drug design.

"Put them into a real situation and see how they act." Oh, such thing is, I'm afraid, not out of the realm of imagination. It's really hard to make such a program: 1. "Real situations" can be aqueous, basic, acidic, etc. 2. There are oceans of molecules. If we try to know how they behave, we should write a program that has them all, literally.

@WilliamBennett As far as building small molecules, that can be done and is regularly done. Where you run into problems is modeling the "real situation". You mention drug discovery. Modeling the interactions of a small molecule in a biological system is a huge field of research. This can be helpful, but it is certainly not as simple as it appears to be. Other applications may have better models.

@WilliamBennett As far as "building your own atomic structure," there are many "molecular editors." I'm partial to mine, Avogadro (http://avogadro.cc/) but there are others. As the comments here indicate, the simulation side is more challenging.

This is a problem beyond the limits of current technology.

You  might want to try BioVia's (formerly Accelrys) Discovery Studio Visualizer. This is free software and the "gateway drug" to their sophisticated suite of molecular modeling and chem/bio informatics software. It is a step above the various drawing programs out there and lets the user generate realistic 3-D models of an infinite variety of chemical structures. http://accelrys.com/products/discovery-studio/visualization-download.php

[…] create your own models, molecular structures, etc., then put them into a real situation and see how they act. This could maybe be used to create new medicines […]

Imagine a huge protein that switches a malfunction (on or off) to be like a lock, while small organic compounds may be seen as possible keys.
You could either

synthesize a huge library of potential keys and test them in the lab or
build a library of keys on the computer and calculate their geometries, charge distribution, etc. using quantumchemical methods and then again use other programmes to examine the interaction between the keys and the lock.

The latter method, figuring out how good the keys fit, is known as docking.
You might want to have a look at the CADDSuite, developed by Oliver Kohlbacher at the University of Tübingen, but please note that these tools tend to be a tad above the entry level of simple moelcule editors.
For an online service, have a look at SwissDock.


text:
is the adsorption of resveratrol within wine or grapes linked with the time it was kept in your mouth?
This question is about resveratrol in wine or grapes and not in some special delivery system.
If one is drinking some wine or eating grapes, does it matter if they keep it in their mouth a bit so the resveratrol within the wine or grapes get absorbed better?
I know now all nutrients can be gotten into onces body via buccal method so I am asking the experts what is the case with resveratrol from wine or grapes.

This is a good question!  I suspect that buccal delivery depends on the molecule in question being membrane permeable, or else on the creation of microlesions that give direct access to the bloodstream.  Wine doesn't create microlesions so membrane permeability is probably the issue.  You could look up the octanol-water partition coefficient of resveratrol to get an idea of its membrane permeability.

This article on the bioavailability or resveratrol seems to partly support your assumption: Keeping it in the mouth for a bit longer rather than swallowing directly might help to facilitate the uptake.
However, the solubility of resveratrol in water is apparently as low as 30 mg/L.


text:
Determine if the following change is an oxidation, a reduction, or neither

The following reaction is not a complete reaction. Determine if the change is an oxidation, a reduction, or neither:
  $$\ce{CrO4^{-2} -> Cr2O7^{-2}}$$

I know that the answer to this problem is neither, but I don't have any idea as to how that answer is reached. Could anyone give me a hint? I am familiar with oxidation, reduction, and oxidation numbers.

If you are familiar with oxidation numbers, what are the oxidation numbers of the chromium atoms in each compound?

@jerepierre in the first compound +6, in the second compound +6 as well.

@jerepierre Does this mean that since there was no change in oxidation numbers, it is neither?

Find the oxidation numbers of the substances in each compound.
You should find the oxidation number of Cr in each compound to be +6, and the oxidation number of O to be -2. Since there is no change in the oxidation numbers, this reaction is neither an oxidation or reduction.

As pointed out in the comments and the other answer, it is neither an oxidation, nor a reduction.
Q: So, what is it?
A: It is a condensation: two molecules of an ortho acid combine under loss of one molecule of water to yield a diacid (pyro acid).
This is frequently found in oxoacids, see

$\ce{H3PO4}$ (phosphoric acid) and $\ce{H4P2O7}$ (diphosphoric acid)
$\ce{H3AsO4}$ (arsenic acid) and $\ce{H4AsO7}$ (diarsenic acid)
$\ce{H2SO4}$ (sulfuric acid) and $\ce{H2S2O7}$ (disulfuric acid)

Two ways you can go about this.
One, recall that the oxidation number of oxygen is always -2 beyond hydrogen peroxide and the elemental forms of oxygen (monoatomic, diatomic, and ozone). Then you can show that the oxidation number for chromium is the same.
If you want to be clever (though not as efficient or just want another way to check yourself), you can attempt to balance the half-reaction like a half-reaction occurring in acidic solution. Why does that attempt show that no oxidation/reduction exists?


text:
What is the difference between K and Kp in the equilibrium equation?
In this equilibrium equation, $$K = K_p\cdot (RT)^{-\Delta n}$$ what does $K$ represent in comparison to $K_p$? It seems to me that they both are equilibrium constants, so how are they different?

Is Kp just reaction quotient possibly?

The equilibrium constant is defined by the expression
$$K_x=\prod_{\ce{B}} x_{\ce{B}}^{\nu_{\ce{B}}}.\tag1$$
For $x$ you can substitute a number of quantities, most commonly when working with solutions is (amount) concentration $c$ and when working with gasses often the partial pressure $p$ is used. In the latter case the equilibrium constants can be related via the ideal gas law $$pV = n\mathcal{R}T.\tag2$$
Substituting this into $(1)$ gives you
$$K_p=\prod_{\ce{B}} \left(\frac{n_{\ce{B}}}{V}\mathcal{R}T\right)^{\nu_{\ce{B}}}.\tag3$$
This can be transformed into
$$K_p=\prod_{\ce{B}} \left(\frac{n_{\ce{B}}}{V}\right)^{\nu_{\ce{B}}} \prod_{\ce{B}}\left(\mathcal{R}T\right)^{\nu_{\ce{B}}}.\tag{3a}$$
We know that the concentration is defined as
$$c_{\ce{B}}=\frac{n_{\ce{B}}}{V}\tag4$$
and from $(1)$ we can set up the concentration based equilibrium constant
$$K_c=\prod_{\ce{B}} c_{\ce{B}}^{\nu_{\ce{B}}}
     =\prod_{\ce{B}} \left(\frac{n_{\ce{B}}}{V}\right)^{\nu_{\ce{B}}}.\tag5$$
Substituting $5$ in $3$ we find
$$K_p=K_c\prod_{\ce{B}}\left(\mathcal{R}T\right)^{\nu_{\ce{B}}}.\tag{6}$$
From the associativity of multiplication it follows therefore
$$K_p=K_c\cdot\left(\mathcal{R}T\right)^{\sum_\ce{B}\nu_{\ce{B}}}.\tag{7}$$
This is often abbreviated with $K=K_c$ and $\sum_\ce{B}\nu_{\ce{B}}=\Delta n$ and therefore
$$K = K_p\cdot (\mathcal{R}T)^{-\Delta n}.\tag{7a}$$

Associative law of multiplications for exponents:
\begin{align}
  x^{n+m} &= x^n\cdot x^m\\
  x^{\sum d} &= \prod_d x^d
\end{align}

K is for concentrations in Molarity. Kp is for what you're using pressures for reactions with all gases.


text:
How do you calculate a unit cell's dimensions from an XRD spectra?
My inorganic lab had us do an XRD measurement, but I've never been explained how to interpret the data.
Question:

Calculate the unit cell dimensions $a$, $b$, and $c$, for $\ce{YBa2Cu3O7}$ from the indexed X-ray powder pattern provided in [the textbook]. Explain why the crystals are nearly tetragonal in terms of the atomic structure of the compound. The following formula is useful:
$$\sin^2{\theta} = \frac{\lambda^2}{4}\left(\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}\right)$$
  The three numbers, $hkl$, called Miller indices, indicate the direction of the scattering plane. The Miller indices can be read from the diffractogram provided. For simple reflections of the type $h00$, $0k0$, and $00l$, the value of $h$, $k$, or $l$ corresponds to n in Bragg's law.

Bragg's law:
$$n\lambda = 2d\sin{\theta}$$
The spectrum:

  (This is a scan from my textbook, while working on it I circled some items. Just ignore that.)

My Attempt:
$$\lambda = 0.154~\mathrm{nm}$$
Scouring the internet I found the formula for tetragonal atomic structure: the axes $a = b \neq c$, and the axis angles $\alpha = \beta = \gamma = 90^\circ$.
Solving for $c$:
$$\begin{align}
\sin^2{(22.9^\circ)} &= \frac{\lambda^2}{4} \left(\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}\right) \\
&=\frac{0.154^2}{4} \left(\frac{0^2}{a^2}+\frac{0^2}{b^2}+\frac{3^2}{c^2}\right) \\
0.151 &= 0.0237\left(\frac{3^2}{c^2}\right) \\
c &= 1.188
\end{align}$$
For $a$ and $b$:
$$\begin{align}
\sin^2{(22.9^\circ)} &= \frac{\lambda^2}{4} \left(\frac{h^2}{a^2} + \frac{k^2}{b^2} + \frac{l^2}{c^2}\right) \\
&= \frac{0.154^2}{4} \left(\frac{1^2}{a^2} + \frac{0^2}{b^2} + \frac{0^2}{c^2} \right) \\
0.151 &= 0.0237 \left(\frac{1^2}{a^2}\right) \\
a &= 0.156
\end{align}$$
Is this the right approach? If I have the right answer, what should the units be?

You are doing it right. For the units since you are only using nm in the calculation (via the lambda), your results will be in nm too. Howerer, crystallographs are used to express cell parameters in Angstrœm (0,1nm) although it's not an official unit.

your are doing it right!

@john sorry for this late comment but if you still want to check whether your calculated unit cell values are correct i have found this small software useful

Diffraction is not a spectroscopic technique, so you don't obtain a spectrum, but rather a pattern, as correctly repotred in the caption. Moreover, indexed Bragg reflections should be indicated without parentheses (as wrongly indicated in the figure); a generic (hkl) notation is used for indexing a crystal face or a single
net plane.

You should not really have assumed that the compound was tetragonal. If it was indeed tetragonal, then the peaks corresponding to the $(200)$ and $(020)$ planes would overlap. Instead, you can see that one occurs at $2\theta = 46.7^\circ$ and the other at $2\theta = 47.6^\circ$. The question may be considered a little misleading, but it does hint at this; it says that the compound is "nearly tetragonal", and it gives you the general formula for an orthorhombic crystal, which is one with $a \neq b \neq c \,(\neq a)$ and $\alpha = \beta = \gamma = 90^\circ$.
The next problem is that you are given values of $2\theta$ (which happens to be the convention), but the formula reads $\sin^2{\theta}$. As such, you need to make the appropriate adjustments.
This question is relatively easy because the peaks have already been indexed for you (i.e. the values of $h$, $k$, and $l$ are already found out). Therefore, all that is left is to plug the numbers into the formula you are given. Which peaks you use is entirely up to you, although it is obviously easier to use $(h00)$, $(0k0)$ and $(00l)$ peaks.
The reflections of the $(200)$ plane occur at $2\theta = 47.6^\circ$. So:
$$\begin{align}
\theta &= 23.8^\circ \\
\sin^2{(23.8^\circ)} &= \frac{(1.54~Å)^2}{4} \cdot \frac{2^2}{a^2} \\
a &= 3.81~Å
\end{align}$$
Similarly, you can show using the $(020)$ and $(003)$ peaks that $b = 3.89~Å$ and $c = 11.64~Å$. This is actually pretty good. The reported values of $a$, $b$ and $c$ are $3.8231~Å$, $3.8864~Å$, and $11.6807~Å$ respectively.
Regarding units, if you make sure that you leave your units inside the intermediate workings (as they should be), you should never run into confusion about "what unit should the answer be". In your question, you converted $\lambda$ into units of $\mathrm{nm}$. $\sin^2\theta$, $h$, $k$, and $l$ are all dimensionless. Therefore, your calculated values of $a$, $b$, and $c$ must match the units of $\lambda$ you used, i.e. $\mathrm{nm}$.
In my answer, I used ångströms throughout. It is not wrong to use $\mathrm{nm}$, but as Babounet noted in the comment section, ångströms are more commonly used in crystallography.

A great explanation. May I ask - can I assume that whenever a diffraction peak (no matter the atomic structure, e.g. tetrahedral, hexagonal etc.) shifts to lower degrees $2\theta$, it means that the cell size has actually decreased as lower Bragg $2\theta$ angle corresponds to lower value of $\sin^2(2\theta)$? Thank you!

Note to my previous post - there was a mistake: it means that the cell size has actually expanded...

@JoshE., I'm afraid that in the last seven years, I've forgotten everything I ever knew about diffraction. You might want to consider asking a new question.


text:
How to calculate the concentration of NO at equilibrium?

$\ce{N2(g) + O2(g) <=> 2NO(g)}$, $K_c = 4.1 \cdot 10^{-4} (T=2000~^\circ\mathrm{C})$
  What is $\ce{[NO]}$ when a mixture of $0.20~\mathrm{mol}$ $\ce{N2(g)}$ and $0.15~\mathrm{mol}$ $\ce{O2(g)}$ reach equilibrium in a $1.0~\mathrm{L}$ container at $T=2000~^\circ\mathrm{C}$?

When K is $\cdot10^{-4}$, you can assume the change is negligible, and just using the concentrations you can get from the given mole amount and the container volume, I get the right answer. When plugging back into, to see if it matches the give $K_c$, it's right:
$$((0.20~\mathrm{M}~\ce{N2})(0.15~\mathrm{M}~\ce{O2}) \cdot 4.1\cdot10^{-4} K_c) = x^2.$$ Ans: $\ce{[NO]} = 0.0035~\mathrm{M}$.
From what I've read and heard you should be able to do it 'the hard' way, using the quadratic formula, like you would if $K_c$ wasn't to $10^{-4}$, and small enough. But I can't seem to get the right answer trying to do it that way.
Rearranging everything, the equation I get to plug into the quadratic formula is
$$x^2 - 0.35x + 0.03,$$ then I get $0.2$ and $0.15$ as the answers. $0.2$ is too high, subtracting it from the $0.15~\mathrm{M}$ $\ce{O2}$ initial wouldn't be right. So using the $4.15$ as $x$, the change in $\mathrm{M}$, and subtraction that from the initial concentrations I have, then plugging that into the equation for $K_c$, even tried rearranging it in case I have got something mixed up, but I don't get anything close to the given $K_c$.

In your quadratic formula, something is missing...

You mean what I have with X^2, 0.35, and 0.03 is wrong, or after that something I'm doing with the quad. formula is wrong? Or?

It is a term not an equation, the = is missing.

I know the =0 at the end is part of that, but but what I have posted above still gives A, B, and C to plug into the quadratic formula, which is what I'm concerned with.

$K_c = \frac{[\ce{NO}]^2}{[\ce{N2}][\ce{O2}]}$
Let x = [NO]
$\frac{x^2}{(0.20M-0.5x)(0.15M-0.5x)} = 0.00041$
$x^2 = 0.00041(0.25x^2 -0.175xM + 0.030M^2)$
$x^2 + 0.000072xM -0.0000123M^2 = 0$


text:
How can iron(III) sulfate be formed?
What is the salt formation for iron(III) sulfate?
Wikipedia tells me that the production of iron(III) sulfate is:
$$\ce{2 FeSO4 + H2SO4 + H2O2 → Fe2(SO4)3 + 2 H2O}$$
I was wondering if there was a difference between salt formation equation and production, because I thought it would've went something more like:
$$\ce{2 Fe(OH)3 + 3 H2SO4 → Fe2(SO4)3 + 6 H2O}$$

I was wondering if there was a difference between salt formation equation and production.

Absolutely! In this case, the latter means that you're trying to turn a side product (waste) of a technical product into another marketable product.
In order to understand, why the oxidation of $\ce{FeSO4}$ is interesting, you have to realize that it is a side product in the synthesis of titanium dioxide ($\ce{TiO2}$). Here, ilmenite ($\ce{FeTiO3}$) is processed with concentrated sulfuric acid to yield $\ce{TiOSO4}$ as the valuable intermediate and $\ce{FeSO4}$ as "waste".


text:
How to understand the concept of moles?
I have trouble relating to the mole concept. For example in a question like the following:

How many moles of $\ce{K}$ are in 1 mole of $\ce{KOH}$?

How is it that in 1 mol of $\ce{KOH}$ there is 1 mol of $\ce{K}$ and x moles of $\ce{OH}$? How is it that 1 mol of $\ce{KOH}$ has 6.02×1023 atoms, and then $\ce{K}$ has 6.02×1023 atoms as well. In addition, $\ce{OH}$ will have 6.02×1023 atoms, but if we add these two it's greater than the original number of atoms in 1 mol of $\ce{KOH}$?

1 Mol of KOH has 36.0210^23 atoms, because there are three atoms in KOH.

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hmm, ok so if I'm looking at say 1 mol of H2SO4, then the total atoms would be 36.0210^23 as well? or would it be *7 instead of 3? | why is it that people say 1 mol = 6.0210^23 if actually the KOH would contain 3 * as many total atoms in 1 mol?

I had the same problem as well, until I sketched out the actual molecules.
When you have one mole of $\ce{KOH}$, you have one mole of the collective "molecule".
If you were to break it down into $\ce{K}$ and $\ce{OH}$, you would get 1 mole of $\ce{K}$, and 1 mole of $\ce{OH}$. This is because you need 1 mole of each to make 1 mole of $\ce{KOH}$.
It's kind of like if you have one hundred keyboards, you would get one hundred 'Enter' keys, one hundred 'Up arrow key' and so on...

ah ok, an interesting analogy; so when someone says i have 1 mol of KOH, it's not actually 6.0210^23 total but 3avg constant, and each atom is 1 mol = 6.02*10^23 combining towards the total?

The concept of mole is just a number. When we say we have 1 mol of KOH, it means exactly that, there are 1 mole of KOH, not 1 mole of atoms. (100 keyboards, not 100 keys).

Ok, this is starting to make a bit more sense.
1 mol KOH = 1 mol K 1 mol OH (1 keyboard, 1 f key, 1 g key)
1 mol H2SO4 = 2 mol H 1 Mol SO4 (3 moles of ions)
(1 custom keyboard = 2 space bars, 1 escape key)
I could then take 2 mol of H and say there will be 2*(6.02*10^23) H atoms, etc. correct?
If so, thank you very much!

I think you've got it! For the H2SO4 example, you would have 2 mol of H, and 1 mol of the ion SO4, which you could count as 1 mol of S, and 4 mol of O. Just understand that the concept of mole is no different then the idea of "dozen". It is just a unit of measurement.

One mole of $\ce{KOH}$ means that there are Avogadro number of (6.023 × 10²³ molecules) $\ce{KOH}$ molecules and we need not consider the sum of number of constituent atoms.
For example,
1 mole of argon contains 6.023 × 10²³ argon atoms.
1 mole of ozone contains 6.023 × 10²³ ozone molecules (and also consists of 3 moles of oxygen atoms).

Fantastic. This is all coming together now and I have a general idea of it all; enough to stop questioning it so much.


text:
How to I identify the diene and dienophile through retrosynthesis of a given compound?
I'm stuck while finding what to use to make this compound:

My first attempt was to use a triple bond as dienophile and a ring as my diene, as shown below -

But I'm not sure if this is right.

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Also your product lacks some charges or protons.

Thats the way it was given to me by a friend but yeah there would be a -2 on the co3 It would be trivial if the bottom double bond was on the bond that connects the two rings but using the diels-alder mechanism I can't really see how to get that structure. Why guess so far is I need a triple bond dieophile and a ring as my diene

It would be great if you could would your retrosynthesis (the one from your comment) into your question! It would be a waste if the question would get closed despite the fact that you have definitely thought on it and came to a reasonable approach!

The compound as drawn is impossible. You're alluding to carbonate [CO3(2-)] in the comment, but that cannot be attached to the structure. Draw it out completely to convince yourself. One mistake in a structure makes me skeptical about the accuracy of the rest.

The $\ce{CO3}$ substituents are questionable. Let's assume that this is $\ce{CO2Me}$ instead!
Since this is homework, I won't provide a full solution but:

Note that double bond patterns may isomerize (in the direction of conjugation).
Alkynes can serve as dienophiles too!

   This was my first attempt is this on the right track? Not quite right yet.

@Loststudent22 Diels-Alder reaction looks fine to me. Remember #1 in my answer. What about adding a strong, bulky base and subsequent quenching?

I kind of see what your saying, so the reaction couldn't be done in one concerted mechanism?

@Loststudent22 I just don't see that, but maybe I'm missing something and I'm preoccupied with the same retrosynthesis that you had in mind too.


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Volatile nature due to hydrogen bonding
Why does intramolecular hydrogen bonding make an organic (or possibly inorganic as well(?)) compound volatile.
What I think is that it might be due to decrease in the solubility of the compound as intermolecular hydrogen bonding cannot be done after that.

possible duplicate of Intermolecular Hydrogen Bonding

@IlmariKaronen No my q is  related to volatility but the one u reffered here.

"Volatile" usually refers to ease of evaporation, high vapor pressure, so it is a property of a pure substance while solubility is a property of the combination of two or more substances.
An intramolecular hydrogen bond makes a compound more volatile because the charge imbalances are offset internally instead of by forming an interaction with another molecule.
Interactions between different molecules of the substance would lower volatility, but intramolecular hydrogen bonds prevent the hydrogen bond donor and acceptor of one molecule from interacting with another molecule.

Charge imbalances are offset internally ?? Plz elaborate.

The group that is the H-bond donor, for example an O atom, would have a partial negative charge, that is offset or partially offset by the intramolecular hydrogen bond.

Related: how is charge balance being offset affect volatility/evaporation?

If partial charges in the molecule don't result in a dipole moment, the only interaction between like molecules will be very weak London focres


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Amplitudes and phases of transform of an image
I have recorded an image of a large macromolecule and after several types of image analysis, have obtained a rotationally averaged image, that is not quite perfect, but appears to show 6-fold symmetry. With the origin on the axis of rotation, I compute a transform and unlike most of my colleagues, if I actually look at the amplitudes and phases of the transform. I have only one image of this molecule and no more specimen left. So my questions are:

What feature(s) I should see in the transform’s amplitudes and phases?
If the amplitudes and phases do not conform exactly to my expectations, what would that mean?
What are the factors that degrade my image?
What should I do to make the transform conform to my expectations?

Any ideas are welcome. Thanks in advance!

Can you describe more about the chemistry behind your data? What technique did you use to acquire data?  X-rays or something? I haven't been around a long time, but I don't think chem.se gets referred many questions from photo.se.


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Why is the specific heat of water high?
How is the specific heat of water so high?
I was reading the hyperphysics article on it but could not understand it properly.

Higher specific heat of water is due to Presence of intermolecular hydrogen bonding in it The amout of heat supplied is used initially to break the bonding Due to this reason water has high specific heat capacity

Water has hydrogen bonding in it. Hydrogen bonding is some kind of intermolecular force (a tutorial and the wikipedia page) that is usually seen in molecules that have $\ce{OH}$, $\ce{NH}$ or $\ce{FH}$ somewhere in their structure.
How does it happen?
Hydrogen atom is really small (atomic radius: About 37 pm) When it bonds with some very electronegative species, like O atom, (in the case of water), the electron is highly absorbed by the more electronegative. That results in a relatively high concentration of positive charge. The other O atom in water is attracted to H and this is the basis of hydrogen bonding.
So...?
Hydrogen bonding in water is far too weak to be compared with intramolecular bonds like covalent or ionic bonds, but it's strong enough that to be broken, it will require lots of energy. That's why water boils at $100 ° C$.
Many know the reason of water's high specific heat to be hydrogen bonding. Partially, it's related to that: In giving heat to water, some heat is "spent" on loosening the hydrogen bonds rather than increasing water's kinetic energy. The argument is valid, and I haven't seen anyone bringing evidence as to its being wrong, but another reason, which was the one hyperphysics gave, relies on the motion of water.
Expanding the topic isn't necessary, and it will add up to confusion. In short, the water molecule is nonlinear. That will result in three rotational degrees of freedom that allows it to have an awful lots of rotational motions! (More types of motions are a cause of an additional bonus to the specific heat)
Liquid ammonia has a higher specific heat, just for the same reason.
Even a very easier reason to understand is the water's relatively low molar mass: $\frac{18 g}{mol}$. What should that mean? More moles of water per one kilogram.
As a summary, the reasons are:

The heated water will contribute much of the heat to loosening, bending or breaking the hydrogen bonds.
Water had three rotational degrees of freedom. In addition to vibration, rotations happen a lot to water molecules. This will result in a higher heat capacity.
Specific heat capacity is defined as the amount of heat required per unit mass to increase the temperature by a degrees celsius. The relatively low molar mass of water allows more moles of it to be there in a mass unit (either kg or g)

As a side note, the exceptional high specific heat is not the only weird property of water. See here for example.

Generally, an increase in the absorbtion of heat energy will result in an increase in the kinetic energy of the molecules and indeed, the individual atoms. The same can be said about water. However, since water has constantly forming H bonds, some of the heat energy will be used to break the water bonds. The more bonds that need to be broken and kept apart (not easy in water) the more heat the water can absorb. When the outside temperature cools down, then the bonds broken by the heat will reform and release energy into the surroundings.

But liquid ammonia's specific heat is higher than water. Are you suggesting that is merely because H-bonding is stronger in ammonia?