text:
How do zeolites catalyse the cracking of hydrocarbons?
Been learning about catalytic cracking of long chain hydrocarbons into shorter chain hydrocarbons, in which the catalyst zeolite is used so that the temperatures needed for the cracking are much lower (about $500\,^\circ\text{C}$ instead of $700\text{-}1000\,^\circ\text{C}$). The question is, how exactly does the zeolite help with the cracking process? Everywhere I look doesn't give a detailed answer, just that zeolite is porous, riddled with holes that can trap molecules. The only clue I have to the process I suppose is this video
https://www.youtube.com/watch?v=nTse5_wXup4
which shows an animation of some hydrocarbons whacking against a tunnel-like zeolite into smaller chain carbons. But to me the video can't be that accurate, because the whole thing seems out of proportion, the hydrocarbons are made from hydrogen and carbon which are pretty small atoms, whereas the zeolite is made from aluminium or silicon which are much larger. Even if this is the actual way it works, why not just have a plane surface to hit the hydrocarbons off to break them then? (Perhaps the rate of collision is much smaller? I'm just pulling at straws here.) Any light shed will be helpful, thanks.

The video you reference to does not tell the whole story.  Yes, zeolites are minerals with open pores, similar to a filter, occurring in nature and are produced commercially, too. Equally to their structure, their composition is interesting -- as both are used to tailor the reactivity and selectivity of this material.  ZSM-5 is one example of designed remarkably high Brønsted acidity.  Hydrocarbons in contact with these sites split into fragments and rearrange because of the intermediate generation of carbocations.  It is advantageous, as these reactions are heterogeneously catalysed by these "solid state acids".


text:
Why do double bonds break
So today we performed an experiment, where we mixed cyclohexene with Bromine water. The equation was

which seems to make sense. The Bromine water we were using was decoloured as evidence that this equation took place, whereas when we mixed cyclohexane with Bromine water nothing happened (to show how double bonds are more 'reactive'). But what I don't understand is why the double bond was broken. I mean, wasn't it happy just being the way it was? Doesn't it still take energy to break the double bond, even if it is less than the energy required to break a single bond? It isn't like it WANTS to take the bromine? (or is it?) The only place where this energy could have come from in my mind is either the vibrational energy due to temperature, or else the kinetic energy from pouring it in? (may be taking too much of a physics approach). If so, doesn't that mean that if I got the cyclohexane, and poured it in from a great height, will it react with the Bromine water? Or is the energy difference between double bonds and single bonds so great that it can only perform under special circumstances? (i.e. under UV light)

BTW, I've never heard about Br-O-H. When I look up "bromine water," I get descriptions of $\ce{Br2}$ in $\ce{H2O}$ not Br-O-H. Are you sure Br-O-H is reacting?

A second BTW: what you have drawn is a mono-organylhydroxyboronium ion (it’s lacking a third bonding partner). Bromine is written with a lowercase r, else it becomes organic-residue + boron.

Keep in mind, that the reason reactions occur spontaneously is because there's an overall driving force: $\Delta G = \Delta H - T\Delta S$. Since there are fewer molecules on the right hand, we can guess that $\Delta S$ isn't driving this reaction. ($G$ is Gibbs free energy, $S$ is entropy, $H$ is enthalpy, and $T$ is temperature.)
So then your question boils down to "why is this reaction favorable for $\Delta H$?"
Even before I look up the bond enthalpies for each of the bonds involved, I have an intuition. Yes, we're breaking the double bond in cyclohexene. We're also breaking the $\ce{Br-O}$ bond in $\ce{Br-O-H}$. So that's two bonds that are broken.
You're right -- that costs energy, so there's clearly something more energetically favorable on the other side. Well, in the product, I'm forming a $\ce{C-Br}$ and a $\ce{C-OH}$ bond.
So the net reaction breaks two bonds, and forms two bonds. As you say, the general intuition is that the "double" part of a double bond isn't quite as strong as the standard $\sigma$ single bond.
Let's look up the bond enthalpies also through Wikipedia:

$\ce{C=C}$ (vs. $\ce{C-C}$ which remains in the product): ~146 kcal/mol vs. ~84 kcal/mol = ~62 kcal/mol
$\ce{Br-O}$ = ~56 kcal/mol (I had to look this one up, so it's a bit iffy)
$\ce{C-Br}$ = ~69 kcal/mol
$\ce{C-O}$ = ~85 kcal/mol

Net $\Delta H \approx$ -36 kcal/mol.
Even if some of the bond enthalpies are off a bit, it's clearly energetically favorable. It has nothing to do with how the species are mixed, or really the temperature. (Well, my guess is that $\Delta S$ is not favorable here, since two molecules form one, so high temperature might make the reaction less favorable from a $\Delta G$ standpoint.) Nor do you need UV light or anything like that -- the difference in bond strengths will push this reaction.

Could you please clarify the symbols? I'm guessing G is change in energy, S is entropy, H is enthalpy, and T is temperature?

Yes, that's correct.

Forget about energy due to pouring in from a great height or kinetic energy of mixing, it is not important, all it does is minutely heat the solution.
The thermodynamic answer from @Geoff Hutchinson is a fine explanation but you can look at it in another more molecular way if you want to.
In any reaction there is a potential energy barrier between reactants and products otherwise most stuff would just react instantly and we would not be here :( . The barrier is overcome only slowly in many reactions and this is because it turns out that most barriers are far greater in energy than thermal energy (average 3RT/2 kJ/mol). When the reactant molecules collide in solution most of the times that this happens they just bounce apart again without reacting. Just occasionally by random collisions with the solvent the reactants have enough energy $^*$to surmount the barrier and produce products (your 'vibrational energy due to temperature'). If the products are at lower energy (as in your example) the barrier back to reactants is even larger than the barrier forward to produce products and so the product is produced in large amounts. If the barrier from products to reactants is small then reaction favours the reactants and little product is formed.
($^*$ I'm assuming, for simplicity, that the molecules orientations are unimportant.)

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text:
What makes something biodegradable?
Learning about biodegradable polymers and non-biodegradable polymers. What exactly sets them apart though? Here's the formula from wikipedia for the biodegradable PHB:

and here for polypropylene, it's non-biodegrable counterpart

As far as I understand, biodegradable just means that bacteria will degrade it into other constituent parts. Is it just that bacteria is a picky eater and won't eat certain polymers? But they seem to be made of similar elements, and the bonds between are generally similar. Any help is helpful.

The former contains oxygen and the bonds are not very similar at all. The polar nature of PHB suggests that it is more reactive, at least in a biochemical sense.

The difference between biodegradable polymers and 'regular' polymers is the bonds that make up the backbone. The backbone is the part of the polymer that extents the brackets. Anything attached to that chain is a side group and is less important.
The two examples you show here are very dissimilar. The polypropylene is only made up of Carbon-Carbon bonds. These bonds are very stable: that holds for chemical and 'biological' stability. The PHB how ever bonds monomers with an Ester functional group (Ester). This group is far more reactive that a C-C bond.
The reactivity of the groups the polymer is made up of, is very, very important. That being said, I cannot claim that all poly-esters are biodegradable. Depending on the groups close to the esters and the enzymes (probably from bacteria) involved, the bond might be cleaved, or might be inaccessible.
Hope this helped!

It is also useful to see that terms like "biodegradable" are not chemical or biological terms which are quantifiable. They are terms used to describe the relation of a chemical in the context of the world. Often "non-biodegradable" plastics could be broken down, it is just that these species are in relatively low frequencies.

An all-carbon backbone is generally less biodegradable, but what really makes polymers badly degradable are high crystallinity and/or hydrophobicity and/or high glass transition temperature. Wood for example is highly crystalline, and can only be degraded by a few specialists (which have had half a billion years to evolve), unless some small animal gives mechanical help.
Same with many polyesters, polyamides and other fiber-forming polymers. They hardly swell in cold water, and so any enzyme would have to break the crystalline order and the backbone bonds at the same time. That kind of energy is usually not available in cold water.
Polystyrene is hydrophobic and has a high Tg, sunlight, oxygen and water have to make the surface a bit more polar chemically before anything biological can attack. The good thing is PS is most often foamed and has a huge surface, plus it is so brittle that it cracks easily, further increasing the surface.

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text:
Why concentrated sulfuric acid for dehydration and dilute for hydration?
To dehydrate ethanol one uses concentrated sulfuric acid:
$$\ce{C2H5OH ->[\text{conc.} H2SO4] C2H4 + H2O}$$
but to go in the reverse direction, dilute sulfuric acid is used:
$$\ce{C2H4 + H2O ->[\text{dil.} H2SO4] C2H5OH}$$
Why is one concentrated, and one dilute? For that matter, how does the sulfuric acid even help in the first place?

Dilute sulfuric acid implies the presence of a lot of water ... and dehydration conditions call for the removal of water. Having water around when one is trying to remove water doesn't help with the removal of water.

"For that matter, how does the sulfuric acid even help in the first place?" Dehydration again implies the loss of water; note that in your dehydration reaction there is an hydroxyl functional group (-OH) which resembles water in that the -OH is one hydrogen away from H2O (water). An acid catalyst helps tack a H+ onto the -OH and this makes a good, stable leaving group.

Think of the reaction as an equilibrium equation:
$$\ce{CH3CH2OH <=>[H2SO4] CH2=CH2 + H2O}$$
Now, using Le Châtelier's Principle, we can deduce Dissenter's comments.
In concentrated sulfuric acid there is little to no water (note that we also start off with little to no ethylene). The equilibrium shifts to produce products.
In dilute sulfuric acid there is lots of water (and little to no ethanol). The equilibrium shifts to produce reactants.
As for the role of sulfuric acid, it acts as a proton source to enable the loss of leaving group in the forward direction and the electrophilic addition in the reverse direction.

Does the hydration of ethylene happen in a significant amount? I'm just curious, but it is rather different from adding $\ce{HBr}$ across the double bond since the bromide can immediately attack the carbocation (in the transition state), whereas $\ce{HSO4-}$ is such a bad nucleophile that it will not add. But this would yield a primary substituted, very unstable and unfavorable, carbocation intermediate, right? Or does $\ce{HSO4-}$ add and is later attacked by $\ce{H2O}$ in an $\ce{S_{N}2}$ type mechanism?

In the hydration in dilute $\ce{H2SO4}$, water molecules far outnumber sulfuric acid molecules. Water is also a far better nucleophile than $\ce{HSO4-}$.

Yes, but what is the mechanism precisely?

@Jori $\ce{H2SO4}$ protonates the double bond, and the carbocation is captured by $\ce{H2O}$. Subsequent deprotonation of the $\ce{OH2^{+}}$ group yields the alcohol.

@JannisAndreska Thank-you. I seem to have completely forgotten that $\ce{H2O}$ is the solvent and therefore is very available (that confused me since primary carbocations normally do not form).

Interestingly, I learnt in my syllabus that cold conc. sulfuric acid should be used for hydration while hot conc. sulfuric acid is used for dehydration. The temperature difference is due to entropy considerations but I don't get the rationale behind using concentration acids. Is it to increase the rate of reaction?


text:
Definition of relative atomic mass
I found out there are two definitions of relative atomic mass $A_\mathrm r$.
First definition is $A_\mathrm r$ is the mass of 1 atom of an element relative to 1/12 the mass of carbon-12 atom. It can be found in any chemistry school textbook.
Second definition is what I just found and dont understand. It states $A_\mathrm r$ is the mass of 1 mole of atoms relative to 1 mole of carbon-12 atoms.
Can you kindly explain second definition with examples?

These are all mathematically identical. Here’s why:
$$\frac{\text{mass of one atom}}{\frac1{12}\text{ mass of one atom of }\ce{^12C}}\\=\frac{\text{mass of one atom}\times N_\mathrm{A}}{\frac1{12}\text{ mass of one atom of }\ce{^12C}\times N_\mathrm{A}}\\=\frac{\text{mass of one mole of substance}}{\frac1{12}\text{ mass of one mole of }\ce{^12C}}$$


text:
How can -G/T and -A/T be thought of as 'disguised' entropies?
I understand that entropy is defined as $\mathrm{d}S=\mathrm{d}q/T$ for a reversible change but I fail to see why $-G/T$ and $-A/T$ can be thought of as 'disguised' entropies. This is a question on my problem sheet at university and my tutor has hinted: "You need to think about the total entropy change for a process".

Do you really mean $-G/A$ and not $-G/T$? I ask because I think you are talking about the Massieu Potential and the Planck Potential (see here).

are you talking about Gibbs and Helmholz free energy?

@RobChem So is it or is it not a question about free entropy?

The fundamental equation of (chemical) thermodynamics $\ce{\Delta G} = \Delta H – {T\Delta S}$ may be divided by  (-T):
$$\ce{-\frac{\Delta G}{T}} = -\frac{\Delta H_{sys}}{T} + {\Delta S}_{sys}$$
In this equation $\frac{\Delta H_{sys}}{T}$ is the entropy change of the system by “heat” exchange with the surroundings, hence $\frac{\Delta H_{sys}}{T} = -\Delta S_{surroundings}$. On this basis, the above equation may be identified as:
$$\Delta S_{total} ={\Delta S_{surroundings} + \Delta S}_{sys}$$
$\Delta S_{sys}$ and $\Delta S_{surroundings}$ strive mutually to a maximum of $\Delta S_{total}$.
$\ce{\Delta G}$ is a disguised entropy change, because $\ce{\Delta H}$ is intrinsically an entropy change too, as explained above.


text:
Find the valences of Z=26
I've tried this:

But its hows that the valence of Fe in normal state is 4, when in the periodic table it isn't. Why does this happen? Can you show me the answer with configuration formula in the states with energy?

You have the electron configuration of iron in your post. Just writing that out is only the first step. What have you done with it to attempt to figure out valences?

Hint:
$$_{26}Fe\equiv1s^22s^22p^63s^23p^64s^23d^6\equiv [Ar]\ce{\underbrace{ ^ v }_{4s^2} \underbrace{ ^ v ,\;^  ,\;^  ,\;^  ,\;^   }_{3d^6} }$$
Think now what will be the configuration after losing electrons:
Spoiler:

! $$Fe^{2+}\to[Ar]\ce{\underbrace{ - }_{4s^2} \underbrace{ ^ v ,\;^  ,\;^  ,\;^  ,\;^ }_{3d^6}=[Ar]\underbrace{ ^ }_{4s^2} \underbrace{ ^  ,\;^  ,\;^  ,\;^  ,\;^   }_{3d^6} }$$
! $$Fe^{3+}\to[Ar]\ce{\underbrace{ - }_{4s^2} \underbrace{ ^  ,\;^  ,\;^  ,\;^  ,\;^   }_{3d^6} }
$$


text:
Why V can't have the valence 7 by taking an electron from the 3p orbital and putting it into 3d?
Does that happen because 3p already has a bond?

Vanadium has a valency of 5 to start with, so transferring an electron from a 3p orbital into a 3d orbital would lead to a valency of 6 not 7.

Vanadium has a stable valency of 5.
Electrons cannot be transferred from $3p$ orbital to $3d$ orbital as the $3p$ orbital is completely filled and has attained stability.
It requires high energy so as to transfer electrons from $3p$ orbital to $3d$ orbital.
Hence,Vanadium has a valency of 5 only and not of 7 or 6.


text:
Differences between Urushibara nickel and Raney nickel
The Wikipedia article about Urushibara nickel catalyst is rather short and vague. What is the mechanism for its preparation, and furthermore, in what sense does it differ from Raney nickel (i.e. when should you prefer the one over the other)?

If you read the Wikipedia article on Raney Nickel, you would have a good start. https://en.wikipedia.org/wiki/Raney_nickel

@LDC3 Urushibara nickel is not mentioned in that article at all.

But it does tell you how to make Raney Nickel, and with the article you posted, you can tell how each is made.

@LDC3 That still does not address the question at all. Wikipedia does not talk about differences in reactivity and uses, other than "It can be used for most hydrogenations where Raney nickel can be used.", which is not very useful.

Both of these statements are from Wikipedia: U-Ni: First nickel is precipitated in metallic form by reacting a solution of a nickel salt with zinc. R-Ni Raney nickel is a fine-grained solid composed mostly of nickel derived from a nickel-aluminium alloy. So you see, R-Ni starts as an alloy and U-Ni starts as a solution. If you were looking for the method of preparation, why didn't you ask for that.

@LDC3 That is about preparation, not reactivity or uses. I was looking for a detailed comparison, so I started a bounty. I appreciate that you are trying to help me out, but the Wikipedia page is just not extensive enough.

So your question was not clearly stated as to what you desired for an answer. I am not a mind reader.

@LDC3 I see, the sentence had accidentally contracted a bit vaguely. Edited that now!

You can find information about the preparation of Urushibara nickel (U-Ni) in the original paper by Urushibara[1].
Regarding the differences with Raney nickel (R-Ni), surface characterization by means of XPS[2] had shown that R-Ni catalysts contain metallic aluminium in the activated phase as well as oxidized aluminium, while U-Ni catalysts contain metallic zinc in the activated phase in addition to oxidized zinc. Also, U-Ni catalysts can be described as supported nickel catalysts, while R-Ni catalysts are shown to have the well known skeletal structure.
The activity of U-Ni catalysts is very similar to R-Ni catalysts, both in nature of the reaction catalysed and in the rate of catalysis. One of the advantage of U-Ni is their simplicity of preparation, while R-Ni catalytic synthesis requires a long time. Despite this, U-Ni catalysts are also not very sensitive to impurities, so they can be prepared from commercial chemicals of ordinary grade, and they can be stored for a long time without decreasing the activity. Another advantage is the safer handling, since U-Ni catalysts are not inflammable upon contact with air, in contrast to Raney catalysts which are pyrophoric. This allows the catalyst to be recovered and regenerated. However, it should be noted that U-Ni catalyst durability is not satisfactory during the reaction, but this is also the case of R-Ni catalysts, so there is not really a big difference when it comes to practical use, maybe slightly supporting the use of U-Ni vs R-Ni as the former is less expensive and easy to produce and regenerate. The only real disadvantage of U-Ni catalysts is that for some particular reductions, they show less activity than R-Ni catalysts, favouring one reaction over another, for example U-Ni catalysts particularly favours the partial hydrogenation of acetylenic compounds to ethylenes[3].
     1 Urushibara, Yoshiyuki. "A New Method of Catalytic Hydrogenation". Bulletin of the Chemical Society of Japan 25 (4): 280–280
     2 Y. Okamoto, Y. Nitta, T. Imanaka and S. Teranishi, J. Chem. Soc., Faraday Trans. 1, 1979, 75, 2027
     3 Kazuo Hata, “New Hydrogenation Catalysts: Urushibara Catalysts” (1971), pp 235-237


text:
What is the hybridization of the nitrogen in the azide ion?
In the azide ion, there is a symmetrical arrangement of three nitrogen atoms. If I've been told to figure out the hybridization of each nitrogen atom, what's the best way to go about doing this? My initial guess was that the central nitrogen was sp hybridized and then the other two were sp2, but I can't seem to justify my guess.

Here are the resonance structures we can draw for the azide anion.

Resonance structure A has an octet on each nitrogen and just one unit of formal
charge on each nitrogen
Resonance structures B and C also have an octet around each nitrogen, but there is a rather large formal charge of -2 on one of the nitrogens.  Such a large charge on a small atom like nitrogen makes structures B and C less favorable and therefore we would expect them to count less to the "true" representation of the azide anion, than resonance structure A
Resonance structures D and E do not have an octet around the singly-bound, terminal nitrogen, these resonance structures will count even less.  In fact, since the azide ion is linear and these two resonance structures are bent, we can say that they don't really contribute to the true picture of the azide ion.

So if we consider only resonance structures A and B (C is equivalent to B) we can see that the central nitrogen is $\ce{sp}$ hybridized in both cases (like in allene for A and like in acetylene for B).  The $\ce{sp}$ hybridization of the central nitrogen is consistent with the linear structure of the azide ion.
In A, the terminal nitrogens may both be $\ce{sp^2}$ hybridized; after all the drawing looks similar to the double bond in ethylene. However, there are other options.  For example, both nitrogens could be $\ce{sp}$ hybridized, with an $\ce{sp}$ single bond, one lone pair in a $\ce{sp}$ orbital and the other lone pair in a $\ce{p}$ orbital (just like the 2 lone pairs on oxygen in water). Without more data there is no way to tell for sure, but in an introductory class $\ce{sp^2}$ hybridization would be a reasonable answer.
In B, the terminal nitrogen with the triple bond appears to be $\ce{sp}$ hybridized just like in acetylene.  The hybridization of the other terminal nitrogen in resonance structure B is anybody's guess; there are many possibilities and since there is only one ligand attached (the central nitrogen), we don't know what direction the 2 electron lone pairs are pointing in.  This nitrogen might be unhybridized, $\ce{sp}$ hybridized, $\ce{sp^2}$ hybridized - again, we just don't know.  Since the true hybridization of the terminal nitrogen is an equal mixture of these two hybridizations (e.g. a blend of B and C), we really don't know how the terminal nitrogens in B and C are hybridized.
In an introductory class, the answer would be that the central nitrogen is $\ce{sp}$ hybridized, and, since resonance structure A is the major resonance contributor, we might say that the terminal nitrogen is $\ce{sp^2}$ hybridized (but, from the above discussion, we know that is an oversimplification).

The structures A,B and C are resonance structures with each other, so are D and E. However, the two sets cannot be in in the same set, since the principle requirement is violated - resonance structures all must have the same atomic coordinates. In general, I would say, that all nitrogen atoms are sp hybridised, as there is no external field, that would necessitate further hybridisation.

I have said it a couple of times already on our site,[1] but I'm not getting tired of saying it again:

Hybridisation always follows the (spatial) geometry of the molecule, never the other way around. As such it is a way to interpret bonding, not to predict it. Following from that, knowing the molecular structure may always give you the ability to suggest a hybridisation scheme.
As a rule of thumb, terminal atoms (excluding hydrogen) have almost always (maximally) approximately sp hybridised orbitals. While other hybridisation schemes can be applied, they are usually not a good representation.[2]
As another rule of thumb, the lesser the mixing between s and  p orbitals, the (probably) more correct the bonding will be described.[3]
Any scheme using d-orbitals or higher (in main group chemistry) is likely to be outdated and needs a more refined approach to bonding.

Using the above it is fairly simple to derive the hybridisation scheme of all the atoms. Since the molecule is linear, the central nitrogen is constrained to have at most sp orbitals. With the rule of thumb, the terminal nitrogen also only have sp orbitals. These orbitals will form the σ bonds and the σ lone pairs. The remaining two p orbitals at every nitrogen will form the π bonds. If you further think about it, the molecule is isoelectronic to carbon dioxide. I have written about the electronic structure before: π Bonding in Carbon Dioxide.
Because of the high symmetry of the molecule, D∞h, automatic interpretation (with natural bond orbital theory) of this bonding situation is very difficult. A natural resonance theory analysis on the DF-BP86/def2-QZVPP level of theory gives us the following structures:
$$\ce{
  \underset{24\%}{[N-N#N]-}
  <->
  \underset{49\%}{[N=N=N]^-}
  <->
  \underset{24\%}{[N#N-N]-}
}$$
Unfortunately the generated hybrid orbitals don't look much like we would expect them to look, because they are also constrained to symmetry. (Occupied orbitals in blue/orange.) You can see that the central nitrogen basically has two lone pairs, which have a very low occupancy (1.29 el.), while the (antibonding) π bond is between the the terminal nitrogen. A linear combination of these two would lead to the expected bonding picture. The virtual combinations of the π bond (red/yellow) on the other hand have a quite high occupancy (0.71 el.). This is often an indicator that a hybridisation scheme is not going to work very well. Resonance effects are not described very well with hybrid orbitals.

Notes

See for example:
What is the hybridization of terminal fluorine atoms in molecules like boron trifluoride?
What is the hybridization of chlorine in vinyl chloride?
Hybridisation of terminal nitrogen in diazomethane
It is very important to understand that atoms are never hybridised. (Even though organic chemists tend to use this jargon - even in textbooks.) The orbitals at an atom can be hybridised. Hybridisation is just one way of looking at the bonding situation, it is one model. As such, this interpretation is correct, whether or not your teacher might agree with it. There is no way to probe an hybridisation scheme, there are only instances where one explanation fits the experimental observations better than another. For example see below.
One most popular example of misapplying hybridisation schemes is water. In many texts the central oxygen is described as having (approximately) four sp3 orbitals, which essentially means that the lone pairs are equivalent. This interpretation is not in agreement with the photoelectron spectrum, which clearly shows that the lone pairs are not equivalent. For more on this I'd like to refer you to Michael Laing's article: "No rabbit ears on water. The structure of the water molecule: What should we tell the students?" (J. Chem. Educ. 1987, 64 (2), 124. DOI: 10.1021/ed064p124).


text:
Why is there a doublet instead of a singlet in this Wittig reaction salt configuration NMR?
I did an NMR sample to my salt p-carboxybenzyl triphenylphosphonium bromide. I got a doublet at 5.26, and according to every article it belongs to the $CH_2$ next to $P^+$, but how come I didn't get a singlet for that?

Phosphorous exists entirely as $\ce{^31P}$, and $\ce{^31P}$ has a magnetic moment of 1/2 (just like a proton). Therefore, a single phosphorous will split any protons coupled to it into a doublet - as you observed. If you looked at the corresponding $\ce{^31P}$ NMR spectrum you would see that the phosphorous signal would be split into a triplet by the two protons.

and another question if i may, i got a very large peak at 3.38 in dmso, could it be water?

The water impurity signal can move around a bit depending on the water concentration, but I'd bet your signal is due to water. This reference says the water impurity in dmso absorbs at 3.3.  If the above answer was helpful, please mark it as accepted. Thanks!

im having trouble finiding out another two multiplets in 7.08 and another in 7.56 at the same reaction ,my solvent was tolouene..

For peak identification it would probably be best if you opened a new question and included a copy of the nmr spectrum along with your experimental procedure.

authors: Tárùñ MâÑdál


text:
Proportion of gases in Ideal Gas Mixture?
Suppose methane ($CH_4$) and ethane ($C_2H_6$) in a container. The specific gravity is $r_g=0.6$. I think this means $0.6=\frac{\rho_{CH_4}}{\rho_{C_2H_6}}$. Now I am trying to understand the question.

Question: what is the proportion of gases?

Solution according to my teacher: $C_{CH_4}=0.1$ and $C_{C_2H_6}=0.9$

My Solution: $C_{CH_4}=24.2\%$ and $C_{C_2H_6}=75.8\%$

Values (Curtis Phase Behaviour book p.162)

$M_{CH_4}=16.04 \frac{lbm}{lbm-mol}$
$M_{C_2H_6}=30.07 \frac{lbm}{lbm-mol}$

What is $x_{CH_4}=\frac{n_{CH_4}}{n_{CH_4}+n_{C_2C_6}}=\frac{n_{CH_4}}{n}$?

By Amagat's law and Dalton's law here,
    I know $\frac{n_j}{n}=\frac{P_j}{P}=\frac{V_j}{V}$.
Specific gravity $0.6=\frac{\rho_{CH_4}}{\rho_{C_2H_6}}$ where $\rho=\frac{m}{V}$ and constant $V$ so
    $0.6=\frac{m_{CH_4}}{m_{C_2H_6}}$ so $m_{CH_4}=0.6m_{C_2H_6}$.
$PV=nRT\rightarrow n=\frac{PV}{RT}$ where constant $\frac{V}{RT}$ so
    $x_{CH_4}=\frac{P_{CH_4}}{P_{CH_4}+P_{C_2C_6}}=\frac{P_{CH_4}}{P}=_{(1)}\frac{n_{CH_4}}{n}=\frac{V_{CH_4}}{V}$.
By $M_i=\frac{n_i}{m_i}$ so $n_i=M_i m_i$ where $n_{CH_4}=M_{CH_4}m_{CH_4}$

so
$\frac{n_{CH_4}}{n}=\frac{M_{CH_4}m_{CH_4}}{M_{CH_4}m_{CH_4}+M_{C_2H_6}m_{C_2H_6}}=_{(2)}\frac{M_{CH_4}0.6m_{C_2H_6}}{M_{CH_4}0.6m_{C_2H_6}+M_{C_2H_6}m_{C_2H_6}}=\frac{0.6M_{CH_4}}{M_{CH_4}0.6+M_{C_2H_6}}=0.242455$
$\rightarrow\frac{n_{C_2H_6}}{n}=1-\frac{n_{CH_4}}{n}=0.757545$
where the results are not 0.1 and 0.9.

Your interpretation of specific gravity is incorrect:

Specific gravity (SG) refers to the ratio between a given density and that of a reference.  Usually it is defined for liquids with respect to water and gasses with respect to air so in this case

$0.6=\frac{n_{CH_4}\rho_{CH_4}+n_{C_2H_6}\rho_{C_2H_6}}{\text{air}}$

where $n_i$ are proportions by mole factor. This is the incorrect assumption

$0.6=\frac{\rho_{CH_4}}{\rho_{C_2H_6}}$.

Gasses are always a little dicier since the reference is much more important with changes in temperature and pressure affecting the resulting density.

Example

The density form of the ideal gas laws goes like:

$\rho = \text{mass / volume = molecular weight * pressure / constant / temperature}$.

Therefore for constant temperature and pressure, the specific gravity
  is equal to the ratio of molecular weights.  This follows because
  $\frac{\rho_1}{\rho_2} = \frac{MW_1}{MW_2}$ at constant conditions.

Equations and Solution

$M_{\text{air}} = 28.97$ (I got $28.84$ using the simple $0.21*32+0.79*28$ and this agrees with  this)
$\text{SG}=0.6=\frac{n_{CH_4}\rho_{CH_4}+n_{C_2H_6}\rho_{C_2H_6}}{\text{air}}$
$1=n_{CH_4}+n_{C_2H_6}$

so by the first equation and the second equation the density of the mixture is $0.6 * 28.97 = 17.38$. The third equation holds only if $0.9*16 + 0.1*30 = 17.4$ so

methane is 90% and ethane is 10%.

Thank you! I made it easier to read with LaTex/structuring keeping the original content the same. I hope people will accept my edits so easier to read the key points in the future +1. Welcome to SE!

I would remove the part Example and leave only the Equations and Solution part -- even though I read it many times I cannot understand how it is related to this question. Or perhaps you can clarify it?

I was trying to emphasize how to get from the specific gravity to the ratio of molecular weights via the ideal gas law.


text:
Since the formation of diamond from Graphite is endothermic and the entropy decreases for this process, how is it possible to make diamond?
Surely G would always be positive and thus would make it impossible to make diamond. Clearly this is not the case. But how? Would an increase or decrease in temperature aid it's formation? I think a decrease in temperature would favor diamond but wouldn't this contradict le chatilier's principle?

Diamond is significantly denser than graphite.
atomic densities:

$1.14  \times 10^{23}$ cm$^{-3}$ for graphite
$1.77  \times 10^{23}$ cm$^{-3}$ for diamond

This suggests that higher pressure would favor the formation of diamond.  Indeed, as the following phase diagram shows, diamond is the most stable allotrope of solid carbon at high pressures.  Temperature has little effect, as long as the pressure is high, diamond is favored over graphite at all temperatures up to the point of liquefication.

If you have a look at this phase diagram, you can see the main condition at which diamond is the favourable structure: high pressure. The Gibbs free energy is dependant on pressure, in addition to temperature: $$dG = -SdT + Vdp$$
With a little math and assuming the temperature is constant, the gibbs energy moving from one pressure to another: $$G(p_2)=G(p_1)+\int_{p_1}^{p_2}Vdp$$
In practice, one method of making synthetic diamonds relies on high pressure and high temperature: carbon dissolved in a molten metal precipitates out onto small seed diamonds.
It's not really necessary to calculate anything to rationalize why diamond would be the favoured allotrope at high pressure—diamond has a much higher density than graphite. At high pressure, the side of the equation with less volume will be favoured by Le Chatelier's principle.

if you take the definition G=H-TS how is it possible for the reaction to happen because surely G is always positive?

Enthalpy is also dependant on pressure. ∆H of formation at standard state (which is what you'll find in a table) is not the same as at very high pressure.

Thermodynamics gives two possible solutions: you make $\Delta G$ positive through appropiate choice of reaction conditions (i.e. high pressure), or you couple the diamond-formation reaction with another reaction, so that the total $\Delta G$ becomes positive.  There are processes for gas-phase deposition of diamond films on surfaces, and those don't depend on high pressure.

You're forgetting the key - enthalpy is defined at constant pressure. You need to be thinking about Helmholtz, not Gibbs. If we say that the reaction from graphite to diamond is caused by pressure, chances are the pressure will not be constant throughout.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help]. At the moment this reads more like a comment than an actual answer - could you elaborate a little more. With a bit more rep, you will be able to post comments on any question/answer.


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Determine # of molecules in an unknown sample
How might one determine the # of molecules in an unknown sample? If the sample is a gas,  we might use $PV=nRT$ to estimate # of molecules. If the sample is a solid or liquid and the chemical formula is known, we simply need to measure the mass. But can we determine the # of molecules without first determining the chemical formula?

There are attempts to redefine the mol through measuring spheres of monocrystalline silicon.  If you have the macroscopic dimension of your sample and the dimensions of your unit cell (from x-ray diffraction) you know how many atoms if silicon are there.

What if the substance does not exhibit crystalline structure?

Freezing-point depression

Are you telling me there is no general way to determine the number of molecules in a sample? That would be strange that something so simple requires complicated and case-specific methods to deal with.

Freezing-point depression actually works quite nicely.

That is for solutions, though. Perhaps I should ask which method experiments use, or whether they care about calculations like this at all.


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Density of Ni vs Cu using unit cell
How might one determine whether Ni is more dense than Cu using only the atomic masses of Ni and Cu, and the fact that Ni has a simple cubic unit cell, and Cu has a face-centered unit cell. You are not given the radii of each element. I feel like this question cannot be answered with the given information only, considering as Ni and Cu have very close densities (0.06 g/cm^3 difference).

A quick look through the web suggests that both Ni and Cu crystallize in a ccp structure.  The space group is the same: Fm-3m.  Considering trends across the periodic table, you should have an answer now.

This was a question given on a test. You are supposed to use only the information given. You cannot use density trends in the periodic table. Also the problems states that Ni has simple cubic, Cu has face-centered. You must use these assumptions. This question is hypothetical.

If what you say is true your teacher should be fired for gross incompetence.  Instructions for an orgo class I was involved in were that if you saw a reaction or reagents that weren't in the answer key you had to give the student the opportunity to give a literature reference in their support.  Chemistry is the real world, which is much larger than what is taught in a particular class.  Your teacher seems intent on killing student independence and initiative.

This is the problem with high school - its not about learning, its about getting the answers correct. I encounter scenarios like this all the time. Of course, to answer the question with the given information only is a bit weird (how are we supposed to predict a difference if the densities are so close). Now, if we are given that Ni and Cu have ccp structure, the answer is obvious by examining relative radii of the two elements. Nonetheless, I believe the question is not totally invalid. Suppose ithe assumption about the unit cell structures was true, can the question be answered?

In a sc packing there is much more empty space than in a fcc packing. That's why of the metals only polonium rystallizes in an sc structure, and it's a semi-metal.  For a real world-example, compare the densities of lead (fcc) and polonium (sc).

I think you need to make at least one more assumption along the way to determining the relative densities. But let’s take it stepwise:

We are given a primitive cubic cell for nickel and a face-centred cube for copper.
We therefore conclude that per unit cell one atom of nickel and three atoms of copper are present.
We know from the periodic table that copper is a bit heavier than nickel ($63.5$ versus $58.7$)
Therefore, it is clear that copper’s unit cell is more than four times as heavy as nickel’s. But we don’t know their sizes.
From mathematical considerations we can also calculate how much of each unit cell is filled with atoms ($74~\%$ for fcc; somewhat less which I am too lazy to calculate now for primitive cubic) — this again means that there should be more weight per volume in the copper unit cell.
What seems to be lacking in my eyes is a way to compare the unit cell sizes, whether by atom size or unit cell dimensions. However, since nickel and copper are right next to each other in the periodic table, it may be safe to assume their sizes to be almost identical.
If that assumption is safe, we can use either route to clearly show how copper’s density is higher. But note that if copper’s atoms were significantly larger than nickel’s, this reasoning would break down extremely quickly.

The density of a crystal is given by the equation
$\rho=\frac{Z\times M}{a^3\times N_0 \times 10^{-30}}~\mathrm{ g/cm^3}$, where $Z$ denotes the effective number of metal atoms per unit cell.
This means
$\frac{\rho_\ce{Cu}}{\rho_\ce{Ni}}=\frac{Z_\ce{Cu}\times M_\ce{Cu} \times {a_\ce{Ni}}^3}{Z_\ce{Ni}\times M_\ce{Ni}\times {a_\ce{Cu}}^3}$.
Just substitute the values to get the answer.


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Ideal gas law derivation from kinetic theory
Consider the derivation of the ideal gas law from kinetic theory presented here: http://en.wikipedia.org/wiki/Kinetic_theory
I have some questions

This derivation assumes the container is a cube, but the law holds for containers of any shape, right?
This derivation assumes each particle has the same mass, but the law holds for a mixture of gasses of different masses, like air, right?
The derivation assumes that the particle intercepts a give side with frequency $2L/v_x$. Why can we make this assumption?

This derivation assumes the container is a cube, but the law holds for containers of any shape, right?

The resulting law holds for any shape of container. However, the details of the derivation depend on the geometry. A cube is easy, so we use it.

This derivation assumes each particle has the same mass, but the law holds for a mixture of gasses of different masses, like air, right?

Yes, it holds for a mixture of any ideal gases - the key is that only the kinetic energy of the gas molecules matters. Heavier molecules would be moving slower at the same temperature, but they would exert the same force (and therefore pressure) on the walls.

The derivation assumes that the particle intercepts a give side with frequency 2L/vx. Why can we make this assumption?

This assumption works for only one particular case. If you notice the third sentence at the beginning of the pressure derivation (emphasis is mine):

When a gas molecule collides with the wall of the container perpendicular to the x coordinate axis and bounces off in the opposite direction with the same speed

They are looking at the special case of a collision at a 90 degree angle, which bounces off the opposite wall. In that case, it will take the particle t = L/V time to reach the opposite wall, and t = L/V time to bounce back - so the total is 2L/V. This illustration may help explain it better:

Now this might sound like an unreasonably strict condition, but the reason that it works is that the component velocity distribution for each particle is symmetrical (it's a normal distribution), which means that for every particle that impacts the wall at 95 degrees (for example) there is another that impacts 85 degrees. As a result, the average angle of impact is 90 degrees for each wall, and therefore the average collision time is given by looking at 90 degree impacts for each component of the velocity vector.

1.So how do we prove the law holds for any container? 2.If the law holds for a mixture of gasses of different masses (still acting as an ideal gas), how do we prove the law in this case? 3.I believe you are misunderstanding that quote from the article. It should read like: "When a gas molecule collides with [a wall of the container that perpendicular to the x-axis]". 4. How do you prove that the velocity component distribution for each particle is the normal distribution?

Also can someone link me to a rigourous proof of the ideal gas law using kinetic theory?

@JoshuaBenabou If you only want a proof that the ideal gas law holds for any container size it might be easier to simply show general applicability via thermodynamics. But if you want a general proof from kinetic theory I would imagine that this might not be easy and possibly quite lengthy.

@JoshuaBenabou -

There are lots of ways, one way is to show that it holds empirically for many container shapes, then derive it from first principles for an arbitrary container (a cube in this case). I am not sure if there is way to rigorously show that it holds for all container geometries.
You would replace the kinetic energy terms with summations over the different masses.
Possibly. But, who is to say the x-axis isn't vertical?
You don't, that's empirically known. At any rate, it doesn't have to be normal, just symmetrical about a mean.

There is a nice general proof using statistical mechanics here:  http://en.wikipedia.org/wiki/Ideal_gas_law.

JoshuaBenabou - yes, the general proof usually is done using stat mech - but that's not kinetic molecular theory. The link you posted is an example of what @Philipp was referring to.


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Can there be more than one combination of reactant/product concentration for a specific equilibrium constant?
For an elementary reaction like $$\ce{A + B -> AB},$$ $$K_{c} = \frac{\ce{[AB]}}{\ce{[A][B]}}$$ and $$\frac{\mathrm{d}\ce{[AB]}}{\mathrm{d}t} = k\ce{[A][B]}.$$ If I let the reaction reach equilibrium, and at this point, without altering temperature, suck away $\ce{A}$ and add $\ce{B}$ at the same rate (so $-\mathrm{d}\ce{[A]}/\mathrm{d}t = \mathrm{d}\ce{[B]}/\mathrm{d}t$) and alter both concentrations slightly, once I am finished, will the system still be in equilibrium or try to shift?
What if the reaction is not an elementary reaction such that $\frac{\mathrm{d}\ce{[AB]}}{\mathrm{d}t} = k\ce{[A]}^n\ce{[B]}^m$?

A single "elementary reaction" does not involve an equilibrium.
There must be a forward reaction and a reverse reaction to have an equilibrium.  This is at least two elementary reactions.
Considering an equilbrium reaction: Just as a square encloses the most area for a rectangle of given perimeter, if the sum [A] + [B] is kept constant,  [A][B] is maximal when [A] = [B].  Deviation from this maximum will shift the equilibrium away from the product and toward the reactants.  In other words, if the system is at equilibrium with [A] = [B], and [A] is decreased and [B] increased, keeping [A] + [B] constant, there will be net reverse reaction until equilibrium is again reached, and [AB] at equilibrium will be less than orginally.

What if I maintain the product [A][B] to be constant while decreasing [A] and increasing [B]. After I stop, will the final state be at equilibrium? In both cases (maintaining [A][B] and [A]+[B] constant), does the new equilibrium state differ than the original?

[A][B]=constant1; [A] + [B] = constant2.  There are two equations and two unknowns, only two solutions at most (the intersections of a hyperbola and a line).  In other words, you cannot decrease [A] and increase [B] while maintaining [A][B] and [A]+[B] constant.  It is mathematically impossible.

@Yandle Instead, if you just keep [A][B] constant, decrease [A] and increase [B], without keeping [A]+[B] constant, the system will still be in equilibrium as long as [AB] is not changed.  The decrease in [A] and increase in [B] would need to be in a manner that avoids changing the volume of the system, because changing the volume of the system with a constant amount of AB would change [AB].


text:
Aqueous Chemistry: Ranking Ion Concentration
I am trying to rank the concentration of $\ce{PO4^3-}$, $\ce{NO3^-}$, and $\ce{Na+}$ that remain in a reaction. The reaction occurs when 100mL of 1.0M $\ce{Na3PO4}$ solution is mixed with 100mL of 1.0M $\ce{AgNO3}$ soln. I have written the equation as:
$\ce{Na3PO4(aq) + 3AgNO3(aq) -> 3NaNO3(aq) + Ag3PO4(s)}$
From this equation I find that there are 0.3mol of $\ce{Na+}$ by 0.1L * 1.0M * 3mol Na, and
0.1mol of $\ce{PO4^3-}$ by: 0.1L * 1.0M * 1mol $\ce{PO4^3-}$ ,
and finally 0.3mol of $\ce{NO3-}$ by 0.1L * 1.0M * 1mol $\ce{NO3-}$.
By these calculations, the concentration of the nitrate ion and sodium ion should by equal. However, the correct answer is that the most concentrated ion is sodium followed by nitrate followed by phosphate. Could someone please explain where I'm going wrong.
Thanks!

What's the concentration of Na+ in 1 M trisodium phosphate, and what's the concentration of (NO3)- in 1M silver nitrate?

The concentration of Na+ in 1M trisodium phosphate should be 3.0M. I found this by finding the number of moles of trisodium phosphate in 100mL of 1.0M solution: 0.1L and then multiplying that by 1M to get 0.1mol trisodium phosphate and then multiplying by 3mol to account to get the number of Na ions. Then I divided by 0.1L to get 3.0M. With similar steps I found the [NO3-] to be 1.0M. I'm not sure why this doesn't work when I try to find concentration after the reaction is complete. Thanks for your reply.

You seem to have overlooked the fact that the volume of the solution has doubled after combining them both. The concentration of phosphate ions could be calculated from the solubility product of silver phosphate, but should be very low.

Let's look at the overall reaction:
$\ce{Na3PO4(aq) + 3AgNO3(aq) -> 3NaNO3(aq) + Ag3PO4(s)}$
As you can see $\ce{Na^+}$ and  $\ce{NO_3^-}$ are spectators ions, they don't participate in the main reaction. Their concentrations, taking into account the dilution, are:
$\ce{[Na^+]= \frac {3 \times 1.0 \times 100}{200}}= 1.5 \ce{M}$.
${[\ce{NO_3^-}]= \frac { 1.0 \times 100}{200}}= 0.5 \ce{M}$.
The net ionic equation is:
$\ce{PO4^{3-}(aq) + 3Ag^+(aq) ->  Ag3PO4(s)}$
The initial concentration of phosphate and silver ions taking into account the dilution are:
${[\ce{PO_4^3-}]= \frac { 1.0 \times 100}{200}}= 0.5 \ce{M}$.
${[\ce{Ag^+}]= \frac { 1.0 \times 100}{200}}= 0.5 \ce{M}$.
According to the stoichiometry of the reaction: 3 moles of silver ions react with one mole of phosphate ion. This means that silver ion is a limiting reactant. It will completely react to form the precipitate of silver phosphate. On the other hand, the concentration of phosphate at the end of the reaction will be:
${[\ce{PO_4^3-}]= 0.5-\frac{0.5}{3}=\frac{2\times 0.5}{3}=0.33 \ce{M}}$.
In order to determine exactly the concentration of silver ion, we have to use the solubility product of silver phosphate:
${[\ce{Ag^+}]= \frac{K_{sp}}{[\ce{PO_4^3-}]^3}=\frac{8.89\times10^{-17}}{(0.5/3)^3}}=0.04\times10^{-17}\ce{M}$


text:
Is purple paint more susceptible to UV damage than other colors?
I was recently sorting some Mardi Gras beads. These are cheap plastic beads on strings, coated in cheap metallic paint. While sorting the beads into different colors, I noticed that the purple beads seem to have suffered more fading than the other colors. This is an effect I have noticed in the past: if the beads sit in the sun for a day or two, the colors fade rapidly. I'm guessing that the cause of that is that the UV radiation from the sun is breaking the molecules used for the pigments in the paints.
My first thought was that I had a pile of beads exposed to sunlight, and the purple beads in that pile had been at the top of the pile. But I noticed this happening in a few different piles that I had stored in different places and for different lengths of time. I have also noticed that the red beads and the purple/pink beads show some fading, though not as much as the purple beads. The yellow, silver, and green beads show almost no fading.
My current hypothesis is that the manufacturers use a pigment in the red and purple beads that is more susceptible to UV damage than the other pigments. So my question is: what is the chemical in these purple and red paints that is breaking down? (I guess there's also an implicit question here: is my thinking here chemically plausible?)

My first guess would be that they use organic dyes, instead of inorganic pigments. Dyes are in general less stable and may decompose via UV decomposition, chemically (UV assisted oxidation) or thermally. Without knowing the exact dye used, it is difficult to sketch a reaction, but in principle I fully agree with your reasoning.

I would also agree with you and the comment from ssavec. The only point that I would make is that it shouldn't be because of the colour of the pigment only the stability of the particular dye it contains (almost certainly organic). In fact the colour of the paint should make it less susceptible to fading as it is reflecting colours at the violet end of the spectrum (subject to your definition of purple) rather than absorbing them.


text:
How thermochromism happen?
I learned that when $PbO$ is hot, it is orange. When $PbO$ is cold, it is yellow. I know this's called thermochromism. But how do that happen? Is it because the ions inside doing something?

In the case of lead oxide and many other inorganic thermochromic materials, the color change is a result of a phase change that slightly alters the distance separating the inorganic ions.
In ordered materials, the energy levels present in atoms and molecules are replaced by energy bands.  Said differently, in materials with extended order there are so many energy levels clustered around one another at a certain energy point and separated from one another by such small amounts of energy, that the distinct energy levels of the molecules "fuzz" out into a band.  Instead of a highest occupied and lowest unoccupied molecular orbital (HOMO and LUMO), a valence band and a conduction band results.

Just as the HOMO-LUMO separation will determine where in the energy spectrum an organic molecule will absorb light, the energy separation between the valence band and the conduction band (called the band gap) will determine where a material with extended order will absorb light.
In the case of lead oxide and other crystalline materials, slight changes in the crystal structure (phase change) that can be brought about by heating or cooling the material, will change the size of the band gap.  This change in the energy separation between the valence band and conduction band will result in a change in the optical (and electronic) properties of the material.  If the change in optical absorption occurs in the visible region, then the thermochromic properties of the material are apparent to the eye.
Here is a link to a nice paper that starts simply, but then goes into greater depth, on this subject.


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Could you suggest me a good coupling method for amide synthesis between tetrahydroisoquinoline and a carboxylic acid?
I want to synthesize an amide using tetrahydroisoquinoline and an unsaturated carboxylic acid which is light sensitive. I tried using DCC as a coupling agent and DCM as the solvent at room temperature, but it didn't work for me. The tetrahydroisoquinoline is in hydrochlorated form.
Do I have to do acid extraction prior to synthesis, or I can use the hydrochlorated form itself? Please suggest a good method along with the conditions for the reaction to be carried out.

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site.

A SciFinder search gave 200 or so examples of coupling tetrahydroisoquinoline with an unsaturated carboxylic acid (including benzoic acids and the like). Surprisingly dicyclohexylcarbodiimide (DCC) was not one of the reagents used, although N-(3-dimethylaminopropyl)-N′-ethylcarbodiimide (EDC) was used in more than 50 cases. Other common amide coupling agents including PyBOP, BOP, HBTU, and HATU were reported. Another related strategy through a mixed carbonate/anhydride by activating the acid with isobutylchloroformate.
I'm not sure if you have to isolate the free base of your amine first, but you should certainly include a base to liberate the active nucleophile. If you haven't tried it yet, add an equivalent of triethylamine or diisopropylethylamine.
Aapptec, a supplier of reagents and building blocks for peptide synthesis, has a nice compilation of the different amide coupling reagents that are used:
Aapptec coupling reagents.

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


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Nuclear spin multiplicity of radicals
How do you calculate the nuclear spin multiplicity of radicals such as $\ce{N2H+}$? How does the net charge on the molecular species effect its overall nuclear spin?

The short answer is that the nuclear spin and electronic spin are separate effects. So an electronic radical ($\ce{N2H+}$) does not affect the nuclear spin of the protons or the nitrogen ($\ce{^14N}$ has nuclear spin 1).
So on a basic level, the net charge does not affect the overall nuclear spins.
The somewhat longer answer is that the nuclear spin states depend on the local chemical environment. This means that we can "see" different proton nuclear spins based on the molecular structure.
If there's are unpaired electrons, paramagnetic NMR requires understanding the coupling of the electron spin and the nuclear spin. It's possible, but not as straight-forward as typical diamagnetic NMR (for example $\ce{^1H}$ peaks may spread over 200-250 ppm, unlike ~8-12 ppm for normal protons).
Two main effects in paramagnetic NMR are pseudo-contact and contact shifts.

The effect of the contact term arises from transfer of spin polarization to the observed nucleus. Spin polarization is a consequence of the very strong electron-nuclear (NMR detected) interaction. This coupling, also known by EPR spectroscopists as hyperfine coupling, is on the order of MHz, vs the usual internuclear (J) coupling observed in conventional NMR spectra, which are on the order of a few Hz. This difference reflects the large magnetic moment of an electron (−1.00 μB), which is much greater than any nuclear magnetic moment (e.g. for $\ce{^1H}$: 1.52×10−3 μB).


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Relativistic effects and lanthanide contraction
So I understand than the lanthanide contraction is due to poor shielding of the 4f electrons which decreases the radius.
However, if Im not mistaken the relativistic effects lead to a contraction of the core orbitals (s,p) while the f and d orbitals expand. So what I do not understand how do they match, one of them decreases the radius and the other one should expand it as the f orbitals expand right?
They usually say relativistic effects are one of the reasons for the lanthanide contraction
Thanks

The lanthanide contraction is caused by two effects

The 4f orbitals are very diffuse and therefore result in poor screening of the electrons further out, those in the n=5 and n=6 orbitals
relativistic effects

This Wikipedia article presents a nice discussion of the lanthanide contraction.  The article estimates that the first effect, the screening effect, is the major factor, with relativistic effects playing only a minor role (~10%) in the observed contraction.

what I do not understand how do they match, one of them decreases the
  radius and the other one should expand it as the f orbitals expand
  right?

That's correct, the outer s- and p-orbitals contract due to the lanthanide effect, while d- and f-orbitals expand (see this SE Chem post for a more detailed description).  In the lanthanide series, the 6s electrons are further from the nucleus than the 4f electrons, therefore they (the 6s electrons) determine the radius.  A contraction of the 6s orbital radius results in a contraction of the observed lanthanide electron radius, even though the 4f electrons have moved slightly further away from the nucleus.

Both are relevant, but IMHO, the poor screening is an important part. The effective nuclear charge (Zeff) increases across the lanthanides, but the added 4f electrons don't do a good job screening that larger nuclear charge. Thus, the atomic radius contracts. A similar trend is seen across other rows.

@GeoffHutchison  Yes, like I said the relativistic effect is estimated to account for only 10% of the contraction, the poor screening by the 4f electrons being the major factor.

Only orbitals with l=0 (s orbitals) have radial components that are non-zero at the nucleus:

s-shells, including and inparticular the 6s shell, are much more sensitive to the relativistic effects (compared to l>0 shells) due to s-orbitals allowing electronc density AT the nucleus.  If s-shell electron density is non-zero at the nucleus, it's reasonable to suspect that s-shell electron density is contracted more by relativistic effects.


text:
How are phenols made?
What is the equation and conditions for the synthesis of phenols? Would it work if we added $\ce{NaOH}$ to benzene and then refluxed?

Please also look up articles by Dr. Jeffrey S. Plotkin (unfortunately some of his articles are not accessible by mouse clicks from google search...)
And, Here, Date last amended: 11th January 2017

That reaction will definitely not yield a phenol.
Phenols are not usually prepared directly from benzene, but from benzene derivatives, such as toluene or chlorobenzene:
Oxidation of toluene, as developed by Dow Chemical:
$\ce{C6H5CH3 + 2 O2 → C6H5OH + CO2 + H2O}$
Hydrolysis of chlorobenzene, using base or steam (Raschig–Hooker process):
$\ce{C6H5Cl + H2O → C6H5OH + HCl}$
Actually, there are processes that allow us to obtain it directly from benzene, but these are rarely used:
Direct oxidation of benzene with nitrous oxide, a potentially "green" process:
$\ce{C6H6 + N2O → C6H5OH + N2}$
The most widely used process for industrial production (95%) involves the partial oxidation of cumene (isopropylbenzene) via the Hock rearrangement:
$\ce{C6H5CH(CH3)2 + O2 → C6H5OH + (CH3)2CO}$

Source of the reactions: Wikipedia English page for phenol

You could say that the cumene process begins with benzene because it's prepared from benzene and propylene on an acidic catalyst.  Benzene is available from the aromatic stream of your petroleum refinery, and propylene is plentifully available from the top of the cracking unit.

There are many ways to synthesize phenols.  The "best" way depends on what aromatic compound you have available as a starting material.  This web page lists many different methods for preparing phenols; if you find a method that looks interesting, you can click on the example and obtain more detailed information.
A very nice, general method for the synthesis of substituted phenols is the Sandmeyer reaction.  In this reaction you start with an aromatic amine.  The amine is diazotized to produce a diazonium salt.  When the salt is mildly heated it eliminates nitrogen producing a reactive phenyl cation.  When the diazonium salt is decomposed in the presence of water, phenols are produced.

To add to the other answer: there is an other method to synthesize phenols from benzene, namely through the benzyne intermediate. First you can react benzene with $\ce{Br2}$, in the presence of a strong Lewis acid like $\ce{AlCl3}$, to form bromobenzene. $\ce{NaNH2}$, a really strong base ($\ce{pK_{a}}=38$; compared to $\ce{NaOH}$ with $\ce{pK_{a}}=13$), can be used to remove a proton ortho to the bromo-group, where the inductive effect is largest. Then bromide is eliminated, yielding benzyne. Benzyne is incredibly unstable (very deformed "$\pi$ bond") and is electrophilic enough for $\ce{OH-}$ to attack, giving the phenol after protonation by the formed $\ce{NH3}$.
Note that this only happens under the most vigorous conditions where $\ce{NaOH}$ and $\ce{NaNH2}$ are fused together.
Direct electrophilic addition of an alcohol group to benzene is not possible, because there is no reliable source of $\ce{OH+}$.

Phenol is C6H6O, essentially adding one oxygen atom to benzene. The cumene process allows the sharing of two oxygen atoms between benzene and propene and make phenol and acetone. Similarly the similar process by Exxon-Mobil allows the sharing between benzene and cyclohexane which is produced by hydrogenation of benzene.
Therefore, I think the readers can see the trend; if a new process can effectively allow the sharing of oxygen atoms with phenol, it can be good.
Another trend is looking at phenol as adding oxygen atom to the benzene ring. Here, there are journal articles covering electrosynthesis of adding oxygen atoms to benzene. Also, nitrous oxide was tried for adding the oxygen, and releasing the dinitrogen byproduct.
Another trend is looking at phenol as the combination of the phenyl and hydroxyl substituents. Here, substituted benzenes are converted into hydroxyl-substituted benzene. Chlorobenzene is one example.


text:
Do liquid O₂ and N₂ form ideal solutions?
I suspect they do, but could not find data on this. (I mean at temperatures near those used in the distillation of liquid air.)

It's not perfectly ideal.  Try to get access to Excess enthalpies of the liquid mixtures nitrogen + oxygen... for quantitative data.


text:
Cocrystals vs Eutectic mixtures
I was reading around about the different solid states of materials and I got to a point where I have reached the multi-component solids. I was not able to find a major difference between the cocrystal and the eutectic mixture from their definitions. Do you know of any fundamental difference?
The question in another way: if you had a mixture of 2 solids that bind to each other somehow, how can you know whether this is a cocrystal or a eutectic mixture?
Many thanks,

That explains a eutectic but does not answer the question, which is a good one for solid-state chemists.
A cocrystal is a crystalline, multiple component solid with a distinct stoichiometry (molar ratio) between components where a single component is neutral overall(can exist alone without a counter ion, typically non-ionic but a zwitterion counts as a single neutral component) and the components interact by non-covalent interactions that can be partially ionic and has distinctive crystalline characterization data, i.e. to be crystalline a solid the mixed solid phase needs to have X-Ray or electron diffraction data that distinguishes it from its starting components.
So, the eutectic phase of 2 solids, or the solid mixture with the lowest melting pt, can be a cocrystal, but a cocrystal is not necessarily a eutectic. A solid mixture with no regular repeating crystal lattice and no consistent molar ratio (stoichiometry) is called a solid solution. So a eutectic phase could be a cocrystal, or a salt, or a mix of those, or physical mix (solid solution).

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site.

Simple: consider a system whose components are fully miscible in solution and upon solidifying separates into two phases.  At the eutectic point both solid phases melt at the same temperature.  At other compositions you'd see liquid phase and one solid phase.


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Define "reactivity"
Does reactivity of a compound depend on kinetic factors, thermodynamic factors, or both? i.e. how is reactivity defined? For example, out of a) 2-bromopropane and b) 4-bromobutan-2-one, which is more reactive to alcoholic KOH? Does the answer account for kinetic factors, or the fact that b) forms a conjugated ketone, and is hence more reactive?

Does reactivity of a compound depend on kinetic factors, thermodynamic
  factors, or both?

The reactivity of a compound depends on kinetic factors, but... The "but" arises because thermodynamic factors can be a part of the kinetic factors.  Let me explain.
The following figure shows the conversion of a "Reactant" to a "Product".  The product is more stable so the reaction is exothermic, heat is released.  The rate of the reaction is determined by the height of the barrier (the activation energy, $E_{a}$) that a reactant molecule must pass over in order to become the product.  The higher the barrier, the slower the reaction.  Since the barrier height is contained in the kinetic equation describing the speed of a reaction  (for example, the Arrhenius equation)  we say that reactivity is determined by kinetics.

But now look at the reverse reaction.  Suppose you had a bottle of "Product" and heated it to convert it to "Reactant".  This reaction is endothermic, heat must be supplied.  The free energy difference between product and reactant ($\Delta G^o)$ is a thermodynamic factor telling us whether the product or reactant is more stable.  In the reverse reaction, the barrier that we must surmount is higher than it was for the forward reaction.  The barrier for the back reaction is given as $E_{a} + \Delta G^o$.
So while the rate or reactivity depends on the barrier height which is a kinetic phenomenon, we can see how the free energy difference ($\Delta G^o$), a thermodynamic quantity, can be viewed as a factor influencing the barrier height.

out of a) 2-bromopropane and b) 4-bromobutan-2-one, which is more
  reactive to alcoholic KOH? Does the answer account for kinetic
  factors, or the fact that b) forms a conjugated ketone, and is hence
  more reactive?

1-bromobutane might make for a better comparison, in this case the bromine is attached to a terminal carbon as in 4-bromobutan-2-one, and both molecules contain 4 carbons - the only difference is that one is a ketone.  In any case, the answer is based on kinetic factors.  I don't know which one will react faster, but here's my thinking. They should have similar activation energies, but since the ketone reacts to produce a more stabilized conjugated $\alpha,\beta$-unsaturated ketone, the reaction of the bromo-ketone will be thermodynamically more exothermic than the bromo-alkane.  This thermodynamic difference will reduce the height of the kinetic barrier for the bromo-ketone and make this reaction faster.

Small nitpick to an otherwise very good answer: The depicted graph shows an exergonic reaction as it is based on $\Delta G$ It does not make any assumptions if the reaction occurring is also exotherm.|| Another note, the bigger $\Delta G^o$ is, the more energy is provided to following reactions, therefore fueling itself and speeding up the process. This is often the case for explosions.


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Why do the d orbitals have these notations?
Why do the d orbitals have the following notations: $xy, yz, xz, z^2$ and $ x^2-y^2$? What do they represent in their wave-functions?

The notation is shorthand for the hydrogen atom $\ell = 2$ wavefunctions in real form, in cartesian coordinates.  In other words, the Schrodinger equation is solved in spherical coordinates, in terms of spherical harmonics, which involve complex numbers; however, linear combinations of degenerate (equal energy) solutions are also solutions.  This permits the wavefunction solutions to be expressed using real numbers only. These real solutions can be transformed to cartesian coordinates.
$z^2$ represents that the wavefunction is proportional to $(3z^2-r^2)/r^2$.
$xy$ represents that the wavefunction is proportional to $xy/r^2$.
$yz$ represents that the wavefunction is proportional to $yz/r^2$.
$xz$ represents that the wavefunction is proportional to $xz/r^2$.
$x^2 - y^2$ represents that the wavefunction is proportional to $(x^2 - y^2)/r^2$.
See this Table of Real Spherical Harmonic Functions for the a list of functions in both spherical coordinates and cartesian coordinates, for s through f orbitals.

Could you comment on the five versus six d-Orbital problem, as in pure versus cartesian orbitals? Compare http://www.chemissian.com/ch5

@Martin AFAIK, solution of the Schrodinger  equation gives 5 orbitals, as in it is enough to consider 5 orbitals to produce all possible wavefunctions. However, calculations using 5 orbitals basis set for some properties require arithmetic on complex numbers, while moving to 6-orbital representation allows to avoid complex numbers in expense of the need to deal with excessive basis set.

@permeakra The use of six functions is an older trick because it significantly reduced computational cost and because they are essentially equal to the real solutions of the Schrödinger equation (there is one s-type orbital more).

http://www.ccl.net/chemistry/resources/messages/2003/12/12.006-dir/index.html

other references on the 6-function basis: http://chem.wayne.edu/schlegel/Pub_folder/174.pdf ; http://theochem.github.io/horton/tut_gaussian_basis.html

@DavePhD The second link appears broken. I tend to archive (https://web.archive.org/) such links just to be sure, though this might not always be an option.


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Reduction of molality of methanol

How many grams of water would you add to $\pu{1.00 kg}$ of $\pu{1.44m}$ $\ce{CH3OH (aq)}$ to reduce the molality to $\pu{1.00 m}$ $\ce{CH3OH}$?

I have tried $M_1V_1=M_2V_2$. I have tried converting it to kg of $\ce{CH3OH}$ and subtracting it from the $\pu{1.00kg}$ to get kg of $\ce{H2O}$. All of which have worked to no avail.

Do they say the density of $CH_3OH$?

I am a little confused with your units here. What do you mean with 1.44m? Is it an solution ethanol in water with a concentration of $1.44~\mathrm{\frac{mol}{L}}$? or is this a given molality? Or does it refer to a pure solution of ethanol and you have $1.44~\mathrm{mol}$ of it? Getting the units right would be the first step of finding the solution to your problem.

As far as I am aware it is molality. So moles of solute per kg of solvent. There is also no given density."How many grams of water would you add to 1.00 kg of 1.44m CH3OH(aq) to reduce the molality to 1.00 m CH3OH?" Is the whole question that I am given.

If the molality is $1.44 \ce{m}$, it means that, there is 1.44 moles of methanol in $1 \ce{kg}$ of water.
To prepare this solution, we add the mass of  $1.44\times 32= 46.08\ce{g}$ of methanol to $1000\ce{g}$ of water.
The mass of the final solution is $1000+46.08= 1046.08\ce{g} $.
Here, you have 1 kg of this solution (water + methanol). So, it contains $\frac {46.08 \times 1000}{1046.08}=44.05\ce{g}$ of methanol.
The mass of water in this one kg-solution is $1000-44.05=955.95\ce{g}$
The objective of the problem is to prepare a solution of molality $1 \ce{m}$,
i.e, it is prepared by adding $32\ce{g}$ of methaonl to $1000\ce{g}$ water.
It can also be prepared by adding $44.05\ce{g}$ of methanol to $\frac {44.05 \times 1000}{32}=1376.56
\ce{g}$ of water.
Our solution already contains $955.95\ce{g}$ of water.
So, we need to add only $1376.56-955.95=420.61\ce{g}$ of water.

Firstly, 1.44 m of CH3OH yields a solution of 1.046 L, (where 1.44 m means (1.44*32=46.08) grams of CH3OH in 1 Kg of H2o, so solution weight will be 1.046 Kg) but we have a 1 Kg solution, so we say:

1.046 Kg solution contains 46.08 g of CH3OH

1.000 Kg solution contains    x  g of CH3OH

x=CH3OH grams in your solution = 46.08×1000/1046.08 = 44.05 g

and water weight in your solution is 1000−44.05=955.95 g

now, to prepare the new solution (1 m) we say:

1 kg of water contains 32 g of CH3OH

y Kg of water contains 44.05 g of CH3OH

y= weight of water required for new solution =44.05×1000/32=1376.56g

then 1376.56 - 955.95 = 420.6 g  is water that should be added to the 1.44 m solution to have the 1 m solution .

Note: 1 Kg almost equals 1 L (for water)


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Is the refractive index of a solution of four substances the median of the four substances' refractive index?
I need to determine the purity of a solution and we need to discuss about the  'theoretical' refractive index of the substance that we are supposed to have.

There is no single equation that accurately predicts the refractive index of a mixture.  This reference discusses 9 approaches to estimating the refractive index of mixtures and provides the formula for each method. The simplest method is the Arago-Biot approach.  It is reasonably reliable and is described by the following formula
$$n = n_1 \phi_1 + n_2\phi_2$$
where $n$ is the predicted refractive index of the mixture, $n_x$ is the refractive index of the $x^{th}$ component, and $\phi_x$ is the volume fraction of the $x^{th}$ component.  The volume fraction of the first component is given by
$$\phi_1 = x_1 V_1 / \Sigma~ x_i V_i$$
Here $x_i$ is the mole fraction of the $i^{th}$ component and $V_i$ is the molar volume (the molecular weight divided by the density) of the $i^{th}$ component. The more ideal the liquids, the more accurate the predictions.

The answer by ron is quite correct, multiple methods exist for binary mixtures, as his reference details quite nicely.  However, the question posed was for a mixture of four substances. I am unaware of work in refractive index mixing rules beyond binary mixtures, a June 2018 search in SciFinder (American Chemical Society literature search engine) found 23 articles on refractive index mixing, but all seemed limited to binary mixtures. Hence, the question of which rule might be reasonably applicable to a quaternary mixture is unanswered in the literature, as far as I can see.
Naturally, one could try extending the existing rules to ternary or quaternary mixtures in some situations, the Arago-Biot rule mentioned above would be easy to change into a four-term sum, (as would Lorentz-Lorenz and several others), but a significant difficulty would be in finding the volume fractions (which requires partial molar volumes) in a quaternary mixture. Determining partial molar volumes is not hard in a binary mixture and can be found in many texts. Extension to ternary or quaternary mixtures is not at all common, but this reference might be of help.
The main point is, there is no apparent "theoretical" refractive index for a mixture of four components.  Even doing this with two components requires that you know the partial molar volume data (these are not constants but vary with composition!), and take your pick of a dizzying variety of mixing rules. What is very certain is that the refractive index is NOT simply the median of the refractive indices of the components in any general sense. I suspect that the original post was from a bewildered student.  This professor is bewildered by the question as well, and would suggest that the person who posed the question didn't appreciate the difficulty of the topic.


text:
What is the difference between nuclear fission and nuclear fusion?
I always mix up the definition of nuclear fission and nuclear fusion. So, what is nuclear fission and nuclear fusion, and an example for each?

Fusion is an increase in the number of protons in the nucleus by fusing two nuclei. You know fusing means joining together, right? So that can help you remember which is which. "Confusion" comes from the same root word, and happens when you mix things up together in your mind - the "fusion" in "confusion" is the bringing things together.
Fission is a decrease in the number of protons by splitting the nucleus: "fission" means a split.
Ron's given you the classic examples for fission and for fusion.
Either process can absorb energy, or release it.
Iron is the key element here.
For elements with a lower proton count (i.e. a lower atomic number) than iron, fusion releases energy, but fission absorbs energy.
Whereas, for elements with a higher proton count (i.e. a higher atomic number) than iron, fusion absorbs energy, and fission releases energy.
(it's not quite that simple, otherwise we might find a way to fuse two lighter-than-iron nuclei to release energy and create a heavier-than-iron nucleus which we then split again back to the two original nuclei and release even more energy. That's impossible)

Some time ago I made this rudimentary infographics to help the students remember which is splitting and which is merging, maybe it can also assist you:

Nuclear fission involves the splitting of a heavy nucleus into two lighter fragments plus the release of energy.  As an example here is one route by which $\ce{^235U}$, a fissile material, can split apart.
$$\ce{^{235}_{92}U + ^{1}_{0}n -> ^{141}_{56}Ba + ^{92}_{36}Kr + 3^{1}_{0}n + 200 MeV}$$
Note the 3 neutrons generated in the fission process, they are part of the chain reaction; they allow the reaction to propagate at an even faster rate as they go on to react with 3 other $\ce{^{235}_{92}U}$ nuclei.
Nuclear fusion involves the fusing together of nuclei to form a heavier element.  Again, energy is released in the process.  Here is an example of a fusion reaction involving the combination of 2 deuterium nuclei.
$$\ce{^{2}_{1}H + ^{2}_{1}H -> ^{4}_{2}He + ^{1}_{0}n + 17$.$6 MeV}$$

Nuclear fusion is where nuclei of atoms fuse, or come together, to make larger atoms and releases energy. Nuclear fission is when atoms split apart, releasing energy, neutrons, and smaller atoms.
Fusion: In the core of the Sun, hydrogen nuclei smash into each other to make deuterium and helium.
Fission: In the reactor core of a nuclear power plant, neutrons hit uranium atoms to make them split and release energy.


text:
Is it relevant to pursue a master degree or a doctorate when applying to a job in the petroleum field?
I have a few more questions of this type for an interview. ( University work )
Thanks in advance.

Hello and welcome to chemistry.se! Please take the [tour] and visit the [help] to acquaint yourself with this page. The question you are asking is outside of our scope and might be a better fit for [academia.se].

Professional careers in the petroleum industry can be very competitive, so a higher qualification such as masters degree could make you a more attractive candidate for a scientific or engineering position. Many companies nowadays specify post-graduate qualifications as necessary or desirable.
A PhD is generally preferred for research positions.

While it is certainly nice of you, that you tried to answer this question, it belongs to the category of questions that should not be asked on this site. The proper action with these is to flag them to be closed/ migrated (or vote to close if you have 500 reputation).


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Why is nitration of veratrole specific in these positions?
Why do we not need $\ce{H2SO4}$ as an acid catalyst for the di-nitration of veratrole?

Why does the 2nd nitro group go to positions 4 or 5, following the directing effects of the methoxy-group? Doesn't the first nitro group, that has been substituted onto the compound, have any effect on its position?

You'll get 3,5 substituted dinitro because of 1st nitro group effect.

When there are activating and deactivating groups on an aromatic ring, the direction of the electrophilic aromatic substitution is generally controlled by the activating groups.
Think that activating groups improve the substitution in certain positions by increasing its electron density. On the other hand, deactivating groups favor one position over another, not by activating it, but deactivating it less than the other positions. So, the directing effect of a deactivating group is masked by the directing effect of the activating ones, which is stronger.
You most likely don't need the sulphuric acid due to the presence of two activating groups. In contrast, as an example, last nitration step of trinitrotoluene is so disfavored that it has to be carried out in boiling sulphuric acid.


text:
What is tryptose
Occasionally, I work with lauryl tryptose broth for growing coliform.  Being a chemical engineer, I suspected at the start that tryptose is a type of carbohydrate because of the -ose syffix.  However, a search for more info on the web did not provide even a hint of what tryptose really is. Anybody there who knows what tryptose is?  I'm particularly interested in its molecular structure.

Tryptose is not a single compound, therefore it does not have a unique molecular structure.  As you are aware, tryptose is a nutrient broth used for growing coliform. It is a mixture of smaller proteins produced by using trypsine as the enzyme to digest  larger proteins. You might compare it to tryptone which also uses the enzyme trypsine in the digesting process, but only casein is digested.  There is also peptone which uses a  mixture of pankreatin and papain to enzymatically digest proteins.

Tryptose is not a sugar or any particular compound, so there is no corresponding molecular structure.
Tryptose is an animal-derived mixture that includes water soluble peptides.
http://www.bdbiosciences.com/documents/Tryptose.pdf
http://www.bd.com/ds/technicalCenter/brochures/br_1_222880_academia.pdf

Thanks guys.  After nine months of wondering, i now realise that tryptose is not a carbohydrate but a mixture of polypeptides (or [nonfunctional] proteins?) of varying length (about 2 to 80 or so, which i estimated using the molecular weight distribution chart from the Bacto Tryptose pdf and using a rough average of 120Da for the amino acids).


text:
Determine the bond angle in a compound
How can we find the bond angle between 3 atoms in a compound?
Please specify an equation which can be used for all compounds.

There is no such equation. The bond angle can be estimated by some simple rules, such as VSEPR theory.

If you know atomic coordinates (say, from a single-crystal diffraction experiment), then the task is basically the same as finding an angle between two vectors in 3D space.

In carbon compounds Coulson's Theorem can be used to relate bond angles to the hybridization indices of the bonds involved.
$$1+\lambda_{i} \lambda_{j} \cos(\theta_{ij})=0$$
where $\ce{\lambda_{i}}$ represents the hybridization index of the $\ce{C-i}$ bond (the hybridization index is the square root of the bond hybridization) and $\ce{\theta_{ij}}$ represents the $\ce{i-C-j}$ bond angle.

Let's look at some examples.

In the case of methane, each $\ce{C-H}$ bond is $\ce{sp^3}$ hybridized and the hybridization index of each $\ce{C-H}$ bond is $\sqrt3$. Using Coulson's theorem we find that the $\ce{H-C-H}$ bond angle is 109.5 degrees.
In the case of the methyl carbocation ($\ce{CH3^{+}}$) the $\ce{C-H}$ bonds are $\ce{sp^2}$ hybridized and the hybridization index of each $\ce{C-H}$ bond is $\sqrt2$.  In this case, using Coulson's theorem we find that the $\ce{H-C-H}$ bond angle is 120 degrees.
Finally, in the case of acetylene the $\ce{C-H}$ bond is $\ce{sp}$ hybridized and the hybridization index of each $\ce{C-H}$ bond is $\sqrt1$.  In this case, using Coulson's theorem we find that the $\ce{C-C-H}$ bond angle is 180 degrees.

Conversely, if the bond angle is known, the hybridization can be determined.


text:
What do the numbers in the preferred IUPAC name for ascorbic acid mean?
If I look up the IUPAC name of ascorbic acid, I find this: (5R)-[(1S)-1,2-Dihydroxyethyl]-3,4-dihydroxyfuran-2(5H)-one.
I am used to numbering of molecules in a way like the 3,4-dihydroxyfuran part in this molecule name. But now what I want to know is: what do the 5R, 1S and 5H parts signify?
For the 5H I came up with this: furan has two double bonds, so the H might signify that this is not the case any more, because if there is an extra H at the fifth position, this would be impossible. This can be seen in the IUPAC name of 2-furanone: 5H-furan-2-one. I don't know though if this assumption is correct, also because in 5H-furan-2-one the 5H is before the furan part and in 3,4-dihydroxyfuran-2(5H)-one it is after (shouldn't it be 2(3,4-dihydroxy-5H)-furanone anyway?)
Fact is that I can't find an explanation for the 5R, nor the 1S (as I see no sulfur). Could anyone explain me these weird molecule numberings?

The numbering of (R)-5-[(S)-1,2-dihydroxyethyl]-3,4-dihydroxyfuran-2(5H)-one has four parts:
1) Numbering of the furan ring

2) Numbering of the ethyl side chain

3) Stereodescriptors R and S

4) Italicized element symbols
The italic element symbol H denotes indicated or added hydrogen.

Using the PIN mentioned in the question would be much nicer ;)


text:
Non stationary effects of molecular orbitals
We have been learning a bit about molecular orbitals in class.
However, we have always considered things under a quasistatic approximation.
What are some effects that can only be explained by not using the quasistatic approximation ?

Is the "quasistatic approximation" the same as the Born-Oppenheimer approximation?

@Philipp I use that term as in electricity physics. I do not think it relates to the BO approximation.

First off, let's define our terms. The quasistatic approximation:

More generally, the quasistatic approximation can be understood through the idea that the sources in the problem change sufficiently slowly that the system can be taken to be in equilibrium at all times.

So the question boils down to "are there ever any quantum mechanical behaviors that are non-equilibrium?" Or. more precisely, are there fast changes where the system is not really in equilibrium.
Well, as noted in the comments, there's the Born-Oppenheimer approximation which, at least in my opinion, reflects a quasi-static view of electronic structure. That is, the nuclei move slowly relative to the electrons, so despite vibrations, etc. the electrons can be considered in equilibrium with nuclear positions.
Obviously the Born-Oppenheimer approximation fails.
The poster suggested the question was about other non-static behavior. In that case, you fall into time-dependent quantum mechanics, e.g. time-dependent Hartree Fock (i.e., RPA) or TD-DFT.

Linear-response TDDFT can be used if the external perturbation is small in the sense that it does not completely destroy the ground-state structure of the system. In this case one can analyze the linear response of the system. This is a great advantage as, to first order, the variation of the system will depend only on the ground-state wave-function.

Are there cases where linear response breaks and the variation in the system doesn't depend only on the ground-state wave-function. Yes. Consider the effects of large perturbations, like large electric fields, nonlinear optical effects (two photo absorption), etc.
As another fairly easy example, consider photo-ionization. That is, remove an electron from a system. In the quasi-static approximation, you'd have something like Koopmans' theorem with frozen ground-state orbitals. In reality, there's orbital relaxation even though the ionization process is very fast.


text:
What would be an appropriate way to clean up equipment contaminated with ethidium bromide?
I want to clean up some old equipment that was used to visualize nucleic acids with ethidium bromide. It's a UV table, built from glass and metal (stainless steel, I assume).
Ethidium bromide is generally considered scary in a biological lab (probably because most stuff there is pretty harmless), but there seems to be a lot of controversy about the actual danger.
What would be an appropriate way to clean this? I'd simply wipe the whole thing with paper towels and some ethanol and dispose of them in the chemical waste. Would that be sufficient?

You should follow proper protocols at your lab. EtBr is mutagenic in nature. The Environmental Protection Agency (EPA) does not regulate EtBr and other regulatory bodies do not classify it as a carcinogen, however, multiple mutagenicity assays were performed. How much EtBr is on the UV table? Is it heavily stained?

@Jun-GooKwak It has been used for years for staining with EtBr, there is no visible residue, though. There is no protocol for this here as the UV table simply never left the EtBr hood while it was used, no decontamination was necessary.

As you stated in the question, there is some controversy surrounding the mutagenicity of ethidium bromide. However, a recently published opinion piece titled "The Myth of Ethidium Bromide" quickly dispels most of the uncertainty:

Ethidium bromide, as far as can be told from the data, is not a human mutagen. It’s not a mouse mutagen or rat mutagen either. Nor apparently a mutagen in cows and other farm animals, where it’s used in veterinary medicine at concentrations one thousand times higher than the red solutions that are so feared in biology labs, seemingly with no bad effects.

This knowledge does however not give us clear directions as to how the clean-up and decontamination should proceed. For this we can turn to the Material Safety Data Sheet (MSDS).
The relevant sections are Hazards Identification, Handling and Storage, and Exposure Controls / Personal Protection, numbered 2, 7, and 8, respectively.
Hazards Identification tells us that it's fatal when inhaled (in its solid form). This means that when it's not in solution, you should be wearing a respiratory mask or have a fume hood panel separate you from the dust. The fume hood should be operational, of course.
Handling and Storage states that you should:

Avoid contact with skin and eyes.
Avoid formation of dust and aerosols.
Provide appropriate exhaust ventilation at places where dust is formed.

Finally, Personal Protection informs us that you should be wearing a lab coat and glasses (duh), but also gives us some data on the breakthrough time of the substance through nitrile gloves, which is 480 minutes. There is also a section on respiratory protection, but it is prefaced with "Where risk assessment shows [respirators] are appropriate," which is something that you must decide for yourself.
Thus, the answer is, as almost always, it depends:
In any case you will want to avoid brushing and dusting off the contaminated surface to reduce the formation of ethidium bromide dust. Personally, I would recommend to wipe all surfaces first with water (according to Wikipedia the solubility is 40 g / L) until clean using paper towels (and gloves, of course). Following that, one final wipe-down with ethanol may be appropriate, but that is left for you to decide.


text:
Curly arrow mechanism for reaction between sodium borohydride and water?
I was asked to compare the reaction between sodium hydride and water and sodium borohydride and water. I know that the former acts more violently than the latter, but apparently, I must use a curly arrow mechanism to show this.
I think that the HOMO for the latter reaction is the oxygen lone pairs, and the LUMO is the B-H bond, but I'm having difficulty moving on from there.
What attacks what first?

I don’t see much point in a curly-arrow mechanism for the reaction of $\ce{NaBH4}$ with water. And I don’t see the point of doing it with $\ce{NaH}$, either. In one case, you have free hydrides, which go ‘Oh my god, I see a proton, I’m going to jump at it!’ and in the other case you have hydrides thinking the same thing, but being pulled back by a boron saying ‘No, stay here, I need your electrons! I’m poor!’

Their mechanisms are easily searchable online.
This is from 2009 from art-xy.com:

Notice that the reaction would be different in the presence of metal catalysts:
$$\ce{NaBH4 + 2H2O ->[metal catalyst] NaBO2 + 4H2}$$
This is from 2007 from chembook.co.uk:

Which reaction is more violent is already explained by Jan:

In one case [$\ce{NaH}$], you have free hydrides, which go ‘Oh my god, I see a proton, I’m going to jump at it!’ and in the other case [$\ce{NaBH4}$] you have hydrides thinking the same thing, but being pulled back by a boron saying ‘No, stay here, I need your electrons! I’m poor!’

Therefore the former is more violent.


text:
Will room-temperature water in a sealed flask evaporate?
Some liquid water is present inside a sealed flask at room temperature. Is water evaporating, and is the system at equilibrium? First of all, can water evaporate in a sealed flask at room temperature? If so, is the reaction happening a reversible reaction? Any help would be appreciated. Thanks.

Even at room temperature, some of the water molecules in the liquid will have sufficient energy to 'escape' and become vapour. Because the flask is sealed however, some of the water vapour molecules will return to the liquid phase.
Eventually, an equilibrium will be reached when the rate at which water molecules which escape the liquid is equal to the rate at which water molecules in the vapour enter the liquid. At this point, the pressure reached by the water vapour is known as the vapour pressure of water (at the given temperature).
The 'reaction' is not completely reversible since some heat is lost to the surrounding environment through the wall of the flask.

Thanks, very helpful. So the system will eventually reach equilibrium, despite the fact that it is not completely reversible?

If heat can transfer between the flask and environment, equilibrium will be reached when the temperature of the flask reaches the temperature of the environment. At 25C, the vapour pressure of water is 0.0313 atm.


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How can I derive the preferred oxidation states of a transition metal from its electronic configuration?
For example $\ce{Fe}\ (Z=26)$ Short formula: $\mathrm{1s^2\ 2s^2p^6\ 3s^2p^6d^6\ 4s^2}$. If we make the electronic formula we get 4 single electrons ($s=1/2$), which means that the valence in normal state is 4, but in the periodic table it is $2/3/6$. How can I find its valence?

Firstly you need to appreciate that the 4s orbital is at higher energy than the 3d orbitals (despite the filling order suggesting the oposite). This means that the two 4s electrons are lost first giving a valency of two (which is common to all the first row transition elements). The next point is that one of the 3d orbitals is fully ocupied so one of these two electrons will be preferentialy released due to the repulsion of its pair. This results in a valency of three. These are by far the most common valencies of iron. In fact compounds of 4, 5, and 6 oxidation states (valencies) are known but are very unstable, so of little importance.

Can you explain me what " two 4s electrons are lost first giving a valency of two" means? How can they be lost? Does that mean they go direclty to 3d?

The simple valence bond model works well for nonmetals and the molecules they form. It fails completely and badly when trying to extend it to transition metal compounds. There is no easy way to explain why iron, which has four unpaired electrons in the ground state ($S = 2$) is typically found in the oxidation states $\mathrm{+II}$ and $\mathrm{+III}$ in compounds in aquaeous solution.
This behaviour can only be explained by complex molecular orbital theory. It involves a lot of delicate balances that typically cannot be explained a priori without extensive stability calculations. However, the general gist is that the orbitals are split into different energy levels depending on the compound’s symmetry and the orbitals’ irreducible representation. At first approximation, we can assume a stable configuration if there are no degenerate orbitals that are unevenly populated, but single or multiple unpaired electrons are not a problem. Indeed, iron(III) compounds, that are often very stable, even have a spin of $5/2$ in high-spin state. The secret thereto is even orbital population.
I am afraid that I can give you no real guidance for transition metals except learning, rote memorisation of typical cases and developing an intuition for the more complex ones.

The electron configuration of iron is 1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2. Here, the 4s shell is the outermost shell. So, iron has 2 valence electrons. While reacting, these 2 electrons can go to any other electronegative element who requires these electrons to complete its outermost shell. So, this gives iron a valency of 2. Now, the one out of 2 orbitals is completely filled. So, due to repulsion this electron can also be given out. This gives iron a valency of either 2 or 3. However, iron is mostly found in valency 3 as it is mostly stable.