text:
Deducing compounds from spectroscopic data for isomers of formula C4H8O

Here's the problem:

I already know the $\pu{3600 cm-1}$ line indicates the $\ce{-OH}$ group and the $\pu{1640 cm-1}$ line indicates the $\ce{C}$-double-bond-$\ce{C}$ alkene.
From the $\ce{^13C}$ NMR data: the two carbons with chemical shifts of 100+ are $\mathrm{sp^2}$ carbons and the remaining two are $\mathrm{sp^3}$ carbons.
I have tried to work out compound D, and suggest this structure:

I am stuck on figuring out compound C. I suggest the structure below, but am unsure whether if it is correct.

What makes you unsure about it?

@PH13 it's just the 1Hproton shifts that are confusing...especially for the Hb that I have labelled in compound C. I have interpreted 'qn' to mean 'quintet' but if it couples to 1x Ha and 3xHc, technically shouldn't it be a 'doublet of quartets'?

That's not a "normal" quintet and thus it should not be named like one ... it's rather a multiplet or as you say at high resolution it would probably be a quarted of a doublet (vice versa). The coupling constants are very close to each other and this makes it a quintet at low resolutions.

@PH13 ok! I see. Thanks for clarifying this. Do you also agree with my proposed structure for Compound D?

Your two proposed structures are correct, although your rationalisation of the 13C spectrum for Cpd D is not quite correct around the double bond. In a nutshell, carbon chemical shifts can be calculated as a function of their α, β and γ contributions. α for what is directly bonded to that carbon, β for substituents one carbon away, and γ for substituents two carbons away. Almost without exception, alkene carbons exhibit a downfield β shift and upfield γ shift for substituted alkenes. So, for both compounds C and D, the carbon closest to the substituents will be shifted furthest downfield. The β and γ contributions for the following substituents are shown (someone more tech savvy than me might tidy this up) :
[Substituent; β ;γ]
[-H; 0 ; 0]
[-CH2O-; +14.2 ; -8.2]
[-CH3 ; +9.0 ; -7.0]
The electronics contributing to β and γ shifts is not well understood, but is more complex than a consideration of just electronegativities/electron withdrawing potential.
Below are the spectra for the two compounds:

Your confusion with the description of the quintet label arises from how people report splitting patterns in the literature. I have discussed this on another question, but essentially the two methods of reporting splittings are to (a) report the observed splitting pattern (here a quintet) and (b) report the calculated/expected splitting pattern (here a doublet of quartets). As you can see, (or least I can), reporting the observed pattern can lead to some confusion, and gives no real information about how this splitting pattern arises.
Further, reporting couplings to the nearest Hz (here 6Hz) might be fine for some occasions where the difference in coupling for the two different partners is less or approximate to the observed linewidth, however, with a good sample and a good operator on a good magnet, what is reported as a quintet may very well appear different, especially once some apodization is applied. For example, below is a expansion of a simulated spectrum for compound C with Jbc=5.8 and Jbe=6.2 (smaller couplings from d and f ignored), and observed linewidth of 0.5Hz. The top spectrum is the normal spectrum, and the bottom spectrum is what it looks like with a gaussian linewidth function applied (gb 0.1, lb -1). They look quite different, and the bottom spectrum is very hard to rationalise as a quintet, but easily recognizable as a doublet of quartets.

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


text:
Why are rates of reaction faster for ionic compounds than for covalent compounds?
Why is it that ionic reactions (i.e., reactions involving ionic compounds) are faster than covalent reactions (reactions involving covalent compounds), even though the ionic bond is stronger than the covalent bond and hence more time will be required to break the bond?

Why do you think so? Reactions between ions in aqueous solutions can be very fast, but ionic bonds are already broken.

As @Mithoron states, many reactions of ionic substances are in solution, in which the compounds are ionized. Also, non-ionic hydrogen peroxide decomposes rather rapidlyl: https://www.youtube.com/watch?v=wXyJgVlHUCw and https://www.youtube.com/watch?v=i06SnRCQN6U

Not all ionic reactions are not faster, i.e. have a larger rate constant than those between neutral species, however, at similar activation energies there are effects that only ionic reactions suffer. There are two considerations, the dielectric constant of the solvent used, water, hexane, ethanol, acetone etc., and the relative charges on the ions, similar or opposite as well as their value, $\pm1,\pm 2 $ etc.
For slow reactions, that is not diffusion controlled, (see below) the 'primary kinetic salt effect' is important. This shows that the rate constant can increase or decrease depending on the ionic strength of the solution and the charge on the ions. It is based on the Debye-Huckel theory of ions in solution which relates activity coefficients in dilute solution to the ionic strength.
The rate constant for a bimolecular reaction  has the form
$$ \log_{10}\left(\frac{k_2}{k_2^0}\right) = 2\alpha z_az_b\sqrt I$$
where I is the ionic strength, the z the charges on the two ions, $(-1,+1),(-1,-1) $ etc. and $\alpha \approx 1.6 \cdot 10^{-2} \pu{ m^{3/2} mol^{-1/2}}$ is a constant originating in Debye-Huckel theory. At low ionic strength ($ \lt 0.5 \pu{mol dm^{-3}}$ ) the rate constants increase or decrease linearly with $\sqrt I$.
The effect of increasing the dielectric constant (relative permittivity) is to lower the rate constant, thus rate constants are lower in water ($\epsilon = 78$) than in cyclohexane ($ \epsilon = 2.2$) for the same reaction. This is due to the fact that, at a given distance, in a high dielectric solvent the electric field of the ion is greatly reduced compared to its value in a low dielectric solvent. Thus in a high dielectric the ions are far more isolated from one another and so rate constants are only affected when ions come in relatively close contact. Conversely in low dielectric solvents the ions 'feel' the presence of one another at a large separation, this can considerably increase or decrease the rate constant depending on relative charges.
The rate constant is given by
$$ \ln(k_2) =\ln(k_2^{\infty}) - \frac{z_az_bq^2 }{4\pi \epsilon _0 \epsilon d_{ab}k_BT}$$
where $k_2^{\infty}$ is the limiting rate constant for a solvent of infinite dielectric constant. The permittivity of free space is $\epsilon _0$, the relative permittivity $\epsilon$, q is the charge on the electron, $d_{ab}$ the separation of the ions at reaction and $k_B$ the Boltzmann constant. A plot of the rate constant vs. $1/\epsilon$ yields a reasonably straight line  for all but the lowest $\epsilon$ values.
When a reaction's activation energy  is low, less than $\approx 10 \pu{kJmol^{-1}}$ the reactants react so rapidly on contact that the rate constant in determined by how quickly they can approach one another in solution; the rate constant then becomes diffusion controlled and thus depends on the solution viscosity.
Using the Stokes-Einstein equation and assuming equal sized molecules the rate constant has the simple form
$$ k_2= \frac{8RT}{3\eta}$$
where $\eta$ is the solvent viscosity. Water has a value $\approx 1\cdot 10^{-3} $ Pa s or $1$ centipoise (cP) at room temperature. In fluid solution such as hexane the rate diffusion controlled constant is typically $10^{10}$ to $10^{11} \pu{dm^3mol^{-1}s^{-1}}$.
The effects of ions changes the diffusion controlled rate constant for the reasons described above and is now
$$ k_2= \frac{8RT}{3\eta}\left(\frac{x}{e^x -1}\right)$$
where
$$ x=\frac{z_az_bq^2 }{4\pi \epsilon _0 \epsilon d_{ab}kT}$$
and is the term used before.
When the ionic charges are the same and the dielectric constant small the rate constant is greatly reduced, and conversely it is enhanced when the charges are opposite in sign to one another. In high dielectric solvents these changes are greatly reduced. These effects are similar to those described above.

The ionic compounds are held together by strong electrostatic forces, their ions get separated during chemical reactions, And reacts fastly while covalent compouds first have to break the bonds and as they are not ionic so they react slow than ionic compounds

Your answer is correct and clear, hence I am not inclined towards downvoting it. However, it could use more details and research. Please find out more on ChemSE's expectations on answers. Welcome to ChemSE!

covalent compounds are formed by sharing of electrons between two atoms . they form very strong bond and they do not contain ions . their reactions are slow because reactions involve breaking of existing bonds and formation  of new bonds. and additional energy require to break bonds therefore reactions are slow                          ionic compounds formed by transfer of electron from one atom to another so they contain ions. they easily ionized they do not to expend energy to break bonds because they are already dissociated so they show faster reaction                                   hope i helped...??

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

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

sort: votes


text:
Orbital correlation diagram for the cleavage of cyclopropylmethylene carbene
In the last exam we were asked to rationalize the mechanism of the cyclopropylmethylene carbene cleavage into acetylene and ethylene.

In the lecture we discussed pericyclic reactions (cycloadditions, electrocyclic reactions, sigmatropic reactions, ene reactions). I don't think it's a con-/disrotatory ring opening, because in my view only a $\sigma$-plane is conserved over the entire reaction coordinate. What type of ring opening is this; is it a concerted or radical based C-C bond cleavage? Because there is highly probable an elongation of the bond length (as drawn in the stretched cyclopropane moiety), I don't assume it to be concerted but rather radical based.
I tried to classify the reaction in terms of symmetry along its entire reaction coordinate. Because from the task, a singlet carbene, I know it's a $sp^2$ hybrid, so I strongly assume the empty p-orbital to be in conjugation with the Walsh orbitals, more precisely with the Möbius type orbitals.

The problem is very similar to cyclopropylcarbinyl cation:

In the exam I was asked to draw the orbital correlation diagram. For me it looks like two radicals forms the double bond of ethylene and the "biradical" like centers recombines with the "biradicaloid" (in $sp^2$ orbital). I call a carbene a diradicaloid or biradicaloid because the two radical electrons share the same orbital.
To correlate reactants with products I need to combine orbitals with the same symmetry. We only draw the orbitals which are involved either in bond braking or in bond making. How should I account for the biradicaloid singlet $sp^2$ orbital in the orbital correlation diagram? (It is forming the triple bond of acetylene, so there must be a change of hybridization.) For me it looks like to have no symmetry (neither A' nor A'' because it lies in the mirror plane, because it is eclipsed with the $\alpha$-carbon-H.
If it's a radical mechanism I assume the Möbius orbital of A'' symmetry (depicted in green) to correlate with the $\pi^*$ of ethylene.
I assume, the cleavage happens preferently with light, because in this way I can put a photon from A'' (green) into the orbital of A'' symmetry on the reactant site and therefore lowering the activation energy for the reaction.
I assume the orbital correlation diagram to look something like this:

There are some points unclear to me:

Has the empty p-orbital A'' symmetry ?
For me it looks somehow ugly to correlate orbitals which don't have symmetry labels (I think this procedure is not allowed and I need to build a proper set.) How can I build a linear combination in this particulate case, so that the orbital correlation from red to red makes sense?
There is a "$\pi^*$ orbital acetylene,2" unused (not correlated). Did I miss an orbital from the reactant site? (because there is one orbital too many on the product site)

What is the type of the ring opening; is it a concerted or radical
  based C-C bond cleavage?

This is a chelotropic reaction, a subset of cycloaddition\fragmentation reactions.  As to whether it proceeds by a concerted or radical mechanism, well, you haven't presented any experimental data that can help us decide.  For example, if the reaction had been run with the cis-2,3-dimethylcyclopropane analogue and both cis- and trans-but-2-ene were found among the products, we could argue for a biradical, two-step (non-concerted) process.  When we create an orbital symmetry diagram, that just allows us to make a prediction as to whether or not we expect the reaction to be concerted.

I strongly assume the empty p-orbital to be in hyperconjugation with
  the walsh orbitals

There are a number of different conformations available to the molecule, but I agree with you that having the carbenic p-orbital aligned with the plane of the cyclopropane ring (the "bisected" conformation) should be favored.  Just as the bent carbon-carbon bonds with high p-character in a cyclopropane ring stabilize the empty p-orbital in a cyclopropylcarbinyl cation (your drawing above), the same thing should happen here with the carbene's empty p-orbital.

I was asked to draw the orbital correlation diagram.

I have difficulty reading your diagram, so I created a new orbital correlation diagram.

Let me say a few things about my figure in order to make it easier to understand.

The rear cyclopropane bond is bisected by the plane of the screen, this plane also contains the cyclopropyl carbon connected to the carbene center.
The carbene p-orbital is perpendicular to the plane of the screen, the carbenic $\ce{sp^2}$ orbital lies in the plane.
The acetylenic sigma bond lies in the plane of the screen just as the sigma bond connecting the cyclopropyl carbon to the carbenic carbon does.
There are two pi bonds in the acetylene, Px is perpendicular to the plane and Pz is contained in the plane of the screen.
This plane of the screen is the plane of symmetry that we will use to determine the symmetry of the various orbitals and bonds involved in this transformation.
I haven't redrawn the Walsh cyclopropane orbitals, we can refer to your drawing up above.
$\mathrm{\sigma_1}$, $\mathrm{\sigma_2}$ and $\mathrm{\sigma_3}$ refer to the 3 bonding Walsh orbitals (cyclopropane) that you've drawn above.
I haven't drawn any of the antibonding orbitals, as we'll see below, we don't need them in this case.
The empty carbenic p-orbital and $\ce{sp^2}$ orbital should be around E=0, but I've drawn them slightly below for clarity.
The "A" and "S" labels tell us whether the orbital is antisymmetric or symmetric with respect to our preserved symmetry element, the plane of the screen.

Now, if you look at the reactant occupied orbitals and their symmetries, we have 3 orbitals with S symmetry and 1 with A symmetry.  In the products we also have 3 occupied orbitals with S symmetry and 1 with A symmetry. So there is a smooth correlation between reactant orbitals and product orbitals.  This analysis suggests that the thermal reaction can proceed in a concerted fashion.
Miscellaneous

Just because a reaction is allowed to proceed in a concerted fashion does not mean that the reaction will occur.  The most common reaction of cyclopropyl carbenes is ring expansion to the corresponding cyclobutene.  Other concerted reactions may be preferred because they happen to be of lower energy.
Of course, if a reaction is permitted in the forward direction, then it is also permitted in the back direction.  But when we heat ethylene and acetylene we don't generate cyclopropyl carbene - why not?  Well, these are all equilibria and the forward reaction is extremely exothermic.  We remove a highly strained cyclopropane and a high energy carbene and replace them with pi bonds.  Therefore we would expect the back reaction (combination of ehtylene and acetylene) to be extremely endothermic.  The products are much more stable than the reactants and we don't expect to see a significant concentration of reactants at equilibria.  It would be interesting to heat ethylene and acetylene to high temperature and see if any cyclobutene (or its ring opened product, buta-1,3-diene) is formed.  If these products are observed they would be suggestive of the presence of cyclopropyl carbene which then underwent the ring expansion reaction to cyclobutene.

Response to OP's comments:

How can we see in general from an orbital symmetry correlation diagram
  if a reaction is concerted or not?

If all of the reactant bonding MO's correlate with product bonding MO's, then the reaction is allowed to proceed in a concerted fashion.  If some reactant bonding MO's correlate with antibonding product MO's, then the reaction is not allowed to proceed in a concerted fashion.
In the current example, the reactant bonding (occupied) MO's $\sigma_1,~ \sigma_2,~ \sigma_3$ and the carbene $\ce{sp^2}$ correlate with product MO's $\sigma$, ethylene $\pi$, acetylene $\pi_x$ and acetylene $\pi_z$ respectively.  All ground state reactant orbitals correlate with ground state product orbitals.  Therefore the reaction is allowed to proceed in a concerted manner.

Is it okay to say biradicaloid instead of carbene?

No, not really. "Biradical" is generally used in cases where the electrons are on different atoms.  Like when we break a bond in cyclopropane we have a 1,3-biradical.

In the orbital correlation diagram I would predict the acetylene a
  little bit lower because of more s character (sp) than ethylene (sp2).
  May be the energy difference is roughly as high as the difference
  between carbene p orbital A and carbene sp2 orbital.

That may be, but in both ethylene and acetylene the pi electrons (which is what we are talking about) are in pure p orbitals.  My guess was that they would have roughly the same energy.

For the retro cheletropic reaction I assume a photolytical pathway

The retro reaction, like the forward reaction, should be thermally allowed and photochemically forbidden.

According to the paper, I've found here...

I can only see the abstract, but from what I read, I don't think this is a reaction involving a photochemically produced carbene in an excited state.  I think that they are photochemically decomposing the diazo compound and produce the same carbene they generate thermally at higher temperatures.  The photochemical route just allows them to generate the same ground state carbene at very low temperatures.

That's a very nice answer!

How can we see in general from an orbital symmetry correlation diagram if a reaction is concerted or not?

Is it okay to say biradicaloid instead of carbene?

In the orbital correlation diagram I would predict the acetylene a little bit lower because of more s character ($\ce{sp}$) than ethylene ($\ce{sp^2}$). May be the energy difference is roughly as high as the difference between carbene p orbital A and carbene  $\ce{sp^2}$ orbital. 4) For the retro cheletropic reaction I assume a photolytical pathway:

According to the paper, I've found here: http://pubs.acs.org/doi/pdf/10.1021/jo00297a057
"The photochemical decomposition of 2 gave only ethylene, acetylene, and cyclobutene as expected from Scheme I." This would make sense for me if I correlate the carbene p orbital A  with acetylene $\ce{\pi_{x}}$ and $\ce{\sigma_3}$ A with an antibonding ethylene $\ce{\pi^{\star}}$ S because this Walsh orbital $\ce{\sigma_3}$ A looks similar as the antibonding $\ce{\pi^{\star}}$ orbital. I promote one electron from A symmetry into carbene p orbital A photolytically and therefore lower the activation energy.

Thank you for your "Response to OP's comments". Conc. 2) I see now that two interacting radicals which are called biradicaloids must be located at two different centers (http://goldbook.iupac.org/DT07358.html). Conc. 4) Because acetylene lies a little bit more down in energy, the two $\Pi$ orbitals of acetylene with A and S symmetry are correlating with $\sigma_2$ and $\sigma_3$. I'm not sure if we can correlate e.g. the sp2 acetylene part (S) from reactant with the ethylene (S) green line in my drawing here: http://imgur.com/W8eKLs2 , because they belong to different molecule parts?

I tried to rationalize the mechanism with a frontier molecular orbital picture. Do you agree with the choosen HOMO/LUMO pairs? I've set acetylene's LUMO or HOMO with one of the two degenerate non-bonding orbitals: http://imgur.com/TdMfNFa

I agree that we cannot correlate the sp2 orbital with ethylene because (like you said) they are in different parts of the molecule and do not directly interact. However, I disagree that the 2 acetylene pi orbitals correlate with $\sigma_2$ and $\sigma_3$, if you did this then you're forced to correlate the carbenic sp2 with the ethylene pi orbital.  As I noted above I think that the carbenic sp2 correlates with the acetylenic pi_z. 5) Yes, I agree with your HOMO-LUMO pairs.

I guess we agree σ1 correlates with σ ethylene. Is there a general rule which orbitals can be correlated and which cannot (in assumption symmetry requirement is fullfilled)? I would start as follow: 1) σ ethylene with σ1 S sounds obvious because same symmetry, lowest lying 2) $\pi$x with sp2 S 3) $\pi$z with p A 4) Next higher lying orbital $\pi$ ethylene with σ2 S sounds logical 5)  Next higher lying orbital from product should be correlating with σ3. Hence I would suggest photochemical reaction from σ3 into A p proving lower activation barrier as drawn in new scheme: http://imgur.com/p77z52p

I guess we agree σ1 correlates with σ ethylene...yes; Is there a general rule...yes, lowest energy orbitals of like symmetry correlate; 1) yes; 2) no, pi_x with sigma_3; 3) pi_z with sp2 (maybe my drawing is not clear?); 4) yes; 5) sigma_3 correlates with pi_x; all ground state reactant orbitals correlate with ground state product orbitals, therefore thermally allowed photochemically forbidden - sigma_3 does not correlate with pi*.

was a type off from me. 6) Even though I see now, that the correlations of the bonding orbitals result in thermal fashion, I don't quite understand how to correlate the antibonding orbitals because we have used the "exception" to correlate σ2 S only with ethylene S instead of acetylene S: Do you agree with my updated scheme?: imgur.com/uHeiagv

yes, that looks correct.

Okay thank you, that looks good, to know that all orbitals from acetylene product correlates only to reactant orbitals which have non-zero coefficients on the acetylene moiety.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
How might ammonia be created in this mechanochemical reaction?
Can mechanical agitation catalyze reactions?  It appears so (thanks to Mithoron for the pointer!).
We observed the following: We were using 10-micron hBN powder to impact-plate copper bearings (e.g., by running them together in a vibratory tumbler, and other similar processes).  Even though the mechanical processes ran in a standard atmosphere and did not exceed 60C we consistently observed that ammonia was released.  As a control:  I have run the same copper with all sorts of other media through the same mechanical processes and never detected ammonia.
Our conclusion is that the BN is decomposing during the mechanical processing.  But hBN is very thermally and chemically stable: hBN doesn't react with Cu below 1000C!
So is it plausible that mechanical abrasion at this low a temperature would break the B-N bond?  If so, what subsequent reactions (again, in a standard atmosphere) might cause the free N to form $\ce{NH3}$ in detectable amounts, instead of just forming $\ce{N2}%edit$?

https://en.wikipedia.org/wiki/Mechanochemistry

How did you determine that the product was ammonia? How much was produced? Where might the hydrogen have come from?

@ericksonla: That the product was ammonia was deduced entirely by smell and mucosal irritation on opening the processing vessels.  One of the core parts of this question is exactly your last: Where could the hydrogen come from?  E.g., does breaking the B-N bond release something other than heat that might ionize water vapor?  I don't know enough to begin making more than wild guesses like that.

I'd guess it is highly unlikely you stumbled upon a low pressure low temperature nitrogen to synthesize a super industrially important chemical from water. What about something like a nitrile o-ring thats getting hotter than your bulk system and decomposing? Many nitrogen compounds smell substantially similar to ammonia and nitrile is chemically more similar than BN.

@ericksonla - Note that I have plenty of control runs in the mechanical processes.  The only time the ammonia (or something that smells and irritates like it) is produced is when BN powder is introduced.  Also, I just recalled further evidence that it is in fact ammonia: We happened to leave one of the plastic (I think HDPE) processing vessels closed for a week or so after running it.  When we opened it up all of the copper was tarnished a dark color.

Fine hexagonal boron nitride powder is hygroscopic, so this implies there might be some water in your system. According to Reaction of Hexagonal Boron Nitride Nano-crystals under Mild
Hydrothermal Conditions "The reaction between water and hBN nano-crystals starts at very low pressure and temperature, i.e. 220 C and 1.0 MPa [~10 atmospheres]."
Those conditions are likely to be met in a ball-mill or vibratory tumbler, at least momentarily at the points of impact (in a similar way, cavitation can heat water to luminescence!).
A simple test would be to preheat the tumbler and its contents to 150 or 200 C for an hour or two, then seal it airtight before running the tumbling cycle. If there's no ammonia odor, then it indicates that adsorbed water in the hBN had been reacting with it, perhaps catalyzed by the copper bearings, as well as by impact heating and pressure.

Let us suppose the hBN is somewhat hydrated.  And let us assume that locally high temp and pressure can be created in the tumblers.  I agree that both are reasonable assumptions.  Then what is the reaction stipulated?  All I get from that paper is that hBN crystals deteriorate to cBN.  (NB: I actually did bake the hBN before several runs to remove moisture, but not certain if I baked long enough to fully dehydrate, and didn't control or note other variables well enough to establish correlation of that with absence of ammonia production.)

Though there might be other products, a simple displacement is $\ce{2BN + 3H2O -> B2O3 + 2 NH3}$. See http://www.webqc.org/balance.php?reaction=BN+%2B+H2O+%3D+B2O3+%2B+NH3+

I can see that equation "balances," and the left side is consistent with our hypothesis.   But what would motivate the double replacement?  The hypothesis is that mechanical action can break the B-N bond of some BN molecules.  For that reaction to occur the water molecules also have to decompose.  Is there any reason to expect or believe that actually happens, other than that it's plausible and consistent with the original observation?  E.g., is the right hand side more stable, or lower energy?  Or does B have a greater affinity for O than does H?  Or anything like that?

See  Reaction of Hexagonal Boron Nitride Nano-crystals under Mild Hydrothermal Conditions

I just re-read the paper.  As I understand it, the paper suggests that "mild hydrothermal conditions" can transform hBN to cBN.  The only other compound mentioned in the paper is ammonium borate, which is produced at 1200C and 5.6 GPa.  Am I missing something else that applies to my questions?

p. 743: "the reaction between water and hBN nano-crystals starts at very low pressure and temperature, i.e. 220◦ C and 1.0 MPa."

But all they describe is a deterioration in the crystalline structure.  Nowhere do they say that anything other than cBN is produced from the reaction.  Granted, I am not skilled in the art, so if you're saying that for those who are it goes without saying that, of course, they react to form something else, and that something else would obviously be $\cf{B2O3 + 2NH3}$, I'll take your word for it.  Bonus points if you can explain why that's obvious ;)

The classic action of water on a nitride is the formation of ammonia. For example, with magnesium nitride:
$$\ce{Mg3N2(s) + 6 H2O(l) → 3 Mg(OH)2(aq) + 2 NH3(g)
}$$
Water vapor may be entering the system, and friction is adding heat. The latter factors may be fostering the expected chemical reaction, resulting in the formation of ammonia gas.


text:
Storage issues with o-phthaldialdehyde (OPA) reagent used to derivatize amino acids
I have made up an o-phthaldialdehyde (OPA) reagent according to Handbook of HPLC page 715 that will be used to derivatize amino acids before an HPLC run. There is no information given on reagent storage, but the OPA crystals themselves must be stored in the fridge. However, storing this reagent in the fridge causes clear crystals to form that cannot be redissolved into solution. Has anyone had similar issues? Is it possible to store this reagent at room temperature without having the reagent lose strength?

Most protocols I have seen call for freshly preparing OPA solutions immediately before use, or perhaps for storing the OPA solution for at most one day.


text:
Is there a soluble salt, or compound that when dissolved in water results in a solution with density less than pure water?
Generally water soluble salts tend to 'fit between' the molecules of water such that the volume of the resulting solution does not increase much relative to the volume and added mass of the salt. So the result is that the solution generally has a higher density than pure water.
But are there any special cases; any salts or compounds that result in an expansion of the resulting solution volume such that its density actually decreases?
Sodium polyacrylate comes to mind, but if I'm not mistaken, the expansion doesn't take place until a significant amount of the compound is added, and then only after the liquid is changed to rather a gel or solid. My question relates to the resulting solution actually being a liquid with density < 1.0 g/cm³.
So then if not for water, are there solutions of other solvents that might behave this way?

You're asking about reversion of contraction of volume?

I think so. Volume is the only thing that can change to reduce density since mass is being added.

You mention "compounds" after salts. Does this refer solely to solids as otherwise methanol, as an example,  would meet your criteria.

@Beerhunter  I'm considering any compound, including a liquid. So if you could explain details as an answer, please do so. Thanks

Beerhunter is correct in his comment.  Water and methanol are mutually soluble in all proportions, and methanol has a lower density than water, so, if you add lots of methanol to the water, the density of the solution will be less than that of water.

No, you can't dissolve a salt to make a lower density solution.
Instead of thinking about salt ions just fitting into water molecules, think about it as if the salt ions are letting everything fit together better. The water molecules are achieving their current density because of the presence of hydrogen bonding. When you add the ions you introduce better bonds between the charged ions and the polar water molecules. These bonds allow everything to fit together tighter leading to a rise in density.
This decrease in overall bond length is a key to the spontaneity of the solvation reaction. If you introduce an ion that increases the bond length you would see no solubility because they system would be more energetically stable as a bulk solid and water with hydrogen bonding.
The polyelectrolyte you bring up raises a good point, but it is not an exception to this rule. Instead, the polyelectrolyte is so large that it "solvates" into water to form a gel. This new material is not really a solution so much as just a new material with water in it. Hydrated salts would be another example.
This should be the case for all solute-solvent systems. However, this is not to say what you want can't be done. You may have to looks at some reversible reaction or cis/trans type switching material. A simple solute-solvent system won't get it done.

Following up with other materials, certainly mixing a less dense fluid like methanol or ammonia would decrease the density. As a first order answer you can take the average density of both fluids based on the molar ratio. This will get you the ideal mixture density. For materials like methanol that are highly interactive with water there is some ideal nature to the mixture density, but it is usually small and can be calculated using excess volume data. A good resource for this info is DDBST: http://www.ddbst.com/ddb-ve.html


text:
The Salt Room, is there salt in the air?
I work in a restaurant where we dry age our uncut steaks in a cooler. Dry aging is a wonderfully complex process that improves the flavor of richly marbled steaks. Our particular type of aging cooler has a wall of himalayan pink salt (NaCl with other minierals and enough iron oxide to make a pink hue). Our process is even patented by a famous chef. The patent claims that the salt improves the flavor of the beef, and that the closer the beef is to the salt wall, the more flavorful the beef becomes.
I have a limited understanding of NaCl's basic properties (this is necessary in my line of work). It seems to me that no appreciable amount of salt is making its way to the beef through the air. Air simply circulates around the cooler keeping the beef dry and mitigating microbial grotwh. There is also moisture control that keeps the humidity within a certain range, somewhere in the 60% ball park (relative humidity, i think).
When I assert to some of my co-workers that the salt does nothing besides pull some moisture from the air (via hygroscopy), they contest. They cite the gradually diminishing volume of the salt and the smell of sea air (by the sea, not in the room) to support the claims that the salt is present in the aging room atmosphere.
I would counter that the salt is simply ending up on the floor as it is dissolved by the water it pulls from the air via hygroscopy (The floor is cleaned daily). And on the sea air: I know that near the ocean, salt, along with other minerals and organic material, can become suspended in the air in microscopic water droplets and these droplets can deposit these salt, minerals, etc on surfaces. However, the formation of these "sea salt aerosols" involves the churning of the sea water, formation of bubbles, and the action of wind, two of which are not present in the salt wall.
I guess the best summary of my question would be the following: Can damp salt with an air current flowing around it transfer any amount of salt, via the air, to another surface?

The simplest way to prove your point would be to convince everyone to do a double-blind taste test, but that might be a bit of a hard sell.

that would involve setting up a chamber to age almost identical cuts of beef in almost identical conditions. completely impractical in the context of our workplace. Also, i forgot to mention that a 1 cm of outer layer of each primal(whole section of meat) is removed before fabrication (cutting the individual steaks) as it has become quite dry and possibly developed fungal growth(which is actually beneficial to the process). HOWEVER, seeing as this is a patented process, i think a series of double blind taste tests would be warranted and even feasible to prove  the merit of the patent's claims.

In short, no. The vapor pressure of salt is trivial compared to the grams of salt you would put on a steak when cooking it. I think your analysis is good- the most likely impact is that the salt modulates the humidity.

I agree with this answer but it is worth noting that since the air is circulating, small dust particles could impact the salt surface, resulting in ejection of microparticles of salt in all directions, some of which may in fact wind up on the beef.  Or more likely, beads of water could land on the salt surface and form a small amount of saturated salt solution, tiny aerosolized droplets of which could also be blown around by the air.

I suppose we'd have to set up an identical aging cooler and analyze multiple pieces of aged meat to discern whether there'd be any difference in the salt and mineral content. My guess is that the difference would probably not be measurable (except possibly with very sensitive equipment) and certainly not be perceptible by taste.

Unless the salt is a very fine powder and the air is being agitated throughout the room at a very high rate, the salt simply would not be prone to becoming airbone. Based on your description, it doesn't sound like it would be anywhere near fine enough to worry about salt dust possibly contributing anything. Even were it fine, the moisture would quickly clump it and make any possibility of it entering the air in the first place pointless.
Basically, it's marketing. It's a dumb idea from a science standpoint, but the power of suggestion is often more potent than scientific principles. I wouldn't question it too hard if you value your job based on this very concept - it's quite likely the chef knows it does nothing.


text:
How to find the concentrations of the acid-base equilibrium between sulfuric acid and sulfur dioxide for a known pH using Octave?
The system is an aqueous solution of sulfuric acid and sulfur dioxide. I know how much sulfur there is and I know the pH.  The system is in equilibrium.  I would like to use Octave to solve the system.
I set up the following equations, ignoring $\ce{SO3^2-}$ because the pH is around 1.5.
By definition:
$$ \mathrm{pH} = -\log [\ce{H+}]$$
Conservation of mass:
$$[\ce{H2SO4}] + [\ce{H2SO3}] + [\ce{HSO4-}] + [\ce{SO4^2-}] + [\ce{HSO3-}] = M $$
Henderson-Hasselbalch equations for the system:
\begin{align}
\mathrm{pH} &= \mathrm{p}K_\mathrm{a}(\ce{HSO3-}) + \log \frac{[\ce{HSO3-}]}{[\ce{H2SO3}]}\\
\mathrm{pH} &= \mathrm{p}K_\mathrm{a}(\ce{SO4^2-}) + \log  \frac{[\ce{SO4^2-}]}{[\ce{HSO4-}]}\\
\mathrm{pH} &= \mathrm{p}K_\mathrm{a}(\ce{HSO4-}) + \log \frac{[\ce{HSO4-}]}{[\ce{H2SO4}]}\\
\end{align}
I then defined the solution vector to have the following order:
$$ x = [ [\ce{H+}], [\ce{HSO4-}], [\ce{SO4^2-}], [\ce{HSO3-}], [\ce{H2SO4}], [\ce{H2SO3}] ]$$
and my code for Octave is attached.  For some reason I always get negative values for some of the results.
function y = phCalc(x)
   pH = 1.87;
   pKa1 = -5;   % this is a guess to drive H2SO4 to ionization in this system
   pKa2 = 1.987;% from a textbook
   pKa3 = 1.857;% from a text book
   M = 0.0012 / 32 * 1000 ; % total sulfur moles per volume
   y(1) = 10^(-pH) - x(1);
   y(2) = x(5) * 10^(pH-pKa1) - x(2);
   y(3) = x(2) * 10^(pH-pKa2) - x(3);
   y(4) = x(1) - x(2) - x(3) - x(4);
   y(5) = x(2) + x(3) + x(4) + x(5) + x(6) - M;
   y(6) = x(4) / 10^(pH-pKa3) - x(6);
endfunction

This question seems to be about programming and not about chemistry. Therefore, it might be considered off-topic here.

If you can separate the chemistry question from the Octave debugging question, that will help us help you work towards solving your problem.

Can you start by listing all the unknowns, e.g. the concentration of each species you would like to know?  My guess is that you should also include water self-ionization, which would introduce one new unknown: [OH-].  I'm also confused how you can be solving for anything without a charge balance.  Also, what is x supposed to represent in your function?

I don't think this question is about programming.  Presenting the code is actually a nice unambiguous way to describe the chemical calculation that is being attempted.  The problem seems to be with the chemistry rather than with the code...

@CurtF. To be honest, I'm torn about this one. On one hand, the question is about the calculation using the Octave language. On the other, if we just replaced the code with a hand-calculation, there's absolutely no question that this would be on-topic. For that reason, I haven't actually flagged to close yet.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical formulae can be formatted using $\LaTeX$ syntax.|| I think this question is well on topic here, it is just like a  homework question, where the attempt is writing out code. However, I might be able to understand the chemistry behind, but I cannot understand the code. I think it is quite necessary to explain all the variables and constants in use. A chemical equation of the equilibrium would also help.

Actually, I thought about separating the two, but it was difficult.  When I set the subroutine to only calculate one of the items (let's say y(2) = x(2)-0; ) then I'd get a result of the pH correctly.  So it looked like Octave was calculating correctly, but then when I went to the chemistry it looked like I was putting in the equations correctly, but it didn't work.  So I was not successful in separating them.  I am sorry.

Actually the charge balance equation I forgot to add, which is y(4).   There's a tiny error there, there should be a factor of 2 before x(3) since so4 requires two H+.  But it doesn't change the result hardly at all.

I think I see the problem now.  For this system to approach this pH, M has to float a bit more, and cannot be solved using that number.  The lowest value for this M is around 2.5.  At this pH, this system only has positive solutions between [0.00031..0.00085]*32/1000.   Apparently there should be expressed another constraint, or maybe the problem is overconstrained.  What do you think?

You are essentially asking a debugging question. "I always get negative values for some of the results." Can you specify what results? Do not expect people start playing with your code just to figure out what is your problem.

I suspect you know the concentration of the two kind of sulfur species , too. Without that your problem is not well determined. 2) Iterative solution for polynomial equations is not necessarily a bright idea: polynomial equations often have several roots. When you hit one (if you are lucky), and if it is not the one you were looking for then you are surprised why it is not the physical solution.

Your code seems to be calculating the differences between a guessed concentration vector x and a calculated concentration vector.  These differences in concentrations between guessed and calculated are y.  For example, the guessed pH is apparently x(1), and the calculated pH is 10^(-pH).  The difference between them is y(1).
Thus, its no surprise that the y values are sometimes negative.  To solve the problem you need to feed your function to a numerical solver such as fsolve.  That function will take your guessed solution x, other parameters such as pH, the various pKas, and M, and your function phCalc as input, and return the value for x which makes all the y variables zero.  That solution is the one you want.
On the chemistry side, you mention sulfur dioxide, but I don't see any equations for the solubility of sulfur dioxide gas in water.  Additionally, sulfur dioxide is reduced relative to sulfuric acid, so there would need to be an additional reaction, and parameter (extent of reaction or degree of reduction or something) that describes the relative amounts of reduced sulfur (as sulfur dioxide, bisulfite, or sulfite) vs. oxidized sulfur (as sulfate, bisulfate, sulfuric acid, or sulfur trioxide).
There is also the issue of charge balance: what concentration of inert cations (e.g. sodium or potassium or whatever) are you assuming?  Charge balance must hold in bulk aqueous solution.  If you don't have any sodium / potassium etc., then you should include the self ionization of water as an equation, hydroxide as a species, and y(7) = x(1) - x(2) - 2*x(3) - x(4) - x(OH) as an equation.  If you do have sodium/potassium/whatever at significant concentration, you can neglect hydroxide and just go with the equation y(7) = x(1) - x(2) - 2*x(3) - x(4) + x(Na) where x(Na) is the cation concentration, which I have assumed is sodium.
#

That's right.  fsolve() gives an x to the user's function in an attempt to minimize the result (the resultant vector should be zero when a solution is found).  That's just the way it's written.  This code works correctly if I try to calculate only [H+], for example, by setting y(2) = x(2); y(3) =x(3); y(4)=x(4); y(5)=x(5); y(6)=x(6), and leaving y(1) as in the code above.    So did I make a mistake in the chemistry?

There are a few things about the chemistry in your equations that I'm not sure of.  Let me edit my answer to expand.

Actually I don't have any sodium or potassium.  The charge balance I forgot to mention but was included in the calculation of y(4).

Thanks for the help.  I think what I did wrong was assume a mass / pH for this system which is nonrealistic.

What you want to solve, is the following system:
$$
\frac{k_W}{\ce{[H+]}} - \ce{[H+]} +
\frac{\ce{[H2SO4]0}\left(\ce{[H+]}~k_{a1}+2~k_{a1}k_{a2}\right)}
     {\ce{[H+]}^2+\ce{[H+]}~k_{a1}+k_{a1}k_{a2}} +
\frac{\ce{[H2SO3]0}\left(\ce{[H+]}~k_{a3}+2~k_{a3}k_{a4}\right)}
     {\ce{[H+]}^2+\ce{[H+]}~k_{a3}+k_{a3}k_{a4}}
= 0
$$
for both concentrations of $\ce{[H2SO4]0}$ and $\ce{[H2SO3]0}$ with the constraint that both concentrations add up to $\mathrm{0.0375~mol/L}$.
I have no idea how to solved this with Octave (see below), but as I am more used to Mathematica, that was my first approach to see where the error came from. The problem that you have with those negative solutions is, that the constrain about the concentrations does not allow two positive concentrations. Have a look at the following picture:
$\hskip3.5cm$
The pH of a pure 0.0375 M sulfuric acid solution would be somewhat around 1.35. Changing the mixture by reducing the amount of sulfuric acid and increasing the amount of sulfurous acid leads to a rising pH. A pure 0.0375 M sulfurous acid solution would then have a pH of about 1.77.
But for concentrations below around $\mathrm{0.025~mol/L}$ the system is solvable, what can be seen in the next image. You could obtain the pH of 1.87 with a mixture of 0.00086859742 M $\ce{H2SO4}$ and 0.024131403 M $\ce{H2SO3}$.
$\hskip3.5cm$

Now doing this with Octave
I never used Octave before, so there might be a better solution. But this is my code:
>> function y = Conc(x)
ka1 = 10^3; % wikipedia, sulfuric_acid
ka2 = 10^(-1.99); % wikipedia, sulfuric_acid
ka3 = 10^(-1.857); % wikipedia, sulfurous_acid
ka4 = 10^(-7.172); % wikipedia, sulfurous_acid
ctot = 0.025; % total sulfur concentration in mol/L
ph = 1.87; h = 10^-(ph);
oh = 10^(-14)/h;
xH2SO4 = (h * ka1 + 2 * ka1 * ka2)/(h^2 + h * ka1 + ka1 * ka2);
xH2SO3 = (h * ka3 + 2 * ka3 * ka4)/(h^2 + h * ka3 + ka3 * ka4);
y(1) = x(1) + x(2) - ctot;
y(2) = oh - h + x(1) * xH2SO4 + x(2) * xH2SO3;
endfunction
>> [x, fval, info] = fsolve(@Conc, [1; 2])
x =
  8.6860e-004
  2.4131e-002
fval =
  6.3704e-011  2.8976e-011
info =  1

And as you can see, it gives exactle the same solutions as Mathematica. (fval is the value for y(1) and y(2) that are near zero, so everything is fine. info = 1 means, that everything the solution has converged.)

Yes, this is what I recognized as well.  1.87 is not accessible from the 0.0375 M system.  Thank you for your excellent answer.


text:
Identifying lone pairs when looking for formal charges
Why are there no formal charges on the 1st compound? I am specifically referring to the nitrogen atom, which I would think should exhibit a +2 formal charge? However, in David Klein's organic chemistry textbook it is said that the 1st compound shows no formal charge anywhere. The second compound, of course, exhibits a +1 formal charge on the nitrogen atom and -1 formal charge on the oxygen atom. This would make more sense if I were to take into account the lone pair for Nitrogen in compound 1 (and the 2 lone pairs for oxygen), but then why would they hide the lone pairs in compound 1 and exhibit them in compound 2? Please, correct my incorrect thinking.

You're right. In compound 1, you are supposed to assume the lone pairs (note, there are lone pairs on nitrogen, oxygen, and chlorine), but the lone pairs in compound 2 are explicitly drawn. I think you'll find that depending on the exact problem, and what they are trying to test, whether lone pairs are implicit/explicit may not be consistent across the entire book. The book should be consistent within a single problem however, which did not happen here.

Thank you for clarifying the whole situation. I will keep my eyes and mind peeled for these things in the future.

I suspect the lone pairs in compound 2 are shown because this electron configuration is unusual for oxygen in organic compounds. It's possible (I don't have that particular book to check) that this is a convention used throughout.

Drawing common electron pairs like they appear in most organic molecules is only then relevant, when

they play a keyrole in the reaction process
they should explain a special molecule structure
they are of an unusual quantity and thus resulting in a (intermolecular) formal charge

Otherwise especially molecules with a lot of atoms with unpaired electrons (e. g. saccharides) can get confusing. And in addition: they take more time to draw ;)
Of course when learning organic chemistry its good to think of them a lot, because of their high reaction potential.
You can be a 100 % sure that when they would not be there in compound 1 that it would be highlighted somehow, because that would result in a very unstable state.


text:
Why do ligands have such a small effect on overall absorption of a complexed ion?
When a metal cation is complexed, there is strong UV-Vis absorption due to the splitting of its $d$ orbitals, thereby allowing electronic transitions.
My understanding is that ligands contribute very little to overall observed absorption. This is evident, for example, in copper solutions. Copper(II) chloride, copper(II) sulfate, and copper(II) nitrate – all solutions of complexed ions – are all very similar in color and have similar absorption spectra.
I think the simple answer is that transition metals have the capacity to allow for electronic transitions while most ligands do not, but I don't believe it is this simple. Also, it is curious that sulfate, chloride, and nitrate are all colorless when dissolved in solution.
My question is: why do ligands have such a small effect on overall absorption of a complexed metal cation?

An excellent question, though I will point out that in a few cases, ligands actually do affect the color significantly. See, for example, the cobalt chloride humidity indicator

@chipbuster I was not aware of this. Thanks for bringing it to my attention.

@chipbuster Interestingly enough, copper II nitrate is trihydrated and copper II sulfate is pentahydrated. In copper, the waters of solvation seemingly have very little color on absorption spectra. Yet cobalt chloride's color varies greatly depending on hydration.

Copper salts solutions aren't colorless and influence of ligands on color is comparable to that of central ion.

@Mithoron You misunderstand. What I was saying is that the ligand has little effect on the overall color (and absorption) of the solution. Copper(II) solutions are primarily blue, regardless of the ligand.

This is absolutely not true. Many ligands can strongly dictate the color of transition metal solutions.
Your example picks three simple salts and then questions whether the color of the solutions (dictated largely by $\ce{[Cu(H2O)6]^{2+}}$) is very different. No, because the resulting majority complex in aqueous solution is likely identical.
Even taking a simple ammonia complex, e.g. $\ce{[Cu(NH3)4(H2O)2]^{2+}}$ you can see a substantial color change.
Many copper acetonitrile compounds are weakly colored or colorless, with UV/Vis optical absorption occurring near the edge of the red to near-IR.
When you get into octahedral complexes, you can find substantial changes in color due to MLCT and LMCT absorptions, as well as significant modulation of the d-d transitions due to the ligand (i.e., high-field and low-field ligands).

Oh, I forgot to comment on your discussion of ligand transitions. Ligands like water, ammonia, etc. don't have visible electronic transitions - that's definitely true. But $\pi-\pi^*$ transitions are important with aromatic or otherwise conjugated ligands. Plus, as I mentioned MCLT and LMCT transitions involve both the metal and the ligand and are often the dominant absorption feature in complexes.

A large part of the colour of transition metal complexes are indeed explained more on the metal’s side of things. To explain this, as I did in a number of previous questions, it is always helpful to recall the molecular orbital scheme of a random transition metal complex; figure 1 shows such a scheme for an octahedral $\ce{[ML6]^n+}$ complex with σ and π interactions.

Figure 1: Molecular orbital scheme of an octahedral complex including σ and π interactions from the ligands to the metal. Based on the schemes given by Professor Klüfers in his coordination chemistry course at the LMU Munich.
Typically, ligands arrive as octet-rule fulfilling compounds, i.e. they have all bonding orbitals (the ones on the right) fully populated and the antibonding orbitals are energetically removed. Equally typically, the metal side includes a nonzero number of d-electrons which is rarely equal to 10 (if it is, the chemistry is much more complex and less ordered) and rarely equates to fully populated or completely unpopulated orbitals.
Therefore, it is typically the metal’s side of a coordination complex where the valence electrons are located. These are typically placed in d-type orbitals which are labelled $\mathrm{t_{2g}^*}$ and $\mathrm{e_g^*}$ in this scheme. They are both antibonding with respect to $\ce{L\bond{->}M}$ bonds. The colour one can observe typically originates from excitations $\mathrm{t_{2g}^*\rightarrow e_g^*}$, which is defined to a significant extent by the metal’s side of things. The greatest difference between transition metals and thus between their hexaaquacomplexes is their d-orbital energy; mainly where they are located with respect to the orbital energies of the ligand group orbitals.
However note that the ligands can also have a significant effect on colours. Most importantly, changing the coordinating atom from e.g. oxygen to chlorine will move the entire set of ligand group orbitals; their new position will result in different recombination energies and thus a different resulting field split. While Geoff mentioned the colour change from $\ce{[Cu(H2O)6]^2+}$ to $\ce{[Cu(NH3)4(H2O)2]^2+}$ I consider that one a mild change (blue to blue) and offer the difference between $\ce{[Fe(CN)6]^3-}$ (red) and $\ce{[FeCl6]^3-}$ (bright yellow).[1]
And finally, sometimes an extreme colour change is due to a structural reorganisation. For example, adding chloride ions to a pink-ish solution of $\ce{[Co(H2O)6]^2+}$ will give the blue $\ce{[CoCl4]^2-}$ solution; however, the former complex is octahedral while the latter is tetrahedral; thus the molecular orbital schemes are very dissimilar and a wildly different colour is expected.

Note:
[1]: I’ll admit that it’s cheating a little bit. The red colour of $\ce{[Fe(CN)6]^3-}$ is a $\mathrm{t_{2g}^*\rightarrow e_g^*}$ transition while the yellow colour of $\ce{[FeCl6]^3-}$ is actually a ligand-to-metal–charge-transfer excitation, corresponding to $\mathrm{t_{1u} \rightarrow t_{2g}^*}$.


text:
Overpotential for photocatalytic water-splitting with TiO2
What is the meaning of overpotential in the context of photocatalytic water-splitting? That is light drives charge separation inside TiO2 and the generated electrons and holes react with water to produce hydrogen and oxygen.
I have seen many papers talk about the high overpotential associated with the oxygen evolution half of the reaction. I understand overpotentials as the potential you apply to (an electrochemical) cell in order to drive the reaction without energy barriers. However, in a photocatalytic system, there are no bias voltages. So, what is the physical significance of overpotential here?
Thanks!

Photon absorption by titania results in an excitation of an electron from the valence band to the conduction band.  The difference in energy of an electron in the valence band and the conduction band is the band gap, which for titanium dioxide is about 3.3 eV.
What matters for electrochemistry, though, is the absolute voltage of (i) the excited electron, and (ii) the absolute voltage of the "hole" left in the valence band, not the just the difference between them (gap).  That is far more difficult.
The effective electrochemical voltages of the excited electron (or hole) can differ from what the desired electrochemical reaction is.  So there could still be an overpotential.  For titania, one paper I found estimates of the absolute energies of the valence band "hole" and the conduction band electron of -7.8 eV and -4.5 eV, respectively (at pH 0).  The thermodynamic potential for water oxidation is -5.8 eV, 2 eV higher than the hole energy.  That energy is lost if the hole is used to power water oxidation.  You could thus view excited titania having an overpotential of 2 eV for the water oxidation reaction.  In contrast, the theoretical potential for proton reduction is -4.5 eV, close to the conduction band electron energy.
Thus the energetics of bulk titania provide lots of overpotential for the water oxidation reaction but the tradeoff is that a high band gap means only high-energy photons can power the electrochemistry.  Lowering the band gap would make more and longer wavelengths of radiation available to power water photolysis, but the band gap can only be lowered in certain ways.  For titania, it would not be very desirable to lower the energy of the conduction band, because that would mean excited electrons would not be able to reduce protons to hydrogen .  Thus, the only way to lower the band gap is by raising the energy of the valence band, which would both lower the overpotential and and make more types of radiation accessible for powering water photolysis.

Hi Curt, I sort of see what you are saying. But I'm still a bit confused. I'm linking a paper here if you don't mind (http://pubs.acs.org/doi/abs/10.1021/ja105340b). According to this paper, there is an overpotential of around 0.7 eV for oxygen evolution - which I understand to be the highest free energy barrier in the free energy profile for oxygen evolution. However, they also use a U value (which I believe is well known in electrochemistry circles) of 1.93 eV.

This they say is the electropotential of a hole at the VBM (that I would interpret as the band gap of the material). What I don't understand though is the relation between this U value and the overpotential. Most of the electrochemistry papers I've read don't explain the connection between them, so I'm wondering if it is well-known in the electrochemistry field.

Specifically if you read the abstract of the paper I linked, it says that the overpotential is 0.7 V (which is the highest free energy barrier in the free energy profile for the OER), which is 1.93 V wrt the SHE (and this enters the equation for calculating the free energy change as the eU term, which I interpret as the bandgap). Thanks for your help!

I'm afraid I'm getting out of my depth pretty quickly...sorry I can't help more and thanks for posting the paper, it was interesting to look at.

Hello. Do you know a modern reference to Theoretical Electrochemistry? I only know the book of Vetter.

@NikolajK: Bard is a classic. Gileadi's physical electrochemistry is also supposed to be good. Disclaimer: I just started working on electrochemistry, so I haven't read Bard's book fully and only just borrowed Gileadi's book.


text:
What happens if you cool water in a container too small for it to freeze?
Freezing a full bottle of water tends to shatter the glass bottle. What if you used something tougher than glass, like diamond? What would happen if you kept dropping the temperature, but restrained the liquid volume so it couldn't freeze and what sort of force would the liquid exert on the container's walls?

You'll want to look into the coefficient of thermal expansion for water. If you integrate the function over the temperature range in question, you will get the volume change. From that (as well as mass of substance and volume of container assumed to be static), you can calculate force (or pressure, if desired).

For the record, we can currently generate pressures of nearly 640 gigapascal (roughly 6 million atmospheres) in specialized diamond anvil cells.

Good question.  Let's assume the container is infinitely strong, non-deformable, and constant in volume.  Let's also assume that cooling the water is an equilibrium process -- that way, we won't have any supercooling.
At equilibrium, the first tiny bit of ice that freezes will take up more volume than the water it froze from.  This will raise the pressure on the rest of the water.  Eventually the pressure may get so high that additional freezing of more water is not thermodynamically favored.
Of course, as the pressure is raised, even the solid ice compresses a bit, freeing up a bit more volume for the liquid water.  According to this paper from 2004, ice is less compressible than water, so as a starting assumption, it may be approximately true to neglect the ice compression effect.
Figure 4 from that same paper gives the freezing point depression of water as a function of pressure:

To fully answer your question, in addition to that data, an equation that gives pressure as a function of ice volume would also be needed.  If we make the assumption I was talking about above -- i.e. that ice is incompressible, then from the data point that water has a constant compressibility of 46.4 ppm per atm we can come up with a very simple version of that equation.
$\frac{\Delta V_{water}}{V_{water}}=46.4 \times 10^{-6} \times P$, where P is the pressure in atmospheres.
Before freezing of a fraction $X$ of the water:
$$V_{ice} = X V_{tot}$$
$$V_{water} = (1-X) V_{tot}$$
After freezing:
$$ V_{ice} = X V_{tot} 1.11 $$
$$ V_{water} = (1-X) V_{tot} - \Delta V_{ice} $$
Combining those equations, you can get
$$0.11 \frac{X}{k(1-X)}= P$$, where $k$ is the compressibility of water.  If even 1% of the water in the container freezes (and all our assumptions are true), then the pressure will be 24 atmospheres!  Freezing 10% of the water would mean a pressure of 260 atmospheres.  Looking at the chart above, reaching this point would require a temperature of only 271 or 272 K, i.e. only -1 °C or -2°C.  Freezing 45% of the water would reach a pressure of 2000 atm, already off the chart above -- but the temperature required to reach that point would only be 253K or -20 °C, the setting of the average home residential freezer!  ((Of course, at these extreme pressures, (i) ice is actually compressible, and (ii) the compressibility of liquid water is not constant but also a function of pressure, so the calculations would get quite a bit more complicated.))
The lesson is that for even moderate degrees of cooling, you'd need a very, very strong container.

I would add that 2000 atm (~200 MPa) we start seeing other phases of water, such as ice III.  Given its density (1.16 g/cm3), I would expect either to see compressed ice I or compressed ice I mixed with ice III.

Wouldn't it be something like instant water freeze experiment?

You probably would ultimately get to situation where you get only ice compressed to starting density of water.

If the container isn't infinitely strong, the container could simply burst. This is exactly why you shouldn't leave beer contained in glass too long in the freezer.

You might want to add a reference to the water phase diagram

Bobson -- that's a good starting point, thanks for adding the link.  The paper I linked to was actually rigorously evaluating methods for calculating the relevant portion of the water phase diagram and comparing to experimental data.  It's a better starting place for experts, but you are right that beginners should start with wikipedia.

But what happens when you go all the way - what is the pressure at 0 K? (or what does it approach in theory)

Pressure is not well defined at 0 K I don't think.

If, at atmospheric pressure, ice is less dense than liquid water (it is) then we have to use pressure to keep it from expanding during the freezing process. The amount of pressure required depends on the compressibility of water and ice in a fairly complex way.
The glass bottle in your example breaks because the pressure exceeds the tensile strength of the bottle. The details of converting pressure to stress in a material is more mechanical engineering than chemistry, but I would refer you to Wikipedia's cylinder stress page if you care to actually work through the calculations. If you made your bottle out of some magical material with infinite strength, the pressure would just keep increasing.
The details of what happens are pretty neat. Water can actually form lots of different kinds of ice with different crystal structures. Check out Wikipedia's phase diagram for more info. If you follow the freezing curve on that line, you'll see that with enough pressure the freeze temperature drops to ~250K!


text:
Is the redox method applicable to all chemical reactions?
Is it possible that some chemical reactions can't be balanced through the redox method? Because it seems to me that the elements on the reactants side have the same oxidation states with the product side.

Of course the method is only applicable for reactions where the oxidation states actually change. A typical example where this would not be the case are acid/base-reactions. To improve your question, it would be good if you could add which "elements on the Reactants side" you are referring to.

Sometimes you aren't able to use the redox method to balance a chemical equation. This is because the redox method is only used to balanced an equation where a element gets oxidised or reduced. However if there is a chemical equation where no element is getting oxidised or reduced, then you won't be able to use the redox method to balance the equation. However, fortunately most of the time these type of equations can be easily balanced. Consider the following unbalanced equation: $$\ce{Ba(OH)2.8H2O (s) + NH4SCN (s) -> Ba(SCN)2 (s) +  H2O (l) + NH3 (g)}$$
Here none of the elements are reduced or oxidised, hence you would not be able to balance this equation using the redox method.


text:
How to produce dry pure oxygen
I'm supposed to produce dry and pure oxygen using the decomposition of hydrogen peroxide with a manganese dioxide catalyst.
$$\ce{2H2O2 (aq) → 2H2O (l) + O2 (g)}$$
The gas will be collected in a bottle by downward displacement of water. How can I safely move the oxygen gas through the drying agent without mixing in atmospheric gas?

Is it for a laboratory experiment? What is the scale of such production?

Since you're collecting the oxygen over water why bother ?!?

Commercial drying agents, such as Drierite (anhydrous calcium sulfate), or other desiccant, should do the job for small quantities of oxygen.
Most inorganic desiccants should not react with diatomic oxygen; as long as the oxygen is produced separately from the desiccant, there should be no issue. (Dropping 90% or stronger $\ce{H2O2}$ on many substances can cause dangerously rapid decomposition, but with 30% or lower concentration, that danger is minimal.)
As far as preventing atmospheric gases from entering, first bleed the apparatus of residual air by discarding the initial output. Seal the later containers underwater, e.g. by sliding a glass plate across the mouth. See Janice Crowleys' demo on YouTube.


text:
Making Cast Iron from Pig Iron
In making cast iron from pig iron - the cruder form of iron - coke and limestone are added to pig iron and scrap iron in a furnace. Cast iron is   more refined form of iron and has lesser carbon content as well. What exactly occurs in this process of making cast iron - what reactions happen in the furnace? Why is coke added as well, when the purpose is to decrease the carbon content?

I have quoted the following answers from the UC Davis ChemWiki page on "The Extraction of Iron."

Why is coke added as well, when the purpose is to decrease the carbon content?

Coke is used primarily as a reducing agent (see below) and comes out as carbon dioxide (and thus does not increase the carbon content of the molten iron).

What exactly occurs in this process of making cast iron - what reactions happen in the furnace?

The reactions are given below in the narrative of the process.
The Blast Furnace
The common ores of iron are both iron oxides, and these can be reduced to iron by heating them with carbon in the form of coke. Coke is produced by heating coal in the absence of air. Coke is cheap and provides both the reducing agent for the reaction and also the heat source. The most commonly used iron ores are hematite, ($\ce{Fe_2O_3}$), and magnetite, ($\ce{Fe_3O_4}$).
The air blown into the bottom of the furnace is heated using the hot waste gases from the top. Heat energy is valuable, and it is important not to waste any. The coke (essentially impure carbon) burns in the blast of hot air to form carbon dioxide - a strongly exothermic reaction. This reaction is the main source of heat in the furnace.
$\ce{C + O_2 \rightarrow CO_2}$
The Reduction of the Ore
At the high temperature at the bottom of the furnace, carbon dioxide reacts with carbon to produce carbon monoxide:
$\ce{C + CO_2 \rightarrow 2CO}$
It is the carbon monoxide which is the main reducing agent in the furnace:
$\ce{Fe_2O_3 +3CO \rightarrow 2Fe + 3CO_2}$
In the hotter parts of the furnace, the carbon itself also acts as a reducing agent. Notice that at these temperatures, the other product of the reaction is carbon monoxide, not carbon dioxide:
$\ce{Fe_2O_3 + 3C \rightarrow 2Fe + 3CO}$
The temperature of the furnace is hot enough to melt the iron which trickles down to the bottom where it can be tapped off.
The Function of the Limestone
Iron ore is not pure iron oxide - it also contains an assortment of rocky material that would not melt at the temperature of the furnace, and would eventually clog it up. The limestone is added to convert this into slag which melts and runs to the bottom. The heat of the furnace decomposes the limestone to give calcium oxide.
$\ce{CaCO_3 \rightarrow CaO + CO_2}$
This is an endothermic reaction, absorbing heat from the furnace. It is therefore important not to add too much limestone because it would otherwise cool the furnace. Calcium oxide is a basic oxide and reacts with acidic oxides such as silicon dioxide present in the rock. Calcium oxide reacts with silicon dioxide to give calcium silicate.
$\ce{CaO + SiO_2 \rightarrow CaSiO_3}$
The calcium silicate melts and runs down through the furnace to form a layer on top of the molten iron. It can be tapped off from time to time as slag. Slag is used in road making and as "slag cement" - a final ground slag which can be used in cement, often mixed with Portland cement.
Cast Iron
The molten iron from the bottom of the furnace can be used as cast iron. Cast iron is very runny when it is molten and doesn't shrink much when it solidifies. It is therefore ideal for making castings - hence its name. However, it is very impure, containing about 4% of carbon. This carbon makes it very hard, but also very brittle. If you hit it hard, it tends to shatter rather than bend or dent. Cast iron is used for things like manhole covers, guttering and drainpipes, cylinder blocks in car engines, Aga-type cookers, and very expensive and very heavy cookware.


text:
Is there a specific chemical property or concept that explains/predicts the triboelectric series?
Is there a chemical property (perhaps electron affinity?) that can be used to predict where a substance would fall in a triboelectric series such as this one? Or would the cause of the charge imbalance vary on a case by case basis?


text:
Why use potassium iodide instead of copper iodide in Sandmeyer reaction?
The Sandmeyer reaction is an extremely useful reaction for the functionalization of aromatic rings through a diazonium intermediate. For conversion of a diazonium intermediate to the corresponding chloride, bromide, or cyanide, the copper(I) salt is used. Surprisingly (to me at least) for iodination, potassium iodide is the reagent of choice, as described in Maitland Jones's Organic Chemistry (and elsewhere I'm sure), but no rationale is given. SciFinder searches also show that potassium iodide is most frequently used.
Is there something special about copper(I) iodide that makes it unsuitable for use in the Sandmeyer reaction?

Sandmeyer reactions using iodides, water, and thiols do not require the copper(I) ion as a catalyst for the reaction to proceed whereas your other nucleophiles do (though not necessarily stoichiometrically).

As you can see the copper(I) ion is important in reducing the diazonium to nitrogen and forming a radical.  Iodine, oxygen, and sulfur must be able to be oxidized by diazonium and therefore do not require copper.

Is  there any evidence that thiols, water (and presumably alcohols), and iodide are oxidized? Oxidation of thiols and iodide are (relatively) facile, but I don't consider oxidation of water to be in the same class. If it has something to do with nucleophilic strength, surely cyanide is a stronger nucleophile than water, yet cyanide requires presence of copper(I). It's probably not appropriate to call copper a catalyst, since it's present in stoichiometric amounts.

The copper is not always added in a stoichiometric amount.  The reaction with just water requires an input of heat to effect the reaction.

The answer says that oxygen must be oxidized by diazonium in order to avoid the need for copper. Is there evidence that water is oxidized in this reaction?

This is incorrect conclusion - only catalytic reaction is with redox steps. It's not like it won't work at all with Cl and Br, but Cu makes an improvement.


text:
Is 1-methoxy-2-methylpropane a positional isomer of 2-methoxy-2-methylpropane?
My book doesn't classify them as positional isomers but the position of the methoxy group is different in both the molecules. So, I guess they should be positional isomers.
I understand that the two compounds are metamers. But can't a metamer be a positional isomer?

As far as my knowledge, I also believe that they should be positional (structural) isomers. Did your textbook specifically state that they weren't positional isomers?

Yes. There was a question to identify positional isomers from four pair of options. This wasn't in the list.

2-methoxy-2-methylpropane is a synonym for MTBE (Methyl tertiary-butyl ether). The name "propane" is a little bit missleading, because it comes from the standardized IUPAC nomenclature system, which names the  ether-oxygen together with the smaller alkyl group an "alkoxy" group with the suffix "-oxy" (in our case: methoxy) and takes the larger alkyl part as its base identifier (here propane) when describing ethers. So MTBE is referred as an ether with all its chemical properties.
Now to the isomerism: Structural isomers have the same chemical formula but different constitutions (arrangement of atoms, bonds, groups). So they definitely must be constitutional isomers.
Constitutional isomerism is divided into 5 subcategories:

Chain isomerism
Positional isomerism
Functional isomerism
metamerism
tautomerism

As you already mentioned: they are metamers. Thats because different alkyl rests are attached to the same functional group (ether). This is the first reason why they are not likely to be positional isomers. And the other reasons might come from the "moving substituent" (alkoxy group), which has a carbon atom (the carbon structure has to be constant within positional isomers, therefore moving the -OCH3 could mean a difference) and is no functional group.


text:
How does the electronic configuration for this acid oxide correspond to its structure?
The following electronic configuration and drawing of the structure are given:
$\mathrm{(1s)^2 (2s)^2 (2p)^6 (3s)^2 (3p)^6 (3d)^{10} (4s)^2 (4p)^4}$

How does this image correspond with the electronic configuration?
My train of thoughts so far:
Since the configuration has 34 electrons, I assume the element Z is actually selenium (atomic number 34). It could also be gallium since Z has no charge and has five covalent bonds to look like krypton. This is mere speculation since I just don't know how to approach the problem.
I am unsure what kind of bonds are present here. The tetrahedral form, even though it consists of a total of five bonds, indicates hybridization, but I am completely unfamiliar with hybridization outside of carbon-like hydbridization (four bonds). I am guessing one double bond leads to sp$^2$ hybridization.

Just to confirm, the atom/group pointed up from Z is supposed to be carbon/methyl?

Alas, that wasn't given in the question.

It might not matter, but just checking to see if there was a possible typo.

I don't really understand the question.

The compound is $\ce{SeO3}$, and here is my reasoning:
From the electronic configuration given in the question, the central atom is selenium (atomic weight 34).
As you can see from its electronic configuration, the valence shell has 6 electrons and it goes on to form 4 covalent bonds with oxygen, of which 3 are $\sigma$ bonds and 1 is a $\pi$ bond. Therefore, it shares 4 electrons and has a lone pair. Its steric number is 4 and hence its hybridisation is $sp^3$.
With the tetrahedral geometry, the lone pair will occupy the uppermost position (which is shown by a dash) and the three oxygen atoms will occupy the base of the distorted tetrahedron (due to the lone pair and arising from VSEPR theory). Further, there would be a negative charge on two of the oxygen atoms.
The central atom is not gallium since its electronic configuration is $(4s)^2(4p)^1$. And krypton doesn't react with oxygen as far as I know.

SeO3 doesn't have lone pair on selenium, also there are two negative charges so it's an anion not oxide like OP suggests.


text:
What is the rationale of the IUPAC name for DDT?
The structure of DDT is

One of the benzene molecules should be the parent chain as it has a greater number of carbon atoms than the straight chain (which has just two carbon atoms). Taking benzene as the parent chain, I think that the name of the compound should be 1-Chloro-4-((1'-(4"-chlorophenyl))-2',2',2'-tricholoroethyl) benzene.
But the Wikipedia article states that the IUPAC name of the compound is
1,1,1-Trichloro-2,2-bis(4-chlorophenyl)ethane.
Why is the ethyl chain taken as the parent chain when it has lesser number of carbon atoms than the benzene ring?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), a structure with two or more identical units that are linked by di- or polyvalent groups can be named in two ways:

by ordinary substitutive nomenclature, in which one of the parent structures is chosen as the senior parent structure and the remainder of the structure is expressed by substituent prefixes
by multiplicative nomenclature, in which two or more parent structures are connected by symmetrical or unsymmetrical single or concatenated substituent groups

Both methods yield correct names, which may be used in general nomenclature. However, the IUPAC recommendations define a preferred IUPAC name (PIN), which is the name preferred among two or more names generated from two or more IUPAC recommendations.

P-45.1 MULTIPLICATION OF IDENTICAL SENIOR PARENT STRUCTURES
P-45.1.1 Multiplicative nomenclature is senior to substitutive nomenclature for generating preferred IUPAC names to express multiple occurrences of identical senior parent structures, other than alkanes, in the name of the parent structure (see P-51.3.1). In most cases multiplicative names are shorter than regular substitutive names. A preferred IUPAC name is generated by multiplicative nomenclature when the following criteria for its use are met (see P-51.2.3).
(1) the linking bonds (single or multiple) between the central substituent group of the multiplicative group and all subsequent structural units are identical; and
(2) the multiplicative groups, other than the central multiplicative group, are symmetrically substituted; and
(3) the locants of all substituent groups on the identical parent structures, including suffix groups, are identical.
When these conditions are not met, substitutive nomenclature generates preferred IUPAC names.

The given structure of the DDT molecule fulfils these requirements; therefore, the preferred IUPAC name is generated by multiplicative nomenclature as follows.

P-15.3.1.3 Multiplicative name formation
Multiplicative names are formed in accordance with the number of occurrences of identical structural units as defined in P-15.3.1.1 and the relationship of the linking multiplicative substituent group to the identical structural units.
When a compound contains identical structural units as defined in P-15.3.1.1 linked by a symmetrical simple, compound, complex, or concatenated multiplicative group (a di- or polyvalent substituent group), it is named by stating successively:
(a) the locants for the positions of substitution of the linking multiplicative substituent atom or group to the identical parent structural unit (the locant 1 is omitted when alone in the name of a mononuclear parent hydride);

Here: 1,1'

(b) the name of the linking multiplicative substituent atom or group;

Here: 2,2,2-trichloroethane-1,1-diyl

(c) the numerical prefix ‘di’, ‘tri’, etc.; and/or ‘bis-‘, ‘tris-‘, etc., with no elision of the final vowel before the name of the identical parent structural unit;

Here: bis

(d) the name of one of the identical structural units including the principal characteristic group and substituents, if any, enclosed in appropriate enclosing marks (see P-16.5).

Here: 4-chlorobenzene

The numbering of the identical parent structural unit is retained and, when there is a choice, the locants of the point of substitution by the linking multiplicative substituent groups on the identical parent structure are as low as possible.

Here: 1,1'-(…)bis(4-chlorobenzene) (not 4,4'-(…)bis(1-chlorobenzene))
Therefore, the preferred IUPAC name is 1,1'-(2,2,2-trichloroethane-1,1-diyl)bis(4-chlorobenzene)

According to pubChem and MarvinSketch (current as of 8/12/2015), the IUPAC name is 1-chloro-4-[2,2,2-trichloro-1-(4-chlorophenyl)ethyl]benzene, not 1,1'-(2,2,2-trichloroethane-1,1-diyl)bis(4-chlorobenzene). I'm not sure why there's this discrepancy in the interpretation of the nomenclature rules, though I'm looking into it.

Hmm, according to the MarvinSketch documentation, "Some aspects of nomenclature are only partially implemented, in particular [...] multiplicative nomenclature. In those cases, a non-preferred but chemically correct name will be generated." It seems that maybe pubChem is using the same or a similar name generating engine as MarvinSketch.

why is the name of the substituent 2,2,2-trichloroethane-1,1-diyl? Where does the 1,1-diyl part come from? According to my knowing shouldn't it be 2,2,2-trichloroethyl? because there is only one ethyl group?

To be honest, I don't see how the last quoted rule justifies it. The numbering of the identical parent structural unit is retained and, when there is a choice, the locants of the point of substitution by the linking multiplicative substituent groups on the identical parent structure are as low as possible. So, we should retain the numbering, but we have no multiple different choices. Isn't there a more descriptive rule in effect? (For now I've made somewhat loose answer elsewhere)

In short
The rule for identifying the parent chain isn't just "longest".
In fact, "longest" is the fallback for when no other means are appropriate. The most important decider is the number of suffix substituent groups, which in this case is (chloro-)phenyl.
It's not a benzene compound, because each benzene ring only has one (n-ethyl), whereas the ethane chain has two.
In long
To understand why this should be so; let's consider the actual precedence list for naming the parent chain (from Wikipedia):

It should have the maximum number of substituents of the suffix functional group. By suffix, it is meant that the parent functional group should have a suffix, unlike halogen substituents. If more than one functional group is present, the one with highest precedence should be used.
It should have the maximum number of multiple bonds
It should have the maximum number of single bonds.
It should have the maximum length.

For DDT, we can use rule one: The ethane in the middle has two phenyls stuck to it, but each phenyl only has one ethyl; so the ethane wins.
For a simpler example of the same principle at work, consider 1,4-dioctylbenzene: A benzene ring with two octyl tails coming off of it. There are again two candidates for the parent chain, one making it a benzene, one making it an octane, 1-(4'-octylphenyl)-octane.
But because choosing the benzene gives a parent chain with two substituents, it becomes 1,4-dioctylbenzene.
Edit:
There seems to be some confusion as to whether the rules presented above are the currently recommended ones. A careful reading of the 2013 Blue Book would be necessary to figure that out, and the text isn't online yet.
However: The priority rules above were the rules used to arrive at 1,1,1-trichloro-2,2-bis(4-chlorophenyl)etane.

According to chemspider, the systematic name is 1,1'-(2,2,2-Trichloro-1,1-ethanediyl)bis(4-chlorobenzene). It would be good if you could cite the version of the recommendations you are referring to as there recently (2013) has been a quite significant change.

@Martin-マーチン Of course, the chemspider page also doesn't cite the version of the recommendations, and they're all "Validated by experts", so… But going in and adding a reference.

@Martin-マーチン And to add to the confusion, dioctylbenzene (where the same principle is applied) is listed only as dioctylbenzene on chemspider. I just don't know what to believe anymore. ;_;

According to pubChem, the systematic name is 1-chloro-4-[2,2,2-trichloro-1-(4-chlorophenyl)ethyl]benzene. Which is right? Haha

I just checked the name using MarvinSketch, which uses the updated IUPAC rules. It also calculated the name to be 1-chloro-4-[2,2,2-trichloro-1-(4-chlorophenyl)ethyl]benzene.

@austinian That seems to be the answer, then. Still, seems like a regression if true; the ethane parent chain name seems clearer to me. Ah well, ah well.

Found that there's a limitation in MarvinSketch WRT multiplicative nomenclature, so the calculated name is not the preferred name. Probably same issue with pubChem, since it specifically says "computed descriptor" in the heading that the IUPAC name is under.


text:
Why is the pKa of the thiol group in N-acetylcysteine higher then the pKa of the thiol group in cysteine?
I understand why cysteine has a $\mathrm pK_{\mathrm a}$ of 8.3. But the $\mathrm pK_{\mathrm a}$ of thiol in N-acetylcysteine is 9.27. Since there is an acetyl group attached to the nitrogen, doesn't this lower the $\mathrm pK_{\mathrm a}$ via the inductive effect?

Shown below are the relevant ionization events of cysteine and N-acetylcysteine. The question is asking about the second ionization of each, the deprotonation of the thiol groups, but we see a similar difference in pKas for the carboxylic acid groups as well. The difference in each case is an ammonium group vs. an acylated nitrogen. It's true that the acetyl group is an electron withdrawing group, but it turns out that the cationic ammonium group is even more inductively electron withdrawing. For both the first and second ionizations, the difference in electron withdrawing power is enough to lower the pKa by >1 pKa unit.

iad22agp's explanation is a good way of rationalizing this. Looking at the first ionization, cysteine is cationic while N-acetyl cysteine is neutral. Since the rest of the structures are similar, we would assume that it's easier to take a positive charge away from a positively charged compound than a neutral compound. Similarly looking at the second ionization, it's easier to take a positive charge away from a neutral compound than a compound that's already negatively charged.
pKa refs: cysteine, N-acetylcysteine

Assume you are titrating the substance with base. You are already at pH 7 and going up. Removing a proton from a zwitterion (cysteine) to form an anion will require less energy than removing a proton from an already anionic species (N-acetylcysteinate anion) to form a dianion.

N-acetylcysteine is a zwitterion, too.


text:
Lithium batteries a reality - why not sodium, pottasium batteries?
Lithium batteries seem to be mainstream now, offering high energy density, high output current capability.
Stepping down the periodic table there are alkali metals with greater electronegativity than Lithium: sodium and potassium, etc.
So why not sodium and potassium batteries. Can't these offer higher energy density, higher current output? Is the reactivity just so great that the chemistry can't be engineered to provide a practical battery? Or are the safety issues just too great?

Beyond the answers, I'd comment that It's not the electronegativity but the oxidation potential that matters in a battery, and Li wins over Na or K. Add the lower mass and it's been very attractive.

The sodium–sulfur battery is already in use for power leveling and other stationary applications, such as a 1.2 MWh installation in WV, USA. The Ford Ecostar did use a Na-S battery, but it did not have sufficient energy for distant travel (and some of the experimental batteries caught fire, but so do Li-ion cells on occasion. Oops!).
Sodium-ion batteries are also in development. In the JACS, the article Na-ion batteries get closer to replacing Li-ion batteries discusses some improvements in maximum current capacity and total energy storage.
That said, for mobile use, lithium has a higher energy/weight ratio than heavier metals, but for other use, sodium is less expensive and is readily available.

There is relatively little research in sodium ion batteries, and as a result, breakthroughs in sodium battery technology are lacking. Recently, that has changed, and more scientists are looking into the potential of sodium ion batteries. However, the Na-ion battery's biggest drawback is its slow charge and discharge rates. This makes them impractical for everyday use. However, due to its lower cost, they are a cost-effective alternative to Li-ion batteries for applications in energy storage for wind and solar power.
With more and more research in sodium-ion batteries, they may one day take over Li-ion batteries, and some companies such as Aquion Energy have already begun manufacturing consumer grade sodium-ion batteries.

I am wondering if sodium ion batteries are used more for energy storage for grid electricity at remote communities and towns. Then sodium ion batteries could provide an use for sodium chloride salt from seawater desalination.

Would they be suitable for domestic (household-scale) battery storage?


text:
What reaction occurs when soaking a piece of meat in apple cider vinegar?
I would like to know what reaction occurs when  soaking a piece of meat in apple cider vinegar. Additionally, how does it happen?
I read a topic which was about the bad reputation of apple cider vinegar, which can actually "cook" your skin texture.

See Texas A&M's page on Meat Tenderization, which indicates that connective tissue can be weakened by marination with salt and vinegar. Also, in Principles of Meat Science, 4th Ed., p. 239.

This "cooking" phenomenon happens because of the basic building blocks of a meat (protein) and how they can be broken down (which is called "denaturation"): heat (traditional cooking), pH changes (your vinegar question) or enzymes (bromelin or papain, which are typical ingredients of meat tenderizer)
When exposed to these factors, the proteins that make the bulk of the meat "break", and the result you can observe is very similar, which is the "cooking". Since it varies with the exposure level to the denaturating agent, this is why you only see the "cooking" happening on the surface of marinaded meats.
pH changes is also an alternative "cooking" method used in some dishes, e.g. ceviche, which is raw fish finely chopped that is soaked in lemon juice to "cook"
Hope it helped :)


text:
Identifying the number of bonding and lone pairs without a dot and cross diagram
I've just learned how to predict the shapes of molecules in class today using VSEPR theory.
I would like to ask is there anyway to find the number of bond pairs and lone pairs without drawing a dot-and-cross diagram?

I assume you're using VSEPR theory to predict the shapes?

Yeap. I'm using VSEPR @bon

@Mathxx  number of bond pairs and lone pairs for atom, molecule or what? Also drawing a dot-and-cross diagram is only for very basic training imo, it's not a real tool or sth imo

There is a simple, four-step calculation that you can perform:

count the valence electrons the atoms you are bonding have.
count the number of valence electrons the atoms would like to have for a noble gas’ valence shell. (i.e. eight for everything main group, two for hydrogen.)
substract $2.-1.$, i.e. the first (existing) from the second (desired). This is your number of bonding electrons. Divide by two for bonding electron pairs.
substract $1.-3.$, i.e. the third (bonding) from the first (existing). This is your number of free electrons. Divide by two for lone pairs.

Using a simple example such as sulphur dioxide:

$6\ (\ce{S}) + 2 \cdot 6\ (\ce{2 O}) = 18$
$8\ (\ce{S}) + 2 \cdot 8\ (\ce{2 O}) = 24$
$24 - 18 = 6$, i.e. three bonding electron pairs.
$18 - 6 = 12$ i.e. six lone pairs.

Unfortunately, that by itself does not allow you to write the structure; you need to know which element is in the centre and how they are connected. Here, we have a central sulphur bonded to two oxygens. The final result is:
$$\ce{O=S+-O-}$$
With two lone pairs on the left-hand oxygen, one on the sulphur and three on the right hand oxygen. The formal charges are important.

Let the number of total valence electrons, including charges (ions) be ($V$).
The number of total electrons required to complete octet/duplet is given by ($T$) = $(2 × \ number \ of \ H \ atoms) + (8 × \ number \ of \ other \ atoms)$
The number of bonding electrons is given by $B = (T-V)$
The number of bonds  is given by =  $\dfrac{B}{2}$ $=$ $\dfrac{(T−V)}{2}$
The number of unshared electrons is given by $N = (V – B)$
The number of lone pairs is given by = $\dfrac{N}{2} = \dfrac{(V – B)}{2}  $

There is a very simple way to do it, but it is not recommended at all.
Also there are a lot of assumptions in this method.

Detect the central atom along with the peripheral atoms.
For Example:- In Methane, the Carbon atom is the central atom.

Count the valance electrons of the central atom and the peripheral atoms.
Exception: For Hydrogen , using "1" won't yield results. So specifically for Hydrogen use "7". This is one of the assumptions we have to make.

Divide the above value obtained by 8. The quotient gives the value of the number of sigma bond pairs and the remainder divided by 2 gives the number of lone pairs.

Explanation:
The number of valance electrons counted divided by 8 will give the number of sigma bonds formed. This is just like counting the number of atoms which are getting complete octets, i.e. forming sigma bonds to get octet complete and becoming stable. The remainder is the no. of non bonded electrons, which when divided by 2 gives the number of lone pairs.
Example:
In Sulfur tetrafluoride:
Number of Valence electrons$$1\times(6)+4\times(7)=34$$
No. Of Bond Pairs$$\frac{34}{8}=4$$
No. Of Lone Pairs$$\frac{34\%8}{2}=1$$
To be very clear, this method is not fool proof. It is just a workaround to save time if you do not want to make the structures as it works on majority of compounds.