What you call bases is in fact a subset thereof. Besides that, there is a whole class of inorganic bases, also known as alkali. These are the hydroxides (sometimes called basic salts) of alkaline metals. Their ionization is somewhat different:
$$\ce{NaOH->Na+ + OH-}$$


What atom would be protonated first in a strongly acidic solution?

In short, none. As Mithoron said, the oxygen atoms can all act as Lewis bases and will all be protonated to an extent. If you rephrase the question to "Which of the atoms has the highest Kb?", well that's another story. Atom D has the lowest Kb, because the carbon in a carbonyl has a slight positive charge. Atom C would be difficult to protonate, the oxygen is already donating a pair to the furan ring to make it aromatic. So the debate is between a carbonyl and a primary alcohol. After consulting some pKa tables I found that protonated ketones have a pKa of around -7.3 while protonated primary alcohols of around -2.4 thus the alcohol is a stronger base (higher pKa means weaker acid thus stronger conjugate base). Finally we can conclude that the primary alcohol -OH group will be the first to accept a proton.

P.S. Fun fact, the quickest to be protonated was none of the ones marked with letter, it would have been the ester's carbonyl group!


Some substances appear white when crushed while others do not. Why?

It depends if the substance in bulk is transparent, highly reflective (metallic) or highly absorptive.

Carbon in its graphite allotrope is highly absorptive, and even finely divided is still black.

Carbon as diamond is transparent. Crushed diamonds are white -- look at a diamond file or abrasive disk. What happens is that incident light is scattered as it encounters particle after particle, refracted or reflected repeatedly, based on particle size, shape and index of refraction (nD). A higher nD increases the amount of light reflected from particles. For that reason, titanium dioxide, $\ce{TiO2}$, which has a comparatively high index of refraction (~2.5), makes an excellent white pigment.

Metals reflect light because the "cloud" of electrons on the surface return a reflected wave out of phase with the incident wave. However, finely-divided metals have fewer electrons to share, and electrons start behaving as discrete, localized charges. Finely divided silver appears black, for example. This is the basis for traditional silver halide photography. Powdered $\ce{AgCl}$ appears white, but where hit by light, it yields metallic silver nanoparticles, which appear black (as a negative image).

If your interested, look up the effect of surface plasmons on metallic reflectivity, which can yield surprising colors, such as gold in ruby glass.


Carbon-Sulfur Bond Lengths; Resonance Effects (Or Lack Thereof)

Quick Summary

The linked paper (from 1975) is probably wrong. A slightly more recent paper (J. Am. Chem. Soc., 1989, 111 (21), pp 8112–8115) suggests the bond lengths are about equal. Some quick calculations (below) using multiple modern methods suggest that, as expected, the C-S bond length in $\ce{CH2SH-}$ is shorter than in $\ce{CH3SH}$.

For both older papers, the computational results (using basic Hartree Fock) would be considered highly suspect under modern peer review, since Hartree Fock is known to have problems with anions. Certainly the 1975 paper doesn't even mention the basis set.

Seen below, it's clear that the results are method dependent, but the best current methods, with reasonable basis sets indicate the bond length is at least slightly shorter due to some partial double-bond character.

More Details

The first thing to consider in this is that there are two forms of $\ce{CH2SH-}$, syn and anti, based on the orientation of the formally C-bound anion and the SH group:

syn form

anti form

These will have different bond lengths, since the anti form will have contribution of resonance structures II and III from Ron's answer. This gives some double-bond character and should likely have a shorter bond, although electrostatic repulsion could also be invoked.

The syn form won't have a clear effect, since the C anion lone pair and the S lone pairs are on the opposite side of the bond. My guess is that the bond length is about the same as $\ce{CH3SH}$.

I think the clear message is that we can explain a lot of things, but it's better to know if our results are accurate before trying to explain them.

What I see with more modern computational methods is that the syn form has similar C-S bond length to $\ce{CH3SH}$ and the anti form is slightly shorter. I suspect the previous result is simply wrong. Calculations involving anions are tricky, since the extra electron is usually highly diffuse. I expect the basis set was simply insufficient.

The difference in energy between the two forms is ~1 kcal/mol, so experimentally, these will interconvert readily, and an average bond length would be observed. As mentioned in the comments, this species would probably be hard to observe, even in the gas phase, since $\ce{CH3S-}$ is more stable.

Computed C-S bond lengths (in Å):

For all methods, full geometry optimization was performed using the methods indicated. I would consider the CCSD calculations to be high quality.


However, there is now that remaining unhybridized p orbital of
  nitrogen, which has 2 electrons, so how can it be involved in π
  bonding?

Everything is more-or-less "right" up to these point. The thing you're missing is that there is only one electron in the unhybridized $\mathrm{p}$ orbital of $\ce{N}$ atom since to get the following usual configuration of $\ce{NO3-}$ ion (image courtesy of Wikipedia)

with two negatively charged $\ce{O-}$ ions one electron has to be moved from $\ce{N}$ atom to $\ce{O}$ atom first. Now note that since all $\ce{O}$ atom are equivalent to each other we need to invoke the concept of the resonance, but for simplicity think first about any single of the resonance structure above.

So, formally we start by having:

one $\ce{N}$ atom with $\mathrm{(1s)^2 (2s)^2 (2p)^3}$ electron configuration;
two $\ce{O}$ atoms with $\mathrm{(1s)^2 (2s)^2 (2p)^4}$ electron configuration;
and one $\ce{O-}$ ion with $\mathrm{(1s)^2 (2s)^2 (2p)^5}$ electron configuration;

In the first step we transfer one of the $p$-electrons from $\ce{N}$ atom to an $\ce{O}$ atom, so we end up with:

one $\ce{N+}$ ion with $\mathrm{(1s)^2 (2s)^2 (2p)^2}$ electron configuration;
one $\ce{O}$ atom with $\mathrm{(1s)^2 (2s)^2 (2p)^4}$ electron configuration;
and two $\ce{O-}$ ions with $\mathrm{(1s)^2 (2s)^2 (2p)^5}$ electron configuration;

At the next step we invoke the $\mathrm{sp^2}$ hybridization of $\ce{N+}$ ion, so that we get:

one $\mathrm{sp^2}$-hybridized $\ce{N+}$ ion with $\mathrm{(1s)^2 (sp^2)^3 (2p)^1}$ electron configuration;
one $\ce{O}$ atom with $\mathrm{(1s)^2 (2s)^2 (2p)^4}$ electron configuration;
and two $\ce{O-}$ ions with $\mathrm{(1s)^2 (2s)^2 (2p)^5}$ electron configuration;

Finally, we form bonds as follows:

One $\sigma$ bond between the $\mathrm{sp^2}$-hybridized $\ce{N+}$ ion and an $\ce{O-}$ ion in forming which we use 1 of the total 3 $\mathrm{sp^2}$-electrons of $\ce{N+}$ ion and the only unpaired $\mathrm{2p}$-electron of $\ce{O-}$ ion;
Another identical $\sigma$ bond between the $\mathrm{sp^2}$-hybridized $\ce{N+}$ and another $\ce{O-}$ ion; here we use 1 of the remaining 2 $\mathrm{sp^2}$-electrons of $\ce{N+}$ ion and the only unpaired $\mathrm{2p}$-electron of $\ce{O-}$ ion;
Yet another $\sigma$ bond, but this time between the $\mathrm{sp^2}$-hybridized $\ce{N+}$ ion and the $\ce{O}$ atom in forming which we use the last remaining $\mathrm{sp^2}$-electron of $\ce{N+}$ ion and 1 out of total 2 unpaired $\mathrm{2p}$-electrons of $\ce{O-}$ ion;
What remains now is one unhybridized $\mathrm{2p}$-electrons of $\ce{N+}$ ion and one unpaired $\mathrm{2p}$-electrons of $\ce{O-}$ ion which are used to form the $\pi$ bond.

Pictorially (image courtesy of Professor Peter Bird):


Find the partial pressure in a container

Since we're assuming these are all ideal gases, we can use Dalton's Law of Partial Pressures to solve this fairly easily; you never need to find moles of anything.  Unless something is strange, the percentages given are by volume or by pressure rather than by mass.  If they are intended to be by mass, what follows would need to be modified.  I suspect that the use of the words "air mass" in the problem are what caused the confusion.

The total pressure, 1.5atm, is equal to the sum of the partial pressures, each of which is proportional to its mole fraction in the container.  Because these are gases that have the same volume, 40L, their mole fractions are their percent of the total amount of gas.

The nitrogen, which is 78.1% of the total, has a partial pressure of:

$ 1.5 \cdot 0.781 = 1.1715\ atm$

That's true whether the oxygen is in the container or not.  To see that this is the case, we can remove the oxygen and do the same calculation.

If we remove 20.9% of the gas (in the form of oxygen) we are left with 81.1% of the original 1.5 atm remaining.

$1.5\cdot 0.811 = 1.1865\ atm$

We can determine the mole fraction of nitrogen in this new mixture of gases by adding up the remaining percentages and finding nitrogen's contribution to it:

$78.1 + 0.964 + 0.0378 =  79.1018\ units\ of\ gas\ total$

$\frac{78.1}{79.1018} \cdot 100\% = 98.73\%\ N_2\ after\ oxygen\ is\ removed$

As before, Dalton's Law of Partial Pressures tells us that the pressure of nitrogen will be its mole fraction times the total pressure, or 98.73% of 1.1865 atm.

$1.1865 \cdot 0.9873 = 1.1714\ atm$


What are Mulliken atomic charges, and how are they computed?

Mulliken atomic charges can be defined as[1]:

$$q_A = Z_A-\sum_{\mu\in A}\left( \mathbf{P\cdot S} \right)_{\mu\mu} \tag{Szabo 3.196}$$

Here I have used the same notation as in Szabo[1], with $\mathbf{P}$ being the density matrix and $\mathbf{S}$ beign the overlap matrix. $Z$ is the nuclear charge, and using the greek letter $\mu$ as indicies indicates that we are working in atomic orbital basis (not in in molecular orbital basis).

The sum runs over $\mu\in A$ meaning, that we only consider atomic orbitals that are centered on the $A$th atom.
we can therefore note that Mulliken charges are only defined when we used atomic-centered basis functions (which we most often do).

[1] : Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory, Attila Szabo and Neil S. Ostlund


It is really easy to solve by the half reaction method, here is my answer:

Half reaction method

Global eq: KCl + MnO2 + H2SO4 -> K2SO4 + MnSO4 + Cl2 + H2O

Cathode: 2Cl-   --> Cl2 + 2e-

Anode: MnO2 + 4H+ + 2e- --> Mn+2 + 2H2O

So the final balanced eq is this one:

2KCl (ac)  + MnO2 (s) + 2H2SO4 (ac) --> K2SO4 (ac) + MnSO4 (ac) + Cl2 (g)  + 2H2O

Variations method

Without half-reaction method will be like this:
You have one Mn+4 reduced to Mn+2
That is 2e x 1 atom (ion indeed) so this is a net variation of 2
On the other hand you have Cl- wich oxidizes to form a Cl2 molecule so this is
1e x 2 atoms wich also is a net variation of 2
Now you use these numbers as the coefficient of the products that contain these atoms and you'll have

KCl (ac)  + MnO2 (s) + H2SO4 (ac) --> K2SO4 (ac) + 2MnSO4 (ac) + 2Cl2 (g)  + H2O

and considering this as valid just left to adjust the rest of the elements
and finally you'll have:
4KCl (ac)  + 2MnO2 (s) + 4H2SO4 (ac) --> 2K2SO4 (ac) + 2MnSO4 (ac) + 2Cl2 (g)  + 4H2O
Wich is also a valid adjust

I always recommend using half-reaction method or undetermined coefficient method instead. Good luck!!


Why is OTs a good leaving group and how does it work?

Tosylate groups ($\ce{CH3C6H4SO3-}$) are good leaving groups because their conjugate base forms are quite stabilized by resonance:


Does mechanotransduction of cells play a role in the biocompatibility of titanium?

Mechanotransduction (mechano-relating to mechanics + transduction-to convert into another form) is indeed the second definition you listed. Mechanoreceptors in somatosensory nerve cells the ability to detect mechanical pressure or distortion. These facilitate our touch, hearing, and proprioception, or the sense of how our body is positioned. Olfaction, however, is actually facilitated through chemoreceptors, and depends on how molecules bind to them.

Cellular response to a foreign object is the result of largely chemoreception. Interaction at the implant-cell surfaces determines how an implant will be perceived by the body. Titanium's biocompatability is a result of the protective oxide film that forms on the surface of the metal in the presence of oxygen. This oxide film is strongly adhered, insoluble in water, and chemically impermeable, preventing reactions between the metal and the surrounding environment. Titanium's high dielectric constant allows it to be easily integrated into cells and bone tissue$^{[1]}$.

From Wikipedia:

  Titanium naturally passivates, forming an oxide film that becomes heterogeneous and polarized as a function of exposure time to bodily environments. This leads to the increased adsorption of hydroxyl groups, lipoproteins, and glycolipids over time. The adsorption of these compounds changes how the material interacts with the body and can improve biocompatibility.

[1] Mishra, B. Review Of Extraction, Processing, Properties, and Applications of Reactive Metals 1999 TMS Annual Meeting, San Diego, CA, February 28 - March 15, 1999; Wiley: Hoboken, 2013.


If you do not have a list of the species that belong in the Pourbaix diagram, then it is not possible to accurately predict what species occur where, unless you have extensive experience with the system in question.  If you have a list of species and need to place them in the appropriate regions of a Pourbaix diagram (fill in the blank style), then the higher oxidation state species occur at the top of the diagram and the more basic (e.g. OH containing) species occur towards the right.

If you have Latimer diagrams of all the species in the Pourbaix diagram (this will require the Latimer diagrams for both acidic and basic conditions, typically) then you can more accurately predict the structure of a Pourbaix diagram by (a) writing balanced half reactions and (b) adjusting them for the appropriate pH using the Nernst equation.


Why doesn't Oxygen form coordinate covalent bonds?

Oxygen very much does form bonds in which both electrons come from the oxygen atom.  Examples include:

1.
The $\ce{H3O^+}$ ion at the center of the solvated proton in aqueous acids, also available as salts of some of the strongest acids such as $\ce{(H3O)(ClO4)}$

2.
Carbon monoxide, with its triple rather than double bond.

3.
Ozone, in which the oxygen atoms form a bent chain and one of the three bonds must have both electrons coming from the middle atom.  The resulting charge separation makes ozone one of the few one-element substances with a nonzero dipole moment.


What is the dipole moment of the most stable conformer of 1,1,2,2-tetrafluoroethane?

The dipole moment is 2.45 Debye. The molecule is dipolar as you have asymmetrical electron distribution around both the carbon atoms.

Source: Complete Structure of Gauche 1,1,2,2-Tetrafluoroethane Determined by Microwave
Spectroscopy, J. Phys. Chem. A, 2000, 104, 9489-9493.


The real reaction is $\ce{Ag+ + e- -> Ag}$ for both cases. Equation written in this way, ${E^0}$ will be equal to the reduction potential of Ag(I) in its 1 M solution.

If it is written as $\ce{AgI + e- -> Ag + I-}$, then ${E^0}$ will be equal to the reduction potential of Ag(I) in 1 M $\ce{I-}$ solution -or in other terms in ${K_{sp}(AgI)}$ Molar Ag(I). solution

Hence, technically, $\ce{Ag+ + e- -> Ag}$ is same with $\ce{AgI + e- -> Ag + I-}$. Latter is more descriptive and practically more relevant, however, because the presence of $\ce{I- /Cl-}$ ions in the medium limits the amount of $\ce{Ag+}$ cations via ${K_{sp}}$ and the notation readily implies this.


Will the volume occupied by a mixture of nitrogen and water be same as the volume occupied by them individually?

The idea behind the question is that you know the partial pressure of N$_2$ is 0.96 atm (= 1 atm - 0.04 atm). And you know that the partial pressure of all the H$_2$ that can be produced from the water is thus (1.86 atm - 0.96 atm)* 2/3 = 0.6 atm. [Ignoring the volume of liquid water, which is only ~1/1000 that of the gas.] From this you can determine the number of moles of H$_2$, which is equal to the number of moles of water = 0.6 atm *3 L/(302 K * 0.0821 atm L / mol K).

Note the number of moles of water given in your answer isn't the total no. of moles of water, it's the number of moles of water vapor prior to the electrolysis.


Why do some chemical reactions require many steps?

There is no fundamental law preventing simple chemical reactions: things are complex because of the combinatorial complexity of chemical compounds

The complexity of many chemical reactions is a byproduct of the fact that there is a very, very large variety of possible chemicals. Much of that complexity happens because of the almost infinite way even some simple elements can be combined together to give complicated structures (carbon being the archetypal example). Theoretically, for example (theoretical because not all of the examples can exist in 3D space) there are 366,319 ways to build different alkane compounds from just 20 carbon atoms and hydrogen atoms (see this question here and this entry in the Encyclopaedia of integer sequences). And this number drastically understates the real complexity as it ignores mirror images and more complicated ways of joining the carbon atoms together (like in rings, for example). The complexity just gets more mind boggling if you start adding other elements to the mix.

No physical law prevent us making any possible compound in one step. But the sheer complexity of the end products makes simple ways to reach many of them extraordinarily unlikely from the laws of probability alone, never mind the specific ways chemical components can be easily joined up to make more complex things.

Here is a simple analogy. Let's say you want to assemble a Lego model of the Star Wars Death Star weapon. There are 4,016 pieces of lego that have to be assembled in the right combination and the right order. There is no physical law that says you couldn't somehow do that in a single step. But no sane person's intuition would assume that this was easy or likely. It isn't physical law that prevents one step assembly: it is combinatorial complexity. Chemistry is, do I really need to say this, more complicated than Lego: not least because atoms can be joined up in many more complex ways than the simple, standard-sized physical pins that join Lego bricks together.

Both nature and synthetic chemists have explored many ways to achieve particular end products from simpler building blocks. Sometimes new chemical Death Star equivalents (like the geometrically beautiful hydrocarbon dodecahedrane, which, incidentally, has 20 carbons but isn't counted in the list of 20 carbon alkanes) are made only after long sequences of reactions. The original synthesis of dodecahedrane took 29 steps but others found better, higher yielding, routes that took only 20. Many important drugs are first synthesised in long sequences of reactions but are later found to be available via much shorter routes (there is nothing like the economics of manufacturing cost to encourage creativity).

So the reason many chemical reactions take multiple steps isn't physical laws but probability theory. There are just too many possible chemicals and too many ways to combine things for single step routes to most given products to be likely to work. Doing one thing at a time (just like you would if building the Lego Death Star) is the way to get what you want.


Why are enthalpies of reaction or Gibbs FE change given in kJ/mol

It makes perfect sense that the units KJ/mol are used.

If you burn 2 cords of wood you get twice the heat as from 1 cord of wood.

Enthaply (and Gibbs free energy) of the reaction is proportional to the number of moles that react.

If the reaction is written as:

3A + 2B => 6C

Then the change per mole would mean: per 3 moles of A = per 2 moles of B = per 6 moles of C.


Axial vs equatorial attack of Grignard reagent on dioxan-5-one and dithian-5-one

I've found the solution in Fleming: Molecular Orbitals-Reference edition:

  The dioxan-5-one is attacked by nucleophiles in the gas
  phase from the equatorial direction, because of electrostatic repulsion from the fraction of negative charge
  carried by the two oxygen atoms in the ring, as Houk had predicted. The selectivity is not particularly high
  (67:33), because the usual torsional strain is still present, enhanced in this case by the shorter C—O bond
  lengths making the ring less puckered, and increasing the dihedral angle by which the carbonyl oxygen atom
  is pushed below the neighbouring C—H bonds. In solution, however, the ketone is attacked
  from the axial direction, just like cyclohexanones. Evidently the solvent insulates the charge on theincoming nucleophile from the electrostatic repulsion, the torsional strain remains high and there are no axial
  substituents hindering axial attack.The sulfur atoms in the dithia analogues are less electronegative than the oxygen atoms in the acetals but the axial lone pairs bulge further up, and are not heavily solvated. However, the carbonyl group is
  displaced even further down by the changes in bond length and bond angles from having two C—S bonds in
  the ring. The one effect makes axial attack less favourable, and the other makes equatorial attack less
  favourable. In the event both solution and gas phase reactions take place with a strong preference for
  equatorial attack.


What changes occur when adding differing amounts of acids & bases?

It is perhaps difficult to comment on the intended semantics of the option 3, but I think that is the most appropriate answer.

Option 1 and 4 are wrong because of the reasons you cite.

To comment upon option 2, we need to know the pH range over which this indicator changes colour. According to Wikipedia, an unequivocal shift from red to yellow occurs from a pH of $3.1$ to a pH of $4.4$. For our case, we would end up with a $\pu{\frac 13 M}$ solution of $\ce{HCl}$, which, assuming complete dissociation and negligible contribution of $\text{H}^+$ from water, we end up with a pH of around $0.48$ which is well in the no-colour-change region of the pH values.

Regarding option 3, the 'more' just refers the water molecules formed as a result of neutralisation reaction, which are additional to the ones already present in either of the mixing solutions. Along with the formation of 'new' $\ce{NaCl}$ molecules (as in they did not exist bonded to one another before), water molecules are also formed.

$\ce{NaOH + HCl -> NaCl + H2O}$, i.e. table salt and water


Why might my limelight experiment have failed and how can I improve it?

Perhaps your flame wasn't hot enough. Real limelights used oxyhydrogen torches, not wimpy natural gas flames.

Or perhaps your limestone chip wasn't just $\ce{CaCO3}$. If it was dolomitic limestone, there was a lot of $\ce{MgCO3}$, too, and $\ce{MgO}$ doesn't do the trick like $\ce{CaO}$ does.


Your text will best explain reversibility.  The short answer is that a reversible process is one whose direction can be reversed by a microscopic change in external conditions.

For example a gas in a cylinder expanding against a piston will do so reversibly as long as the external pressure and the internal pressure are essentially identical.  Of course for expansion the internal pressure must be infinitesimally greater than the external pressure.

Next, entropy is really only defined for systems at equilibrium.

Next, the equation for the entropy above is not correct except at constant temperature.  In reality it is $dS = dq/T$, which generally must be integrated in order to determine the entropy by this method.

For the second question, there is some confusion here. $\Delta H$ is equal to $q$ only at constant pressure.

In fact there are lots of ways to compute the entropy change in a process, but this space is too small to go into them.  They should be in your textbook.


Which gems can be prepared on a small scale basis?

Most likely the only gems you will be able to make at home are plastic gems.
Wikipedia sheds some light on the production of some synthetic gems, Corundum and Cubic Zirconia, both of them not possible at home (or at least my home).


Reactions which confirms the presence of cyclic structure of glucose

Yes, such a reaction exists. In fact, it can not only confirm the presence of ring forms in solution but also how prevalent they are.

In 1962, Walker Jr, Gee and McCready published Complete Methylation of Reducing Carbohydrates, in which they subject solutions of sugars in dry DMF to silver(I) oxide  and methyl iodide. The reaction proceeds according to scheme 1 below.

Scheme 1: Methylation of glucose with $\ce{Ag2O}$ and $\ce{MeI}$ as published by Walker Jr, Gee and McCready.

Before reporting on the products, they note:

  It was somewhat unexpected that methylation of
  aldoses would proceed so smoothly in the presence
  of silver oxide. Glucose, shaken with silver oxide in
  DMF at room temperature, is partially decomposed
  in several hours.

Usually, silver(I) salts in solution are used in the Tollens test in which reducing sugars such as glucose reduce the silver ions to give elemental silver which forms a silver film on the inside of the glass. The presence of methyl iodide completely suppresses this reaction as they state in the next sentence:

  When methyl iodide is present
  in the reaction mixture, however, no silver mirror
  (indicating oxidation-reduction) is formed

As for the two step process that I have indicated, the authors state:

  In this reaction methylation
  of hydroxyl of the hemi-acetal group of glucose is
  essentially complete in one hour (as shown
  by loss of reducing function), but complete
  etherification of the remaining hydroxyls requires
  considerably longer.

With the experimental section detailing that the reactions were run ‘at least $\pu{16h}$’ which permits drawing the mechanism as a two-stage process as 1 can be isolated if the reaction is cut short.

Mechanistically, this reaction process is different from an acid-catalysed acetalisation with methanol. In the latter, an entire $\ce{OMe}$ group is attached to the sugar and the reactive species is an aldehyde form. This could be the open-chain aldehyde form (requires an additional reaction step) but most likely occurs via a cyclic aldehyde as depicted below. When using $\ce{Ag2O/MeI}$, the acetal oxygen is supplied by glucose as methyl iodide does not have any available oxygen. The two competing mechanisms are shown in scheme 2.

Scheme 2: Mechanisms of the acetalisation of glucose. (a) $\ce{H+/MeOH}$ and cyclic aldehyde intermediate; (b) $\ce{H+/MeOH}$ and chain-form aldehyde intermediate; (c) $\ce{Ag2O/MeI}$ single-step mechanism.

As these mechanisms show, acid-catalysed acetalisation need not start from the cyclic form to get acetal 1; it is possible to draw the reaction mechanism as originating from the chain form of glucose. But methylation with $\ce{MeI}$ requires the oxygen to be in place as an $\ce{OH}$ function that is deprotonated and attacks $\ce{MeI}$ which is not possible if the chain form were present in solution.

As to why the $\ce{C{1}}$ oxygen is methylated first, and the explanation is the higher acidity of an acetal hydroxy function when compared to an alcoholic one as the authors state:

  This difference in rate is expected
  from the difference in acidity (and ease of
  anion formation) of the hydrogens on the two different
  types of hydroxyl groups.

Furthermore, it bears stating that this reaction not only proves cyclic forms are present but also that they are prevalent. Methylation of the $\ce{C{1}}$ oxygen is rapid compared to all the other hydroxy groups meaning that they should be present in at least appreaciable ratios. Furthermore as mentioned above, under these conditions the reduction of silver(I) to a silver mirror – which requires the open-chain aldehyde form mechanistically – is practically entirely suppressed. Given that the Tollens test is a standardised method to detect glucose, the complete absence of all traces of silver film serves as a good indicator of the low concentration of open-chain aldehyde in solution.

The original paper is: H. G. Walker Jr, M. Gee, R. M. McCready, J. Org. Chem. 1962, 27, 1000–1002. DOI: 10.1021/jo01053a047.


The question is ambiguous, unfortunately.  There are two possibilities that I can see.  Either the question:

is asking about a reaction in which some quantity of wax undergoes the reaction you give, in which case you are 100% correct.
is asking about a reaction in which part of the paraffin is being oxidized to C and part is being oxidized to CO2, the usual reaction that occurs.  Though that is more likely to occur in real life, in my opinion, it also is impossible to solve without further information.  The best you could do would be to set a lower limit (the value you've already found) and an upper limit, the heat of reaction for $\ce{C21H44 + 32 O2 -> 21 CO2 + 22H2O}$.  (I guess that would be an upper limit and a lower, numerically, since the values are both negative, but this reaction would release far more heat.)


Chemical methods to extract Oxygen from Rust

Good question; unfortunately, no good answers. The problem is that oxygen, pure oxygen, doesn't like to exist as such; it will form a diatom like many gases, but it's very electronegative, and so it will prefer to get the extra electrons from something more willing to donate them, like the iron. This is why "oxidation" has that name; until fluorine was isolated in the mid-1800s, oxygen was the strongest known electron acceptor ("oxidizer"). It's easier to run a reaction that gives the oxygen something else to oxidize, producing the raw metal, than to get the iron to complex with something else and release the oxygen. In at least one circumstance, this is very good for humans and other aerobic organisms; oxygen complexes with the iron in your hemoglobin for transport to your cells. In most other circumstances, we could do without this behavior.

Rust is insoluble in water but will dissolve in strong acids. It will react with hydrochloric acid to form iron chloride and water (sometimes producing hydrogen gas if there isn't enough oxygen to go around). The easiest "household" method to obtain oxygen from iron oxide, therefore, is to dissolve the rust in muriatic acid forming iron chloride and water, neutralize any remaining HCl by adding lye to form sodium (and/or potassium) chloride and water, distill the water from the chloride salts by boiling and then cooling the vapor, then electrolyze the water with a battery to produce oxygen and hydrogen. That's cheating, though. First, water's water, and you're adding far more to the party with the acid and base solutions than you would ever get from the rust. Second, muriatic acid and lye, while available in relatively pure form for pool maintenance and for soapmaking respectively, are not exactly everyday household chemicals in those forms (both are components of other household chemicals, such as drain cleaner and lime scale remover, but are somewhat difficult to isolate from them).

Theoretically, if you heat the rust enough, the components will dissociate and you'll be left with pure iron and oxygen. Unfortunately, even with carbon in the mix (producing elemental iron metal and carbon dioxide, which is the reaction behind iron ore refinement and recycling), the minimum temperature required for this reaction to become favorable is 842K = 569*C ~= 1056.2*F. I can't find data on iron-oxygen bond enthalpy or entropy, which is needed to calculate the spontaneous decomposition temperature of the compound, but logic states it's much higher than with added carbon, because you're not getting any energy release by forming the carbon-oxygen bond, so every Joule of energy needed to break the iron-oxygen bonds has to come from outside the material. Theoretically it would cost you about 2.5kWh/kg of heat added to the iron; in practice the amount of heat you'd have to generate (and lose to various inefficiencies) to get the metal that hot is beyond any household device's capabilities, unless your hobby happens to be glassblowing.

The "thermite reaction" between iron oxides and aluminum under heat is another well-known method of separating the iron and oxygen. The rust and aluminum powder, given some initiating heat, will exchange the oxygen to form alumina (Al2O3) and pure iron, and enough heat to keep the reaction going (with plenty to spare; a thermite reaction is too bright to watch without a welder's mask and in fact is used in some situations to weld iron). Two problems with this; first, it's illegal in the US to make thermite without BATFE approval, and second, the reason it works at all is because aluminum oxide is ridiculously stable, more so than iron oxide, and so it takes even more energy to reduce the oxide back down to the raw metal (about 8.2 kWh/kg for aluminum, theoretical, versus 2.25kWh/kg for iron). This is evidenced by the highly exothermic nature of the thermite reaction itself; if creating aluminum oxide produces far more than enough heat to smelt the iron to get the oxygen, think of how much would be needed to go the other way.

Pretty much the only direct route to releasing oxygen gas from rust without a blast furnace is to introduce an even more powerful oxidizer. There are very few known to modern science, and none of them are "household chemicals" to say the least. The easiest to work with would be elemental fluorine gas, and the experienced chemists in the crowd will know that for the extremely relative statement that it is. Fluorine is the most electronegative element in the known universe, and so pretty much any substance that isn't already highly fluorinated (like some polymers, such as Teflon) will react pretty vigorously with it. That includes the water, proteins, enzymes and other compounds in your own body; a fluorine gas leak is a "drop everything and run" type of emergency. As used on rust, the fluorine will simply barge in and displace the oxygen from the rust to form a white to pale green iron fluoride, liberating the oxygen as a gas. You then have to separate the oxygen from the excess fluorine, and one of the standard ways in industry to catch excess fluorine is to pass the exhaust gas over or through a powdered sodium oxide filter, which will form sodium fluoride and liberate more oxygen in a very similar reaction as you got with the iron. Sodium oxide isn't a pleasant chemical to work with either; add water (such as by breathing in any loose powder) and it forms the strong base sodium hydroxide, which eats flesh about as efficiently as anything else you could think of.


Why are zinc shot and mossy zinc not considered toxic but zinc granules are?

The key to toxicity is concentration, which is related to how rapidly the material disperses (or concentrates) in the environment. Which is also related to water type (fresh or salt) and temperature. MSDS sheets are a bit of a joke. They are worded in a way that fends off lawyers rather than gives detailed useful toxicity data. Consider this MSDS for Sodium Chloride
Bear in mind this useful advice when using it at home: "Eye Protection Wear safety glasses with side shields (or goggles) (European standard - EN 166). Hand Protection Protective gloves"


Why does a Lower Concentration of SO2 call for an increase of equipment dimensions in a Contact Process Plant?

You need higher quality of equipment and larger "chambers" to deal with low concentrations. Besides, in order to meet needs of plant (I mean, each plant has production output) you will need stated amount of raw materials, which you can find in material balance sheet of the applied technological processes. So, when using low-concentrated things as raw materials, even if yield is higher (it's higher usually by less than 10%, not really much) you will need larger volume of raw materials (since needed amount is usually given in mass of pure $SO_2$ in entering gas).


What makes selectivity other than activation energy?

It's not just the activation energy that determines the selectivity of a reaction. There is a thermodynamic factor too. This is the classic kinetic-vs-thermodynamic control that you will find in most chemistry textbooks at a degree level.

Reactions under thermodynamic control have outcomes that depend on the position of the equilibrium and therefore the relative stability of the possible products
Reactions under kinetic control have outcomes that depend on the rate at which the reaction processds, and therefore on the relative energies of the transition states leading to the alternative products.

[Organic Chemistry, 2nd ed., J. Clayden, N. Greeves, S. Warren, OUP] page 264

You are considering only activation energies ($\Delta G^\ddagger$) i.e. the difference in energy between the transition states and the reactants.

If your reaction has two pathways, then the pathway with the lower activation energy is faster. This means that the product of that pathway is formed faster. This is kinetic control.
However, the reaction can be reversible. The activation energy for the reverse reaction is $\Delta G^\ddagger + \Delta_r G$. Therefore, if the reaction energy (i.e. $\Delta_r G$) is not too large, then there would be backward reaction.
In this condition, the products and reactants are in equilibrium, so the side with the lowest Gibbs free energy is favoured. If the product from one pathway is more stable (i.e. more negative $\Delta_r G$) than the other product, then the stable product will the be the one with higher yield when the equilibrium is reached.
This is thermodynamic control.
In the above diagram for example, pathway 1 is faster as TS1 is lower in energy. So product 1 is the kinetic product. Pathway 2 is slower, but the product is lower in energy, so product 2 is the thermodynamic product.
Whether a reaction is under thermodynamic or kinetic control would depend on the reaction conditions, the specific identity of the reactants, and how much time is allowed for the reaction.


How to Convert Sucrose to Glucose and Fructose

Sounds like a great experiment!

As you state, a sucrose solution in water eventually hydrolyzes to fructose and glucose, but without a catalyst, the reaction rate is extremely slow; beverages are stable at room temperature for years. However it is hydrolyzed in hours at room temperature by adding sucrase enzyme (secreted in the small intestine), invertase enzyme (derived from yeast) or even a weak acid such as citric acid, tartaric acid, or even dilute hydrochloric acid.

You may need to test whichever of these are available for compatibility with Benedict's solution, though, and you may need to test some of the other ingredients in the drinks to see if they also cause false positive otr negative effects. You might compare the results from the Benedict's test with those from a blood-sugar glucometer (the solutions may need to be diluted in both tests to stay within a measurable range).

BTW, study of yeast fermentation of sucrose is responsible for the name enzyme, meaning leavened with yeast, so one could say invertase is the epitome of an enzyme.

Please share your experimental results here.


Symmetry element in the coversion of cyclopent-2-enyl anion to pentadienyl anion?

Let's look at the back reaction, the closure of the pentadienyl anion to the cyclopentenyl anion.  Once you understand this you'll see that the exact same reasoning applies to the reverse reaction that you asked about.  So here's a picture of the reaction we're talking about.

And here is a picture of the 5 molecular orbitals for the pentadienyl system

Since the pentadienyl anion has 6 pi electrons, the first 3 molecular orbitals (from the bottom up) are filled, each with two electrons.  That makes the third molecular orbital (the one in the middle of the diagram), let me call it $\ce{\psi}_3$, the highest occupied molecular orbital (HOMO).  The HOMO is so important because it is the orbital that will control our process.  If we look at the symmetry of the $\ce{P}$ orbitals on $\ce{C_1}$ and $\ce{C_5}$ in $\ce{\psi}_3$, we see that in order to get constructive overlap (overlapping the parts of the $\ce{P}$ orbitals with the same sign) as we rotate the $\ce{P}$ orbitals on these two carbons to form the bond that forms the cyclopentenyl ring, we must rotate the tops (or bottoms) of these two $\ce{P}$ orbitals towards each other.  If we were to rotate the top of one and the bottom of the other towards each other we would overlap a plus side of one $\ce{P}$ orbital with the minus side of the other $\ce{P}$ orbital and this would not result in constructive or bonding overlap.  The type of rotation that leads to bonding overlap in this example is termed "disrotatory".  It preserves a plane of symmetry bisecting the pentadienyl system (e.g. it contains $\ce{C_3}$) and is perpendicular to the plane of the screen.  This is the symmetry element that is preserved during the ring closure.  The exact same arguments also explain the ring opening of the cyclopentenyl anion to the pentadienyl anion.


Gibbs Free Energy : What is it trying to say actually?

You can prove that the Gibbs free energy is equal to the maximum amount of work that a certain reaction (or any physical process in general) can perform, i.e. $\Delta G = w_{max}$. Such a proof can be found in any decent Physical Chemistry textbook (such as Atkins'). The remaining energy must be "lost" as heat to basically account for the observation that is essentially stated with the Clausius inequality $\Delta S \geq 0$ (i.e. the entropy of the universe is always increasing) . Of course the definition of entropy $\Delta S = \frac{q_{rev}}{T}$ and the Clausius inequality that is derived from it (and hence the Gibbs free energy) are inherently connected to Kelvins (and Clausius') statement of the second law. You can see from the definition (don't forget the change of entropy of the surroundings!) that (since $\Delta U = q + w$) the entropy is decreasing when heat is completely converted to work.

Ask yourself the following: if heat could be completely converted to work, then the collision of an inelastic ball with the earth (or any other object) would be reversible, have you ever seen a tennis ball suddenly jumping back from the ground?


Is a carbon chiral it two of its groups are cis - trans isomers?

Any time all four groups bonded to carbon are distinct, including any sort of isomeric groups, you have a chiral center.  Comments associated with the related question referenced by @Loong indicate that the cis vs trans scenario is even included in the $R$ vs. $S$ naming convention.  This case is an $S$ configuration as the cis group takes second place and is counterclockwise from the hydroxyl group in the appropriate view.


Interaction between atom or molecule is neglected

True. In an ideal gas the gas particles are unaware of each other's existence.

  The transformation to liquid or solid is observed under appropriate condition

False. As mentioned in the answer to (1), the particles don't sense each other. Since there are no attractive interactions there is nothing to prevent the gas particles from flying apart, there is nothing to encourage association with other particles. This can of course be a bad assumption in reality, particularly if the gas is highly compressed, but you should not be applying the ideal gas law under such conditions anyway!

  The volume of constituent atom or molecule is neglected

True. This is a central tenet in the definition of an ideal gas. A correction for a definite particle volume is included in other gas laws, such as van der Waals'.

  The compressibility, PV/RT is independent of pressure

True. This follows directly from the definition, $PV=nRT$


The Chemistry behind creating Polylactic Acid (PLA)

Polylactic acid polymers are creating a lot of interest these days because they're biodegradable. On the downside, corn starch is often one of the starting materials...and corn is also used as a food source, so there's some controversy over the wisdom of this approach.

I think I've found a site that takes the approach you're basing your question on...but there are a couple of issues with the equation you've written. So, I'll answer this question in two parts. Plastic from starch is what you saw in the video; PLA from starch is what you asked about.

Plastic from starch. The recipe for the plastics-from-starch uses starch, glycerol, acetic acid (the $\ce{C2H4O2}$ in your equation) and water as a solvent. (They do not add lactic acid, the $\ce{C3H8O3}$ in your equation, but lactic acid can be produced by hydrolysis of the starch.) With the starch-acetic acid-glycerol recipe, you do not actually make a PLA polymer, but a starch polymer that is somewhat like it instead. (Ref)

So, starch is a polymer that looks like this:

In water, it tends to ball up into globular clusters due to hydrogen bonding. By adding acetic acid and heating the starch in water, you "denature" the starch and turn it into "disordered chains" that stretch out. When you let these disordered chains dry, they become entangled and form a flexible film, the plastic. In this recipe, the glycerol is included as a  plasticizer, so it is unchanged chemically in the reaction. The glycerol also helps to keep the starch stretched out by stabilizing it with hydrogen bonding. Plasticizers (as the name would indicate) are additives that increase the plasticity of a plastic and make it more durable or flexible. Changing the amount of glycerol in the recipe can have a big effect on the strength of the plastic film you make. This is the process you're seeing in the videos, I think.

Making PLA from starch. This is going to be a bit of a challenge to explain depending on your understanding of organic chemistry, but hopefully the pictures will help. When you make PLA commercially, you digest starch, usually with a bacteria, to turn it into lactic acid (see picture below for the structure of lactic acid). This lactic acid must be purified to make a high quality product. If you just combine a bunch of lactic acid molecules to make PLA, one water molecule would be lost each time they combine...sort of like this. (This is an acid and alcohol combining, not two acids, but the idea is the same.) The water in the polymerization reaction leads to a poor quality plastic.

So, in commercial synthetic schemes, two purified lactic acid molecules are reacted to form lactide...they come together in a controlled condensation reaction, with the loss of water. (Ref for picture.)

Now you can use the lactide, i.e. "lactic acids with the water removed", to do the polymerization to PLA. You add an initiator to open the lactic acid by breaking the bond between the $\ce{O-C}$ in  $\ce{O-C=O}$ and this gives you 2 lactic acid molecules hooked together...which react with another lactide molecule to add two more lactic acid molecules to the growing chain, and so on. And that's how the PLA forms.

So the reaction sequence is strach-> lactic acid-> lactide + $\ce{H2O}$ -> PLA

Probably more than you wanted to know!


I'm going to assume you are in high school, thus talking about Brønsted acids/bases.

A Brønsted acid is a compound that can donate a proton $\ce{H^+}$ whereas the base can accept a proton. For some chemicals it can be easily seen that they can donate a proton, like in the example of $\ce{HNO3}$ you give. There is obviously a proton in there so it makes sense that it is an acid. For chemicals with $\ce{O^-}$ in their structure (like the $\ce{KOH}$ in your example which dissociates into $\ce{K^+}$ and $\ce{OH^-}$) it is also clear that they can accept a proton and thus are obviously a base.

Now the difficult part is when a molecule can do both: these are called amphiprotic substances, the most famous being water $\ce{H2O}$, but there are many more of these, think about $\ce{H2PO4^-}$, it both can accept and donate a proton. In these situations molecules can indeed be considered both a base and an acid and the way they behave depends on the other molecules they come into contact with. If the other chemical is a stronger acid, as measured by the acid dissociation constant then the current molecule will act as base and vice versa.


Why is nitric acid often used to digest any solid sample?

This is pretty much correct. Nitric acid is a convenient oxidizing acid which will dissolve most of the metals (except for $\ce{Au}$, $\ce{Ta}$ and elements of platinum group), forming either oxides or nitrates; sulfur is oxidized to sulfate; phosphorus to phosphate; organics is oxidized and nitrated. This assures consistent predictable products in probe preparation.

Most oxides can be separated by filtering, and the majority of inorganic nitrates is soluble in aqueous solutions, which paves the way for the further classic analytical methods (titration, UV-Vis spectroscopy, atomic emission spectroscopy, etc.). Nitrate-ion is not prone to hydrolysis, and formed nitrates are no longer demonstrating oxidizing properties (unless they are solidified and heated), which is overall quite convenient.

Also, nitric acid is cheap and doesn't react with glass or PTFE, which allows to built various lab apparatus suitable for particular analytical method.


What are good examples of questions that make college students think about chemistry concepts?

Intermolecular forces

Are there contradictions about intermolecular forces in the postulates of kinetic theory?

Why do molecules having a higher relative molecular mass have stronger inter-molecular forces?
Kinetics of chemical reactions

Does a first order reaction really get completed only at infinity?
Equilibrium and direction of reactions

What does it mean to shift equilibrium?

At equilibrium are the rates of the forward and backward reactions equal to zero?
Acid/base reactions

Is the behavior of a dilute strong acid the same as the behavior of a weak acid?
Redox reactions (electrochemistry)

Why is it important to use a salt bridge in a voltaic cell? Can a wire be used?


Yes, this is a very interesting question that till today, not everyone agrees the precise mechanism as to how it works. Here is one with homolytic fission:

For your second query, I don't think it is feasible as the $\ce{C=O}$ carbonyl bond is rather stable.


Why does insufficient data lead to random errors?

More datapoints (if everything is done correctly) lead to more trustworthy results. With less datapoints, the possible (statistical) error grows, until at one point, when you only have as many datapoints as you have parameters, it looks like you have no error at all. (That is of course stupid, you should always have derived the error range of the individual data points.)

Now what is "insufficient"? Is that when the statistical error grows too large for your feeling? Or when you only have two points to draw a straight line through? Or when you only have two points to draw a parabola? That's your prerogative.

(note that I have also above used the word "error" in the same sloppy way your books seem to do. The "errors" are in your data, and they have a systematic component (hopefully small) and a distribution. The derived values (e.g. the slope of a linear regression) have a confidence interval.)


Flame Test Spectrograms Not Lining Up With Reality

The first step you can take to make the spectral bands narrow is to reduce the slit width. It is slightly wide at this moment. We are not sure where you were looking for wavelengths? In the compiled tables in handbooks or schematic diagrams on the web, the emission wavelengths are listed for the excited atoms using very high temperature flames using air acetylene, or nitrous oxide acetylene flames. The ordinary flame temperatures in a laboratory are too low to electronically excite the atoms of any other element other than alkali metals. For Li, Na, K, Cs, you would see a line spectrum for each element (one or two lines). For potassium is extremely difficult to see the far red wavelengths (the resonance wavelengths listed in tables) until and unless your eyes are sensitive to red. The flame colors Ca, Ba, Sr, Cu are usually due to molecular compounds of these metals in the ordinary flames.

The key point is that the flame temperature is too low to see the line spectra you were expecting. However, if you look at discharge tubes, then you can do a good comparison.


Why can dissolution reactions (e.g. dissolution of calcium chloride) be exothermic?

The endothermicity (taking up heat) or exothermicity (giving off heat) depends on a balance of factors including the entropy change, the degree with which the ions become associated with water, and the possibility of actual chemical reactions.

The most notorious example is sulfuric acid, which reacts so strongly with water that it can actually boil the water, and spatter the acid, which is why one always adds the acid to the water(water as solvent) and precautionary measures (gloves and proper clothing) are needed to handle the reaction.


Why is it so difficult to create mock NMR spectra of macromolecules?

Macromolecular NMR spectra are difficult to compute because macromolecular electronic structures are computationally expensive to solve with accuracy. Obtaining precise chemical shifts and splitting requires accurate treatment of electron density and geometry, especially for simulation of NOESY, which I understand is important in macromolecular structural elucidation. This requires good methodology and large basis sets.
An interesting case study by Saielli and Bagno demonstrates the use of DFT (with cc-PVTZ basis) in predicting NMR spectra for two candidate structures and the correlating them with experimental data, however the mean error in chemical shifts is nontrivial1 (0.2 ppm in the best match). The same people do $^{1}\ce{H}$ and $\ce{^{13}C}$ NMR predictions of a number of complex molecules2 (with intriguing discrepancies in shifts) however this is a far cry from predicting, say, a large protein $^{1}\ce{H}$ NMR spectrum.
I have seen EPR spectra of transition metal clusters simulated as abstract spin systems (for instance by EasySpin) however I presume that this is very qualitative.

(1) Saielli, G.; Bagno, A.; Can two molecules have the same NMR spectrum? Hexacyclinol revisited.; Org. Lett.; (2009) 11, pp. 1409-1412
(2) Bagno, A.; Rastrelli, F.; Saielli, G.; Toward the complete prediction of the $\ce{^1H}$ and $\ce{^{13}C}$ NMR spectra of complex organic molecules by DFT methods. Application to natural substances.; Chem. Eur. J.; (2006) 12, pp. 5514-5525


Why do the splittings in mirrored COSY crosspeaks differ?

Let's focus for now on the two circled crosspeaks. In a COSY sequence, the frequency in the indirect dimension (the vertical axis) corresponds to the spin which the magnetisation is on during the $t_1$ period. The frequency in the direct dimension (horizontal axis) corresponds to the spin which you measure the magnetisation of during $t_2$.

It turns out that these crosspeaks are from butanal, so let's also colour-code the protons in butanal as follows: we have $\ce{C\color{red}{H}_3-C\color{blue}{H}_2-C\color{green}{H}_2-CHO}$.

The red crosspeak occurs at ~1.65 ppm in the indirect dimension, and ~0.95 ppm in the direct dimension. This means that the magnetisation during $t_1$ was on the blue $\ce{C\color{blue}{H}_2}$ proton, and was subsequently transferred to the red
$\ce{C\color{red}{H}_3}$ proton during acquisition (in $t_2$). The coupling between these two sets of protons is referred to as an active coupling; it is the coupling that is responsible for generating the crosspeak.

Conversely, the the blue crosspeak arises due to transfer from $\ce{C\color{red}{H}_3}$ (during $t_1$) to $\ce{C\color{blue}{H}_2}$ (during $t_2$). So, it also has an active coupling between the same nuclei. The difference here is that the final $\ce{C\color{blue}{H}_2}$ protons are also coupled to the green $\ce{C\color{green}{H}_2}$ protons! This is what is called a passive coupling: one that does not generate the crosspeak, but is still reflected in the multiplet structure as a splitting.

The differences in the crosspeak structure arise because of differences in passive couplings to other nuclei.

Exactly how this is manifested is not easy to explain in this case, because there are multiple equivalent nuclei. But if you are keen, read on. (I'm also not going to present any quantum mechanical analysis, because that would really take way too long.)

In fact, the red crosspeak also has one passive coupling – the $\ce{C\color{red}{H}_3}$ group is coupled to not just one, but two protons adjacent to it, and the crosspeak resembles a case where one of the two protons is an active coupling partner and the other is passive.

We first deal with passive couplings. These lead to in-phase splittings, i.e. peaks which are both positive and are separated by the coupling constant, approximately 7 Hz in our case. For simplicity's sake we will stick to just the direct dimension for now. Because we only have one effective passive coupling, this leads to a doublet with a peak separation of ~7 Hz.

(The indirect dimension, broadly speaking, has similar behaviour in that it is also affected by passive and active couplings, but the exact source of these couplings can differ. Also, it is sometimes harder to observe multiplet structure in the indirect dimension because of different resolutions in the two dimensions.)

Active couplings, on the other hand, lead to antiphase splitting, where peaks have opposite signs. So, our doublet is further split into two doublets: one positive and one negative, and these two doublets themselves are 7 Hz apart. We find that the middle peak exactly cancels itself out, and we are left with one positive and one negative peak separated by exactly twice the coupling constant. This is what is observed in the red crosspeak.

On the other hand, the blue $\ce{C\color{red}{H}_2}$ protons have one active coupling (to one of the $\ce{C\color{red}{H}_3}$ protons), but the other two $\ce{C\color{red}{H}_3}$ protons are effectively passive couplings. On top of that, it has two further passive couplings to the $\ce{C\color{green}{H}_2}$ protons.

The four passive couplings lead to the formation of a $1:4:6:4:1$ quintet (we're going to assume that all the couplings have similar size, which is quite reasonable). The single active coupling causes an antiphase splitting of this quintet, which is easier to express mathematically:

1   4   6   4   1
   -1  -4  -6  -4  -1
-----------------------
1   3   2  -2  -3  -1

(imagine that there is 7 Hz separation between every column of numbers). This $1:3:2:-2:-3:-1$ pattern exactly matches how the blue crosspeak looks like.

For more information, see e.g. Claridge, High-Resolution NMR Techniques in Organic Chemistry (3rd ed.), Section 6.1.5.2.


Does (2-fluoroethyl)benzene undergo elimination via E1cb or E2?

I think that it is easier for the molecule $\ce{Ph-CH2-CH2-F}$ to undergo E1cb mechanism than E2 mechanism. It can be also possible that the molecule undergoes elimination through a mixture of E1cb and E2, but it can be said  that majority portion of the molecules will prefer the E1cb rather than any other pathway.

First of all, an E2 mechanism goes via simultaneous cleavage of the $\ce{C_{\beta}-H}$ and $\ce{C_{\alpha}-LG}$ bonds (LG being the leaving group). So, the rate of the reaction depends on the ease of breaking these two bonds. In this case the leaving group is fluoride ($\ce{F-}$), which has a very strong bond with the $\alpha$-carbon, making it a bad leaving group. So, due to this reason, the rate of E2 should lower for this molecule.

Now, if we check the plausibility of E1cb reaction pathway, we will find that it is most suitable in this case.

Firstly, the proton abstraction by the base gives a highly resonance stabilised benzylic carbanion intermediate, and this carbanion formation is the slow step. As this intermediate's energy will decrease, the activation energy will decrease resulting in an increase in the rate constant as, $k = A\exp(-E_\mathrm a/RT)$, helping to increase the rate for E1cb.

Secondly, the breaking of $\ce{C-F}$ bond doesn't happen in the rate determining step, which also doesn't alter the rate of the reaction  in this case.

So, observing the above mentioned cases, it can be definitely concluded that  E1cb is the more preferable pathway to undergo $\beta$-elimination to form Styrene.


Why does water weaken ion ion attractions?

Electrostatic field caused by ions or any other external source causes partial orientation of molecular electric dipoles along the field vector. Their orientation causes displacements of center of positive and negative charges. This displacement causes induced electrostatic field, that decreases the primary field intensity by the value of relative permittivity.
By other words, you replace the vacuum permittivity in the Coulomb law by the water permittivity.
$$F = \frac {q_1 \cdot q_2}{4 \cdot \pi \cdot \epsilon  \cdot r^2}$$


Well, the first step is to stop thinking electrons as being very small balls that orbit around the nucleus in a circular path. This is known as the Bohr model. Despite this model being an excellent model at an introductory level, it doesn't tell the entire story.

There are several issues with the notion that electrons travel in a circular orbit around the nucleus, the main one being that they should eventually undergo orbital decay and hence eventually slow down and crash into the nucleus. Of course, this doesn't happen. So how do electrons move around the nucleus?

This is where quantum theory came to the rescue in the early 20th century. It stated that electrons could not be treated as a classical particle and didn't not have a definite position and momentum. So this basically means that we can't exactly know an electron's position and its momentum at the same time. This relationship is given by the Heisenberg Uncertainty Principle.

I won't go into detail about it since to keep this explanation basic, but if you are interested check out this video which provides a really good explanation.

So basically what quantum theory tells us that we actually can't specify where and what a path an electron is going to take. However the Bohr model assumes that you can precisely plot out where an electron is going to be, but we now know that due to the Heisenberg Uncertainty Principle you can’t precisely know the exact position and momentum of an electron. So we actually can't construct an orbit and say that the electron is always going to follow that path.

Therefore, it makes it impossible to plot an orbit for an electron as how can you? If you don’t know where the electron is going to be next, there is no way you can predict its path.

So how can we get around this problem? Well, if we can’t draw the orbit for an electron the next best thing we can do is to plot an electron density probability orbital for it. Also known as atomic orbitals.

An orbital is a well defined region of space. Therefore an atomic orbital is a region of space that shows where the electron will be 95% of the time (usually we take 95% but it could be any number really such as 90% or 75%).

A simple way of thinking of orbitals is that imagine that you have a magic camera that can take a sequence of photos of an electron in a hydrogen atom. The electron appears as a dot. Now if we superimpose all of these images, you will see something like this:

Since we generally take atomic orbitals to be the region of space that shows where the electron will be 95% of the time, the very outer sphere of the above image is said to be the orbital of the electron.

Hopes this helps you understand orbitals. If you have any question, feel free to ask.


Treatment of allylic alcohol with thionyl chloride: why is the product rearranged?

As a disclaimer, it is worth noting that the reaction of an alcohol with thionyl chloride can proceed via multiple different mechanistic pathways (Wikipedia has a pretty good overview), depending on the conditions.
In the $\mathrm{S_Ni}$ mechanism a free carbocation is not formed; instead the key intermediate is an intimate ion pair $\ce{R+...^-OSOCl}$. This is not the same as a typical $\mathrm{S_N1}$ mechanism, where the leaving group is fully dissociated.
Following the formation of the intimate ion pair, the anionic component breaks down to form $\ce{SO2}$ and chloride ion, the latter of which attacks the cationic component $\ce{R+}$ to form $\ce{R-Cl}$. The reaction proceeds with retention of configuration, because the nucleophilic chloride ion originates from the leaving group $\ce{-OSOCl}$, and hence must attack from the same face as the leaving group left from.1

If the carbocation $\ce{R+}$ is allylic, then it is electrophilic at both the α-position (i.e. where the hydroxyl group started) and γ-position (i.e. two carbons away from the hydroxyl group). Therefore, following breakdown of the $\ce{-OSOCl}$ leaving group, the chloride can choose to attack either position. The attack at the γ-position is sometimes called the $\mathrm{S_Ni'}$ mechanism.

While the example above proceeds with extremely high selectivity for the rearranged product, this is not always the case. The above reaction was reported by Young in 1962,2 and carried out in dilute ether solution, which strongly favours the $\mathrm{S_Ni'}$ pathway. According to Gu and Zakarian in Comprehensive Organic Synthesis II (Chapter 6.16),3

Thionyl chloride has been used traditionally as the reagent of choice to ensure a high degree of selectivity in favor of allylic transposition, however, the outcome is often less predictable than desired. [...] The reaction, however, is often more complicated and dependent on reaction conditions and, to a greater extent, the structure of substrate. This is illustrated in the total synthesis of codeine reported in 2011. When allopseudocodeine was treated with thionyl chloride, a nearly equimolar mixture of regioisomeric products was produced.

References

Smith, M. B. March's Advanced Organic Chemistry: Reactions, Mechanisms, and Structure, 7th ed.; Wiley: Hoboken, NJ, 2013, pp 408–409.

Young, W. G. Unexpected rearrangement and lack of rearrangement in allylic systems. J. Chem. Educ. 1962, 39 (9), 455–460. DOI: 10.1021/ed039p455.

Gu, Z.; Zakarian, A. Functional Group Transformation via Allyl Rearrangement. In Comprehensive Organic Synthesis II; Knochel, P., Molander, G. A. Eds.; Elsevier, 2014; Vol. 6, pp 636–754. DOI: 10.1016/B978-0-08-097742-3.00624-8.


By definition molecular weight is mass $m$ divided by amount $n$, e.g. for the unknown metal $\ce{M}$:

$$M(\ce{M}) = \frac{m(\ce{M})}{n(\ce{M})} \tag{1} \label{eq:1}$$

From the chemical reaction of hydroxide formation one can deduce that $\pu{1 mol}$ of $\ce{M}$ results in $\pu{1 mol}$ of $\ce{M(OH)2}$. However, the amount of hydroxide-anions $\ce{OH-}$ in the solution (assuming complete dissociation) is twice as high as the amount of $\ce{M(OH)2}$. Hence

$$n(\ce{M}) = n(\ce{M(OH)2}) = \frac{n(\ce{OH-})}{2} = \frac{C(\ce{OH-}) \cdot V}{2} \tag{2} \label{eq:2}$$

where $C$ – concentration and $V$ – volume.

Plugging $n(\ce{M})$ from \eqref{eq:2} in \eqref{eq:1}, a final expression for the molecular weight can be established. Remember, before plugging in the numbers in make sure the units are unified.

   $$M(\ce{M}) = \frac{2 \cdot m(\ce{M})}{C(\ce{OH-}) \cdot V} = \frac{2 \cdot \pu{0.115 g}}{\pu{6.56e-3 mol dm-3} \cdot \pu{0.4 dm3}} = \pu{87.6 g mol-1} \tag{3} \label{eq:3}$$
 The unknown metal appears to be strontium $\ce{Sr}$.


Hydrochloric acid is already dissociated to $\ce{H+}$ and $\ce{Cl-}$

Sodium carbonate as ionic compound dissolves as

$$\ce{Na2CO3(s) <=> 2 Na+(aq) + CO3^2-(aq)}$$

Chloride and sodium ions are in this reaction called the spectator/bystander ions, not being involved in the reaction.

The ongoing reactions are these :

$$\ce{ 2 H+ + CO3^2- <=>> H2CO3 <=>> H2O + CO2(aq)}$$

followed by the dissolution equilibrium:

$$\ce{ CO2(aq) <=> CO2(g)}$$

If solution is lead to crystallization, the spectator ions then meet to form ionic crystals, keeping their charges.

$$\ce{Na+(aq) +  Cl-(aq) <=> NaCl(s)}$$


Why isn't ionization energy called cationization energy?

Atoms can have electrons added to them or removed from them, which in turn ionises the atoms. However, electron addition and removal aren't the same thing in terms of enthalpy changes. The key here is the difference between the
process and the enthalpy changes during the process - as much as an atom is ionised during the production of a cation or an anion, the enthalpy changes are different. Because of this, there are two different terms used for the enthalpy changes during these processes:

'Ionisation energy' is the enthalpy change when one mole of gaseous atoms form one mole of gaseous positive ions (cations), under standard conditions,
'Electron affinity' is the enthalpy change when one mole of gaseous atoms form one mole of gaseous negative ions (anions), under standard conditions.

As seen from the definition for 'ionisation energy', it's already described that there's a production of cations, so it would be redundant to call it 'cationisaion energy'.  The word 'ionisation' in itself is typically used to describe the removal of electrons from an atom, which could also contribute to the term's name.

The reason why there are two different terms is due to how the electrons are dealt with during ionisation. The removal of electrons is always endothermic, whereas the addition of electrons is only sometimes endothermic.

When removing electrons, more energy is required than what's released to remove the electrons from the atom, hence why it's always endothermic. This is due to the Coulombic attraction between the negatively charged electron(s) in the atom's orbit and the positively charged proton(s) in the atom's nucleus. The required energy will increase with each successive removal, since the electrons will get closer to the nucleus and the attraction will increase because of it. This enthalpy change is explicitly called 'ionisation energy'.

When adding electrons, only the first addition is exothermic whilst the rest are endothermic. The first addition is 'easy'; you're just adding an electron to a neutrally charged atom, and the energy released during this process is greater than the energy required. However, with each successive addition, the energy required will increase. This is due to the Coulombic repulsion between the anion and the electrons, so more energy is required than what's released to 'force' the electrons into the atom's orbit, hence why successive electron additions are endothermic. This enthalpy change is explicitly called 'electron affinity'.


Why is SN2 favored greatly over SN2' in this reaction?

While pi-based resonance effects are frequently important, you must always consider inductive effects in every reaction.

Electrostatically, there is a much higher positive charge on the ipso carbon than the beta carbon, because of inductive effects.  The net effect is that there is a lower energy of activation for nucleophile attack at the electrophilic center because the coloumbic forces are more favorable for attack there.  The softer the nucleophile and electrophile become, that less important this effect is, but it is still quite considerable.

Without doing the QM calcs, I would rate the phosphonium electrophile as quite hard.


Double alkylation of benzene to form tetralin

You've got it right, double acylation and double reduction - but the first reduction occurs after the first acylation.  FWIW, the name of the reaction is the Haworth reaction and is commonly used to synthesize substituted tetralins, naphthalenes and higher aromatic polycyclics.

In your particular case, here is a diagram of the first 3 steps in the reaction

Friedel-Crafts acylation of benzene with succinic anhydride
reduction of the ketone using a Wolff-Kishner or Clemmensen reduction
a second Friedel-Crafts acylation using strong acid; it is an internal cyclization and forms the needed ring
and finally (not shown in the above diagram) a second Wolff-Kishner or Clemmensen reduction of the resultant tetralone to produce the desired tetralin


Calculating the total uncertainty of the final product

General approach

\begin{equation}
    \sigma_{F}^2 = \sum \left( \frac{\partial F}{\partial x_i} \right)^2 \mathrm{d}x_i^2
\end{equation}

Here we neglect all correlation terms formed by squaring what is essentially the first order Taylor expansion with respect to the variables about their uncertainty. $x_i$ is a variable or parameter which contains uncertainty and $F$ is the function which depends on the variables and parameters $x_i$. $\mathrm{d}x_i$ is the uncertainty in $x_i,$ i.e., you measure the mass of your solution 10 times and it is 1 kg ± 0.1 kg. Your $\mathrm{d}x_i$ for mass is 0.1 kg. If you want you can think of $\mathrm{d}x_i$ as $\Delta x_i$ instead.

For a given calculation $f$ you take its derivative $F$ wrt each variable/parameter, hence you know its rate of change, and $\mathrm{d}x$ is the known/measured uncertainty of that variable/parameters, i.e., how much the function might be out for a given error in a specific parameter.

So, in your original expression you only accounted for the uncertainty in the parameters, but not how their error would propagate into each and every function/equation/calculation they are used. If a function is linear in a given $x_i$ then at most it will be out by the uncertainties magnitude multiplies by some constant in the equation. However, it may be that the function contains $x_i^{10}$ in which case the error in $x_i$ will be to the 10th power which could be huge! So, use the above equation, with the appropriate derivatives to "propagate" error.

Some background derivation

The change in a function due to a variable/parameter can follow from its differentiation

\begin{equation}
\mathrm{d}f = \left( \frac{\partial f}{\partial x} \right) \mathrm{d}x + \left( \frac{\partial f}{\partial y} \right) \mathrm{d}y
\end{equation}

It is possible that the error due to variable $x$ is equal but opposite to $y,$ and the errors cancel. However, when we square it, which gives us

\begin{equation}
\sigma_f^2 = \left( \frac{\partial f}{\partial x} \right)^2 \mathrm{d}x^2 + \left( \frac{\partial f}{\partial y} \right)^2 \mathrm{d}y^2 + 2\left( \frac{\partial f}{\partial x} \right)\left( \frac{\partial f}{\partial y} \right) \mathrm{d}xy,
\end{equation}

this way errors don't cancel. Also, we almost always assume the correlation term $\mathrm{d}xy$ is negligible. Neglecting the correlation terms, we find ourselves left with the top equation.

More info can be found here: Wikipedia uncertainty propagation. Also, don't forget to donate to Wikipedia. :)


can an alloy be used to make another alloy?

An alloy is purely a mixture of metals (and sometimes non-metals) which are non-chemically bonded.

So yes, you could mix steel and silver to create an alloy. There would be nothing stopping you from doing this, but I cannot say whether or not it would produce any desirable results. Mixing alloys with other materials is merely creating another alloy. Stainless steel is actually just regular steel combined with Chromium.

An alloy is usually created to increase certain properties of its constituents. For example, iron is mixed with carbon in order to create steel, which is far stronger and can be used to manufacture items such as girders. Or, in other cases, Chromium may  be added, if the steel is to be used in, say, cutlery.


Why is tert-butoxide often used in elimination reactions when it is not necessary?

Several reasons:

t-BuOK is especially known as a strong base, and a poor nucleophile. Its large, bulky structure causes it to perform exceptionally poorly in substitution, literally eliminating any side reactions when the desired product is the elimination product.
It is easily available, like Raphaël insists.
It's basic strength depends on the medium. It is very strong in DMSO, where the solvent complexates well with $\ce{K+}$ ions. However it is relatively weaker in benzene. This flexibility gives it inevitable popularity.
Many condensation reactions (Stobe's, Darzen's) are seen to have greater yield when t-BuOK is used.


What signal is measured at the detector in atomic absorption spectroscopy?

You raised an excellent point, which is not as simple a UV-Vis spectrophotometry because most molecules do no fluoresce or emit light, whereas atoms do. This issue makes an atomic absorption experiment slightly more challenging (at least in principle). The critical question is "what" signal reaches the detector. How does one distinguish the contribution of flame emission from the hollow cathode lamp (HCL) beam? Last but not least, is there is a way to distinguish atomic emission due to HCL?
Have a look at a typical AAS instrument. The critical component is the chopper, which is nothing but a rotating mirror with holes. It merely blocks the light from reaching the flame while rotating at an exact frequency.

You can see only two signals are needed in an atomic absorption spectrophotometry experiment: a) Light beam intensity when there are no atoms in the flame and b) light beam intensity when analyte atoms are present in the flame.

You aspirate a blank (no analyte solution) into the flame- this sets the 100% transmission or absorbance = 0 at the chosen wavelength. Ignore the reference beam; it is for ensuring that the lamp is stable. We are only interested in the beam passing through the flame.

Next, you would aspirate your analyte solution into the flame. Let us take Na as the most straightforward example. There is light absorption by Na atoms in the flame, leading to a reduction in the HCL beam's light intensity after passing through the flame. The ratio of light intensity in steps 2 and 1 gives the transmittance, which is easily converted into absorbance.

However, there is an almost continuous emission spectrum from the flame too.
Here is the smart idea. I do not know who came up with it. The detector continuously sees the background emission from the flame. However, the light beam from the hollow cathode lamp reaches the detector alternately at a particular frequency due to the rotating chopper. The electronics are tuned to register only the light, which is oscillating due to the chopper

The critical but no so critical problem: When atoms absorb a specific wavelength, they emit precisely the same wavelength as you noticed in your post in the flame (called atomic fluorescence). the chopper cannot stop that light. Hence it is part of the transmitted light beam.

The good news is that atomic fluorescence is not directional, unlike the light beam from the HCL. Secondly, the intensity of atomic fluorescence light is so small compared to the HCL beam intensity that it is practically negligible. There is no way to prevent atomic fluorescence from reaching the detector in an atomic absorption experiment, to the best of my knowledge.
Can we see stars during daylight? Stars are always there and "part" of the sunlight. However, stars contribute very little to the daylight. In the same way, atomic fluorescence contribution in an atomic absorption experiment is almost null because the hollow cathode lamp beam is quite strong as compared to the atomic fluorescence in random directions.


How did early chemists measure concentrations and purity?

Not everything comes as a solution. Rather, the historical chemist would have started of with solid substances. These were purified according to ‘established procedures’ (distillation, recrystallisation, dissolution and precipitation) which worked quite well. Then, one of these substances would have been used to make a stock solution — and because it is solid you can simply weigh it.

Now you might assume that the historical chemist had no knowledge on how two substances would react — which is likely true. However, once he knew the composition of one type of solution, he was able to express the other in equivalents: ‘this solution is able to reduce two equivalents of the permanganate solution’ (using the first example that jumped to my head; not sure whether that is actually a likely thought).

Going via the diversion of analytical chemistry (is this a compound or an element?), it was established that a law of multiple proportions existed: if two elements form more than one compound, the mass ratio of one of those elements is always a simple integer ratio. This was easily extended to reactions: if something reacted with 2 equivalents of A but 1.5 equivalents of B, then maybe it wasn’t the concentration I assumed but I should multiply by two?

Over time and more data points, the current system would have been established.


Actually in $\ce{CH4}$ with slightly more electronegtaive carbon it is in $-4$ state. With $\ce{CO2}$ the oxygen is more electronegative oxygen carbon is $+4$. Therefore oxidation is also defined as addition/replacement of a more electronegative element.

Electronegativities $$\begin{array}{}\bf H&2.2\\\bf C&2.55\\\bf O&3.44\end{array}$$


Keeping the electronic configuration in the ground state is expensive in Ab-initio MD?

Let us first note that the calculation of the first derivative of the total DFT or HF ground state energy of a molecule with respect to its nuclear coordinates (called the gradient) depends on a converged self-consistent field (SCF) solution. The solution is obtained in an iterative fashion and is represented as a collection of molecular orbital (MO) coefficients. In some sense, they are unique for the nuclear coordinates. However, if we move the nuclei only very slightly, repeat the calculation, and compare the new set of MO coefficients with the previous ones, they will be almost identical.

Almost, that's the key here. The gradient will become more of an approximation as you move away from the nuclear coordinates for which the underlying SCF solution was originally calculated (and merely recalculate the gradient at new coordinates using old MO coefficients). One way of dealing with this problem is to take the old SCF solution as a starting point for a new SCF solution, which should converge in just a few steps. But those few SCF iterations can easily dominate the cost, given that a gradient calculation after obtaining the SCF solution typically has a computational cost on the order of 1-3 SCF iterations.$^{1}$

To summarize: I understand the sentence "However, the computational cost of dynamical simulation appears to be mostly due to keeping the electronic configuration in the ground state." in the quoted section of the question to mean: recalculation of the SCF solution is necessary at least every few steps of the molecular dynamics (MD) simulation and that recalculation is the most expensive part.

$^{1}$ A specialized quantum chemistry program, that would keep certain integrals and other intermediates across the different SCF procedures, would help somewhat with the computational cost at the expense of more on-disk storage and more uncertainty, since all those quantities implicitly depend on the nuclear coordinates.


The required HLB doesn't depend on the concentration of water. The required HLB of mineral oil is ca. from 9 to 12. You have to find out the required HLB of your mineral oil at first.


You can call it a monomer, although your audience will expect that you are about to present its oligomerization or polymerization. And according to IUPAC you should put parentheses if in the brackets you have the repeat unit of the polymer: e.g. poly(styrene).


Wouldn't this mean that the products have more enthalpy?

The products have less enthalpy, maybe the following diagram remediates misconception :

$ΔH=Σproducts-Σreactants = {-965.5 -(-74.87)}=-890.63$


Why are actinides not commonly included in rare earth metals?

It comes down to the history of their discovery, natural occurrence and applications. It's a fascinating story that you can read about in here:

http://www.vanderkrogt.net/elements/rareearths.php

The rare earths (i.e. lanthanides) were discovered over a few decades in the late 1800s. They were laboriously separated from "rare earths" - the name used back then to describe the rare minerals (silicates, oxides, phosphates, etc) that included these elements. These are all naturally occurring rocks. You can literally pick up from the earth a mineral that has rare earths in it. They are also stable (apart from Pm). Therefore, they are used in industry. There are rare earth mines, rare earth magnets, rare earth lasers. You get the stuff from the earth.

Actinides, were mostly discovered in the early to mid 1900s, after rare earths were a "thing". Other than U and Th, you do not find them in nature. You do not mine them from the earth, and they do not exist in "earths". Their radioactivity precludes their use in industrial applications other than a few niche uses. Although their chemical properties are similar to the lanthanides, there are many differences. They are simply not rare earths. Almost, but not quite.

I'll also add that the definition of "rare earths" is a loose one. The broad chemical definition includes the lanthanides La–Lu, Y and Sc. As far as geologists care, Sc is not a rare earth element, and Y is only "honorary". Industry and mining usually consider the lanthanide oxides as REO (rare earth oxides) and use REY (rare earths + yttrium) when they are talking also about Y, because it's not obvious to them that it is a rare earth.


Is it so because it is impossible to solve mathematically?

Yes, in both classical and quantum mechanics there is no known general closed form analytical solution for the three-body problem. To be more precise, for the case of classical mechanics, an analytical solution in the form of a convergent power series exist, but is useless in practice due to very slow convergence.

  If the answer could be explained in terms that a new master student in
  theoretical chemistry would understand, that would be great!

I doubt that is possible. For the case of classical mechanics, Poincaré had a simple argument: one has more degrees of freedom than integrals of motion, and thus, could not solve the equation of motion through the usual integration method. But, as I mentioned, that's not the whole story, and secondly, I'm not sure how to deal with quantum mechanics anyway. So, as Ivan Neretin mentioned in his comment, things are the way they are: for all practical purposes the three-body problems could not be solved analytically.

  Why cannot the Schrödinger equation be solved exactly for systems in
  which more than two particles interact?

And besides, you have to be careful the wording here. I mean, you can easily have a two-particle system for which there is no known analytical solution. This is because the procedure to obtain analytical solution for a two-particle problem relies on the center-of-mass separation trick which works if the potential energy depends only on the relative positions of the particles. More formally, it works only if interactions in the two-particle system are invariant to the translation of the system as a whole. This is the case of two isolated particles that interact just with each other, for instance, an electron and a proton in an isolated hydrogen atom. For this system you can indeed use the center-of-mass separation, reduce the two-body problem to a set of two one-body problems, and get the well-known analytical solutions.

But what if you put that hydrogen atom in a inhomogeneous electric field? Center-of-mass separation won't work than, since potential energy terms due to interaction of both the electron and the proton with the external field depend on their actual position with respect to the field, not just on their relative positions. The same thing happens when considering two electrons in a field of a fixed proton, i.e. just the electronic subsystem of a helium atom. The electronic two-body problem is also unsolvable analytically in that case due to the presence of external (for the two-electron system in question) inhomogeneous electric field created by the proton.


Why is the electronic wavefunction of helium a product of the two 1s wavefunctions when ignoring electron-electron repulsion?

The same mechanics as for the separation of angular and radial parts of the hydrogen electronic wavefunction are in place here: You have a Schrödinger equation that is separable into two (i.e. one for each electron). You solve them and in order to combine them, they must be multiplied.

It so happens in this case that the solution for each equation is known and recognizable: the electronic wavefunction of $\ce{He+}$. The fact that the two wavefunctions are multiplied is not related at all to electronic structure theory per se (no involvement of LCAO etc.), but merely a result of the separability of the Schrödinger equation.


What is the difference between volatility and evaporation?

Evaporate is a phase change from liquid to gas without changing the molecule itself (or the atoms is the liquid is atomic).

Volatility is essentially the tendency of a liquid to evaporate. So a highly volatile liquid would show a great tendency to evaporate.

A measure of volatility is the vapor pressure. There is a very related term partial pressure. So if we think of air as being nitrogen and oxygen, then there is a partial pressure for nitrogen and a partial pressure for oxygen, and the sum of the two partial pressure is the actual pressure (the absolute pressure).

To make things a bit more confusing chemists think of absolute pressure. So because we are in earth's atmosphere there is always 1 atmosphere of gas pressure. So when you pump a car tire up to 30 lbs/sq inch, there is already 14.7 lbs /sq inch air pressure due to the atmosphere. So the absolute pressure inside the tire is 44.7 lbs/sq inch.


Proton-decoupled phosphorus spectrum of a ruthenium complex

Let’s call the isomers mer-trans, mer-cis and fac from left to right for easier addressing.

What you didn’t indicate yet is which symmetry elements the different ones have and if they are chiral; this is helpful in determining the look of the NMR spectra because it helps us identify enantiotopic and homotopic atoms which give identical spins. The mer-trans isomer is clearly the one with the highest degree of symmetry: I judge its point group to be $C_\mathrm{2v}$. The other two should be $C_\mathrm{s}$, if I did the looking correctly.

Once we have identified the symmetry, we need to find out which atoms are chemically and/or magnetically equivalent. For the two $C_\mathrm{s}$ compounds it is pretty easy: If the plane of symmetry transforms one atom onto the other, they are chemically equivalent but enantiotopic (so magnetically non-equivalent). If the plane of symmetry contains the atoms in question, they must be chemically non-equivalent, thus have different spins. For the mer-trans isomer it may look slightly more tricky, but note that spinning around the $C_2$ axis will always place one atom of the same kind in question on top of the other thus again making them chemically equivalent and homotopic. (Also note that the three phosphorus atoms can never all be chemically equivalent to each other since one always occupies a special position by symmetry.)

With our sets of chemically equivalent or non-equivalent atoms in hand — to check: I arrived at sets of 1, 1 and 2 hydrogens and always two phosphorus; in one phosphorus set we have a pair chemically equivalent atoms, the other two contain a pair of enantiotopic ones — we can predict the look of the spectra and what will couple to what. Note that the $\ce{^{31}P\{^1H\}}$ spectrum will not contain any couplings to hydrogen atoms.

For mer-trans, your reasoning is sound and I also expect the result to be a doublet and a triplet; the coupling constants must be identical. If I had to guess, I would say that those closer to $\ce{CO}$ should have a higher chemical shift due to $\ce{CO}$’s electron withdrawing nature; but it could be due to a push-pull mechanism that its influence is greater in the para-position.
mer-cis should also be a doublet and a triplet for more or less the same reasons. The chemical shifts must be non-identical but could be closer to each other due to every phosphorus being a neighbour of $\ce{CO}$ (which is still the single most differentiating species in the compound).
fac cannot be a doublet of doublets since that would be a single signal for all three; the three are, however, chemically non-equivalent. The single special phosphorus will again display a triplet because it is coupling to two distinct phosphorus atoms. The other two should show up as a doublet. I’ll let you do the reasoning.

When proceeding to the hydrogen spectra, remember that $\ce{^{31}P}$ is also a magnetically active, spin ½, $100~\%$ abundance nucleus. Thus, you must consider the coupling of the protons with each other (if applicable) plus the coupling of the protons with the phosphorus atoms (the magnetically non-equivalent ones). In no particular order you should arrive at:

One spectrum containing two distinct proton shifts, each with two doublet couplings and a triplet coupling. I cannot say a priori which one should be larger or smaller.
One spectrum containing a single proton shift which is split up into a doublet and a triplet. Again, I cannot say which coupling is stronger.
One spectrum containing a single proton shift, split into a doublet of doublets of doublets (ddd). At least here I don’t have to say which coupling is stronger because they are all distinct.

Retroactively assinging proton spin systems, I get $\ce{AX, A2}$ and $\ce{AA'}$; for phosphorus I arrive at $\ce{AX2, AA'X}$ and $\ce{AX2}$, again in no particular order for all six spin systems (i.e. you would still need to assign them to the correct structures). If you wish to combine the spin systems into one entire one per structure, use $\ce{M}$ rather than $\ce{A}$ in the hydrogen ones (or $\ce{K}$ and $\ce{R}$ instead of the hydrogen $\ce{AX}$). Note however that I am terrible at spin systems and that you should take those with a grain of salt.


The term "fused" in chemistry generally refers

to become physically joined together

as by melting, adherence or other means of consolidation [1].
At an industrial scale solutions containing $\ce{CaCl2}$ are obtained either as by-products in the production of soda (by the ammonia method) and Berthelot's salt, or by the adding $\ce{HCl}$ to limewater. Solutions are then evaporated and $\ce{CaCl2}$ is isolated as a white or greyish porous mass, which is called fused calcium chloride and is generally water-free or contains trace amounts of water.
To make it clear, in this case "fused" ≠ "melted", as it's not about melting $\ce{CaCl2}$ (m.p. = $\pu{772^\circ C}$), but rather a physical adhesion as a result of a set of steps applied to calcium chloride solution (or its deliquescent form) of gradually increasing the temperature from $\pu{70 .. 80^\circ C}$ to approx. $\pu{300^\circ C}$. This results in formation of hydrated forms ($\ce{CaCl2 * x H2O}$) with subsequent complete dehydration and solidifying above $\pu{250^\circ C}$ alongside with minor partial hydrolysis to form $\ce{CaO}$ and $\ce{HCl}$.
Fused refers to the fact that this porous mass is granulated, but not dispersed, in order to increase the surface area and improve its dehumidifying ability and at the same time to prevent the salt from deliquescing swiftly.
Reference

fused. (n.d.) American Heritage® Dictionary of the English Language, Fifth Edition. 2011.


How does toluene react at higher temperatures and why?

Assuming you're talking about Friedel-Crafts alkylation of toluene, this can be explained by the reaction being reversible and thermodynamic control taking precedence at sufficiently high temperatures. The directing ability of substituents in electrophilic aromatic substitution reactions is a function of the reaction being under kinetic control at lower temperatures. According to Fleming's book on MO theory, a single electron-donating substituent on benzene conjugated to the ring's $\pi$ system disrupts the normal degeneracy of the $\psi_2$ and $\psi_3$ molecular orbitals of unsubstituted benzene. Comparing the intermediates of an EDG-substituted benzene undergoing electrophilic attack, the HOMO of the $\pi$ system, $\psi_3$, is higher in energy when attack occurs at the meta- position than when it occurs at the ortho- or para- positions. Using the concept of resonance, this can be illustrated in an abstract sense by considering the resonance structures of the intermediates resulting from attack at the various positions; an additional resonance structure exists for ortho- and para- attack if the substituent is conjugated to the $\pi$ electron system of the ring.

With alkyl groups, the electrons are donated by hyperconjugation, which is a weaker effect (and not apparent from resonance structures), but nonetheless has a similar impact on regioselectivity and the $\pi$ molecular orbitals. Additionally, the coefficients on the atoms of the HOMO of the $\pi$ system are lowest at the meta positions when an electron donating group is present (including alkyl groups), indicating electron density is higher at the ortho- and para- positions.

(I should clarify, of course, that resonance is not an actual physical process/phenomenon, merely an abstraction which provides a qualitative picture of the extent of electron delocalization and which often correlates well with the actual distribution of electron density in the molecular orbitals.)

Ultimately, at low temperatures, when the reaction is kinetically controlled, the favored pathways are the ones in which the transition states and intermediates are stabilized, leading to ortho-/para- substitution.

At higher temperatures, the preferred product is the thermodynamic one, and the advantage of meta- substitution is that there is less crowding and reduced steric hindrance in, e.g., a 1,3,5-trisubstituted benzene by comparison to a 1,2,4-trisubstituted benzene. Additionally, Friedel-Crafts is actually reversible at high temperatures. Even if the kinetic product forms initially, then it can decompose by dealkylation in the presence of Lewis-Acid catalysts and acid at high temperatures, at which point equilibrium can be established with the thermodynamic product.

According to March, experimental evidence suggests that methyl groups will mainly migrate intramolecularly in the isomerization reaction to give the thermodynamically preferred meta-substituted benzenes, while both intramolecular rearrangements and complete dealkylation with subsequent intermolecular reactions are possible with other types of alkyl groups.