Color changes of a solution of acetone and sodium hydroxide

Under basic condtions, acetone does condense with itself, first to form mesityl oxide, which is colorless, and then to form isophorone, which has a yellow color.

Both mesityl oxide and isophorone are still reactive carbonyl compounds likely to be attacked by deprotonated acetone carbanions.  That leads to bigger and more conjugated condensation products, which leads to darker and darker colors.

I found a paper which looks in some detail at the base-catalyzed polymerization of acetone, albeit with a different base and at high-temperature conditions in the gas phase, but I suspect a similar pathway happens, more slowly, at ambient temperatures in the liquid.  In the liquid, even larger molecules can likely form because they will stay solvated and can keep reacting (as opposed to in the gas phase where big molecules "crash" out of the phase as soot/gunk/etc.)

The color is likely in part from isophorone but much moreso from large polyconjugated molecules derived from isophorone containing many consecutive sp2 hybridized carbons. Such compounds are very often colored due to intense absorption of visible light.


Boiling means dynamic evaporation of liquid.

Infinite boiling would mean to have infinite amount of that liquid and therefore the egg would have infinite mass.

Such an egg would collapse into a black hole and we would not be able to continue in our observation.

In reality, in inert atmosphere, with overlapping stages:

protein coagulation
water boiling
organic matter thermal dehydration
general organic matter decomposion.
carbon stage

If we kept temperature just above $\pu{100 ^\circ C}$ to keep boiling,  stage 3-4 would be very slow and we would not reach carbon.


How to breakdown wax, preferably beeswax

As far as I know beeswax does not decompose. If you just want to melt it, use heat. If you want to break it down using a solvent, try turpentine (paint thinner).


Can a molecule be neither gerade nor ungerade?

g/u is not a property of a molecule; I assume you meant molecular orbital.
All functions can be written as a linear combination of even + odd functions: there is a short explanation on Wikipedia. Basically, you have a function $f(x)$; now define

$$\begin{align}
f_\mathrm e(x) &= \frac{1}{2}[f(x) + f(-x)] \\
f_\mathrm o(x) &= \frac{1}{2}[f(x) - f(-x)] \\
\end{align}$$

It is clear from the definition of parity that $f_\mathrm e$ is even (since $f_\mathrm e(x) = f_\mathrm e(-x)$) and likewise that $f_\mathrm o$ is odd. Now since $f(x) = f_\mathrm e(x) + f_\mathrm o(x)$, we have shown that any function $f(x)$ can be expressed as a linear combination of even and odd functions.

In three dimensions where you have $\psi(x,y,z)$ you simply need to define

$$\begin{align}
\psi_\mathrm g(x) &= \frac{1}{2}[\psi(x,y,z) + \psi(-x,-y,-z)] \\
\psi_\mathrm u(x) &= \frac{1}{2}[\psi(x,y,z) - \psi(-x,-y,-z)] \\
\end{align}$$

and analogously to above one can see that every function $\psi(x,y,z)$ can be expressed as the sum of a gerade component $\psi_\mathrm g$ and an ungerade component $\psi_\mathrm u$. These components don't have any physical meaning (unless $\psi$ itself is either g or u, in which case one component is simply $\psi$ and the other is zero), but that's somewhat beside the point.

If a molecule does not possess a centre of inversion then its (canonical) molecular orbitals will not possess g/u symmetry.

So the answer is yes, and yes. There is no "or".


Why is the product of the reaction "Bi + O2" Bi2O3 instead of Bi2O5?

Many elements have multiple valencies. The most common oxidation state of Bi is +3. Bismuth (V) is also possible. Read more here on Bismuth oxides.


Skipping the defatting stage in extraction of alkaloids

One big risk with the method you propose is foaming and gellation turning your extractions into a huge mess.  Under basic conditions, the fats (e.g. triacylglycerols or TAGs) are likely to hydrolyze, forming carboxylate salts of fatty acids.  "Carboxylate salts of fatty acids" is a long way of saying soap.  These soap-like compounds could lead to excessive foaming or gellation of the sample, which will hinder phase separation and gunk up your apparatus.


XSteam IF97 tables (MS Excel version) lack of inputs

If you have an initial estimate for the pressure, $p_0$, then you can quickly solve iteritively for the pressure that matches the desired entropy $s=s^*$ using modified Newton's method:  $$p_{n+1}=p_n-\frac{s(p_n,T)-s*}{\left(\frac{\partial s}{\partial p}\right)_T}$$The partial derivative of s with respect to p can be obtained by finite difference at the initial guess for p.


Ions leave the salt bridge and move further, without accumulating at the bridge end. For all salt bridges where there is no other reactions involved, so all simple salts in solution or molten salts.
Ions move all the way through. There is some concentration difference in salt bridge and main half cell volume. It is true for simple salt solutions and molten salts. Same ions move through half cell and salt bridge.
The speed of a wave is on the order of 100 000 km/s. The speed of ions moving is on the order of 0.001 mm/s. You can estimate the ion speed from calculating amount of ions participating in a reaction per second and dimensions of the bridge and ion density in the solution or molten salt.
Yes. But it is important to remember that there are two cancelling each other currents in the salt bridge. Thus magnetic field is close to zero.
Most important is a number of ion types. If there are just two types, such as in solutions and molten salts, then this intuition applies. If more ion types are present, and especially if ions can convert between types, things will get more complex. Another option to consider is a fuel cells where only one type of ions passes through, and electrons use outside route to complete the loop. So, cases with one negative and one positive ion types are considered here. All other collections of ions will behave very differently.
No. Ions are constantly being replenished by the bigger volume of the both half cells.

Electroneutrality strictly speaking is not conserved when the bigger volume than an atom is considered and precise measurments are available. There is a gradient of potential due to different concentration of ions. Concentration of ions is the reason for the effect you are interested in. Important parts to consider are electrodes, where ions react, thus dissapearing from what we calculate, and narrow areas, where ions' speed increase, such as in salt bridge.


Why does ozone have an electrostatic potential map similar to that of SO2?

Resonance structures of $\ce{SO2}$ and $\ce{O3}$.

As you can see from the diagram $\ce{O3}$ and $\ce{SO2}$ are isoelectric and have the same shape and this accounts for their very similar electrostatic potential maps. Both molecules are polar although $\ce{SO2}$, with a dipole moment of 1.62D, is more polar than $\ce{O3}$ (0.53D). As the diagram shows, the polarity of both molecules is due to the their trigonal planar shape with a formal positive charge on the central atom.


How Does Benzyl Alcohol Produce a Peak at m/z = 79?

Interesting question.

The cycloheptatriene - norcaradiene tautomerization is a common equilibium.  Usually the norcaradiene is present in small amounts, but as long as some is present the reaction can rapidly proceed through this intermediate.

In the diagram below, the norcaradiene can undergo hydrogen shift with concomitant decarbonylation to produce the m/z = 79 peak in the mass spectrum.


How can we rationalise the amount of crystal water in coordination complexes?

I have given two answers: one regarding the hydration of metal cations in solution and one regarding salts of metal cations and their hydration extents.

When in solution, the answer lies with size. Think of a BB gun pellet and a basketball. How many basketballs can you stick to the sides of a BB gun pellet? 2, maximum. That's why some metal cations are hydrated with fewer waters than others; these metal cations tend to be the smaller ones.

Also, it is not $\ce{FeSO4}$ that is hydrated but rather just $\ce{Fe^{2+}}$. Iron sulfate is soluble in water so there shouldn't be unionized iron sulfate in solution; in addition, if the iron sulfate were unionized, it would be hard to coordinate water molecules to the metal cation - it's charge density is diluted, so to speak, by the presence of the doubly negative sulfate anion. So the correct formula should be $\ce{[Fe(OH_{2})_6]^{2+}}$.

When not in solution - i.e about the salt examples you posted: outside of water these salts can coordinate that many water molecules likely because there is no water to hydrolyze the metal-sulfate bond. By having the sulfate part of the molecule stick around, this likely facilitates more hydrogen bonding with water molecules - note that the sulfate anion is polarized by the water molecules; the negative charge density resides overwhelmingly on the oxygen molecules.

Consider the above picture of ferric sulfate. Note that there are four oxygens with partial negative charges on the sulfate ion. At least some of these partial negative charges facilitate hydrogen bonding in the solid-state. Also note the charge on the iron atom; this positive charge density also facilitates hydrogen bonding to water molecules.

In addition, we can consider calcium sulfate dihydrate - gyspum, drywall, sheet rock - the stuff your walls are likely made of. From here we clearly see that two of the oxygens on the sulfate anion are available for association with water molecules. Hence the existence of $\ce{CaSO_4*2H_2O}$


Why are solids and liquids not included in the equilibrium constant? What about in a reaction rate calculation?

It very much depends on what definition of the equilibrium constant you are looking. The most common usage of the same has quite a variety of possible setups, see goldbook:

  Equilibrium Constant
  Quantity characterizing the equilibrium of a chemical reaction and defined by an expression of the type $$K_x = \Pi_B x_B^{\nu_B},$$ where $\nu_B$ is the stoichiometric number of a reactant (negative) or product (positive) for the reaction and $x$ stands for a quantity which can be the equilibrium value either of pressure, fugacity, amount concentration, amount fraction, molality, relative activity or reciprocal absolute activity defining the pressure based, fugacity based, concentration based, amount fraction based, molality based, relative activity based or standard equilibrium constant (then denoted $K^\circ$ ), respectively.

The standard equilibrium constant is always unitless, as it is defined differently (goldbook)

  Standard Equilibrium Constant $K$, $K^\circ$ (Synonym: thermodynamic equilibrium constant)
  Quantity defined by $$K^\circ = \exp\left\{-\frac{\Delta_rG^\circ}{RT}\right\}$$  where $\Delta_rG^\circ$  is the standard reaction Gibbs energy, $R$  the gas constant and $T$  the thermodynamic temperature. Some chemists prefer the name thermodynamic equilibrium constant and the symbol $K$.

It is worthwhile to note, that the first definition is always an approximation to the standard definition and in the standard definition all compounds regardless of their state are included in the equilibrium.

If you are looking at a reaction that takes place in gas phase, solid materials will play a constant role, since their partial pressure will solely depend on their vapour pressure and can therefore be regarded as constant. Therefore it can be seen as a part of the equilibrium constant. (The same applies to liquids while deriving for gas phase.) If you would increase the amount of solid in the system you would still not change it's concentration in gas phase.

For the reaction $$\ce{H2 (g) + I2 (s) <=> 2HI (g)}$$ you can form the standard equilibrium constant with activities/ fugacities
$$K^\circ = \frac{a^2(\ce{HI})}{a(\ce{H2})\cdot{}a(\ce{I2})},$$
with $$a=\frac{f}{p^\circ}.$$
The activity for a pure solid is normally defined as one ($a(\ce{I2})=1$) and therefore
$$K^\circ\approx K = \frac{a^2(\ce{HI})}{a(\ce{H2})}$$

For concentration dependent equilibrium constants, the following assumption is to be considered: $c(\ce{H2O})\approx55.6~\mathrm{mol/L}$ and is usually always much larger than that of any other component of the system in the range where the equilibrium approximation can be used. Also in most solution based reactions, the solute itself does not directly influence the reaction. Its concentration will therefore not change (significantly) and can be included in the equilibrium constant.

You are correct in assuming that the kinetics of a surface reaction will only depend of the actual area of the surface (and of course the reactants forming the products). But here again this area is most likely to be considered as constant and will result in a scalar for the reaction rate (in the area where an accurate description of the reaction is possible).

As of Greg's comment, He is of course right. In an equilibrium process the surface area will affect both directions of the reaction, hence it will not show up in the reaction rate at all. (The only effect it might have is, that the equilibrium state will be reached faster.)


Not all bonds are ionic bonds. While saying that "Carbon is 4+ valency and is a $\ce{C^{4+}}$ ion" works for carbides like $\ce{Be2C}$, it will not work for a covalent compound like $\ce{HCOOH}$.

Instead of ionic bonds, here we have covalent bonds. Whereas in an ionic bond one atom donates an electron to the other, in a covalent bond, two atoms each contribute some electrons and share them.

For example, with $\ce{O2}$, each atom shares two electrons with the other. Neither "give up" any electrons, they are just shared and contribute to the octets of both atoms:

The formate ($\ce{HCOO-}$) ion is like this:

The central carbon shares one electron with a hydrogen, and two electrons with an oxygen. It also shares one electron with a negatively charged oxygen. This completes the octets (or doublets in the case of hydrogen) of all four atoms.

With electrons drawn on:

Note that while the charges are not of the $1\times(\ce{C^{4+}}) + 2\times(\ce{O^{2-}}) + 1\times(\ce{H-})$ form, the oxidation numbers are.


Crystallization in a solution of 2 or more solutes

When the solution contains several substances, it is very often possible to achieve their separate sequential crystallization due to the different nature of the solubility of the substances as a function of temperature. In a certain temperature range, the solution will always be saturated with respect to one substance and unsaturated with respect to the other. This leads to the isolation of only the first substance from the solution, while the other remains in the mother liquor.

The difference in the compositions of the liquid (vapor) and solid phases formed during the partial crystallization of the solution, melt, and gas phase always leads to the fractions enriched with one component, and sometimes even to practically pure components. Possible degree of separation during crystallization is determined by the type of the phase diagram (composition vs temperature) in the solid-liquid(gas) system.

Obviously, salt and sugar do not form solid solutions, therefore depending on the concentration and temperature one component will crystallize first, and, when water is fully evaporated, a mechanical mixture of crystals is left.

This is a basis of various crystallization methods of mixtures (binary or multicomponent) separation, such as:

Fractional crystallization;
Crystallization in the presence of auxiliary substances;
Fractional melting and sublimation;
Multiple recrystallization.


The general formula is $\Delta H=\Delta U + \Delta (pV)$.

Since $\Delta U=q_v$ (heat at constant volume), then $\Delta H=q_V+V\Delta p$, so no, heat at constant volume does not equal enthalpy, unless p is constant, which just brings you back to the original relation $\Delta H = q_p$.


How to determine which p-type GTO to choose in HF method?

Have a look in the Gaussian manual if something is unclear.

If you want to use an extern basis set in Gaussian you have to respect the gaussian input conventions for the basis set, which you can find here (gaussian.com/gen/). Since you are using the EMSL Basis Set Exchange the server gives you a file that satisfies these conventions. Like:

Type  NGauss  Sc
$α_1\quad d_{1μ}$
$α_2\quad d_{2μ}$
…
$α_N\quad d_{Nμ}$

Type stands for the shell, then you have the number of primitives with NGauss and scaling factor Sc. Afterwards you put your exponents and coeffients of the primitive gaussian functions which then will be contracted to the respective contracted gaussian functions. You might notice that the scaling factor is set to unity for almost all basis sets—I checked some Dunning, Pople and Ahlrichs basis sets and always found it set to unity, which seems resonable. In the TURBOMOLE format the scaling factor is omitted totally, so it seems to be a Gaussian specific hack.

Now to your second question. You are specifing the shell, not a single function, a shell contains always all spherical harmonics belonging to the given quantum number or all cartesian gaussian functions. The spherical harmonics are the solution of the Schrödinger equation for an electron on a surface. You have generally $2l+1$ spherical harmonics where $l$ is the azimudal quantum number specified as 0/s, 1/p, 2/d, 3/f, 4/g, 5/h and so on. There exists also the possibilty to use cartesian gaussians with $(l+1)(l+2)/2$ degenerated functions.

Why am I telling you that? You always choose a whole shell with a azimudal quantum number, how many function belong to this shell depends on the quantum chemistry code you are using. ASAIK, Gaussian supports both.

In case of a p-shell it does not matter, both expressions evaluate to three p-functions (as expected). So you get three p-functions with one p-shell.

The basis set you have chosen would be evaluted as:

(9s4p1d) → [3s2p1d]

Having nine primitive s type gaussians (note that Dunning basis sets use a general contraction scheme), four primitive p type gaussians and one primitive d type gaussian which form a basis set out of three s shells, two p shells and one d shell with 14 contracted spherical harmonic gaussian functions or 15 contracted cartesian gaussian functions for carbon.


What makes some metals melt at higher temperature?

Some factors were hinted, but let me put them in an order of importance and mention some more:

metals generally have a high melting point, because metallic interatomic bonding by delocalized electrons ($\ce{Li}$ having only a few electrons for this "electron sea") between core atoms is pretty effective in those pure element solids compared to alternative bonding types (ionic $\pu{6-20 eV/atom}$ bond energy, covalent 1-7, metallic 1-5, van-der-Waals much lower). Also, ionic lattices like $\ce{NaCl}$ have a higher lattice and bonding energy, they have weak interatomic long-range bonding, unlike most metals. They break apart or are easily solvable, metals are malleable but don't break, the electron sea is the reason for their welding ability.
the crystal structure and mass play an inferior role among your filtered elements (just look up the crystal structure of those elements), as metallic bonding is not directional unlike covalent bonding (orbital symmetry). Metals often have half filled $\mathrm{s}$ and $\mathrm{p}$ bands (stronger delocalized than $\mathrm{d}$ and $\mathrm{f}$) at the Fermi-edge (meaning high conductivity) and therefore many delocalised electrons which can move into unoccupied energy states yielding the biggest electron sea with half or less fill bands.
noble metals like $\ce{Au,Ag}$ have a full $\mathrm{d}$ orbital, therefore low reactivity/electronegativity and are often used as contact materials (high conductivity because of "very fluid" electron sea consisting only of $\mathrm{s}$-orbital electrons. Unlike tungsten with half or less occupied $\mathrm{d}$-orbitals they show no interatomic $\mathrm{d-d}$ bonding by delocalized $\mathrm{d}$-electrons, and more importantly, a half filled $\mathrm{d}$-orbital contributes 5 electrons to the energy band, while a $\mathrm{s}$ only 1, $\mathrm{p}$ only 3, the electron sea is bigger among the $\mathrm{d}$-group.
The "packaging" of core atoms in the lattice (interatomic distance) among the high $Z$ atoms (compared to e.g. $\ce{Li}$) is denser (more protons, stronger attraction of shell electrons, smaller interatomic radius), means stronger interatomic bonding transmitted by the electron sea:

You can see here that in each series ($\ce{Li,\ Na,\ K}$) the melting points rise to a maximum and then decrease with increasing atomic number (lacking unoccupied energy states for delocalized $\mathrm{d}$-electrons), bigger electron sea being here a stronger factor than a bit more dense packaging.

Boron as a semi-metal shows metallic and covalent bonding, Carbon strong directional covalent bonding and is able to build a network of bonds unlike other non-metal elements showing covalent intramolecular bonding, e.g., in diatomic molecules but not strong intermolecular bonding in macromolecules because of lacking unpaired electrons.

So there are some bigger trends for melting points explaining the high melting points of $\mathrm{d}$-metals, but also some minor exceptions to the rule like $\ce{Mn}$.


Why does pOH increase when pH decreases?

At a particular temperature, the $K_{\text{eq}}$ for the following reaction (yes, it's the auto-protolysis of water):

$$\ce{H2O(l) <=> H+(aq) + OH-(aq)}$$

will be constant. Note that $K_\mathrm{w}=K_{\text{eq}}\times[\ce{H2O}]=\ce{[H+(aq)][OH-(aq)]}$ is called the auto-protolysis constant of water. It is considered a constant because concentration of water $([\ce{H2O}])$ is assumed to be constant.

So, you can directly observe from here, that since $\ce{[H+(aq)][OH-(aq)]}=K_\mathrm{w}=\text{constant}$ at a particular temperature, if $\ce{[H+(aq)]}$ increases (i.e. $\mathrm{pH}$ decreases), $[\ce{OH-(aq)}]$ must decrease (i.e. $\mathrm{pOH}$ must increase).

Your interpretation of the Le Chatelier's principle is also correct. In fact, it is actually what is happening behind the scenes. When you add more acid (assuming a 100% ionized acid), the $Q$ value of the auto-protolysis of water increases, causing the reaction to shift backward. However, the backward shift is unable to decrease the concentration of $\ce{H+}$ ions as much as it decreases the concentration of $\ce{OH-}$ ions. See explanation below:

Consider adding $\pu{0.01M}$ $\ce{HCl}$ to pure water (at $\pu{25^\circ C}$) at $t=0$. This is what happens till achieving equilibrium at $t=t_0$:

$$
\begin{array}{ccc}
&\ce{H2O(l) &<=>& H+(aq)&+&OH-(aq)}\\\hline
-&-&&\pu{10^-7M}&&\pu{10^-7M}\\
t=0&-&&\pu{(10^-7+10^-2)M}&&\pu{10^-7M}\\
t=t_0&-&&(10^{-7}+10^{-2}-x)~\pu{M}&&(10^{-7}-x)~\pu{M}\\
\end{array}
$$

Now, if you actually solve for $x$ at $t=t_0$, you'll find $x$ to be of the order $10^{-8}$ or below. Notice that, for the $\ce{H+}$ ions, this means a decreases in concentration from $\approx\pu{0.01M}$ at $t=0$ to $\pu{0.0099999M}$ at $t=t_0$. This a hardly noticeable change. However, it is a catastrophe for the $\ce{OH-}$ ions, whose concentration would probably be reduced, by a factor of more than ten thousand, from the initial $\pu{10^-7 M}$! The change in $\mathrm{pOH}$ is, hence, much more pronounced.


Supposed low temperature of calcium carbide synthesis

Liu et al. in 2012 [1] showed experimental data that lead me to believe this to be inaccurate. Two relevant informations from their findings:

$\ce{SiO2}$ and $\ce{Al2O3}$ do not react with coke at temperatures lower than $\pu{1460 °C}$, but they do react with $\ce{CaO}$ to form $\ce{Ca3SiO5}$ and $\ce{Ca2Al2O5}$, respectively […]
Note that the behaviors of $\ce{SiO2}$ and $\ce{Al2O3}$ observed in this work do not agree with some literature, which reported that both $\ce{SiO2}$ and  $\ce{Al2O3}$ reacted with coke at temperatures of $\pu{∼ 1200 °C}$.

Also, in Fundamentals of Metallurgical Processes: International Series on Materials Science and Technology [2, p. 174] it is written that calcium carbide is not stable in the presence of silicon and undergoes the reaction:

$$\ce{2 SiO2 + 2 C + CaC2 <=> Ca + 2 Si + 4 CO}$$

References

Li, G.; Liu, Q.; Liu, Z. $\ce{CaC2}$ Production from Pulverized Coke and $\ce{CaO}$ at Low Temperatures—Reaction Mechanisms. Ind. Eng. Chem. Res. 2012, 51 (33), 10742–10747. DOI: 10.1021/ie300671w.
Coudurier, L.; Hopkins, D. W.; Wilkomirsky, I. Fundamentals of Metallurgical Processes, 2nd ed.; Hopkins, D. W., Ed.; International series on materials science and technology; Pergamon Press: Oxford [Oxfordshire]; New York, 1985; Vol. 27. ISBN: 978-0-08-032536-1, eBook ISBN: 9781483162409, DOI: 10.1016/C2013-0-03773-5.


Arrange the following compounds in the increasing order of equilibrium constant for hydration

To my knowledge, comparing the steric hindrance of the carbonyl group is not the path you want to follow to answering this question, because the trigonal planar configuration of the carbonyl carbon provides plenty of room in order for the inductive effect to outweigh the steric hindrance. We have to examine the partial positive charge of the carbonyl carbon resulted from the C=O and the other substituents. The C=O bond has a great dipole moment resulting in a partially positive carbonyl carbon. This charge can be stabilized or increased by alkyl substituents (hyperconjugation) or electronegative atoms (inductive effect) respectively. The greater the positive charge on the carbonyl carbon the higher the equilibrium constant for hydration (due to the mechanism). Therefore, we can easily say that (A) < all, because of the two alkyl substituents. C > A, because the chlorine reduces the hyperconjugation effect. B > C (only 1 alkyl substituent). E > B (because of the same reason as the C > A scenario. And lastly D > E, because of the abscense of any alkyl groups (same reason as the B > C scenario). Now, why is B > C and D > E? Why is less alkyl substitution increasing the charge to a greater extent than 1 more alkyl group with an electronegative atom? If the chlorine was attached to the carbonyl carbon itself (attach the chlorine in E to the C=O), it would greatly icrease the charge and in fact this molecule would have the highest Constant of all. However, we have to keep in mind that the inductive effect greatly deminshes the further away you go from the chlorine atom. The carbonyl carbon is 2 σ bonds away from the chlorine (as in C and E), so yes it will feel the induction but the alkyl group will still donate elenctron density via hyperconjugation to a lower extent. In conclusion, the chlorine in C and E is reducing the hyperconjugation effect but it doesn't neutralize it completely. Therefore, B > C and D > E


Radial pair distribution function (VMD). How to define it for a water NaCl system with multiple Na and Cl?

I haven't done it in vmd, but a rdf is really kind of like a histogram. For a particle of interest, you count how many particles are between r and $\Delta r$ from it and record that number in a bin, then move outwards, recording numbers in each shell into a bin for that shell. That tells you the rdf of that particle interacting with everything else. If there are three other types of particles in the simulation, you will get 3 rdfs from that one rdf calculation centered at the particle of interest.Repeat for other particles once you are done so you can get a nice average.

So all you need to do is count particles in $\Delta r$'s (and from this you get the volume too!, that is trivial though. I would do this for all Na's and Cl's and then average each of their bins to get one rdf for Cl-Cl, one for Cl-Na, one for Cl-O, one for Cl-H.

So that is how it should work, but I haven't used vmd to do it myself :S

RDF's don't have to be atom-atom, they could be molecule-molecule, or molecule-atom... anything really.


A Question on the Role of Triplet Oxygen in the Formation of Artemisinin

I offer the following thoughts on the paper you cite regarding the formation of artemisinin 9 from carboxylic acid 1. You are correct that the allylic hydroperoxide 2 is formed by an ene reaction with singlet oxygen (1O2). The stereochemistry of the hydroperoxide group is the result of the ene reaction occurring on the convex face of the unsaturated cis-decalin 1. The crude hydroperoxide 2 was exposed to an atmosphere of triplet oxygen (3O2) in a vigorously stirred mixture of petroleum ether and catalytic trifluoroacetic acid. The initial steps are 2 --> 6 via radicals 3, 4 and 5. Triplet oxygen forms hydroperoxide radical 3 which undergoes cyclization to 1,2-dioxetanyl radical 4. It is likely that radical 4 reacts with triplet oxygen prior to dioxetane fragmentation owing to the stereochemistry of the hydroperoxy radical in 5. The chain reaction is completed by the reaction of radical 5 with hydroperoxide 2. In the transformation 2 --> 3 the radical HOO was formed and can also intercede as a source of a hydrogen atom regenerating oxygen. Finally, acid-catalyzed cyclization of 6 leads to artemisinin (qinghaosu) 9. There is no issue of "forbidden" reactions.


If all intermolecular forces are electrostatic in nature, why don’t large non polar molecule dissolve in water?

Dipoles can be induced, but they are not permanent. You’re basically asking why a spring doesn’t like to stay compressed when you push on it—it's because that’s not its most stable position. The electron density around  hydrogen in a $\ce{C-H}$ bond is much higher than the hydrogen in a polar $\ce{O-H}$ bond, so electrostatic repulsions overwhelm any attractions that may be induced. See the diagram below$[1]$:

$[1]$ Bond polarity diagram - Iverson Group UT Austin


Vibrational Spectrums of HI, HBr, HF, and HCl

The fine structure in the diatomic vibrational spectra is based on various population of the rotational levels.

The best demonstration of isotopic effects in vibrational spectra is to switch hydrogen for deuterium.

The frequency of vibration (from Hooke's Law adapted to a harmonic oscillator) is given by:

$$\nu = \frac{1}{2\pi}\sqrt{\dfrac{k}{\mu}}$$

where $k$ is the force constant of the oscillator and $\mu$ is the reduced mass:

$$\mu=\dfrac{M_1M_2}{M_1+ M_2}$$

Switching $\ce{HCl}$ to $\ce{DCl}$ has a much larger effect than the two chlorine isotopes (35 and 37):

$\ce{H^{35}Cl}, \ \mu\approx\frac{35}{36}=0.972$
$\ce{H^{37}Cl}, \ \mu\approx\frac{37}{38}=0.974$
$\ce{D^{35}Cl}, \ \mu\approx\frac{70}{37}=1.892$
$\ce{D^{37}Cl}, \ \mu\approx\frac{74}{39}=1.898$

The difference between $\ce{^{35}Cl}$ and $\ce{^{37}Cl}$ is in the third decimal place.

Here is a nice Google Image search for the vibrational spectra of HCl and DCl


UV radiation on IBr (Radical Substitution)

The presumed stability of $\ce{Br-}$ is overcompensated by the instability of $\ce{I+}$. While some polyatomic iodine cations like $\ce{I2+}$ exist, $\ce{I+}$ is found only to be stable in the gas phase and in the absence of a counterion. Due to the low difference of electronegativities (0,5) of Br and I, the I-Br bond is predominantly covalent, which also makes heterolysis unlikely. Homolytic cleavage of the I-Br bond is therefore energetically more favorable.


First thing to say, I don't know much about electrochemistry. But concluding from thermodynamics, rewriting your equation a bit and making dimensional analysis there are some things to say:

It seems that you interchange the words dimension and unit. So an example: Length is a dimension and meter, inches... are units that measure length. You are probably asking for the physical interpretation of the electrolyte diffusional conductivity and not for the specific unit it is measured in.
The logarithm can be applied only to dimensionless numbers. In order to calculate it, you have to divide by some $c_0$ usually this is given in mol/l. In addition both terms in brackets have to be dimensionless, because there is a sum with a dimensionless number and in sums the dimensions have to match. So the only term relevant for the dimension, is the fraction.
The resulting dimension of the fraction is in SI units [A / m] and not [A m$^2$ / mol]


Why was anhydrous magnesium carbonate “upsalite” discovered so late?

The paper describing the synthesis of Upsalite has a clear introduction explaining the problem:

  Magnesium is the eighth most abundant element in the earth’s crust and essential to most living species. It can form several structures of hydrated carbonates such as nesquehonite ($\ce{MgCO3·3H2O}$), and lansfordite ($\ce{MgCO3·5H2O}$), a number of basic carbonates such as hydromagnesite ($\ce{4 MgCO3·Mg(OH)2·4 H2O}$), and dypingite ($\ce{4 MgCO3·Mg(OH)2·5 H2O}$), as well as the anhydrous and rarely encountered magnesite (MgCO3) [5]. In contrast to other alkali earth metal carbonates, chemists have found anhydrous magnesium carbonate difficult to produce, particularly at low temperatures. Above 100°C, magnesite (crystalline $\ce{MgCO3}$) can be obtained from $\ce{Mg(HCO3)2}$ solutions by precipitation. However, at lower temperatures, hydrated magnesium carbonates tend to form, giving rise to what has been referred to as “the magnesite problem” [6].Yet, not only chemists have been intrigued by magnesium carbonates. Although abundant in nature, where crystalline forms exist as traces in most geological structures, pure magnesium carbonate is seldom found on its own in larger deposits, a fact that has puzzled geologist for more than a century [7].

As you can see, anhydrous crystalline magnesium carbonate has been synthesized for a long time (and occurs in nature), in the form of a dense phase: magnesite. The new material reported under the name of upsalite is simply an porous amorphous polymorph of magnesite. New polymorphs of inorganic structures are regularly discovered/synthesized by researchers, and reported in the literature. The most striking example may be the family of zeolites, with more than 200 polymorphs of aluminosilicate.


How can I choose the correct mechanism for nucleophilic substitution of aryl halides?

Before comparing the three main mechanisms, I will first explain why aryl halides typically do not undergo SN2-type reactions. Take bromobenzene as a typical example of an aryl halide. A nucleophile, like hydroxide ion, cannot simply displace the bromide from bromobenzene. Quoting from Organic Chemistry (1st ed., p 589) by Clayden et al.,

  In the aromatic compound, the C–Br bond is in the plane of the ring as the carbon atom is trigonal. To attack from the back, the nucleophile would have to appear inside the benzene ring and invert the carbon atom in an absurd way. This reaction is not possible!

In rare cases, SN1 reactions can take place; these characteristically occur when there is an excellent leaving group on the arene. One of the few examples is in the case of an aryldiazonium ion, where the corresponding leaving group is nitrogen gas.

Note that the presence of a diazonium ion does not necessarily imply a SN1 mechanism; there are also other possible mechanistic pathways, mainly involving radical chemistry.

The next possibility is that of an addition-elimination reaction, commonly termed SNAr. This is made possible by electron-withdrawing groups (e.g. nitro or carbonyl groups) when they are ortho and/or para to the halogen, as the negative charge in the intermediate can be delocalised into these electron-withdrawing groups:

Finally, the benzyne mechanism. Consider bromobenzene: it cannot undergo SN2 reactions (as explained above), SN1 reactions (bromide is not a sufficiently good leaving group), or SNAr reactions (no electron-withdrawing group ortho or para to bromine). However, under strongly basic conditions, it is possible to eliminate HBr by abstracting a proton ortho to the halide. Examples of such strong bases include sodium amide (NaNH2), LDA/lithium diisopropylamide (LiNi-Pr2), or potassium tert-butoxide (KOt-Bu).

To summarise the three categories:

If you have a very good leaving group - primarily nitrogen gas - on the aromatic compound, then a SN1 mechanism is possible.
If you have good electron withdrawing groups ortho or para to the halide, then it is likely an addition-elimination or SNAr reaction.
If the above two are not possible, and if you have an extremely strong base such as NaNH2 or KOt-Bu, then the benzyne mechanism may be applicable.


A stereospecific reaction is one in which there is only one possible stereochemical outcome for the product and that outcome is controlled by the stereochemistry of the reactant.

One of the earliest examples studied by most students of organic chemistry is nucleophilic substitution by the $\mathrm{S_N 2}$ mechanism. This reaction in its simplest form has a concerted mechanism. Reactions with concerted mechanisms often have stereospecific outcomes because the concerted mechanism demands a certain geometric arrangement of the reactants. In the case of the $\mathrm{S_N 2}$ mechanism, the nucleophile must approach from the opposite side of the molecule from the leaving group for steric and electronic reasons.

Likewise, the $\mathrm{E2}$ elimination mechanism leads to a stereospecific outcome. Stereospecificity does not require a one-to-one mapping of stereochemical inputs to outcomes. Rather, it requires a predictable mapping of inputs to outcomes.

For example, consider the $\mathrm{E2}$ elimination of 2-chloro-3-methylpentane with hydroxide. The reactant has four possible stereoisomers and the major Zaitsev product has two possible stereoisomers. Each stereoisomer leads to only one possible stereoisomeric product even though each stereoisomer of the product can be prepared from two stereoisomers of the reactant.


For atoms in the first 2 rows of the periodic table the octet rule arises as a consequence of filling orbitals four orbitals which each hold 2 electrons - each of which appears as a pair of dots in your Lewis structure.

Were you to construct the Lewis structure for neutral $\ce{NH2}$ you would indeed have 5 electrons from the nitrogen, and two from hydrogen, 7 i.e. 3 pairs and 1 extra electron on its own in a orbital.  Electrons in unpaired orbitals are however very reactive, the pairing of electrons lowers the total energy of the system.  As such, it is more favourable for the $\ce{NH2}$ molecule to remove an electron from some other molecule or atom near by and form ions e.g. $\ce{Na}$ to create $\ce{Na+NH2-}$.

Also by creating the charges, you create an electrostatic attraction between the ions which also lowers the energy and makes the ion more stable than the neutral molecule.


Fractional index in a chemical formula of metal oxide

There is nothing special about fractional indices in chemistry, it is just a pointer signifying we are dealing with a non-stoichiometric compound.
Group 6 metal are known not only for their regular oxides existing in numerous polymorphic modifications, but also for forming homologous series of nonstoichiometric oxides $\ce{\ce{M^{VI}_nO_{3n-1}}}$ and $\ce{\ce{M^{VI}_nO_{3n-2}}}$ with the general formula $\ce{M^{VI}O_{3-x}}$ $(\ce{M} = \ce{Mo}, \ce{W};$ $x \in (0; 1)).$

Now, to make things clear, notation "$\ce{WO_{2.9}}$" does not refer to a molecule; $\ce{WO_{2.9}}$ is a formula unit showing composition of a network solid.
To further illustrate the origin of the fractional coefficients here, let's refer to the crystal structures of the stoichiometric and non-stoichiometric tungsten(VI) oxides $(\ce{WO3}$ and $\ce{WO_{2.9}},$ respectively).

Depending on the temperature, $\ce{WO3}$ adopts several polymorphs with various crystal structures; at the room temperature the most common structure it adopts is the one of distorted $\ce{ReO3}$ type, being made of tilted $\ce{[WO6]}$ octahedra cross-linked via the corners (for the illustration, see e.g. Wikipedia).

On the other hand, $\ce{WO_{2.9}}$ is an oxygen-deficient compound in comparison.
To account for that and eliminate the presence of point defects, crystal structure undergoes a transformation resulting in increased number of edge-sharing groups of $[\ce{WO6}]$ polyhedra and reduced number of the corner-sharing ones, e.g. the structure gets more "packed", and, more importantly, a crystallographic shear is introduced (which is not an impurity).
Let's have a look at the asymmetric unit of $\ce{WO_{2.9}}$ of the crystal structure determined by Magnéli [1]:

Figure 1. Crystal structure of $\ce{WO_{2.9}}$ [1, ICSD-24736] showing an asymmetric unit $\ce{W10O29}.$
Color code: $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#2194D6}{\Large\bullet}~\ce{W}$.

Due to the asymmetric unit $\ce{W10O29},$ the chemical formula (formula unit) is reduced to $\ce{WO_{2.9}}.$
One can also fill the entire unit cell knowing the number of formula units $(Z)$ per unit cell; here, $Z = 2$ and this results in equal formula $\ce{W20O58},$ often seen in the literature:

Figure 2. Crystal structure of $\ce{WO_{2.9}}$ [1, ICSD-24736] showing packed unit cell.
Color code: $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#2194D6}{\Large\bullet}~\ce{W}$.

Finally, to illustrate the location of the aforementioned shear plane, let's complete coordination shells on all tungsten atoms and view the structure along the $b$ axis:

Figure 3. Crystal structure of $\ce{WO_{2.9}}$ [1, ICSD-24736] showing polyhedral representation ($[\ce{WO6}]$ octahedra) with the shear plane along $a$ axis denoted with red arrow.
Color code: $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#2194D6}{\Large\bullet}~\ce{W}$.

As for the assignment of the formal oxidation numbers for the non-stoichiometric compounds like this, this is typically considered a non-trivial task of a questionable value.
Also, see the answer of mine to the related question Oxidation state of tungsten in $\ce{W3O8}$.

References

Magnéli, A. Structure of β-Tungsten Oxide. Nature 1950, 165 (4192), 356–357. DOI: 10.1038/165356b0.


Yes, It will.
It's a Nucleophilic Substitution reaction,

Chlorobenzene + Sodium Methoxide --> (Under extreme conditions ) Anisole

Extreme conditions because Aryl halides don't go Nucleophilic Substitution easily.


What type of organic compounds are used in OLED technology?

The short answer is a variety of pi-conjugated materials.

A huge variety of compounds, both small molecules, oligomers, polymers, and organometallic species have been used. Key requirements for the emissive material is a high efficiency emission process, either fluorescence or phosphorescence.

Some of the first compounds were simply conjugated aromatic molecules and dyes. Martin Pope at NYU observed luminescence from anthracene (e.g. Wikipedia) in the 60s.

A wide variety of conjugated dyes have been tried with varying success.

In general, there are 2-3 different types of compounds used in an OLED:

Hole-transport materials: Easily oxidized and conduct positive charges (holes).
Electron-transport materials: Easily reduced and conduct negative charges (electrons).
Emissive materials: Holes and electrons recombine and emit light.

In many cases, these are made in a sandwich device: ITO anode with hole transport layer on top, emissive layer, then electron transport layer, then cathode. (I'm leaving out various interface agents, "blocking layers," adhesion layers, etc.)

You're asking about the emissive component. Basically, you want materials that will efficiently emit (fluoresce or phosphoresce) at different wavelengths / colors.

The biggest real breakthrough, IMHO was from $\ce{Alq3}$ which is an aluminum organometallic compound with remarkably bright emission. But emits green, so for displays, you need other things.

I've been out of the field for a while, but here are some examples. Chemical companies like lumtec sell various components.

Polyfluorenes: different copolymers have tailored emission colors. Usually the "bridge" carbon is substituted with two alkyl substituents for solubility and processability (and to avoid aggregation).

  In OLEDs, polyfluorenes are desirable because they are the only family of conjugated polymers that can emit colors spanning the entire visible range with high efficiency and low operating voltage. Furthermore, polyfluorenes are relatively soluble in most solvents, making them ideal for general applications[...]

Various vinyl carbazoles: Similar reasons to the polyfluorenes.
Phosphorescent metal complexes (e.g., iridium): If you work out statistics, you get a lot of electron-hole combinations that are triplets, so phosphorescent emission can be very efficient, particularly with transition metals with a high spin-orbit coupling.

I think most leading emissive materials still largely fall into those categories. Small molecule purely organic targets exist, but aren't as efficient.

A bigger question is usually how you intend to deposit the compounds. Vapor deposition (i.e., evaporation or sputtering) can be done with almost anything that sublimates, but if you want to do drop-casting, spin-coating, or inkjet printing, you'll want soluble compounds.


Difference between Lindlar and Rosenmund catalysts

This is a rather interesting question because these names actually refer to classes of reactions (specific to certain reagents and products), and aren't constrained by specific proportions of substances or even the identity of these substances.

$\hspace{4cm}$

A Rosenmund catalyst is used to reduce acyl chlorides to their corresponding aldehydes, and is typically composed of palladium supported on $\ce{BaSO4}$, and poisoned by sodium acetate, N,N-dimethylaniline, thiourea, thiophene, dibenzothiophene, ethyldiisopropyl amine, or, most commonly,
quinoline.

A Lindlar catalyst is used to reduce alkynes to cis alkenes, and is typically composed of palladium supported on $\ce{BaCO3}$, $\ce{CaCO3}$, or carbon, and poisoned by lead acetate and quinoline.

The broad list of poisons which cause nearly identical catalytic activity of the supported palladium in both Lindlar and Rosenmund catalysts tells that the poisons are not involved in the catalytic process itself, but rather are affecting the surface of the catalyst. Interestingly, Lindlar-active palladium shows no activity for the conversion of benzoyl chloride to benzaldehyde under various conditions. This demonstrates that despite similar catalyst preparation procedures, the two reactions are catalyzed by different active sites.$^{[1]}$

$[1]$ Mcewen, A. B.; Guttieri, M. J.; Maier, W. F.; Laine, R. M.; Shvo, Y. The Journal of Organic Chemistry 1983, 48 (23), 4436–4438.


Is there a (carbon) heterocyclic compounds named from number three to infinity?

To my knowledge, the Hantzsch-Widman nomenclature defines stems for ring sizes up to 10 (ecine for unsaturated, ecane for saturated heterocycles).

Oxepane is not the general stem for a seven-membered heterocyclic ring, but the suffix for one containing an oxygen atom.


How to merge PDB-files to build trimer from monomers

The PDB-file only contains the asymmetric unit, and no information on a potential biological relevant multimer. So you will need to get information about the state in solution experimentally.

That being said, one can often deduce from the size of the contact surface whether you have a potential quarternary structure. You can do this by hand, but the PISA-server is quite helpful there, and it will also output the multimer PDB-file:

http://www.ebi.ac.uk/pdbe/pisa/

If you just want to apply symmetry operators to your asymmetric unit in order to create symmetry mates, you can do so in

Pymol:
- click Action/generate/symmetry mates/within X Å
- from the command line:
        symexp prefix, selection, cutoff
        , e.g. symexp sym,1GVF,(1GVF),5
   see also http://pymolwiki.org/index.php/Symexp

Coot:
- if you just want to look at the symmetry mates, activate View/Cell&Symmetry with an appropriate radius and/or select "Symmetry by molecule"
- if you would like to actually create the symmetry mates, try Extensions/Modelling/New Molecule from Symmetry Op  - however you will have to specify the SymOp manually.

You will need to have the CRYST1-record in your PDB-file with the correct space group.


How is the association/ dissociation constant derived?

This looks a lot like a normal equilibrium constant. There is very little background information provided, but I assume that this formula applies to an experimental procedure in which the same total amount of the solute is dissolved into each solvent and the concentrations of free (dissociated) solute are measured. In this case that $C_1$ and $C_2$ should be the other way round. In the following, I assume that $C_1$ is the measured concentration in benzene and $C_2$ is the measured concentration in water.

In order for the formula to work we also need to assume that in benzene (or whatever the other solvent is), the solute associates almost completely.

With these assumptions the derivation of the formula is practically identical to the standard derivation of the formula for an equilibrium constant.

Let's call a single molecule of the solute A and a complex of $n$ associated molecules B.

The association reaction then is

$$n\ce{A <=>[k_1][k_2] B},$$

where $k_1$ and $k_2$ are the rates of association and dissociation reaction respectively.

The speed of the dissociation reaction is proportional to the concentration of B, so we can write $k_2=b\ce{[B]}$.

The association reaction requires $n$ units of A coming together and from probability theory we can tell that the probability of that happening is proportional to the concentration of A multiplied by itself $n$ times, so we can write $k_1=a\ce{[A]}^n$.

At equilibrium the rates of association and dissociation reactions are equal, so we can write:

$$k_1=k_2 \Rightarrow a\ce{[A]}^n=b\ce{[B]} \Rightarrow \frac{b}{a}=\frac{\ce{[A]}^n}{\ce{[B]}}$$

Now we raise both sides to the power of $\frac{1}{n}$ in order to get

$$\left(\frac{b}{a}\right)^{\frac1n}=\frac{\ce{[A]}}{\ce{[B]}^{\frac1n}}$$

Now, $\ce{[A]}$ is the concantration of dissociated solvent in benzene, so we can substitute $\ce{[A]}=C_1$. On the other hand $C_2$ is the concentration of the solute in water, which is the same as the concentration of total concentration of both associated and dissociated forms in benzene, so we can write $C_2=\ce{[A]}+n\ce{[B]}\approx n\ce{[B]}$ if $\ce{[A]}$ is much smaller than $\ce{[B]}$ as we assumed in the beginning.

By substituting we can write

$$\left(\frac{b}{a}\right)^{\frac1n}=\frac{C_1}{\left(\frac1n C_2\right)^{\frac1n}} \Leftrightarrow \left(\frac{b}{na}\right)^{\frac1n}=\frac{C_1}{C_2^{\frac1n}}$$

Because $n$ is independent of $C_1$ and $C_2$, we can substitute $K=\left(\frac{b}{na}\right)^{\frac1n}$ and we get

$$K=\frac{C_1}{C_2^{\frac1n}}.$$


In Lassaigne's test, the organic compound is first fused with Sodium Metal instead of any other Group 1 Metal, Why is this so?

You asked this historical question before (check Sci math history which is pretty good). For historical searches, Google scholar works the best. https://scholar.google.com/scholar?hl=en&as_sdt=0%2C44&q=lassaigne+test&btnG=&oq=lassaign

The very first article is very useful on the history of sodium fusion test.
"A lost centenary: Lassaigne's test for nitrogen. The identification of nitrogen, sulfur, and halogens in organic compounds" J. Chem. Educ., 1945, 22 (5), p 212

The first few lines of the article clearly mention that Lassaigne used potassium rather than sodium. He must have the same question his mind which you are asking now. Surprisingly, it was another person named Jacobsen who suggested using sodium. The original test indeed used potassium.

Why sodium is used today in some teaching labs? Lower price and less reactivity than potassium. Essentially nobody uses this historical test in analytical laboratories anymore. Today, an very small quantity of organic compound is burned and the gaseous products are analyzed by gas chromatography.


Carbonates are salts and esters of carbonic acid $\ce{H2CO3}$.

In generic nomenclature you may therefore consider a hydrogen carbonate also a subset of all carbonates. They are sometimes also referred to as primary carbonates.

Another considerable point is, that the structural subunit $\ce{O=C(-O-)2}$is still conserved.

In a stricter sense a hydrogencarbonate is only all forms involving $\ce{HCO3-}$, so it appears to be not a generic carbonate, but a class of its own. This refers to its one $\ce{O-H }$ bond and therefore it being a different structure.

However, another point of view should always consider aqueous solution equilibrium.
\begin{aligned}
\ce{HCO3- + H2O &<=> H3+O + \color{\red}{CO3^{2-}}}\\
\ce{HCO3- + H2O &<=> H2CO3 + {}^{-}OH}
\end{aligned}

You can see that carbonate ions will be present to a certain extend in solution. The chemistry will therefore also to some extend resemble the chemistry of the carbonate.

Chemical properties of bicarbonate compound might differ from the pure carbonates - incorporation of water, solubility, ... .

On another page, in organic chemistry the compound class carbonates refers to structures of the type $\ce{\color{\red}{R}-{CO3}-\color{\navy}{R'}}$, where $\ce{\color{\red}{R},\color{\navy}{R'}$\in$ Alkyl, Aryl}$ - hence esters.

To sum it all up, since a more stringent definition of carbonate is not present to me, I would always consider a bicarbonate a carbonate.


Normal Phase TLC using ethyl acetate as eluent

TLC is only about polarity.
Since all three compounds share the benzoic acid functionality, we only have to consider the rest.
Then we have three compounds: $\ce{R-OCOCH3}$, $\ce{R-OH}$ and $\ce{R-H}$.

In general the polarity is:

alkane < ester < alcohol

That is what we have here also: 3 < 1 < 2.

The one with the lowest polarity will have the highest R$_f$.
To show that it works, I have run them on a silica TLC plate with AcOEt:

AcSA: acetylsalicylic acid; SA: salicylic acid; BA: benzoic acid.


How to increase the concentration of HCl

Add Calcium Chloride to destroy the azeotrope.  Then distill and capture the vapor phase.

See the second method "Extractive Distillation" here:

http://www.jacobs.com/uploadedFiles/wwwjacobscom/20_Learn_About_Us/25_Products/252_Chemetics/Hydrochloric%20Acid%20Concentration.pdf

(Obviously, proper equipment and safety precautions are essential.)


Is it possible to efficiently convert hydrogen to methane?

No. You'll end up putting more energy into the process than you'll get out of burning the methane less the energy of burning the hydrogen that you started from. So although you can make methane from hydrogen, it isn't "efficient."


How exactly is energy created by oxidative phosphorylation?

Energy is "created" by the syntesis of ATP. The ATP molecule is considered an energy carrier, because it is used in other reactions as a source of energy. The released energy comes from the loss of one or two phosphate gropus to form ADP or AMP.

Remember that energy is not created, but simply transferred. The oxidative phosphorylation is a mechanism to recreate the ATP using the energy from other sources (O2, nutrients, etc).

Also why do they need an electron acceptor? What would happen if there was no O2 to accept the electrons?

I'm not sure I can give you a precise answer, but I'll try. In chemistry reactions, the system is always working towards equilibria. In general, if you take away one reagent of a reaction, there won't be a reaction at all. If there's no oxygen available, ATP won't be formed, and as a consequence the body resorts to other sources of energy, or cells will die.

Wouldn't they just bind to something else?

I don't think so. The electrons aren't free particles moving around the cell, the transfer only happens if all the reagents for the reaction are complete. Enzymes are, generally speaking, very specific, and what could happen is a "saturation" of NADH's which wouldn't be converted back to NAD and hence the cycle would stop.

But there's this talk of electrons being in "lower in energy" at the end. What does that mean? How is an electron lower in energy?

I don't remember hearing or reading that. Can you post some context or source?


Both the answers are correct but the most appropriate will be B as the chelation occurs because of hydrogen bonding which makes it more stable.
H-bonding is  possible in your structure but the ring is less stable
.


Beside the resources for learning chemistry Poutnik indicated, you may learn some of the chemical names and concepts from wikipedia, too. Frequently, articles are available in more than one language and sometimes a version in an other language offers additional details, too.  You find the switch at the left side of the pages:

Typically, you find some references at the bottom of these pages which may be accessible by you without going to a university library.

As side note: You mention that you did not yet start to study chemistry at a university at all.  Hence, it is too early for you to judge if a Slovak (or Czech, given the language) university is good or bad for your aspiration of a PhD in chemistry; because you likely do not know yet which specialty will attract your interest, nor potential co-workers and supervisors.  Remain active to obtain a solid chemical education first -- a chemistry department with classes about 30--60 students per year may be more helpful for laying such a foundation, than a larger institution where you become "just a number".  Do not exclude smaller departments (and countries) just because they are smaller, or less frequently in the scientific media, than others when thinking about a move to an other institution is an opportunity.


Alkene reactions with sodium tert-butoxide and CHBr₃

Strong bases react with chloroform $(\ce{CHCl3})$ and bromoform $(\ce{CHBr3})$ to form dihalocarbenes by alpha elimination.

The net reaction is as follows:

$$\ce{RO- + CHCl3 -> ROH + Cl- + CCl2}$$

The base removes the hydrogen atom from chloroform or bromoform to form an anion. Remember, when you deprotonate something, the electrons stay behind.

$$\ce{RO- + Cl3CH <<=> ROH + Cl3C:^{-}}$$

The trihalomethyl anion is unstable and fragments to lose a halide ion and generate the carbene.

$$\ce{Cl3C:^{-} <=>> Cl- + Cl2C:}$$

The mechanism using arrow-pushing formalism can be found at this website from the University of Liverpool.


How is the alpha carbon defined in alkyl halides?

The α carbon is the one directly attached to the functional group. There is no contradiction to your book’s claim that the α carbon of 2-bromobutane is the one directly bound to bromine.

This is due to the different definitions of functional groups and what they include. The keto functional group is defined as $\ce{-C(=O)-{}}$, i.e. it includes the carbon atom doubly bonded to oxygen. Halogen functional groups, for example the bromo functional group are defined as only the heteroatom, so in this case $\ce{-Br}$. Note that because the carbon is not included in the definition of this functional group, it is instead the α carbon.

This means that reducing the ester functionality of a β-halo ester will result in a γ-halo alcohol even though no group migrated.


Constructing an equation for the precipitation of a salt

Precipitation is usually a single or double displacement reaction, and refers to the formation of an insoluble, solid salt from soluble ions. For instance, silver bromide, $\ce{AgBr}$, is insoluble in water. However, it is made up of $\ce{Ag+}$ and $\ce{Br-}$ ions, which we can introduce in the form of soluble salts:

$$\ce{LiBr(aq) + AgNO3(aq) -> LiNO3(aq) + AgBr(s)}$$

In an aqueous solution, lithium bromide and silver nitrate underwent a reaction, forming lithium nitrate and silver bromide. In the products, silver bromide is denoted with the state symbol (s), which stands for "solid". This reflects its insolubility in water and hence its tendency to "precipitate".

The net ionic equation is when we only write the ions that participate in the reaction. The ions that do not participate are called spectator ions. For example, in the reaction above, only $\ce{Ag+}$ and $\ce{Br-}$ participated in precipitation. Thus, the net ionic equation for the reaction above is:

$$\ce{Ag+(aq) + Br-(aq) -> AgBr(s)}$$

One way of deriving this net ionic equation is to realise that salts such as $\ce{LiBr}$ and $\ce{AgNO3}$ do not exist as molecules in water, but rather dissociate to give their constituent ions:

$$\begin{align}
\ce{LiBr(s) &->[dissolution] Li+(aq) + Br-(aq)} \\
\ce{AgNO3(s) &->[dissolution] Ag+(aq) + NO3-(aq)}
\end{align}$$

However, $\ce{AgBr}$ is insoluble and does not dissociate into ions. If we go back to the first equation and replace all the soluble salts with their constituent ions, we get

$$\ce{\underbrace{Li+(aq) + Br-(aq)}_{from LiBr} + \underbrace{Ag+(aq) + NO3-(aq)}_{from AgNO3} -> \underbrace{Li+(aq) + NO3-(aq)}_{from LiNO3} + \underbrace{AgBr(s)}_{does not dissociate}}$$

The spectator ions which do not take part in the reaction are $\ce{Li+}$ and $\ce{NO3-}$, as they are unchanged in this reaction. Subtracting them from both the left- and right-hand sides, we obtain the net ionic equation:

$$\ce{Ag+(aq) + Br-(aq) -> AgBr(s)}$$

If we take this idea and apply it back to $\ce{CaCO3}$, we first need to come up with soluble sources of the ions $\ce{Ca^2+}$ and $\ce{CO3^2-}$. A possible choice is $\ce{CaCl2}$ and $\ce{Na2CO3}$, although that is hardly the only option.

The overall equation may be written as

$$\ce{CaCl2(aq) + Na2CO3(aq) -> 2NaCl(aq) + CaCO3(s)}$$

and the net ionic equation as

$$\ce{Ca^2+(aq) + CO3^2-(aq) -> CaCO3 (s)}$$

where $\ce{Na+}$ and $\ce{Cl-}$ are the spectator ions.

Finally, here are some web pages you might find helpful:

http://www.occc.edu/KMBailey/Chem1115Tutorials/Net_Ionic_Eqns.htm

http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Reactions_in_Aqueous_Solutions/Precipitation_Reactions


Finding the Ka of a weak acid after addition of a strong base

When you used the ICE method the second time, you forgot that HA does not have an initial concentration of 0. Instead, the initial concentration of HA is 0.0091 as you calculated previously. For simplicity's sake, you can just use Henderson-Hasselbalch equation after your first ICE method to determine the answer.


How can molecularity be the order of the rate determining step?

What your book, and most other elementary books present, is at best a generalization, not applicable to all cases.

Molecularity, as per the collision theory, is defined only for elementary reactions. One of the things that define an elementary reaction is the fact that the entire transformation from reactants to products occurs simultaneously, and follows from a properly oriented, appropriately energetic collision of the reactant species. If the reaction is not a direct consequence of a single collision, it is not an elementary reaction, and hence molecularity is meaningless.

Now, what usually happens is that out of several elementary steps constituting a reaction, one is very slow as compared to others. So whatever the rate of that elementary reaction, becomes practically the rate of the entire reaction. And since it is an elementary reaction, its order is its molecularity, and to an approximation, the molecularity of the slowest step then becomes the rate of the equation.

But, in most cases, the slowest step is not slow enough for it to entirely determine the rate. It is the slowest step, and it is an elementary reaction and therefore its molecularity is its order. But now, its order is not necessarily the order of the entire reaction, since that was an approximation valid only in cases where the slowest step was so slow that it entirely determines the rate. Then, the order of the entire reaction, due to the influence of the other steps (elementary in itself), can be fractional, while the order of the slowest step, is still its integer molecularity. ( here the term slowest step is preferred over the misnomer rate determining step, as it does not entirely determine the rate).

Also keep in mind, that many a times what people refer to as the "rate determining step" or the "slowest step" is not an elementary reaction. then its order need not be an integer, even if it is slow enough for the order of the net reaction to be equal to the order of that step.


Why should a system try to reverse any change that has been done to it?

Well, lets first write out the general equation of a chemical reaction:

$$\ce{aA + bB <=> cC + dD}$$

As you obviously know, since this is at equilibrium, not all the reactants react to form the products. Hence in a solution which is at equilibrium, all species of $\ce{A, B, C, D}$ will be present. However the special thing is that the ratio of the concentration of these species will usually be constant, no matter how much of each species you start with. This ratio is known as the equilibrium constant. The only thing that effects this ratio is temperature. The equilibrium constant is given by:

$$K_\mathrm{c} = \ce{\frac{[A]^a[B]^b}{[C]^c[D]^d}}$$

One more equation you need to know is the reaction quotient which gives the ratio of products to reactants at a certain time during a reaction and is given by:

$$Q_\mathrm{c} = \ce{\frac{[A]^a[B]^b}{[C]^c[D]^d}}$$ Note how this is similar to the equilibrium constant, but here, $\ce{[A]}$, $\ce{[B]}$, $\ce{[C]}$ and $\ce{[D]}$ are the concentration of the substance at a specific time during the reaction and doesn't necessary have to be when the system has reached equilibrium. This means that if $Q$ is smaller than $K$, the reactants will react to form more products so that $Q$ will increase so that it will eventually equal $K$ and when $Q$ is larger than $K$, vice versa. When $Q$ equals $K$, the system is at equilibrium.

So now lets apply this concept to help us explain Le Chatelier's Principle by considering different scenarios. Lets consider a mixture of hydrogen and nitrogen gas in a container. They react to form ammonia gas. The equation of this reaction is:

$$\ce{N2(g) + 3H2(g) <=> 2NH3(g) + heat}$$

Also the equilbrium constant is:

$$K_\mathrm{c} = \ce{\frac{[NH3]^2}{[N2][H2]^3}}$$

Scenario 1: Taking or adding more reactants or products

Lets say that this mixture of gases have finally reached equilibrium, so $K = Q$. Now I suddenly take some hydrogen has out.This would decrease the concentration of hydrogen gas and hence increase $Q$. So now $Q>K$. Therefore for the system to adjust itself, it needs to decrease the value of $Q$. It does this by creating more products and decreasing the amount of reactants. So in short, the equilibrium shifts to the left. This is what Le Chatelier's Principle predicts. Now by using the exact same logic, we can predict what happens if I were to change the concentration of ammonia, nitrogen, etc.

Scenario 2: Changing pressure or volume

Pressure and volume are inversely proportional to each other. Meaning that an increase is pressure will result in a decrease in volume and vice versa. This is due to the ideal gas law:

$$pV = nRT$$

Therefore, increasing the pressure of a system has the effect on the equilibrium  as decreasing the volume.

Now, back to example of hydrogen, nitrogen and ammonia gas in a container. If I were to suddenly decrease the volume of the container by two (equivalent to doubling the pressure) this would mean that the concentration of each species would double. Now, how would this effect our $Q$ value, would it be larger or smaller than $K$? Well the numerator would increase by fourfold while the denominator would increase by sixteenfold. Hence $Q<K$. Therefore as mentioned above, the equilibrium will shift to the right.

This can be applied to if the pressure was decreased, volume was decreased, etc.

Scenario 3: Temperature is changed

Now this one isn't really explained by using equilibrium constant. Instead will use a more intuitive approach.

For endothermic reactions, heat is required for the reaction to go forward. So by increasing the temperature, there is an increase in heat energy available to the system. Therefore the forward reaction will occur more readily. Hence it is expected that the equilibrium will shift to the right. Meanwhile, a decrease in temperature means less heat energy, hence equilibrium will shift to the left. The same logic can be applied to exothermic reactions.


Why would a ~1 cm thick layer of argon be a significantly poorer conductor of heat than air?

The thermal conductivity coefficients of some gasses in units of W/(m•K) are helium 0.142, argon 0.016, air 0.026, methane 0.03, propane 0.015, bromine 0.004, and steam 0.018. These are all rather similar except for helium and bromine.

Using the kinetic theory of gasses the thermal conductivity coefficient is given by $ \kappa \sim n\bar c\lambda C_V$ where $n$ is concentration, $\bar c$ the mean speed, $\lambda$ the mean free path and $C_V$ the heat capacity. Using definitions $n\lambda \sim 1/\sigma^2$ where $\sigma$ is the diameter of a gas atom or molecule; $\bar c \sim 1/\sqrt{m}$ and the heat capacity is $C_V \sim sk_\mathrm B$ where $k_\mathrm B$ is Boltzmann constant and $s$ is a multiplier depending on the type of species, atom, diatomic molecule, etc.

The conductivity coefficient at constant temperature is therefore $\displaystyle \kappa \sim \frac{sk_\mathrm B}{\sigma^2\sqrt{m}} $.

Thus we can see that helium will have high $\kappa$ as it has a small diameter and mass, and bromine will have a small value of $\kappa$; large diameter and mass. The other gasses are similar to one another because although the heat capacity increases between an atom and molecule and increases further as the molecule has more vibrational modes, the diameter and mass also increase when the molecule has more atoms and these terms cancel out to some extent.

So it is not just about speed, but also size, mass and number of vibrational modes. Using argon over dry air would seem to give a small advantage, but it may not be cost effective, i.e. heat saved vs. initial cost and energy needed to use argon.

(The molecular heat capacity increases above that for translational motion with the number of squared terms in the energy, potential and kinetic, each term contributes $k_\mathrm BT/2$ to the energy and so $k_\mathrm B/2$ to the heat capacity. Each vibration populated thus contributes $2k_\mathrm B/2$ and rotations the same for a linear molecule, but $3k_\mathrm B/2$ for a not linear one.)


How to obtain reactant lifetime using transition state theory?

As far as I understand, there are two things that I should answer for you.

Calculating rate constants with any trajectory methods is different than using TST. In fact, the whole point of using QCT is so we can account for nonstatistical effects. These nonstatistical effects are usually extremely important, as it is also mentioned in the article

  ... This dynamic matching leads to pronounced nonstatistical effects
  on the lifetimes...

Any time you are using TST, you are assuming there are no nonstatistical effects whatsoever. This approximation is probably never true, but it leads to surprisingly good numerical results. The way you calculate reaction rates in QCT is by integrating the reaction probability using a Monte Carlo integration formula, cleverly sampling the initial conditions of the reactants. I usually recommend a good book of W. Miller. The book itself is a bit outdated, but the derivation by Porter and Raff in the first chapter is beautiful.

Now you asked about the determination of the lifetime. The article, as I understand it, talks mostly about the lifetime of the activated complex. This can be done because in QCT we know the coordinates of atoms in every single timestep, so we can compare them to some given values to decide whether we are in the activated complex region or not. From the article:

  Transition zones were defined for the C–C forming (3‡) and the C–N breaking (5‡) steps (see Supporting Information for details). [...] The lifetime of
   intermediate 4b was defined as the elapsed time between departure from the transition zone 3b‡ and entrance in transition zones 5b‡

So, long story short: QCT determines lifetimes differently from TST, this is because TST is inherently statistical, while QCT also considers nonstatistical effects. Instead, QCT calculations rely on defining geometries for different states, and simply calculating how many integration steps the system spends there.


Why does NaCl act as a catalyst for the reaction of CuSO4 + H2O + Al?

Aluminum resists corrosion in neutral or slightly off-neutral water because of the very insoluble Al2O3 film on the metal. If you break this film, it will corrode the bare metal and reform.

But if you scratch the Al and attach a cathode (a less active metal), you have a galvanic cell, and H2 can be evolved from this cathode as Al dissolves. The simplest example of this is when you put Al in H2O in contact with a drop of mercury. Scratch the Al thru the Hg; the Hg amalgamates with the Al (attaching itself as cathode), allowing H2 to be evolved as Al dissolves elsewhere.

The experiment with CuSO4 is similar: scratch the Al; some Cu will be deposited and act as a cathode and the rest of the Al will eventually dissolve in the H2O. If you are too lazy to scratch the Al (I'm being facetious!), you can add a little Cl- ion to the H2O, which will begin corrosion of the Al, deposition of Cu, galvanic cell production, dissolution of all the Al.

There is an electrochemical example posed as a test question: would it be better to make a ship out of aluminum with copper rivets or out of copper with aluminum rivets? An aluminum boat? Nah! It would corrode in seawater. Copper would survive longer. But NO! The copper boat would have a huge cathode and tiny aluminum anodes (the rivets), which would corrode rapidly and the copper plates would fall apart. On the other hand, an aluminum boat would not last forever in seawater, but the tiny cathodes would limit corrosion current, and the large aluminum anodes would have corrosion distributed over the whole ship, so it would survive longer than the other way. Of course, any intelligent person would use copper rivets on copper plates, and aluminum rivets on aluminum, but it's just to make a point.


Choices B and C are two different ways of saying the same thing. Say $E_{MO}(A)$ and $E_{MO}(AB)$ are the energies of electrons in bonding and anti-bonding orbitals, respectively, relative to the electrons in the original atomic orbitals from which the molecular orbitals are formed, or, to make matters simple, set the energy of electrons in the original atomic orbitals to zero, that is, set as a reference $E_{AO}(1)$ and $E_{AO}(2)=0$ (assuming for simplicity that there are two MOs formed from two AOs).
To say bonding occurs is equivalent to saying that the energy of electrons in the separate atomic orbitals is higher than the energy of the electrons in the bonding molecular orbital. Using that reference system, positive energies (>0) disrupt bonding, and negative lead to bonding. Then saying that the effect of the antibonding electrons is greater than that of the bonding ones is equivalent to the statement
$$0\lt E_{MO}(B) + E_{MO}(AB)$$
or equivalently
$$0\lt  \frac{E_{MO}(B) + E_{MO}(AB)}{2} = E_{MO,\text{avg}} $$
but this is just what choice C says, since the average energy of the atomic orbitals is just zero, that is
$$  \frac{E_{MO}(B) + E_{MO}(AB)}{2} = E_{MO,\text{avg}} \gt 0 = E_{AO,\text{avg}}  $$


Is the solution isotonic, hypertonic, or hypotonic?

Osmotic pressure for non-electrolytic solutes is given by

$$\pi = CRT$$

where $C$ is the effective concentration of all the solutes. In our case, with multiple solutes, we simply add all their concentrations to obtain the effective concentration. This gives us

$$
\begin{align}
\pi_\mathrm{cell} &= 0.05RT\\
\pi_\mathrm{environment} &= 0.03RT
\end{align}
$$

Since $\pi_\mathrm{environment} < \pi_\mathrm{cell},$ The environment solution is hypotonic.


Is the cathode and anode on the cell notation of an Electrolytic cell on the same side as if how you will write it for a Galvanic cell?

A good mnemonic to remember which is the anode or the cathode is to remember that Right side represents Reduction in a cell diagram. This is fixed by those who invented cell notation.. Also, by definition, reduction occurs at the cathode. Whether it is a spontaneous cell or not, is another question.

Now as EdV said, calculate the cell potential as written i.e., consider cadmium half cell as the cathode, and the other silver electrode as the anode. If the Ecell=Ecathode-Eanode turns out to be positive, everything is fine, all that means that the cell, as written, is a spontaneous cell.

If Ecell is negative, then the cell is non-spontaneous.


How to separate paracetamol and caffeine?

The answer that my teachers had in mind was column chromatography. Essentially we use the van der Waals force of silicon gel to do an elution of the caffeine / paracetamol solution dissolved in water.

More details can be found here:
http://www.chemguide.co.uk/analysis/chromatography/column.html


Ammonium sulphide is $\ce{(NH_4)_2S}$; it is colorless. Yellow ammonium sulphide contains some sulfur $\ce{S}$ dissolved in the preceding solution. So the anion  $\ce{ S^{2-}}$ is mixed with some disulphide $\ce{S_2^{2-}}$, which is yellow. If you add an acid to this solution, you produce the reaction $$\ce{S_2^{2-} + 2 H^+ -> H_2S_2}$$ or$$\ce{S_2^{2-} + 2 H_3O^+ -> H_2S_2  + H_2O}$$
But $\ce{H_2S_2}$ is thermodynamically unstable with respect to disproportionation. It gets slowly decomposed according to $$\ce{H_2S_2 ->  H_2S + S}$$The sulfur S produced here is not soluble into water, and will make the solution cloudy. If there is enough disulphide, the sulphur will make a precipitate of elementary sulphur, which may be pale yellow or white.

Ref.: N.N. Greenwood A. Earnshaw, Chemistry of the Elements, A. Wheaton & Co, Exeter, 1986, p.805 - 807.


Predict the standard potential at 310 K for the cell constructed from the hydrogen electrode and metal-insoluble salt electrode $\ce{Ag(s)|AgCl(s)|Cl−}$

The reactions would be:
$$\ce{AgCl +e- ->Ag +Cl-}\tag{$E_1$ V}$$
$$\ce{2H+ +e- ->H2}\tag{0 V}$$
Adding:
$$\ce{AgCl +H_2->Ag +Cl- +2H+}\tag{$E_1 V$}$$

The standard potential would be:
$$E=E^\circ -\frac{RT}{\nu F}\log\ce{\frac{[Cl- ][H+]}{P_{H2}^2}}\quad (E^\circ =E_1,\nu=1)$$
H
Now you have $E_1$, also you must have $\ce{[Cl- ],[H+ ]}$ beforehand, there's no other way, other than considering the fact that (standard) Hydrogen electrode is maintained at $\text{1M}$ and $\text{1 atm}$, and to maintain charge $\ce{[Cl- ] =[H+ ]}$.In that case,
$$E=0.222-\frac{8.314\times 310}{96500}\ln(1)=0.222$$

Just for info (from Wikipedia):
$$\begin{array}{c|c}\text{Temperature}&\text{Potential} E^\circ\\
^\circ\text{K} &\text{V versus SHE at the same temperature}\\\hline
298 &0.22233\\
333 &0.1968\\
398 &0.1330\\
423 &0.1032\\
\end{array}$$
Also when you consider $\Delta S\ne\text{ constant}$:(A.J. Bard, R. Parson, J. Jordan, "Standard Potentials in Aqueous Solution", Marcel Dekker, Inc., 1985.)
$$E^\circ=0.23695 - 4.8564\times10^{−4}t - 3.4205\times10^{−6}t^2 - 5.869 \times 10^{−9}t^3$$


How can placing a lone pair in axial position differs from that in equatorial in number of electron domains?

Here's a drawing of the situation.

In the example with the axial lone pair of electrons, when they say, "90 degree to three other electron domains", they just mean that the axial lone pair is situated 90° relative to 3 other substituent bonds - the 3 equatorial substituent bonds in this case.  When the lone pair is located in an equatorial position, then it has a 90° interaction with only 2 other bonds - the 2 axial substituent bonds in this case; the lone pair is 120° away from the 2 other equatorial substituents.


How to calculate the cation exchange resin capacity?

First of all, you will need more sulfonic groups than sodium ions since sodium doesn't have a very good affinity for sulfonic groups.

Anyway, you can find some specs here — figure 2. This source is using sulfonated polystyrene with various crosslink percentages and expresses the capacity in $\mathrm{mmol}_\text{sulfonic groups}/\mathrm{g}_\text{resin}$ ($1\ \mathrm{meq} = 1\ \mathrm{mmol}$).


Naming of bicyclo[2.2.1]hept-5-ene-1,4,5,6-tetramethyl-3-bromo-2-ethyl carboxylate

The individual parts of the proposed name are correct; however, their order of citation in the name is wrong.

The name of the parent hydride is ‘bicyclo[2.2.1]hept-2-ene’. Note that the ending ‘ene’ receives the lowest locant possible (here: ‘2’ since the bicyclic ring system is numbered starting with one of the bridgeheads).

The principal characteristic group is expressed by means of a suffix, which yields the name ‘bicyclo[2.2.1]hept-5-ene-2-carboxylate’. Note that now the characteristic group receives the lowest locant possible (here: ‘2’), which
is cited immediately in front of the suffix. Thus, the locant for the ending ‘ene’ in the functionalized parent hydride is changed to ‘5’.

The other substituents are cited as prefixes in substitutive nomenclature: ‘3-bromo-1,4,5,6-tetramethylbicyclo[2.2.1]hept-5-ene-2-carboxylate’

Finally, the name of the ester is formed according to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), by placing the ‘hydroxylic’ component in front of the name.

  P-65.6.3.3.1 Monoesters

  Monoesters formed from a monobasic acid and a ‘monohydroxylic’ component are named systematically by placing the ‘hydroxylic’ component denoted by an organyl group (alkyl, aryl, etc.) in front of the name of the acid component expressed as an anion derived from the appropriate acid (…).

Therefore, the preferred IUPAC name (PIN) of the compound given in the question is
‘ethyl 3-bromo-1,4,5,6-tetramethylbicyclo[2.2.1]hept-5-ene-2-carboxylate’.


How exactly is ammonium sulfate formed in a Kjeldahl analysis?

Ammonia is basic so:
$$\ce{NH3 +H+<=>NH4+}$$
Sulphuric acid is acidic(since it's name is itself so acidic!):
$$\ce{H2SO4<=>H+ +HSO4- <=>2H+ +SO4^2-}$$
Now the final reaction would be like:
$$\ce{2NH3 +H2SO4<=>NH3 +NH4+ +HSO4- <=>2NH4+ +SO4^2- <=>(NH4)2SO4}$$
And remeber since proteins are polymers of peptides which itself are molecules consisiting of both a carboxylic ($\ce{-COOH/-COO-}$) and amino ($\ce{-NH3/-NH4+}$) group, therefore with strong heating ammonium sulfate is expected.


What prevents the anode from corroding in electrolytic organic reactions?

One would control the composition of the electrodes (mercury, platinum, Ebonex, boron-doped diamond coating) and the potential of the cell so that the organic reaction went forward while the electrodes remained inert.

Metal oxidation is facilitated by Lewis base ligands that drive reaction by stabilizing the ionic products.  Don't have them in there.


Authors may be sloppy about notation in this matter. I recommend considering $R_\ce{H} \approx \pu{10973 cm-1}$ and $Ry \approx \pu{2.18e-18 J}$, noting $Ry = hc \cdot R_\ce{H}$. Units of wavenumbers $(\pu{cm-1})$ and energy are often considered interchangeable in practice because they are proportional to each other by the constant value $hc$.

In my notes, I would always be sure to write $R_\ce{H}$ or $Ry$ to explicitly remind myself "which" Rydberg constant I was using (in fact I merged the R and y into a single symbol because I didn't like the suggestion of multiplication.)

Note also that there is a unit of energy known as a Rydberg, with $\pu{1 Ry} = Ry = hc \cdot R_\ce{H}$.


Reaction mechanism for the synthesis of 9,10-phenanthrenequinone

Here's what I was thinking.

  This probably happens via a single electron oxidation of that double bond to a radical cation as a first step. The cation can trap water and then we can oxidize the radical to a cation. Eventually, you get to a diol which you can oxidize to a diketone.

I ommited the steps for chromium oxidation from the alcohol to the ketone. Sounds like you are already familiar with those.


How is traditional thermite powder stored such that it doesn’t corrode?

Thermite powder is a heterogeneous mix and consists of fuel and oxidant, in you case aluminium $\ce{Al}$ and rust $\ce{Fe3O4}$. Separately, those can be corroded over time all right: aluminium can be oxidized and form aluminium(III) oxide, hydroxides and even some salts depending on what atmosphere thermite has been kept in. As you probably know, aluminium, being a reactive metal, doesn't corrode under ambient conditions as it's surface is protected by the thin layer of $\ce{Al2O3}$. Iron(II) is also capable of further oxidation, although rust is already the most stable form under ambient conditions.

However, it seems like you are using the term "to corrode" quite liberally to denote the act of burning of the thermite, e.g. the primary exothermic reaction

$$\ce{8 Al + 3 Fe3O4 → 4 Al2O3 + 9 Fe}$$

which only occurs when enough external energy (say, in a form of heat) is supplied. The thing is, not every reaction that is permitted by thermodynamics is also permitted by kinetics. Burning of a thermite is a heterogeneous process requiring to overcome activation barrier. Simply put, there are numerous solid particles or two phases being barely in contact, plus there is a protective layer on one phase ($\ce{Al2O3}$). Here, you need energy at least to increase thermal motion (promoting mass transfer between the phases) and compact the mix (ideally, melt to increase contact area).

That's why thermite won't easily go off, and it's a good thing making its storage safe and the working process predictable. There are only a few documented accidents [1, p. 62] resulting in ignition of thermite-alike mixtures, but none of them were spontaneous; instead, they were caused by the supply of substantial amount of external energy (heat, electricity, sparks, kinetic energy etc.), or third-party chemicals acting like catalysts:

  Aluminium-painted rusty surfaces are only hazardous if the coating has
  been heated to 150 °C: on cooling, the hazard remains (Eckart et al., 1971).
  Paints with cellulose nitrate as an additive are always hazardous. [...]

  Hawkins Colliery, 1950. The impact of a hand-held drilling machine cased
  in Elektron C alloy (94% Mg) upon rusty steel scaffolding ignited
  firedamp, burning 10 men, one fatally. This incident led to the initiation
  of a major research programme into the dangers of light alloys by the
  Safety in Mines Research Establishment. [...]

  SS Esso Durham, 1961. An explosion in one of the ship's tanks on 29th January was attributed to the impact of a brass object, of weight 361 b, onto a magnesium anode on the wall of the tank after a fall of 16 ft. From simulations of possible alternatives, it was considered likely that a grazing impact on the rough, dirty magnesium surface would have smeared the brass, which would then have reimpacted the anode a fraction of a second later, causing a thermite reaction and an explosion. The oxide necessary was probably provided by dirt in the area (ignitions were not obtained when impacting a clean magnesium surface). The possibility of ignition from impact onto hardened steel after a 40 ft fall was eliminated. [...]

  An aluminium anodizing works, Bletchley, Buckinghamshire, 1971. A ferrous metal object caused a spark on an aluminium grinding and polishing machine; the abrasive belts were lubricated by mineral oil, which caught light. The fire spread to the oil tank and extract ducting, causing £65000 worth of damage. [...]

  Horden Colliery, 1955. A steel weight was dropped on a friction prop with
  aluminium frictional surfaces, leading to ignition of firedamp.

  Glyncorrwg Colliery, 1955. A rusty steel wedge was struck from an aluminium roof bar, leading to ignition.

  United Kingdom, 1971. Propane was vented into a workshop during the filling of 140 lb propane cylinders when one was accidentally overfilled. The propane was ignited, and two people were injured in the subsequent explosion. The ignition source was believed to have been a spark from rusty steel cylinders being loaded outside, which, on examination, were
  shown to have aluminium smears on them. An aluminium roller conveyor was being used in the loading operation. A very light impact on such a smear is capable of generating a spark through a thermite reaction. [...]

  Ruhr Refinery, 1976. Two explosions within two hours of each other occurred on the 2nd December whilst switch-loading diesel into two tankers of carbon steel fabrication using an aluminium loading arm. The explosions were attributed either to a static electricity discharge or to a thermite reaction. [...]

  SS British Industry, 1963. The tanker had cathodic protection by incorporating a series of magnesium anodes in seven tanks. The cause of an explosion on the tanker was attributed to the impact of loose anode materials onto rusty deposits in the tank in which the explosion occurred. On examination after the explosion, many of the anodes were seen to be almost detached from their angle-iron bases, and bits had broken away. It
  was deemed that detachment could have been effected by a washing machine whipping violently in the tank, and that there would have been flammable concentrations of crude oil vapour in the tank.

References

Bond, J. Sources of Ignition: Flammability Characteristics of Chemicals and Products; Butterworth-Heinemann: Oxford; Boston, 1991. ISBN 978-0-7506-1180-0.


How to report result to the nearest 0.01 mg/L?

Reporting to the nearest quantity means that you report as many decimal places of the quantity as suggested. This is a common analytical chemistry expression given in procedures or analytical protocols. For example, if we say weigh  $\pu{5 g} \; \ce{KOH}$ to the nearest $\pu{0.1 g}$, one would record the weight to one decimal place, e.g., $\pu{5.1 g}$ or $\pu{5.3 g}$, to one decimal place but you will not report $\pu{5.1176 g} \; \ce{KOH}$ even if your balance is capable of doing so.

Your test report should include at least two decimal places, the report should appear as $\pu{7.24 mg/L}$ or something like that number.


Are all deliquescent salts highly soluble?

As already pointed out, dissolution and deliquesence are different phenomenon.

For dissolution to take place, the energy produced by hydration of the salt's constituent ions should compensate for the energy required to break the crystal lattice.

Solubility then depends on temperature, pH and also the amount of salt already present in the solution. (Recall: definitions of solubility products)

Hygroscopic characteristics in general can be attributed to the both absorption and adsorption (at the surface).

Particularly, deliquescence occurs when the vapour pressure of the saturated aqueous solution of a substance is less than the vapour pressure of water in the ambient air.

When water vapour is collected by the pure solid compound, a mixture of the solid and liquid or an aqueous solution of the compound forms until the substance is dissolved and is in equilibrium with its environment; when this happens, the vapour pressure of water over the aqueous solution will equal the partial pressure of water in the atmosphere in contact with it.

A crystalline salt aerosol particle will deliquesce in the atmosphere when the relative humidity surpasses a characteristic value, i.e deliquescence point.


What is the use/function of lithium chloride as an excipient in sofradex eardrops?

Probably the main reason $\ce{LiCl}$ is used in liquid to soften ear wax is that $\ce{LiCl}$ is quite hygroscopic and pharmacologicically inactive under the conditions used (small amount used to retain moisture).

As you state, taken internally, in contact with the eyes, or left in contact with skin for extended time, there are some biological effects, but I'd not be concerned about those with a tested commercial product.


Purification of primary amines using Schiff base immobilization

Let's take your questions one at a time.

Are there any alternative ways to purify primary amines from a complex mixture that I am unaware of?

Your end application is LC-MS.  There are plenty of papers on amine-specific derivatization methods for metabolome analysis. For example, there is one by Zhao et al. [1]. Instead of purifying the primary amine metabolites and isolating the purified mixture as a separate step, they just rely on the analytical instrumentation to be able to identify primary amines from their derivatization pattern.
Alternately, as suggested in the comments by @Waylander, you could acidify the sample to form ammonium ions, and then pass the sample through a cation exchange column, where only positively charged ions will bind.  As you noted, this will not be perfectly selective for primary amines, because species like quaternary ammonium salts etc. will also be retained.
You could combine the derivatization approach with the SCX approach, of course.  If you chose a primary-amine-selective derivatization, you might get a selective method.  At some point, however, the complexity of the method will become a drawback.

Have anybody else, published or unpublished, used Schiff base chemistry to purify primary amines?

Well this question is impossible to answer for the unpublished case.  I couldn't find any published methods in a quick search, which makes me suspect it is not the best approach to your problem.  But I could be wrong of course!

What would be your odds that this could work? Is it worth pursuing?

Unanswerable without data.  I suspect that secondary amines, at least some times, will also react with the aldehydes to form Schiff-base-like iminium cations, and may also be selected by your proposed method.

Any ideas for improvements?

I think derivatization, without separation or isolation of the amines, would be an attractive direction for you to consider.  The paper I linked, its references, and the papers that reference it might be good starting places.
Reference

Zhao, S.; Li, H.; Han, W.; Chan, W.; Li, L. Metabolomic Coverage of Chemical-Group-Submetabolome Analysis: Group Classification and Four-Channel Chemical Isotope Labeling LC-MS. Anal. Chem. 2019, 91 (18), 12108–12115. DOI: 10.1021/acs.analchem.9b03431.


What is the effect of polar protic solvent on nucleophilicty?

Protic solvents affect nucleophilicity because they form a shell around the nucleophile via their hydrogen which has a positive partial charge and therefore interacts with it, thus the nucleophilicity will decrease. This effect is stronger for smaller nucleophiles as they can be "trapped" more easily. So taking the classical nucleophiles you obtain this order:

$$\ce{I^-} > \ce{Br^-} > \ce{Cl^-} > \ce{F^-}$$

Aprotic solvents affect nucleophilicity in a more complex way. You can sometimes read that the previous order of nucleophile strength is inversed and small nucleophiles are favored:

$$\ce{F^-} > \ce{Cl^-} > \ce{Br^-} > \ce{I^-}$$

But be careful, this is not always true. There is more consideration to take.

A "soft" anion is highly polarizable and has more electron density to give, making it a good nucleophile compared to "hard" ones which are tightly holding their electrons.
This order is not respected by some polar aprotic solvents, acetone is the only one coming in my mind.


Why is change in entropy negative and change in enthalpy negative for the reaction of Magnesium and Hydrochloric acid?

For starters, you should be suspicious when the reference page is formatted as it is.

The real issue I see is that the page lists $\Delta S_{f}$. This is where you should immediately call BS.

We don't use this quantity because entropy is not an energy, therefore, you cannot set the standard state entropy to 0. That means that entropy of formation is pretty useless because you need to know the actual entropy of the standard state of components to know what the entropy of the species is. For elements, the standard molar entropy is positive. However, for solutions (and this is what I got wrong in my now-deleted comment), the standard molar entropy can be negative thanks to solvation (specifically, solvation shells of water can be highly "ordered").

In order to compute the entropy of this reaction, you need to know the standard molar entropy of each species. The easiest thing to do is to look it up in a trustworthy reference source, for example a NIST database or CRC Handbook of Chemistry and Physics.

$$
\begin{array}{|c||c|}
\hline\hline
\text{substance} & S^{\circ} (\mathrm{J}\ \mathrm{mol}^{-1}\ \mathrm{K}^{-1}) \\
\hline
\ce{H2(g)} & 130.68 \\
\ce{Mg(s)} & 32.67 \\
\ce{HCl(aq, 1 M)} & 56.6 \\
\ce{MgCl2(aq, 1 M)} & -25.1 \\
\hline
\end{array}
$$

Sources: hydrogen, magnesium, hydrochloric acid CRC handbook under "Thermodynamic Properties of Aqueous Ions", magensium chloride CRC handbook under "Thermodynamic Properties of Aqueous Ions"

These are the same values in your reference, but I had trouble believing those data. So, basically, the answer is correct. This reaction has negative entropy change.

Why? Because solutions are kind of evil. Solubility and other solution phenomena can be really hard to predict because they depend so much on how the solvent interacts with the solute. In this case, it's probably driven by some idiosyncratic size of the chloride anion, which somehow allows water molecules to form very nice solvation spheres around the anion. I'm rationalizing this because the standard molar entropy for solutions of magnesium iodide (84.5) and magnesium bromide (26.8) are all positive. These ions are probably of a size that prevents the water molecular from fitting together nicely into a single shell. The imperfections in the solvation shell increase entropy.

That means that the large entropy of hydrogen gas is balanced by some stupid solvent effect with negative entropy, so the overall change in entropy is negative.