Does the protective layer of Aluminum oxide form on the surface of aluminium alloys (such as Al-Zr) as well?

Short answer: Yes.

Long answer: Any alloy containing 'passivating' elements such as aluminum, which have the tendency to form a protective oxide as you describe, has the potential to also form the same sort of passivating oxide layer as the pure material.  It's not guaranteed, though.

Pure zirconium also forms a passivating oxide $(\ce{ZrO2})$$^{(1)}$, so the AlZr alloy you mention is almost certain to form a similar protective oxide.

More broadly, a major area of research at present is in the 'alumina-forming' alloys ("alumina" is another word for aluminum oxide, $\ce{Al2O3}$), where aluminum is included in, e.g., steels or nickel superalloys, to try to achieve this passive aluminum oxide layer on the surface in high-temperature, corrosive environments$^{(2-5)}$.  With materials like a stainless steel-plus-aluminum, the passive oxide scale only forms if the alloy has the right percent range of aluminum -- otherwise, the aluminum oxide ends up staying spread throughout the metal and not concentrating at the surface.  If the alumina doesn't end up spread evenly over the surface, you don't get nearly as good of a protective effect.

(1) http://www.mtialbany.com/metals/zirconium/
  (2) http://topaz.ethz.ch/function/web-het-secured/pdfs/Prescott-92.pdf
  (3) http://www.metalle.uni-bayreuth.de/de/download/publications_downloads/Bensch__Sato__Warnken__Affeldt_______Reed__Glatzel__Acta_Mat_.pdf
  (4) http://www.netl.doe.gov/File%20Library/Events/2010/ods/Gordon_Tatlock_Corrosion.pdf
  (5) http://www.ornl.gov/science-discovery/advanced-materials/science-to-energy-solutions/alumina-forming-austenitic-alloys-licensed


Rate of nitration of C6H6, C6D6 and C6T6

The rate-determining step is the attack of the benzene aromatic ring onto the electrophile, and the loss of aromaticity is what causes the instability of the transition state, resulting in a higher activation energy and thus a slower rate of reaction. The subsequent deprotonation is after the rate-determining step and thus will not affect the rate of reaction, despite C-T > C-D > C-H bond strength.


How does the surface tension of water decrease with an increase in the concentration of NaCl?

The surface excess of a salt (NaCl) is negative because salts prefer to be hydrated.  So there is less salt at the inferface than in the bulk phase.  The Gibbs adsorption equation says the change is surface tension is the opposite of the surface excess - so for a salt solution, the change is positive and the surface tension increases.


When to include or exclude lone pair counting electrons according to Hückel's rules?

As written on Chemistry Libre Texts,

  Perhaps the toughest part of Hückel's Rule is figuring out which electrons in the compound are actually π electrons. Once this is figured out, the rule is quite straightforward.  π  electrons lie in p orbitals and  sp2 hybridized atoms have 1 p orbital each. So if every molecule in the cyclic compound is sp2 hybridized, this means the molecule is fully conjugated (has 1 p orbital at each atom), and the electrons in these p orbitals are the π electrons.

In furan, there are two lone pairs, one in the sp2-orbital, and the other in the p-orbital. The latter is hence the set of π electrons.

In the cyclopentadienyl anion, the lone pair is present in the p-orbital, as all the sp2-orbitals are involved in bonding.

Thus, π electrons and not just lone pairs are to be counted in Huckel's Rule.


Using vapor mole fraction and pressure to determine liquid mole fraction

Raoult's law tells us the relationship between the composition of the liquid and the vapor pressure above it:

$$P_A=\chi_A P^\circ_A$$

Where:

$P_A$ is the partial pressure of $A$ in the vapor above the solution
$P^\circ_A$ is the vapor pressure of pure $A$
$\chi_A$ is the mole fraction of $A$ in the solution

For a two component solution $A$ and $B$ (I don't want to use $x$ to avoid confusion with $\chi$) we can then use Dalton's Law to get:

$$P_T=P_A + P_B = \chi_A P^\circ_A +\chi_B P^\circ_B= \chi_A P^\circ_A +(1-\chi_A)P^\circ_B$$

However, from the definition of mole fraction we can derive a relation between the mole fraction in the vapor (from here on out $\phi$ to avoid confusion between vapor and solution, which is what your mistake is).

$$\phi_A=\frac{n_A}{n_T}$$

Where $n$ is the number of moles, and since at constant temperature and volume, $n\propto P$...
$$\phi_A = \frac{n_A}{n_T}=\frac{P_A}{P_T}$$

So, we can rewrite $P_A = \phi_A P_T$ and $P_B=(1-\phi_A)P_T$

We are given $\phi_A=0.3$. Thus $\phi_B=0.7$

Here we go. Substitute $P_A=\chi_A P^\circ_A$ and $P_T=\chi_A P^\circ_A +(1-\chi_A)P^\circ_B$ into $P_A=0.3 P_T$:

$$\chi_A P^\circ_A=0.3\left(\chi_A P^\circ_A +(1-\chi_A)P^\circ_B\right)$$
$$0.7\chi_A P^\circ_A=0.3(1-\chi_A)P^\circ_B$$
$$0.7\chi_A P^\circ_A + 0.3\chi_A P^\circ_B=0.3P^\circ_B$$

I think you can get it from here.


Bohr's theory of the atom precedes quantum theory but does describe the Balmer and other lines in atomic hydrogen.  I've outlined the calculation so that you can see where the Bohr radius comes from.

He  assumed that the electron orbits the nucleus in a circle with a radius r and is determined by the balance of centripetal acceleration and Coulomb attraction towards the nucleus. For an electron of mass $m_e$ and speed $v$ this is
$$\frac{m_ev^2}{r}=\frac{e^2/(4\pi\epsilon _0)}{r^2}$$
 where e is the charge on the electron and $4\pi\epsilon_0$ puts us into SI units, ( $\epsilon_0$ is the permittivity of free space).

The total energy is the sum of kinetic and potential parts $$E=\frac{m_ev^2}{2}-\frac{(e^2/4\pi\epsilon _0)}{r}$$
and using the first equation gives
$$E=\frac{e^2/(4\pi\epsilon _0)}{2r}$$

Bohr had to make the assumption that electrons circling in an orbit do not radiate energy, which is contrary to classical electrodynamics, but he then needed to determine which orbits were allowed. He assumed that the angular momentum is in quantised units of Plancks constant h and a (principal) quantum number n. This gives
$$ m_evr = n\frac{h}{2\pi}$$

which when combined with the first equation gives

$$ r=a_0n^2$$

which is the equation you ask about and $a_0$ is conventionally the symbol for the Bohr radius and is 0.0529 nm.
It is calculated as
$$a_0 = \frac{\hbar^2}{e^2/(4\pi\epsilon_0)m_e}$$

where $\hbar = h/2\pi$.
Re-expressing the  energy using $a_0$ gives
$$E=-\frac{e^2/(4\pi\epsilon_0)}{2a_0}\frac{1}{n^2}$$
which is the famous Bohr equation and explains the energy levels in hydrogen atoms and so the transitions between them.


How many grams of sodium sulfate and water in 100ml of 0.5M solution

As Jason indicated your calculations are corrected.

370 grams of 15% solution of sodium sulfate does have 314.5g $\ce{H2O}$ and 55.5g $\ce{Na2SO4}$.
Your calculation for grams of solution are correct. 47.3g of solution is required.

However the density would be expected to be somewhat above 1.00 g/ml not less. So a little less than 47.3 ml of solution would be required. However the density of the solution is nowhere near 10 g/ml so the book answer is horribly wrong.

To solve for ml of solution, I'd have just stated an assumption that the density of the solution was 1.00 g/ml since you weren't given that information. There is absolutely no way to calculate the density given the information in the problem.


Applying Raoult's Law to a condensed gas mixture

You have confused the vapor-phase mole fraction with the liquid-phase mole fraction.  The liquid-phase is mole fraction in usually denoted as $x_i$ for compound $i$.  The vapor-phase mole fraction of $i$ is $y_i$.

Raoult's law is that the partial pressure over an ideal solution is equal to the pure-component vapor pressure times the liquid mole fraction, $p_i=x_i*p^°_i$.

Separate from Raoult's law we have the ideal gas law. It says that that says that partial pressures are proportional to mole fraction in the vapor phase, i.e. that $p_i=y_iP$.  So when you took the ratios of $p_{benz}/P$ and $p_{tol}/P$, you were solving for the vapor-phase mole fractions, i.e. for $y_{benz}$ and $y_{tol}$.

Your mistake comes in taking those $y_i$ values and treating them as $x_i$ values.  That's a mis-formulation of Raoult's law.

You can read more about Raoult's law and the ideal gas law for binary ideal mixtures in my answer to a previous chem.se question.


Direction of mass transfer in two film theory

Just keep in mind: the driving force of mass transfer is a concentration gradient. Nature tries to 'equilibriate' a difference in concentration between two phases.

As such, the mass transfer in that sense will be from the phase of high concentration (S2) to the phase of lower concentration (S1).

Two film theory is only a simplification of the mass transport directly at the interface, in which we assume there are two boundary layers on each side of the interface in which the concentration gradient is constant, giving us linear concentration profiles, and the concentration directly at the interface is determined by Henrys Law. This makes dealing with such problems much easier.

Mass and Heat Transfer are often taught in one course as they have many similiarities, but Heat Transfer offers a more intuitive introduction to the principles of transport phenomena in my view:

In heat transfer the driving force is the temperature gradient. We intuitively know that heat transfer will be from the hotter system to the colder system, and formally know that from the 2nd law of thermodynamics. Can you see the similarity to mass transport?


How is the magnetic property of an element affected when the element is part of a compound?

Magnetism appears in many very different forms that have very different origins.

All elements show some diamagnetism, even though often overshadowed by other, stronger, forms of magnetism, because it is a property of the bound electrons.
The diamagnetic contribution caused by the inner electrons that are not part of a chemical bond will therefore remain unchanged if the compound bonds chemically. The diamagnetic contribution from the electrons participating in the bond will of course change, so even for the most simple form of magnetism, it is not possible to predict the diamagnetic behavior of a compound from that of the elements.

Paramagnetic materials have a permanent magnetic moment from uncompensated magnetic moments, typically electron spins (there are exceptions where a paramagnet is formed from larger units, like magnetic particles).
In a very simplified way, you will find unpaired electrons typically on the outer levels of elements, where they are influenced and participate in chemical bonds. This is even the case for the strongly localized d and f electrons that are characteristic for the strongest paramagnets like $\ce{Gd}$.
So clearly, paramagnetism is influenced by the chemistry.

Both diamagnetism and paramagnetism are extremely weak effects that normally don't play a role in everyday life.

Now for the 'real' forms of magnetism, ferromagnetism (ferri- antiferro and ferri etc.): All of these types of magnetism are collective phenomena, i.e. they are the result of the internal interactions of a large system, through conduction electrons (itinerant), superexchange, double exchange, etc.

Due to the collective nature, it is not possible to predict the behavior of a compound from that of the physical parameters of the elements. (Of course, you can simulate the magnetic behavior of more or less any material, but I think that was not the question)

You could not call $\ce{Fe}$ itself a ferromagnet, the ferromagnetism is a property of a huge number of $\ce{Fe}$ atoms that are acting collectively. This ensemble is very sensitive to changes, often even a small temperature change (e.g. at 20°C for $\ce{Gd}$) will completely change the magnetic properties. A chemical reaction or even simple alloying, will completely change the magnetic properties.

The macroscopic magnetic properties of a material (susceptibilitiy, coercivity, remanence, saturation magnetization, frequency dependence of parameters etc.) do not allow to predict the properties of a compound that contains this material. So you can't predict the coercivity of  $\ce{NdFeB}$ given the coercivities of  $\ce{Nd}$, $\ce{Fe}$ and $\ce B$.

It is possible to simulate the complete electronic structure of a compound with Density Functional Theory. This simulation will tell you the magnetic parameters of the material, including the ordering temperatures in the case of ferromagnetic materials, to a varying degree of accuracy. It is a very computationally intensive approach, to compute the properties of a new material is a research project, requires some ad-hoc assumptions and will need to be experimentally verified as it is not yet 100% reliable.

This computational approach does not make any use of the macroscopic physical properties of the elements in a compound, it directly solves the many body problem of electrons and nuclei.

In summary, it is not possible to predict the change in magnetic property of a compound given the knowledge of the magnetic properties of its participating elements, but you can simulate the electronic structure of a compound, and that will tell you how the compound will behave magnetically.

Note: Neodymium magnets aren't made out of $\ce{Nd}$ metal. They are made from sintered $\ce{Nd2Fe14B}$. $\ce{Nd}$ is only a small fraction of the entire material. This is a very complex material. The exceptionally high coercivity, i.e. the resistance against a change in the magnetization, is the result of high magnetocrystalline anisotropy of $\ce{Nd2Fe14B}$ and the interfaces between the grains that pin the magnetization. This is something you would not think about given just the magnetic parameters of $\ce{Nd}$, $\ce{Fe}$ and $\ce{B}$.


Is there a notion of generalized acidity for lithium ions?

A complex of lithium ion with lithium hudroxide is known as the salt $\ce{[Li2OH]^+[ClO4]^-}$ [1].  As with the hydronium salt, this requires an essentially nonbasic counterion; with a chloride counterion a double salt is obtained with no evidence of a complex cation.  From the abstract (some evident faults in translation from Russian have been corrected):

An effort is made to obtain salts of Li2OH+ and Li3O+ cations, belonging to a series of simple oxonium compound cations H3O+, LiOH2+, Li2OH+, Li3O+, where proton and lithium replace each other. Binary compounds of lithium hydroxide with lithium chloride and perchlorate are obtained: LiOH·LiClO4, LiOD·LiClO4, LiOH·LiCl, LiOD·LiCl and LiClO4·2LiOH. A conclusion is made by IR absorption spectra, that an adduct with perchlorate can be built of ClO4 anions and complex [Li2OH]+ cations. For adducts with chloride, equivalence of hydroxide and chloride anions with regular alternation of the anions in the crystal lattice is most probable.

Reference
1.
Nikitina, Z. K., Nikitina, N. I., Rosolovskii, V. Y., "Complexing Lithium Salts and Lithium Oxide or Hydroxide: Synthesis and Characterization of a New Cation-Lithiumhydroxonium Li~2OH^+", Russian Journal of Inorganic Chemistry 42(9), 1295-1298  (1997).


If no oxygen is present, will cut avocados stop browning?

Though I'm sure that it's possible for them to brown due to substances other than oxygen, in general, yes, if you remove oxygen from the cut surfaces of an avocado, it will not turn brown, and yes, if you reduce the quantity of oxygen, browning will slow.  The same works for apples, bananas, and pretty much any other fruit that browns upon exposure to air.

It's not particularly practical for intact avocados, but one method for preventing guacamole from browning is to put plastic wrap right against the surface of the dip, preventing air contact.  Additionally, if you look at packaged guacamole, in addition to the browning resistance caused by acidity, the amount of oxygen in any given container is limited to a few small bubbles.  The oxygen in those bubbles is quickly consumed and the remaining gas does not oxidize the guacamole further.

You could conceivably test this with a vacuum chamber and keep normal slices of avocado green for a prolonged time.  You also could use a food vacuum sealer that bags and evacuates food for long term storage, though ripe avocados are delicate enough that they will probably be crushed into paste by the process.


Why does silver oxide form a coordination complex when treated with ammonia?

It's quite elementary (perhaps). Silver (and gold) are known to show two main coordination numbers in its compounds, namely, $2$ and $4$. To start, we can consider $\ce{[Ag(OH)2]-}$ to be our initial complex as in aqueous solutions it can be thought of to be existing in an equilibrium with $\ce{Ag2O}$. See footnote [1].

A quick glance at the order for the strength of ligands tells us that $\ce{NH3}$ is a stronger ligand than $\ce{OH-}$. What I now intend to do is to compare the thermodynamic stability of $\ce{[Ag(OH)2]-}$ with that of the diammine complex $\ce{[Ag(NH3)2]+}$.

The strength of ligands is directly proportional to the thermal stability of a complex so yes, the diammine complex is more stable.

A similar case is seen with copper(I). Cyanide complexes of copper(I) are known to be extremely stable and hence it prefers forming a complex with cyanide over other ligands.

[1] Nevertheless, saying this may be highly controversial because Ellingham diagrams tell us that $\ce{Ag(OH)}$ being unstable gets converted into $\ce{Ag2O}$ which is a much more stable compound for silver to be in. However in aqueous solutions, an equilibrium of such a kind can always exist.


Why do nitro groups confer explosive tendencies?

Explosives chemistry is a rather complex topic. I've heard that this book is a good source of information about it (I haven't read it).

In a nutshell, your intuition about the nitro group is accurate. Formation of $\ce{N2}$ is highly energetically favourable. To get an explosive, what we need is a rapid reaction that produces a lot of heat and gas to both cause the obvious effects of an explosion and to propagate the reaction to other molecules of explosive. The rapid part is what separates something like TNT from toluene. The combustion of toluene is also energetically favourable, but in TNT, to create $\ce{N2}$ is not dependant on mass transport of any other species and can thus happen very rapidly, whereas the combustion of toluene is limited by how quickly oxygen is transported to it. If one were to vaporize toluene in the correct concentration in air, this transport problem goes away and an explosion can occur. TNT also has an advantage in this regard because the nitro groups provide a source of oxygen to react with the carbon and nitrogen remaining (not enough for all of it, but it helps). Explosives are often mixed with fuels or oxidizing agents to produce a more oxygen-balanced mixture for a more efficient explosion.

As for what happens first in an explosion, many different reactions can occur during an explosion, but it may be helpful to consider what it takes to actually detonate TNT and think about the timescale of the reactions. TNT is a solid at room temperature and has a flash point of 163 °C making it difficult to even ignite, and while it will burn in a fire, there is no risk of explosion. For an explosion to occur, enough gas and heat has to be produced to propagate the reaction through the bulk of the material. In practice this is done using a much more sensitive explosive (other explosives like lead azide or nitroglycerine are unstable enough to be set off by heat or pressure) to produce a small shockwave that provides the activation energy to initiate a reaction as it travels through the explosive which then sustains the shockwave through the rest of the material. In a normal explosive (not a fuel-air explosive or the like), the reactions that contribute to the bulk of the explosion are limited to what the explosive is made of because the speed of the explosion is too fast for air to play much of a role initially.

In the case of TNT, the experimentally-measured time it takes for a shockwave to pass through is 100–200 fs (no idea how one measures that), so any oxidation of the carbons seem unlikely to contribute much to the the initial explosion, given the only readily available source of oxygen is from the nitro groups which must presumably decompose first. This group proposed a few decomposition pathways for TNT, including homolytic cleavage of the $\ce{C-NO2}$ bond, rearrangement from $\ce{C-NO2}$ to $\ce{C-ONO}$ followed by homolytic $\ce{O-NO}$ cleavage, and $\ce{C-H}$ attack from an adjacent nitro group to the methyl ring substituent, but they found that only the first was fast enough to occur during detonation, the others possible only for lower temperature thermal decomposition. This initial decomposition step is the only thing fast enough to contribute to the shockwave that sets off the rest of the TNT, while the reactions that produce the final products occur (relatively) long after the initial blast has initiated the rest of the explosive.


It basically comes down to scaling with the size of the system, which will correspond to the number of electrons (that is, occupied valence orbitals) and the size of the basis set (which must have a sufficient quality, that is, size to yield meaningful results).

Edit: I should note that not all quantum chemists consider DFT to be ab-initio. Given the prevalence of such calculations, I have included it in my answer.

Let's consider density-functional theory (DFT) and Hartree-Fock (HF) theory first. These are a ground-state methods that describe a system with a single "wave function" called a Kohn-Sham or Slater determinant, respectively. This is valid for many, but not all systems - especially when there are several d-block metals closely interacting, this can be a bad approximation. For DFT, the cheapest useful specific methods (called functionals) in the most common implementations (i.e. programs) will scale as $\mathcal{O}(N^3)$ (where $N$ is some measure of the system size) due to the calculation of the electron-electron interaction.$^1$ At some point, a matrix diagonalization is performed, which has the same scaling. For HF, the scaling is worse, because of the calculation of a term called exchange (that arises from the specific form of the wave function ansatz), which scales as $\mathcal{O}(N^4)$.

At this point, I should mention that efforts to lower the prefactor of the computational cost function of these workhorses of quantum chemistry are ongoing and some have reached maturity. For "small" systems, the cost is then often dominated by terms other than the highest scaling ones. However, you will eventually hit the scaling wall. There are also efforts to lower the scaling behavior by introducing more approximations, smarter discarding of near-zero terms etc. Again, you may find out the hard way, that in the system you wish to calculate, the approximations do not hold or are not effective.

For excited states, one can assume a $+1$ scaling behavior and a worse prefactor. When considering the ground state with more accuracy, one then turns to post-DFT methods such as double-hybrid density functionals or post-HF methods such as Configuration Interaction, Møller–Plesset perturbation theory, Coupled-Cluster theory (e.g. CISD, MP$n$, CCSD(T)). In the standard implementations, they scale at least as $\mathcal{O}(N^5)$, but can go up to 6 and 7. Again, smart selection of the involved terms can lead to significant speed-ups, leading to an empirical linear or near-linear scaling at the expense of increased prefactors and less certainty.

The comments to the question mention classical dynamics of proteins. The scaling wall even for a single-point energy is hit well before approaching 100 $\ce{CNO}$ atoms (in my personal experience, which is a bit dated. Make it 200, then). The necessary derivatives with respect to the nuclear coordinates also tend to add to the prefactor or the scaling in a significant fashion.

$^1$ This scaling applies if a particular additional approximation called density fitting is made, which is usually a very good one. Note that better DFT methods will have worse scaling.


Why does it require so much pressure to create diamonds?

There are two allotropes of carbon crystal, graphite and diamond.   Both
are stable, so the formation of diamond structure is competitive with
the other structure (or structures, if you include buckyballs
and tubes).   The important consideration, therefore, is crystal growth rate,
which will depend on solvent and seed material, as well
as temperature and
pressure.

For manufacture of large artificial diamonds, liquid iron 'solvent' is
used, which is why high temperature is important; carbon has to diffuse through the liquid solvent. At high temperature,
the (relative) instability of diamond means that
a pressurized environment is required: diamond has to expand (diamond is 3.53 g/cc, but graphite is 2.23 g/cc) in order to become graphite, and pressure prevents that.   Pressure alone might
become so high that graphite recrystallizes as denser diamond, in a shock wave (as long as you only want to make small diamonds) made with explosives.  At elevated temperatures and low pressure, diamond
recrystallizes as graphite, however.

It is
possible to grow diamond films (depending on a substrate that
seeds the crystal form of interest) at low pressures as well, from
vapor.   Pyrolyzing CH4 in the presence of a cold surface that doesn't
seed the graphite form is the general approach.

To summarize: in growth of a carbon crystal, the environment determines
the dominant crystal structure in a number of ways.  The Diamond Makers by Robert Hazen, is a good treatment of this subject.


How to determine the vapor pressure of a glycerine and propylene glycol mixture?

Your best bet for this will be to use the Clausius-Clapeyron relation. This is an exponential function that describes the vapor pressure of a substance as a function of temperature, and uses the enthalpy of vaporization (latent heat) and a constant as parameters.

The constant is dependent on the particular substance, but there is a way to eliminate it from the equation if you can get one data point: the vapor pressure at a specific temperature. If you can do that, then you can use the two-point form to predict vapor pressure at another temperature:

$\ln\frac{p_2}{p_1} = \frac{-\Delta H_{vap}}{R}(\frac{1}{T_2}-\frac{1}{T_1})$

This version only has one parameter - the enthalpy of vaporization ($\Delta H_{vap}$). $p_1$ and $T_1$ are the vapor pressure and temperature at point 1 (an arbitrary point), and $p_2$ and $T_2$ are the vapor pressure and temperature at a different arbitrary point.

For a pure substance, if you know the enthalpy of vaporization and the boiling point, you an calculate the vapor pressure at any other temperature. That is because the boiling point is the temperature at which the vapor pressure equals one atmosphere. In other words, you would set:

$p_1 = 1 \space \mathrm{atm}$

$T_1 = T_b$,

Then solve for $p_2$. The equation would be:
$p_2 = p_1\exp([\frac{-\Delta H_{vap}}{R}](\frac{1}{T_2}-\frac{1}{T_1})) \tag{1}$

When you use this equation, make sure you first convert the temperatures to Kelvin. This only works with an absolute temperature scale. Pressure can be in whatever units you want, as long as both pressures have the same units. In other words, convert 1 atm to whatever units you want to use for vapor pressure.

The data you listed didn't have the enthalpy of vaporization for either component, but they are available from NIST:

$\Delta H_{vap} = 91.7 \space \mathrm{\frac{kJ}{mol}}$ - glycerol

$\Delta H_{vap} = 71.2 \space \mathrm{\frac{kJ}{mol}}$ - ppg

Incidentally, NIST lists different values for the boiling point than you have. I would double check those before plugging numbers in.

Now, you don't have a pure substance; you have a mixture. Depending on how non-ideal it is (which basically means how much the molecules of one component will affect the other components) you might have to account for that if you want to get very accurate results. On the other hand, if you just want an estimate and are planning to experimentally adjust the power output from there, you can get a good starting point by assuming that the mixture and the vapor are both ideal. Under those conditions, the total vapor pressure of the mixture will be equal to the sum of the partial pressures...

$p_t = p_{gly} + p_{ppg}$

... and you can use the Clausius Clapeyron equation from above for each component individually to find the respective vapor pressures. Then you can add them together if you want the total vapor pressure. However, you may be more interested in the vapor pressure of each component individually, since (from what I have heard) glycerol is more responsible for the "cloud" that forms, while propylene glycol is the better organic solvent. It would be helpful to know how much vapor of each component a given mixture would produce at a given temperature.

That's how you can predict vapor pressure from temperature. To relate this to the amount of heat you need, you will need to do another step. Temperature is a function of heat supplied and the specific heat of the substance:

$q = mC\Delta T \tag{2}$

where $q$ is the amount of heat transferred, $m$ is mass, and $C$ is the specific heat. You know the specific heat of the mixture, and the mass is an independent variable since you will control the composition and amount of the mixture. If you assume that volumes and heat capacities are additive, that mixing is fast, and that heat transfer is fast (at least locally), you can estimate the temperature (or rather, the maximum temperature) by connecting this to the power supplied by your device. Power in watts is the same as Joules/second. Therefore, you can multiply the power output by the time to get a total (maximum) amount of heat transfer (in Joules):

$q = P*t \tag{3}$

We can combine equations $(3)$ and $(2)$ and solve for the operating temperature:

$T=\frac{Pt}{mC}+T_i \tag{4}$

where $T_i$ is the starting temperature of the mixture (before you apply power).

Now we can substitute this back into $(1)$ to come up with an estimate of the vapor pressure of each component as a function of the mass of the liquid, the power applied, and the length of time that it is applied for:

$p_{vap} = (1 \space \mathrm{atm}) \exp([\frac{-\Delta H_{vap}}{R}](\frac{1}{\frac{Pt}{mC}+T_i}-\frac{1}{T_b})) \tag{5}$

The real vapor pressure will likely be lower, but I think this would get you to a good starting point.

A couple of things to remember:

$m$ refers to the mass of each component that is in close proximity to the heating element.
$Delta H_{vap}$ is in kJ/mol - make sure that the $R$ you use has the right units.
$T$ must be in Kelvin (both of them)
The units of $P$, $m$, and $C$ have to cancel out to give K.

If you want to get a (rough) upper estimate of the mass that will be vaporized, you could assume an ideal gas and use

$m=\frac{pVM}{RT}$

where $V$ is the volume of the vaporizer chamber, and $M$ is the molar mass of the component.


Why does the exchange interaction in Hartree-Fock theory lower the total energy?

Can anyone explain why the exchange contribution to the total energy is negative?

I find it misleading that exchange interaction is treated as something that changes total energy of the system. This lowering of energy is actually due to the Hartree-Fock scheme being in principle inexact, and is not really specific to indistinguishable particles.

Let me elaborate. Consider a hypothetical system of distinguishable particles with exactly the same masses, charges and spins (and all their other properties entering the Schrödinger's equation). Let these parameters be equal to those of an electron. Now, if you solve the Schrödinger's equation for this system in the electromagnetic field of an atomic nucleus, you'll get some (spinor-valued) wavefunctions $\Psi_n$, where $n$ enumerates the states. These wavefunctions won't in general be antisymmetric.

But the symmetry of the Hamiltonian, namely its invariance with respect to exchange of a pair of particles (since all their relevant properties are the same), tells us that there's a large degeneracy. If you linearly combine the degenerate states corresponding to the same energy so as to get symmetric $\Psi_k^{\mathrm s}$ and antisymmetric $\Psi_k^{\mathrm a}$ linear combinations, these will still be solutions of the Schrödinger's equation we started with.

If you throw away the symmetric solutions and only leave the antisymmetric ones, these will actually be the solutions for the case where our particles are indistinguishable. You see? We've just arrived at exchange interaction between indistinguishable particles, without any change in energy. What we got is just a possible increase in ground state energy due to losing a bunch of states (and thus renaming one of the formerly excited states to the ground one).

Now, why do we get lowering of states in inexact schemes when we introduce such antisymmetrization? That's directly related (at least for the antisymmetrized ground state) to the variational principle, which tells us that an approximation of a ground state will always have mean energy $E_{\text{approx}}$ (which is the Rayleigh quotient) obeying

$$E_{\text{approx}}\ge E_0,$$

where $E_0$ is the exact ground state energy.

Before introducing the Slater determinant, the trial wavefunction was smooth and generally nonzero at the loci of electron-electron collisions. But the exact (even non-symmetric) eigenfunction should have a local minimum (a cusp, see here for an example) or a (smooth) zero at such loci due to the Coulomb repulsion. After antisymmetrization, many of such loci of collisions will get zeros. This will make the approximation better, and the mean energy lower, closer to the exact eigenenergy.

If instead of Hartree-Fock method we used some other approximation that had begun with exact treatment of electron collisions, sacrificing something other, then such antisymmetrization might not lead to any improvement in estimation of eigenenergies (limit case being very precise numerical treatment yet disregarding Pauli principle, which is basically equivalent to an exact solution for distinguishable particles).


Why is density fitting not used to reduce four-center integrals further down to two-center integrals?

I try to answer your question, although I'm not completely sure if I understand what you are asking for.

You want a decomposition like for example:
$$
 (pq|rs) = \sum_P A^p_P A^q_P A^r_P A^s_P
$$
am I correct?

In the framework of density fitting this type of decomposition is not possible without further steps/tricks.
If we look at the two electron integrals
\begin{equation}
(pq|rs)=\int d\pmb{r}_1 \int d\pmb{r}_2 \phi_p(\pmb{r}_1) \phi_q(\pmb{r}_1) \frac{1}{\pmb{r}_{12}} \phi_r(\pmb{r}_2) \phi_s(\pmb{r}_2) \quad,
\end{equation}
we can consider this electron repulsion integral as repulsion between two generalized electron densities
\begin{equation}
(pq|rs)=\int d\pmb{r}_1 \int d\pmb{r}_2 \rho_{pq}(\pmb{r}_1)  \frac{1}{\pmb{r}_{12}} \rho_{rs}(\pmb{r}_2) \quad,
\end{equation}
with $\rho_{pq}(\pmb{r}) = \phi_p(\pmb{r}) \phi_q(\pmb{r})$ and $\rho_{rs}(\pmb{r}) = \phi_r(\pmb{r}) \phi_s(\pmb{r})$.
This density maybe approximated with the aid of an auxiliary basis set:
\begin{equation}
 \tilde{\rho}_{pq}(\pmb{r}) = \sum^{N_{aux}}_P d_P^{pq} \chi_P(\pmb{r})
\end{equation}
Hence the name "density fitting". The coefficients $d_P^{pq}$ are determined by minimizing the functional
\begin{equation}
  \Delta_{pq} = \int d\pmb{r}_1 \int d\pmb{r}_2 \frac{\left[ {\rho}_{pq}(\pmb{r_1}) - \tilde{\rho}_{pq}(\pmb{r_1}) \right] \left[ {\rho}_{rs}(\pmb{r_2}) - \tilde{\rho}_{rs}(\pmb{r_2}) \right]}{r_{12}}
\end{equation}
and leads according to Dunlap [1] to:
\begin{equation}
 d_Q^{pq} = \sum_{P} \left(pq|P\right) \left[\pmb{J}^{-1}\right]_{PQ} \quad,
\end{equation}
with
\begin{equation}
 \left(pq|P\right) = \int d\pmb{r}_1 \int d\pmb{r}_2 \phi_p(\pmb{r}_1) \phi_q(\pmb{r}_1) \frac{1}{\pmb{r}_{12}} \chi_P(\pmb{r}_2)
\end{equation}
\begin{equation}
 J_{PQ} = \int d\pmb{r}_1 \int d\pmb{r}_2 \chi_P(\pmb{r}_1) \frac{1}{\pmb{r}_{12}} \chi_Q(\pmb{r}_2)
\end{equation}
Then we can re-write the electron repulsion integral as
\begin{equation}
 (pq|rs)=\sum_{PQ} \left(pq|P\right) \left[\pmb{J}^{-1}\right]_{PQ} \left(Q|rs\right)
\end{equation}
The framework of density fitting does not give us the possibility to decompose $\left(pq|P\right)$ further,
but it is possible in combination with more general tensor decomposition techniques like the canonical decomposition (CP)[2],
in which we approximate a tensor in the format you were fishing for (as long as I understood you correctly)
\begin{equation}
 X_{i,j,...d} = \sum_{r}^{R} \omega_r  a_{ir}^{(1)} a_{jr}^{(2)} ... a_{dr}^{(d)} \quad,
\end{equation}
where $R$ is the rank of the tensor. Without losing generality, we will in the following absorb the weighting factor $\omega_r$
in one of the factors.

We can apply this general tensor decomposition to the density fitting representation of the two electron integral.
For the $[J^{-1}]_{PQ}$ matrix we can simply us a SVD
\begin{equation}
 [J^{-1}]_{PQ}  \approx \sum_{r_x}^{R_x} \bar{U}^{P}_{r_x} \bar{V}^{Q}_{r_x}  \quad,
\end{equation}
where the singular values $\sigma^{1/2}$ are absorbed into the vectors
For the 3-index integrals we do a CP decomposition
\begin{equation}
  \left(pq|P\right) \approx \sum_{r}^{R} A_{r}^{p} A_{r}^{q} A_{r}^{P}
\end{equation}
Thus the 4-index integrals can be written as
\begin{equation}
 \left(pq|rs\right)  = \sum_{PQ} \left( \sum_{r_1}^{R_1} A_{r_1}^{p} A_{r_1}^{q} A_{r_1}^{P} \right)
                                 \left( \sum_{r_x}^{R_x} \bar{U}^{P}_{r_x} \bar{V}^{Q}_{r_x} \right)
                                 \left( \sum_{r_2}^{R_2} A_{r_2}^{r} A_{r_2}^{s} A_{r_2}^{Q} \right) \quad.
\end{equation}
In some sense we are done since only two index quantities occur, but we can simplify the expressions a bit.
The sums can be rearranged to
\begin{equation}
  \left(pq|rs\right) = \sum_{r_1}^{R_1}\sum_{r_2}^{R_2}\sum_{r_x}^{R_x} \left( \sum_{P} A_{r_1}^{P}  \bar{U}_{r_x}^P  \right) \left( \sum_{Q} \bar{V}_{r_x}^Q   A_{r_2}^{Q} \right)
                         A_{r_1}^{p} A_{r_1}^{q} A_{r_2}^{r} A_{r_2}^{s} \quad,
\end{equation}
and a summation over $P$ and $Q$ yields
\begin{equation}
  \label{eqn:thc1}
  \left(pq|rs\right) = \sum_{r_1}^{R_1}\sum_{r_2}^{R_2}\sum_{r_x}^{R_x} P_{r_1 r_x} Q_{r_x r_2}
                         A_{r_1}^{p} A_{r_1}^{q} A_{r_2}^{r}  A_{r_2}^{s} \quad.
\end{equation}
Only $P_{r_1 r_x}$ and $Q_{r_x r_2}$ depend on $r_x$ and via summation we introduce a new coefficient matrix $B_{r_1 r_2}$, which
combines the expansions length $R_1$ and $R_2$.
\begin{equation}
  \left(pq|rs\right) = \sum_{r_1}^{R_1}\sum_{r_2}^{R_2} B_{r_1 r_2} A_{r_1}^{p} A_{r_1}^{q} A_{r_2}^{r} A_{r_2}^{s} \quad,
\end{equation}
which is quite similar to the THC format introduced by Martinez and co-workers.[3]
In one additional step we obtain a CP factorization for the full electron repulsion integral
\begin{equation}
  \left(pq|rs\right) = \sum_{P}^{R_1\cdot R_2} \bar{A}_{P}^{p}  \bar{A}_{P}^{q} \bar{A}_{P}^{r} \bar{A}_{P}^{s}
\end{equation}
So finally we arrived at a decomposition in only two index quantities and there are several groups[4,3,5], who looked
at this integral format(s). It is not of widespread use since so far the available algorithms for the required CP decomposition
are quite costly. In combination with density fitting the costs will scale with $\mathcal O(N^4)$, which is not that bad,
but the prefactor is so large that for a long range of molecular sizes a CCSD(T) calculation will finish quicker than the decomposition
of the two electron integral tensor.

The story is a bit different for the intermediate step, which can be identified as the THC format. Martinez and co-workers
are heavily working on using this format and enabled $\mathcal O(N^4)$ scaling implementations for MP2 and CCSD. However,
for MP2 it is to the best of my knowledge not really compatible with DF-MP2 implementations.

So after this long answer the short conclusion: Such integral formats exist in form of the THC format or CP decomposed integrals,
but further work is necessary to increase the efficiency and resolve some other problems.

References:

Dunlap, B. I., Connolly, J. W. D. & Sabin, J. R. On Some Approximations in Applications of X Alpha Theory. The Journal of Chemical Physics 71, 3396–3402 (1979).
Carroll, J. D. & Chang, J.-J.   Analysis of individual differences in multi-dimensional scaling via an n-way generalization of “eckartyoung” decomposition. Psychometrika 35, 283–319 (1970). URL http://dx.doi.org/10.1007/BF02310791
Hohenstein, E. G., Parrish, R. M. & Martínez, T. J. Tensorhypercontraction density fitting. I. quartic scaling second- and third-order Møller-plesset perturbation theory. The Journal of Chemical Physics 137 (2012). URL http://scitation.aip.org/content/aip/journal/jcp/137/4/10.1063/1.4732310.
Benedikt, U., Auer, A. A., Espig, M. & Hackbusch, W. Tensor decomposition in post-Hartree Fock methods. I. Two-electron integrals and MP2. The Journal of Chemical Physics 134 (2011). URL http://scitation.aip.org/content/aip/journal/jcp/134/5/10.1063/1.3514201.
Schmitz, G., Madsen, N. K. & Christiansen, O. Atomic-batched tensor decomposed two-electron repulsion integrals. The Journal of Chemical Physics 146, 134112 (2017). URL https://doi.org/10.1063/1.4979571. https://doi.org/10.1063/1.4979571


How can one determine if all reagents will react in the reaction?

That is correct, the two numbered reagent and solvent pairs refer to two reactions, that will be done after one another.
There is an intermediate product that can be isolated, or at least identified.


If A reacts with B (aq), does that mean it is soluble in it?

Summary, TL;DR:
No, A reacting with B does not mean that A is soluble in B. Solvation and chemical reaction are two different processes, although the distinction can in some cases seem less than perfectly clear. All of the example compounds you discussed are soluble or miscible with both water and hydrochloric acid. The IUPAC definition of a chemical reaction is "A process that results in the interconversion of chemical species", which is fundamentally different from solvation, and does not apply to any of the examples in the question.

As you suggest, solvation and chemical reactions are two fundamentally different concepts. For example, if compound A reacts with water to form a homogeneous liquid solution, that is a different processes than if compound A were to simply be dissolved by water.

Solvation is defined by the IUPAC Gold Book as follows:

  Any stabilizing interaction of a solute (or solute moiety) and the solvent or a similar interaction of solvent with groups of an insoluble material (i.e. the ionic groups of an ion-exchange resin). Such interactions generally involve electrostatic forces and van der Waals forces, as well as chemically more specific effects such as hydrogen bond formation.

Whereas the IUPAC Gold Book defines a chemical reaction as:

  A process that results in the interconversion of chemical species.

One aspect of solvation vs. reaction that may seem confusing is the solubility of ionic species. Even though $\ce{NaCl}$ becomes $\ce{Na+}$ and $\ce{Cl-}$ in an aqueous solution, this does not constitute a chemical reaction as defined above, and we say that $\ce{NaCl}$ is soluble in water. The same is true of your example of $\ce{H2SO4}$; even though it dissociates in water, it is not converted to a new chemical compound. One way to think of this is if you remove the solvent, the solute should typically resume to it's original form. From our previous example, if we evaporate the water from our aqueous solution of $\ce{Na+}$ and $\ce{Cl-}$, we just get the solid $\ce{NaCl}$ back.

Your examples of solubility in hydrochloric acid are a bit complex because that is a two-component system of water and $\ce{HCl}$. All of the compounds you discuss are water soluble and it doesn't matter if the $\ce{HCl}$ is there or not. One slight exception in your examples is ethylamine. Ethylamine itself is miscible with water, but many higher molecular weight amines are not water soluble, but are soluble in hydrochloric acid. In this case, the $\ce{H+}$ from $\ce{HCl}$ protonates the amine, making the hydrochloride salt. This is still an example of solubility however, as once you remove the solvent, you are left with the original compound, in this case the amine.


Molecular Imaging - Any surprises to be had?

I don't think this is going to put basic quantum chemistry upside down. Because chemistry is complicated, and these tools are "inaccurate". Let me illustrate : it was thanks to spectroscopy of atoms that people discovered the spin of the electron. It then explained a lot of fact that puzzled chemists. A molecule is something awfully complicated, that we can only solve with approximations and powerful computers. So to my mind, molecular observables are unlikely to contradict quantum theories. Now you need the LHC to split particles in order to test the theories.

This guy will tell you that there are surprises to be had about large molecules, the ones from natural sources (or plain coal) that are a pain in the ass to study by NMR,  or enzymes (like this one, studied with single molecule diffraction).

There will be always surprises. But if we could predict them, would they still be surprises ? (sorry that was too long for a 'comment')


Conflict Between Organic and Physical Chemists--explaining the intensities of IR absorptions

In considering the C-H stretch, let's consider that the C atom is attached to something else via an "immobile" bond(s).

For $sp^3$ hybridized the C would have 3 other bonds besides the immobile one.

For $sp^2$ hybridized the C would have 2 other bonds besides the immobile one.

For $sp^1$ hybridized the C would have 1 other bonds besides the immobile one.

For a non linear molecule there should be 3N-6 modes of vibration, but a linear on has 3N-5.

   Hybridization  # Atoms  Linear   Vibration Modes
       sp3            4       No         6
       sp2            3       No         3
       sp1            2       Yes        1

The more vibration nodes, the more microstates for the molecule in both its current state and its excited state. Thus having more microstates increases the probability that there are more transitions from the current state to some excited state.

For an analog on the microstate overlap, think about atomic spectra vs UV-VIS. Atomic spectra are line spectra, UV-VIS is band spectra.


Should cyaphide create similar chemical compounds as cyanide?

Methylidynephosphane or phosphaethyne is known and it has been achieved in 1961 by Thurman Gier. He passed phosphine gas at low pressure over an electric arc produced between two carbon electrodes. Condensation of the gaseous products in a –196 °C (–321 °F) trap revealed that the reaction had produced phosphaethyne along with acetylene and ethylene.
$$\ce{PH3 ->[electric arc][carbon electrodes] HCP + H2C=CH2 + HC#CH}$$
The ions $\ce{PCO-}$ and $\ce{PCS-}$ are also known. From the abstract of this paper1:

A range of λ3-phosphaalkynes $\ce{P⋮C−R}$ $\ce{(R = F, Cl, O-, OLi, OH,}$
$\ce{OSiMe3, S-}$, $\ce{SLi}$, $\ce{SH, Me,}$ $\ce{SiMe3, H)}$ has been geometry-optimized at the HF, MP2, and CCSD(T) level of theory. Vibrational frequencies as well as 31P and 13C NMR chemical shifts were calculated and compared to experimental values of the large variety of λ3-phosphaalkynes prepared over several years via differing synthetic routes. With a few obvious exceptions, a good agreement of calculated and experimental parameters was found. This also opens up the possibility to make predictions on
spectroscopic data of compounds not yet synthesized or fully
characterized, e.g., $\ce{PCOSiMe3}$ and $\ce{PCSH}$. The bonding situation
especially of the negatively charged species $\ce{P⋮C−O-}$ and $\ce{P⋮C−S-}$ was
studied via NBO analyses to determine the degree of phosphaalkyne and
heterocumulene contributions to the mesomeric anions.
$$\ce{^{-}X-C#P <-> X=C=P- (X=O,S)}$$

Cyaphogen ($\ce{(CP)2}$) has not been synthesized yet. It is considered a pseudohalogen. Structurally, it is formulated as $\ce{P#C-C#P}$.
Currently, no binary cyaphides are known but numerous organometallic cyaphide complexes are known. Some of them are listed in the Wikipedia article.
Reference

Theoretical Studies of NMR Chemical Shifts and Vibrational Frequencies in λ3-Phosphaalkynes P⋮C−R, Klaus Hübler and Peter Schwerdtfeger, Inorganic Chemistry 1999 38 (1), 157-164, DOI: 10.1021/ic9811291


Sulfates and persulfates as accelerants in black powder?

Sulphates and persulphates cannot be considered as analogues to chlorates and perchlorates.

Perchlorates are frequently used in professional pyrotechnics, being more reactive than nitrates.
Chlorates are too reactive for safe use.
Sulphates are useless as not acting as oxidant.
Persulphates have limited oxidative oxygen content, compared to nitrates or perchlorates, but as chlorates are too reactive for safe operations.

As a general rule, you must know what you are doing. More reactive mixtures may not only lead to faster wearing of the guns, but also to a gun blowing into your face.

Sulphur cannot be replaced by sulphates/persulphates, as it has the opposite role than oxidants. It would be like if you wanted to replace oxygen for breathing by carbon dioxide, as both contain oxygen.

Gun powder

  Chemical reaction

  Gunpowder does not burn as a single reaction, so the byproducts are not easily predicted. One study showed that it produced (in order of descending quantities) 55.91% solid products:
   potassium carbonate, potassium sulfate, potassium sulfide, sulfur, potassium nitrate, potassium thiocyanate, carbon, ammonium carbonate and 42.98% gaseous products: carbon dioxide, nitrogen, carbon monoxide, hydrogen sulfide, hydrogen, methane, 1.11% water.

  However, simplified equations have been cited.

  A simple, commonly cited, chemical equation for the combustion of black powder is
  $$\ce{2 KNO3 + S + 3 C -> +K2S  + N2 + 3 CO2}$$

  A balanced, but still simplified, equation is $$\ce{10 KNO3 + 3 S + 8 C -> 2 K2CO3 + 3 K2SO4 + 6  CO2 + 5N2}$$


What makes certain zeolites better at ion exchange?

Zeolites are selective ion exchangers.

What is usually required for a good zeolite is that it selectively absorbs the ion you care about rather than any old ions in the mixture.

That selectivity is a product of the shape of the holes in the 3D structure. Bigger holes will be less selective than smaller holes. If all the ions in the mix fit in the hole the zeolite won't reject the ones you don't want. So, ideally, the holes will be big enough for one ion but too small for another one. This allows selective exchange.

That is about it. It isn't about general equilibration of ions; it is about selective exchange of ions.


When one say that the electron goes from 1 orbit , say 2S, to another
  (higher) orbital, say 5P, it mean that the electron is excited to that
  orbit by some mean.

First off, there are no orbits. Electrons do not circle the nucleus. Rather they are in particular quantum states, or, in other words, they "occupy" particular orbitals. Note that when we say that an electron "occupies" an orbital, we just mean that it is in a particular quantum state. There is no spatial occupation in a classical sense out there: eletrons do not "sit" or "circle" anywhere, they are just in their quanatum states.

Consequently, when we say that electron "goes" from one orbital to another, what we actually mean is that the quantum state of the electrons changes. And again, an electron transition is not a mechanical motion, thus, electrons upon transitions do not "move" through nodes.


Why weren't the moles of oxygen calculated the same way as C?

You have done combustion analysis, so you have introduced a large amount of oxygen to the ascorbic acid sample.  Because of this, you can't know how much of the oxygen in the $\ce{CO2}$ and $\ce{H2O}$ you produced actually came from the acid sample and how much came from the atmosphere.

But you do know that the mass of your ascorbic acid is the mass of all the carbon plus the mass of all the hydrogen, which you calculated, plus the mass of oxygen.  The difference in mass between what you have calculated and what you started with is the mass of oxygen that was in the ascorbic acid.

If there were any other elements in your sample, such as sulphur or nitrogen, those amounts would also have to be calculated before finally working out how much mass you haven't accounted for which must have been oxygen.


How do I neutralize copper sulfate blue tint on a sofa?

There are two forms of copper sulfate you may encounter, the waterfree form ($\ce{CuSO4}$) which is colourless; and the other, the pentahydrate ($\ce{CuSO4 * 5 H2O}$) which is blue/greenish (ref).  You need to get rid of both forms.

In contrast to application on organic matter, the oxidative action by bleach, regardless if based on hypochlorite or the one that is chlorine free, won't remove the stain, yet may -- depending on the dye (mixture) deployed on your carpet attack the colour. On the other hand, though, both copper salts are fairly soluble in water.

Hence, in several cyles, you might attenuate the problem:
1) Dampen the area with some of hand warm water (e.g., with a spray bottle used to dampen your clothing prior to ironing) and subsequently 2)  rapidly dry it by pressing a fresh, clean cotton tissue against the tainted area. The white tissue is meant to act like a blotting paper, to catch both the water as well as the salt dissolved in it, too.  Hence you have to move on, to expose each time a fresh, clean, and dry section of the clean tissue towards the dampened area.

As for every cleaning of a fabric where you do not know yet how well "it works", try first with a small sector of your stained area.  If the fabric is cotton based, the polar groups in cotton actually may catch the copper salt fairly strong.

PS:  As you witnessed the accidental spill, you may be more sensibilized to see the stain, than a guest not familiar with it would be (especially as the dark fabric was not dyed uniformly).


Can a Birch Reduction be done using gaseous anhydrous ammonia?

A word of caution: pure ammonia, in either its gaseous or liquified form is extremely dangerous to handle, especially in the context of your 'hobby' where, in your own words you "don't have access to a ton of glassware or specialty equipment." Any ammonia that escapes from your reaction and isn't ventilated properly will cause severe burns to anything it comes into contact with, including completely trashing the lining of your nose and lungs etc. In an attempt to relieve your curiosity, a photograph of a birch reduction showing the characteristic blue is shown below:

That said, the question is an interesting one in the context of laboratory chemistry.

  Is there a way to use a gaseous form of ammonia for the birch reduction that doesn't generate sodium amide?

Yes and No. The most common way of performing these reactions on a small scale in a lab context is to essentially pass the gas over your reaction as illustrated below:

Practically, you're not actually making liquid ammonia and then transferring it to your reaction, but rather your passing it over your reaction flask which is fitted with a condenser which will cause in-situ liquefaction of the gaseous ammonia. Even a simple dry ice condenser is sufficient to do this, and the formation of the active reagent is easily observed by the appearance of the characteristic blue/green colour.  Any gas not liquified is then passed to the back of the fume cupboard, or, ideally, bubbled through a solution of something to neutralise it and hence stop it being released.


From your argument I assume the real question you want to ask is "Why HOOH is such a good oxidizer compared to molecular oxygen?".

From thermodynamical point of view, molecular oxygen is a very potent oxidizer. However when you are talking about reactivity, you should always take a look at kinetics, not only at thermodynamics.

One important part of the answer to your question is that not HOOH a good oxidizer, but O2 is a very bad oxidizer from kinetic point of view. Molecular oxygen is in the triplet state, therefore its reactions are often spin-forbidden. Especially the oxygen->superoxide->peroxide steps require often level crossings between different spin surfaces. One very important reason why transition metal complexes catalyse oxidation in biology is this: as they often have electrons to flip and large spin-orbit coupling, the level crossings became easy. There is literally tons of literature on this in bioinorganic chemistry.

Note that in flames and in high temperature burning that reaction is essentially a huge variety of radical reactions, so the mechanism is completely different from the oxidation reactions in aqueous solutions.

Molecular oxygen has also 4 electrons to take when turned to oxides. This sometimes require complicated mechanism where every single step is thermodynamically and kinetically is feasible, and there is always a oxidizable component around.

Also note, that HOOH itself need catalyst to be an effective catalyst, like in Fenton reactions and similar.


Why are these molecular orbitals invalid for hexatriene?

I read up a bit about Huckel theory. There's a pdf that explains this particularly well.

At one point, the following formula is derived at one point:

$$c_s\propto\sin\left(\frac{\pi k s}{n+1}\right)$$

which gives the coefficient (sans a positive normalisation constant) of the atomic orbital of the $s$th atom in the $k$th MO of a conjugated linear or monocyclic system on $n$ atoms.

A quick and dirty method has also been provided.

Using this, one can get a general idea of how the MO should look, and then one can get the exact +/- sequence by considering the number of nodes. Note that this is really the same thing as applying the $c_s$ formula given above, just in a graphical manner.


Case 1: Mixing equal volumes of two solutions with a pH difference of 1.
Initially we have:
\begin{array}{|c|c|c|}
\hline
&[\ce{H+}]& \mathrm{pH} &n_\ce{H+}\\ \hline
a& x & \log x & xV\\
b&\frac{x}{10} & -\log x + 1 & \frac{xV}{10}\\
\hline
\end{array}
\begin{align}
\overline{\mathrm{pH}}=\frac{\mathrm{pH}_a+\mathrm{pH}_b}{2}=\frac{-\log x-\log x+1}{2}=-\log x+0.5
\end{align}
After complete mixing, we obtain:
$$
({n_{\ce{H+}}})_\mathrm{mix}=({n_{\ce{H+}}})_{a}+({n_{\ce{H+}}})_{b}=xV+\frac{xV}{10}=\frac{11xV}{10}\\
[\ce{H+}]_\mathrm{mix}=\frac{1.1xV}{2V}=\frac{11x}{20}\\
\mathrm{pH}_\mathrm{mix}=-\log \frac{11x}{20}=-\log x - \log\frac{11}{20}=\overline{\mathrm{pH}}-0.5+0.2596=\overline{\mathrm{pH}}-0.2404
$$
Case 2: Mixing equal volumes of three solutions with consecutive pH differences of 1.
Before mixing:
\begin{array}{|c|c|c|}
\hline
&[\ce{H+}]& \mathrm{pH} &n_\ce{H+}\\ \hline
a& x & \log x & xV\\
b&\frac{x}{10} & -\log x + 1 & \frac{xV}{10}\\
c&\frac{x}{100} & -\log x + 2 & \frac{xV}{100}\\
\hline
\end{array}
\begin{align}
\overline{\mathrm{pH}}=\frac{\mathrm{pH}_a+\mathrm{pH}_b+\mathrm{pH}_c}{3}=\frac{-\log x-\log x+1-\log x+2}{3}=-\log x+1
\end{align}
After complete mixing, we obtain:
$$
({n_{\ce{H+}}})_\mathrm{mix}=({n_{\ce{H+}}})_{a}+({n_{\ce{H+}}})_{b}+({n_{\ce{H+}}})_{c}=\frac{111xV}{100}\\
[\ce{H+}]_\mathrm{mix}=\frac{1.11xV}{3V}=\frac{111x}{300}\\
\mathrm{pH}_\mathrm{mix}=-\log \frac{111x}{300}=-\log x - \log\frac{111}{300}=\overline{\mathrm{pH}}-1+0.4318=\overline{\mathrm{pH}}-0.5682
$$
(It seems you got the sign wrong on your last equation.)
Caveats:
Technically these calculations only work well if we assume that the activity of the solvated proton is the same as its concentration, which is a poor approximation below $\mathrm{pH}$ of 0 and above $\mathrm{pH}$ of 14.
You also get a slight deviation if you mix solutions that have $\mathrm{pH}$ too close to neutral, because the auto-dissociation of water affects the equilibrium values of $[\ce{H+}]$ and $[\ce{OH-}]$.
For example, if we mix two equal amounts of liquid, one being pure water($\mathrm{pH} = 7$) and the other being aqueous $\ce{HCl}$ solution with $[\ce{HCl}] = \pu{10^{-6}M}$. Considering the auto-dissociation constant for water as $K_\mathrm{w} = 10^{-14}$, the $\mathrm{pH}$ for the last solution is actually approximately $5.9957$ because water auto-dissociation already affects it slightly.
Neglecting that, we mix the liquids creating a solution with $[\ce{HCl}] = \pu{5\times 10^{-7} M}$. Directly using $[\ce{HCl}] = \pu{5\times 10^{-7}M} = [\ce{H+}]$ yields a $\mathrm{pH}$ of $6.3010$.
Using the trick, we get the pH for $[\ce{HCl}] = \pu{5.5\times 10^{-7}M}$ as $6.2596$. However, the real value for $[\ce{H+}]$, taking into account the shifting of the water auto-dissociation equilibrium, is $[\ce{H+}] = \pu{5.19258\times 10^{-7}M}$, such that the real $\mathrm{pH}$ is $6.28461$. A small deviation surely, but it's present.


'Aromatic alcohol' is not a term that appears anywhere in the IUPAC literature - while its meaning is evident, it doesn't make sense in the same way that 'aromatic heterocyclic' does so is probably best avoided.
What is defined by IUPAC is the definition of a phenol, and the definition of an aryl group. The definitions below come from an old IUPAC glossary of  'class names' which, although old, is still referenced in the most recent organic nomenclature guidelines and the most recent IUPAC Gold Book.

  phenols:
  Compounds having one or more hydroxy groups attached to a benzene or other arene ring

  aryl groups: Groups derived from arenes by removal of a hydrogen atom from a ring carbon atom.

  Glossary of Class Names of Organic Compounds and Reactive Intermediates Based on Structure.
  Pure App. Chem. 1995, 67, 1307

To actually answer your question, the answer is (II) - as Jerepierre alluded to, the NCERT books / guides are renowned for being less that accurate, and a quick search around chem.SE will reveal just how many errors there are! In this case, its hard to tell if the error is just due to typesetting (someone typed III instead of II), or it it lies with the ambiguity of the term 'aromatic alcohol' (is it referring to phenols, or is it referring to a compound which contains both an aryl ring and a hydroxyl group).

I think its fair to assume that the question is asking you to identify the phenols, in which case the answer is (II) is because A and D are the only structures in which the hydroxyl group is directly attached to a benzene or other arene ring, which is our working definition of a phenol / aromatic alcohol.


Does IUPAC nomenclature have the ability to name all organic compounds?

Definitely not.

You got yourself in trouble specifying all organic compounds, because there is a truly immense, mind-boggling number of possible compounds. No one even knows how to accurately determine such a quantity. A very rough estimate, making some incredible simplifications such as the use of only carbon, hydrogen, oxygen, nitrogen and sulfur atoms, is that there are some $10^{63}$ "reasonable" unique structures for compounds with a molecular weight below $\rm{500\ g\  mol^{-1}}$. Thanks to combinatorics, chemical space is enormous. We (or all sapient species in our observable volume, for that matter) will never come close to scratching its surface.

Furthermore, IUPAC nomenclature is largely created a posteriori. That is, though there are many rules trying to cover as many bases as possible (in the process becoming quite unwieldy at times), eventually some unexpected compound with unusual connectivity is discovered and becomes of wide interest. Thus standardising its nomenclature and that for closely related structures becomes fundamental to allow communication between scientists. A recent example of this occurred with fullerenes, which quickly jumped to prominence after 1985. IUPAC just had to create an entire new section of nomenclature for this class of compounds, which is not at all uncommon.

The closest thing to an absolute method of describing a compound's structure is to have a table of positional data (XYZ coordinates) giving the relative positions between the atoms determined from X-ray/neutron diffraction. Any attempt at simplifying this data will be lossy, whether the structures are drawn (not too lossy) or named (very lossy).

The structures you show have comparatively simple IUPAC names, in fact. Heme is a type of porphyrin, which is a widely occurring framework in biomolecules. The central framework can have its positions numbered and the substituent in each one read off separately. Regarding Zantac, the Wikipedia page for the compound states its IUPAC name in the "Identifier" section in the box at the right, namely N-(2-[(5-[(dimethylamino)methyl]furan-2-yl)methylthio]ethyl)-N'-methyl-2-nitroethene-1,1-diamine.

As some interesting examples of the relations between available nomenclature and chemical space, consider the following:

1,1,1,2,2,2-Hexaphenylethane : A molecule with a simple structure and a simple IUPAC name which likely cannot exist in reasonable conditions.
Maitotoxin : an awe-inspiring biomolecule with a rather large structure but containing fairly simple connectivity between atoms, whose IUPAC exists but is quite complex - disodium (2S,3R,4R,4aS,5aR,6aS,7aS,8R,9R,10R,11aR,12R,12aR,13aS,14aR)-10-[(2R,3R,4R,4aS,6S,7R,8aS)-6-[(1R,3R)-4-[(2S,3R,4R,4aS,6R,7R,8aS)-6-[(1R,3S,5R,7S,9R,10R,12R,13S,14S,16R,19S,21R,23S,25S,28R,30S)-25-[(1S,3R,5S,7R,9S,11S,14R,16S,18R,20S,21Z,24R,26S,28R,30S,32R,34R,35R,37S,39R,42S,44R)-11-[(1S,2R,4R,5S)-1,2-dihydroxy-4,5-dimethyloct-7-en-1-yl]-35-hydroxy-14,16,18,32,34,39,42,44-octamethyl-2,6,10,15,19,25,29,33,38,43-decaoxadecacyclo[22.21.0.0³,²⁰.0⁵,¹⁸.0⁷,¹⁶.0⁹,¹⁴.0²⁶,⁴⁴.0²⁸,⁴².0³⁰,³⁹.0³²,³⁷]pentatetracont-21-en-34-yl]-9,13-dihydroxy-3,7,14,19,30-pentamethyl-2,6,11,15,20,24,29-heptaoxaheptacyclo[17.12.0.0³,¹⁶.0⁵,¹⁴.0⁷,¹².0²¹,³⁰.0²³,²⁸]hentriacontan-10-yl]-3,4,7-trihydroxy-octahydropyrano[3,2-b]pyran-2-yl]-1,3-dihydroxybutyl]-3,4,7-trihydroxy-octahydropyrano[3,2-b]pyran-2-yl]-2-[(2S,3R)-2,3-dihydroxy-3-[(1S,3R,5S,6S,7R,8S,10R,11R,13S,15R,17S,19R,21R,22S,24S,25S,26R)-6,7,11,21,25-pentahydroxy-13,17-dimethyl-8-[(2R,3R,4R,7S,8R,9R,11R,13E)-3,8,11,15-tetrahydroxy-4,9,13-trimethyl-12-methylidene-7-(sulfonatooxy)pentadec-13-en-2-yl]-4,9,14,18,23,27-hexaoxahexacyclo[13.12.0.0³,¹³.0⁵,¹⁰.0¹⁷,²⁶.0¹⁹,²⁴]heptacosan-22-yl]propyl]-4,8,9,12-tetrahydroxy-hexadecahydro-2H-1,5,7,11,13-pentaoxapentacen-3-yl sulfate
This hydrocarbon : A molecule which likely exists, with a seemingly very simple structure, but with slightly quirky connectivity which makes naming it a challenge. Add a few more bridges and I'm sure you can break any existent nomenclature rules.


Why is an electropositive metal like Zinc capable of plating out?

It's too simple to think of this situation in terms of electronegativity and it's certainly not limited to zinc. If you look at a table of standard electrode potentials, you can see that the reduction of many metal ions like $\ce{Zn^2+,Pb^2+,Cd^2+,}\textrm{ etc.}$ have standard potentials below that of the reduction of hydronium (0 V) and should be much less favourable than making hydrogen gas if this were the only consideration.

However, in real electrochemical cells, it generally takes a certain amount of extra voltage to get a certain reaction to occur, known as the overpotential. Like all chemical reactions, redox reactions have a certain activation energy separate from the energy change between the reactants and products, e.g. elemental gold is far lower in energy than a gold solution, but a gold solution will not immediately reduce spontaneously because there is an activation barrier to the process, an extra potential must be applied to make it happen at a reasonable rate. Similarly, though the standard potential for reducing hydronium is 0 V, in practice it may not be observed until much more negative applied voltages.

This overpotential for hydrogen evolution is strongly dependant on the type of electrode used. For the reaction to occur, hydronium ion must adsorb to the surface. In electrodes like platinum, this interaction is stabilized so the overpotential is relatively small and it's thus quite difficult to deposit metals like zinc onto platinum. For this reason, other electrodes must be used for working with aqueous solutions at very negative potentials. Glassy carbon, mercury, and bismuth electrodes are often used for this as hydronium is not readily adsorbed to their surfaces, giving large overpotentials for hydrogen evolution. These types of electrodes may be useful down to -1 V or so and more exotic materials like boron-doped diamond electrodes, even more negative.

However, no electrode can prevent hydrogen evolution if the potential is negative enough. Zinc is about the most easily oxidized metal that is practical to electroplate from aqueous solution. Some metals are so easily oxidized that they are basically impossible to plate in aqueous solutions which is why one generally cannot electroplate from aqueous aluminum solutions or solutions of alkali metals (things like sodium and potassium are so easily oxidized that they react violently with water to form hydrogen gas).


Why does chlorine replace a hydrogen in propene instead of adding to the double bond?

It's not that the chlorine radical doesn't attack the $\pi$-bond, it just doesn't do it as fast. Alkyl radicals are much less stable than allyl radicals, so hydrogen abstraction will occur much faster than radical addition ($k_1 >> k_2$).


Why does benzoyl peroxide homolytically cleave to give phenyl radical

The mistake you're making (which is an easy one to make) is that you're neglecting to consider the carbon dioxide product. Carbon dioxide is a stable molecule, and its release also increases the entropy of the system.

So even if the phenyl radical is less stable than the benzoyl radical intermediate, the combination of phenyl radical + $\ce{CO2}$ can still have an overall lower free energy than the benzoyl radical.


Why it is hard to rinse off soap from your body using rain water

Rain water is pretty much devoid of the minerals such as calcium salts that are often found in so-called "hard" tap water taken from underground sources. Calcium ions in tap water react with the fatty acid anions in soap to create insoluble soap scums, which (while they may remain on your skin) do not have the slippery feel of soap. Without the calcium ions present, the soap remains in soluble form (sodium salts of fatty acids) and I would propose that the slippery feel of soap on your skin comes from the traces of it that remain even after considerable rinsing.


Calculate the third and fourth nearest neighbours in bcc

With $\sqrt3\over2$ being that close to $1$, BCC packing is better not looked at in terms of coordination spheres. But if you insist...

Say you are sitting in the center of a cell. Then:

Your first neighbours are at the corners of the same cell.
Second neighbours are at the centers of the nearest adjacent cells.
Third neighbours: centers of the next adjacent cells (those which share two corners with your cell).
Fourth neighbours: far corners of the nearest adjacent cells.

On a side note, there are more than just 8 of the latter.


Reactant energy distribution and spontaneity of reactions

Yes; you are on the right path in your second paragraph, but distinguish between activation energy and free energy as they are not connected. I'm sorry but I think that your third paragraph is not all correct as it stands. If the free energy change is positive the reaction is not spontaneous by definition. Perhaps you mean, if the overall enthalpy change is positive can the free energy now be negative if the overall entropy change is also positive? The answer is yes.

The thermodynamic quantities only give us information about initial and final states at equilibrium. The activation energy is obtained from kinetic, that is rate constant measurements usually vs, temperature and not from thermodynamic quantities. In the vast majority of reactions the activation energy is greater than thermal energy (approx kBT, where kB is Boltzmann's constant). These reactions are generally slow because, by the decreasing exponential nature of the Boltzmann distribution with energy, not many reactants have enough energy to react in any given time period, say 1 sec. (This is lucky otherwise we would all have reacted eons ago!) In some special types of unimolecular reactions, e.g. electron transfer, bond dissociation & isomerization, with zero or very small activation energy then clearly these will be very fast possibly with rate constants $ >10^{12}\ /sec $.
In the case where the products are of higher energy than reactants, reaction still occurs but the equilibrium lies on the side of the reactants. Again by the Boltzmann distribution argument products react faster to reactants than the other way round.


What is the difference between an amorphous solid and glass?

While a glass is generally considered to be a supercooled, configurationally frozen liquid, not all amorphous solids are glasses. For example, amorphous silicon is a four-fold coordinated semiconducting solid, much like crystal silicon.  Liquid silicon is 8-12 fold coordinated with metallic bonding. Amorphous silicon has been shown to display a first-order phase transition to both the crystal and the liquid.

So, all glasses are amorphous solids, but not all amorphous solids are glasses.


I would recommend using the functional PySCF cubegen script as a starting point for your own script. (https://sunqm.github.io/pyscf/_modules/pyscf/tools/cubegen.html)

If we have a grid defined with nx ny and nz points along the three axes, generating the cube file is just a nested loop over z, y, and x. (In that order) You should end up with nz lines of output each with nx*ny entries.


Why is mer-AlQ3 more stable than fac-Alq3?

The stability of meridional isomer over facial isomer can be deduced by considering the interactions between atoms of the ligand 8-hydroxyquinoline.
 In considering the interactions if we take the repulsion between the electron rich atoms dominated in 90 degrees and at 180 degrees these interactions become negligible, we will find that in meridional isomer, there are dominating
i) 2 N-N interactions
ii) 2 O-O interactions
iii) 8 N-O interactions.
 But if we look into the facial isomer of tris(8-hydroxyquinolinato)aluminium we will find there are dominating
i) 3 N-N interactions
ii) 3 O-O interactions
iii) 6 N-O interactions.
 So in facial isomer two N-O interactions of mer isomer has turned into 1 N-N interaction and one O-O interaction. Due to high charge density on small sized O atoms O-O repulsions are much much greater than N-O interactions and thus the electronic repulsions are much more greater in fac isomer which decreases its stability than meridional isomer


You can synthesize your desired compound from Benzene in the following way.

First, you can synthesize m-dinitrobenzene, and perform selective reduction to have 3-nitroaniline. Now, you can acylate the ring by first protecting the amine group. After acylation you can convert the amide to amine by acidic hydrolysis, diazotize the amine group and perform Sandmeyer's reaction. Then just reduce the remaining nitro group to amine to have your product.

The synthetic scheme is given below,


What is the mechanism behind the formation of phosgene from tetrachloroethylene

Given that the reaction occurrs at high temperatures, the reactive species is likely singlet oxygen, which reacts much like an alkene$^{[1]}$. Electron rich alkenes without allylic hydrogen, like tetrachloroethylene, react with singlet oxygen to form 1,2-dioxetanes$^{[2]}$.

A $[_\pi2_s + _\pi2_a]$ cycloaddition, though a poor reaction geometry for two alkenes, is made more accessible by the small, linear oxygen molecule. This, as well as peroxides' predisposition to nonplanar conformations, makes the transition state hardly any more strained than the dioxetane itself. Once formed, the dioxetane quickly decomposes to give two molecules of phosgene. To give you an idea of their stability:

  A small sample of the [dioxetane] product, on warming to room temperature, melted and exploded.$^{[3]}$

$^{[1]}$ Wikipedia, Singlet Oxygen, Reactions, Organic Chemistry
$^{[2]}$ Singlet Oxygen: Applications in Biosciences and Nanosciences Volume 1
$^{[3]}$ Stereospecific Formation of 1,2-Dioxetanes from cis- and trans Diethoxyethylenes by Singlet Oxygen


from what I understand a higher pi HOMO is synonymous with more
  nucleophilic.

Yes, that's generally true.  If you have a substituent on a double bond that increases the electron density in the double bond, then the double bond (usually one carbon more than the other) becomes more nucleophilic and the energy of the HOMO and LUMO are raised.  Conversely, electron withdrawing groups attached to a double bond withdraw electron density, making the double bond less nucleophilic and lowering the energies of the HOMO and LUMO.

Comparing cyclohexanone enol (cyclohex-1-en-1-ol) to cyclohex-2-en-1-ol, you can draw a resonance structure for the former that increases electron density in the double bond (especially the $\beta$ carbon).  This electron donating resonance effect overwhelms the electron withdrawing inductive effect of the hydroxyl group.  Therefore, the double bond in the cyclohex-1-en-1-ol will be more nucleophilic and its HOMO and LUMO will be higher in energy than those for cyclohex-2-en-1-ol.


Finding Gibbs energy at different temperatures

You might know that Gibbs free energy is given as:

$$\Delta G=\Delta H-T\Delta S$$

This implies that:

$$\Delta G^\circ=\Delta H^\circ - T \Delta S^\circ$$

Which can be rewritten as:

$$\Delta S^\circ = \frac{\Delta H^\circ - \Delta G^\circ}{T}$$

Often it is a good approximation to assume that $\Delta H$ and $\Delta S$ are temperature independent. Now that you have found the entropy you can calculate Gibbs free energy at different temperatures using the first equation.

If you want to find the equilibrium constants, you might have seen this equation before:

$$\Delta G=-RT\ln\left( K_\text{eq} \right)$$


Good to know that someone is working in DSP and chromatography but there are many software out there like PeakFit, why a new one?

Your first test could be peak detection as function of resolution as well their height ratios. Find out under what thresholds your peak detection fails. This is the most important thing.

Peak shapes: Gaussian, Exponentially modified Gaussians are very very common. Tailing is far more common than fronting. There is a very nice equation that covers all three shapes. Search BI-EMG in Google Scholar and you would find it. "BI" here is bi-directional."

Sharp & wide peaks: It is a relative term. You have to check what maximum efficiencies are achievable on instruments on GC, SFC and LC.

Different levels of skewness: Asymmetry of 2 is usually the maximum allowed number. Asymmetry of 1 is highly desirable. In practice it is possible to obtain 1.0 to 1.1.

Baseline drifts: Look more like linear ramps. You have to familiarize yourself with gradient elution.

Peak resolution: Start from peak resolution of > 0.5

Tailing and fronting: There is a very nice single equation that covers all three. See point no. 1.

Smoothing: I have written a couple of papers on smoothing. You can play with wavelets (most complex), Hamming filter, Savitsky Golay, simple central moving average, and Gaussian weighted moving average is the most common. I cannot disclose which instruments use which method. You can assume Gaussian noise rather than uniform. To the best of my knowledge wavelets or FFT filtering is not used in commercial GC, HPLC etc.


What are the products of the dissociation of sodium bicarbonate in water? What is the relative pH of the solution?

It's not as intuitive as it seems and your questions are all legitimate questions, but sometimes even good arguments can't be used as evidence in chemistry.

  $\ce{NaHCO3 -> Na+ + HCO3-}$, but that one doesn't involve water at all

Look at this reaction:
$$\ce{NaCl <=>Na+ +Cl-}$$

Even this reaction doesn't "involve" water in the schematics but is right, we assume that is a dissociation of a salt in water.

  $\ce{NaHCO3 + H2O -> Na+ + OH- + H2CO3}$, but doesn't $\ce{H2CO3}$ decompose into $\ce{H2O + CO2}$?

Yes, but in fact this is an equilibrium:
$$\ce{H2CO3 <=> H2O + CO2}$$

And then you have to take into account another important equilibrium many time ignored:
$$\ce{CO2 (aq) <=> CO2 (g)}$$
So the amount of $\ce{CO2}$ present in the solution depends on by the type of system and the pressure exerted. Like a Coca-Cola bottle: if the system is closed, the $\ce{CO2}$ remains dissolved otherwise the equilibrium shift to the right and the $\ce{CO2}$ "escapes" until the rate between the $\ce{CO2}$ molecule dissolved and the $\ce{CO2}$ molecule escaped forms the solution is the same.

  $\ce{NaHCO3 + H2O -> Na+ + OH- + H2O + CO2}$, but that has water on both sides of the equation.

Okay, it is not a net ionic equation but the stoichiometry is still right. Maybe water is used to indicate that we are talking about a reaction that takes place in water.

Which reaction is not right?

All the reactions here are right, I think maybe they put some formal inaccuracies to help you, but in fact they are all part of series of equilibria that normally occur and can be described as:
$$\begin{multline}
\ce{NaHCO3 + H2O <=>[A] Na+ + HCO3- + H2O <=>[B] Na+ + H2CO3 + OH-} \\
\ce{ <=>[C] Na+ + OH- + H2O + (CO2(aq)<=>[C_2]CO2 ^ (g))}\end{multline}$$

The question should be more specific asking for the ultimate products of the reaction because all these are products of the dissociation of sodium hydrogen carbonate and because are equilibria is probable that all the species are present in different amounts when the reaction reach the equilibrium. However, the ultimate products are described by the third answer. And this equation can give you a hint about the acidity of the solution. When $\ce{CO2}$ escape from the solution (equilibrium $\mathrm{C_2}$ goes to the right) the concentration of the manly acid species goes down but the hydroxides from the equilibrium $\mathrm{B}$ are still present so at the end the solution in an open system with an atmosphere with a low partial pressure of $\ce{CO2}$ will have a higher pH, will become mildly alkaline.


Converting fractional coordinates into cartesian coordinates for crystallography

You have to use a matrix to convert. This is derived in most textbooks on crystallography, such as McKie & McKie 'Essentials of Crystallography'.
The matrix is
$$ M=\begin{bmatrix} a & 0 & \\b\cos(\gamma)& b\sin(\gamma) & 0\\c\cos(\beta) & cn_2 & c\sqrt{\sin^2(\beta)-n_2^2} \end{bmatrix}$$
where $$n_2=\frac{\cos(\alpha)-\cos(\gamma)\cos(\beta)}{\sin(\gamma)}$$
and $a,b,c $ are the unit cell dimensions and $\alpha,\beta,\gamma$ the angles in radians. The determinant of $M$ is the cell volume.
If $V_1$ is the vector of fractional cell coordinates $x/a,y/b,z/c$ etc. the matrix multiplication calculation $\hat d_{12}=(V_1-V_2)\cdot M $ and then $L_{12}=\sqrt{ \hat d_{12}\cdot \hat d_{12}}$ gives the bond length atom 1 to 2. The angle $\theta$ between two bonds 12 and 23 is given by a dot product $\cos(\theta)=\hat d_{12}\cdot \hat d_{23}/(L_{12}L_{23})$
An example calculation in python is shown below for a triclinic crystal. The @ is matrix multiplication.
import numpy as np
a = 7.55                # cell parameters
b = 4.99
c = 12.50
alpha = 122.5*np.pi/180
beta  = (95+18/60)*np.pi/180
gama  = (118+54/60)*np.pi/180

V1 = np.array( [-0.2812 , -0.0628 , 0.1928 ] ) # x/a, y/a, z/a
V2 = np.array( [-0.2308 , -0.0972 , 0.2931 ] )
V3 = np.array( [-0.3639 , -0.1913 , 0.3521 ] )

n2 = (np.cos(alpha)-np.cos(gama)*np.cos(beta))/np.sin(gama)
M  = np.array([[a,0,0],[b*np.cos(gama),b*np.sin(gama),0],
     [c*np.cos(beta),c*n2,c*np.sqrt(np.sin(beta)**2-n2**2)]])

dcm1 = (V1-V2) @ M             # row x matrix
L12  =  np.sqrt(dcm1 @ dcm1)   # sqrt(dot product)
L12


In nuclear chemistry, how does a neutron split to form a proton and an electron?

One neutron has changed into a proton, that's what has happened. Our convention in chemistry is to identify nuclear species by the proton number (or the "atomic number") and that is why a new nuclide with an increased atomic number is formed. Both the $\beta^+$ and $\beta^-$ decay follow identical paths: in $\beta^-$ decay, a neutron changes into a proton (thus giving no change in mass number and increase in atomic number), whereas in $\beta^+$ decay, a proton decays into a neutron giving no change in the mass number and a decrease in the atomic number.


You need 15 g HCl and 6 g NaI per 100 g of the solution. Treating 31.45% (I don't know much about practice) as exact, there is 15 g HCl in 15 g / 31.45% = 47.69 g of your muriatic acid. So you need 6 g of sodium iodide, 47.69 g of your muriatic acid and the rest, i.e. 46.31 g, of pure water per 100 g of the solution.


What is the difference of molar volume between dissolved 35Cl- and 37Cl-?

I don't have an exact answer for chloride specifically, but consider the theoretical equations in the following two articles:

Isotope Effects on Molar Volume and Surface Tension: Simple Theoretical Model
and Experimental Data for Hydrocarbons J. Chem. Phys. 44, 457 (1966)

  We assume the molecules are held together in a
  condensed phase by van der Waals "bonds" or covalent
  bonds represented by 6-12 or Morse potential-energy
  functions, V(x). For monatomic molecules x is taken
  to represent the distance between molecular centers.

and

Lattice Constants of Separated Lithium Isotopes J. Chem. Phys. 27, 1030 (1957)

which finds atoms of Lithium-6 $4 \times 10^{-4}$ times more separated than those of Lithium-7 and also cited to a reference which considers lithium-6 versus lithium-7 fluoride-19.


I would say no, it is only a hydrogen bond, not a halogen bond.

To be a halogen bond, the halogen atom must accept electron density from the other member of the bond.  If the other member of the bond is a hydrogen atom bonded to a more electronegative element, I don't see have the halogen atom could be an acceptor of electron density.

If you can get access see "Halogen Versus Hydrogen" Science Vol. 321 no. 5891 pp. 918-919.  Footnote 7 there explains: "A halogen bonding donor is a species that contains an electrophilic halogen that can become a member of a halogen bond. In the literature on halogen bonding, the reader should be alert to how the words 'donor' and 'acceptor' are used. In a complex RX---B, RX is the halogen bond donor but the electron acceptor (Lewis acid); B is the electron donor and halogen bond acceptor (Lewis base)."


Mixing two containers of each water and ethanol to varying degrees of purity

$$\begin{array}{c|c|c}
\ & \text{water} & \text{ethanol}\\ \hline
\text{Volume in concern, in L} & 10.00 & 5.00 \\ \hline
\text{Density in kg/L} & 1.000 & 0.789 \\ \hline
\text{Molar mass in g/mol} & 18.015 & 46.069\\ \hline
\text{mass in kg} & 10.000 & 3.945\\ \hline
\text{moles} & 555 & 85.63\\ \hline
\text{mole fraction} & \frac{555}{555+85.63} = 0.866 & \frac{85.6}{555+85.63} = 0.134
\end{array}$$

From the given graph, it looks like we have an excess volume of $\pu{-0.75 mL/mol}$ (at $\chi_\text{EtOH}=0.134$). Since we have total amount of water and ethanol, $555+85.63=\pu{640.63 mol}$, we can calculate the excess volume of the container:

$$\pu{-0.75 mL/mol} \times \pu{640.63 mol} \times \pu{1 \times 10^{-3} L/mL}= \pu{-0.480 L}$$
Thus the actual volume is not $\pu{15 L}$, but $\pu{(10.00+5.00-0.480) L}=\pu{14.52 L}$

Suppose, we transfer $V_1$ of water/ethanol mixture back to the container-2 containing the remaining of $\pu{5.00 L}$ of ethanol. That $V_1$ contains:
$$\frac{\pu{555 mol}\text{ water}}{\pu{14.52 L}}\times V_1 \ \pu{L} = 38.22V_1 \ \pu{mol}\text{ water}$$
and,
$$\frac{\pu{85.63 mol}\text{ EtOH}}{\pu{14.52 L}}\times V_1 \ \pu{L} = 5.90V_1 \ \pu{mol}\text{ EtOH}$$

Thus container-2 has $38.22V_1 \ \pu{mol}$ of water and $\left(5.90V_1 + \pu{85.63 mol}\right)$ of ethanol after the addition of $V_1 \ \pu{L}$ of water/ethanol mixture from container-1.

Suppose we have made both containers have the same purity (by $\%(w/w)$) by this action. Let's calculate the masses of water and ethanol ($m_{w2}$ and $m_{et2}$, respectively) in container-2:

$$m_{w2} = 38.22V_1 \ \pu{mol}\text{ water}\times \frac{\pu{18.015 g}\text{ water}}{\pu{1.0 mol}\text{ water}}= 688.5 V_1 \ \pu{g}\text{ water}$$

Similarly,

$$m_{et2} = \left(5.90V_1 + 85.63\right) \ \pu{mol}\text{ EtOH}\times \frac{\pu{46.069 g}\text{ EtOH}}{\pu{1.0 mol}\text{ EtOH}}= \left(271.8V_1 + 3945\right) \ \pu{g} \text{EtOH}$$

Now, we'll calculate the masses of water and ethanol ($m_{w1}$ and $m_{et1}$, respectively) in container-1. Amounts of water and ethanol in this container is:

$$\frac{\pu{555 mol}\text{ water}}{\pu{14.52 L}}\times (14.52 - V_1) \ \pu{L} = \left(555 - \frac{555V_1}{14.52}\right) \ \pu{mol}\text{ water}$$
and,
$$\frac{\pu{85.63 mol}\text{ EtOH}}{\pu{14.52 L}}\times (14.52 - V_1) \ \pu{L} = \left(85.63 - \frac{85.63V_1}{14.52}\right)  \ \pu{mol}\text{ EtOH}$$

Thus,

$$m_{w1} = \left(555 - \frac{555V_1}{14.52}\right) \ \pu{mol}\text{ water}\times \frac{\pu{18.015 g}\text{ water}}{\pu{1.0 mol}\text{ water}}= \left(555 - \frac{555V_1}{14.52}\right)\times   \pu{18.015 g}\text{ water}$$

Similarly,

$$m_{et1} = \left(85.63 - \frac{85.63V_1}{14.52}\right)  \ \pu{mol}\text{ EtOH}\times \frac{\pu{46.069 g}\text{ EtOH}}{\pu{1.0 mol}\text{ EtOH}}= \left(85.63 - \frac{85.63V_1}{14.52}\right)\times   \pu{46.069 g}\text{ EtOH}$$

Theoretically, the condition for same purity by $w/w$ is:

$$\frac{m_{w1}}{m_{w1}+m_{et1}} = \frac{m_{w2}}{m_{w2}+m_{et2}} $$

or

$$\frac{m_{et1}}{m_{w1}+m_{et1}} = \frac{m_{et2}}{m_{w2}+m_{et2}} $$


Calculating equilibrium concentrations of silver and halide ions

Both chloride and bromide ions are present in 10-fold excess over silver ions. That means that the chloride and bromide concentration in solution will not drop by much (they will remain major species).

The solubility product of AgCl is 200-times higher than that of AgBr. If both AgCl and AgBr precipitate, the chloride solution would be 200-times higher than that of bromide. That is impossible because so little silver ions are available, and bromide and chloride start at equal concentrations. Thus, no silver chloride will precipitate (if this explanation is not convincing, suspend your disbelief until we can check the equilibrium concentrations).

Most of the silver ions will precipitate. As a first approximation, let's say the silver ion drops from 5 mM to zero, so the bromide concentration will drop from 50 mM to 45 mM. Of course, this can't be the equilibrium, so we can dissolve some of the AgBr again.

$$[\ce{Ag+}] = \frac{K}{[\ce{Br-}]} \approx \frac{\pu{1e-11}}{\pu{45e-3}} = \pu{2.2e-10}$$

So the concentration of silver ions is roughly $\pu{2.2e-10 M}$, that of bromide ions roughly $\pu{45 mM}$ and that of chloride ions $\pu{50 mM}$, with no silver chloride precipitating.

  At equilibrium condition, both AgCl and AgBr would have precipitated out.

Now that we have solved the exercise, you can verify that AgCl will not precipitate. The reason is that the silver ion concentration is extremely small due to the presence of 45 mM bromide ions. In fact, if NaCl and AgNO3 had been mixed first, AgCl would have precipitated. However, adding NaBr would have dissolved it again eventually, leading to the same final state.

  However, this doesn't match the answer key. Where am I wrong?

As mentioned in the comments, the set of final concentrations does not make sense from a stoichiometry (charge balance) perspective, and also contradicts the claim that AgCl is saturated. If you assume that both AgBr and AgCl have formed and attempt to solve the problem by brute force, the solutions to the two equations with two unknowns will only yield solutions that are not attainable (negative concentrations, or concentrations higher possible with the available starting materials).


How do I select a sub-freezing temperature for my experiment that will allow the formation of frost, but not supercooled liquid water (rime)

Surprisingly, at a glance, I had to run my experiments with the glass substrate at higher temperatures in order to ensure ice would form and not liquid water.  When you think in terms of the difference between the saturation vapor pressure of liquid water and that of ice at the same temperature, this solution is obvious.  The temperature has to be above the dew point but remain below the frost point, which is always higher than the dew point at temperatures below 0 oC.  The reason for this is that at any given temperature, liquid water has a higher vapor pressure than ice.

As a real-world example of the ice-growth mechanisms observed in my trace-gas uptake experiments, what was a problem for my experimental conditions actually has important implications for ice-particle growth in clouds, in which context it’s called the “Wegener-Bergeron-Findeisen Process”.  Here, when there is a mixture of ice and supercooled water droplets, water vapor condenses onto the ice particles so that the humidity with respect to liquid water decreases enough for the liquid water to further evaporate.  Thus there is a net transfer of water mass from liquid droplets to solid particles and the solid particles may then grow large enough to precipitate (as in fall to the ground in this context).


Corey -House synthesis vs Wurtz reaction

As $\ce{R-Li}$ is very reactive ( even more than $\ce{RMgX}$ ). To decrease the reactivity of $\ce{R-Li}$ , it is reacted with $\ce{CuX}$ to form Gilman Reagent which is less reactive than $\ce{R-Li}$ to produce alkanes.

If  $\ce{R-Li}$ reacts directly with $\ce{R'-X}$, there will no formation of alkanes, but interchange reaction occurs.

$$\ce{RLi + R'X --> RX + R'Li}$$


What does the equilibrium constant mean for reactions with very small concentrations?

"In this case, the equilibrium constant suggests the products are favoured but if you look at how much product we have compared to reactants there's so much more reactants than products."  This is what I take as the heart of your question.  To that point:
I believe you need to compare the concentration of all the reactantS, not just one of the products.  In this case, the concentration of all of the reactants is 5.76 X 10 -8 which is a lot less than the 2.4 X 10 -4 which you might have been comparing to the product in the equation.


Optical activity of (1r,3R,5S)-3,5-dimethyl-4-methylidenecyclohexan-1-ol

The given molecule,(1r,3R,5S)-3,5-dimethyl-4-methylenecyclohexanol , is optically inactive .This molecule has a plane of symmetry (as shown below) and, therefore, is achiral.


Does NaOH have a higher OH- concentration than NH3?

Given that the solutions have equal concentrations, a solution of $\ce{NaOH}$ (a strong base) will be more basic than one of $\ce{NH3}$ (a weak base), since $\ce{NH3}$ only dissociates partially:

$$\ce{NH3 + H2O <<=> NH4+ + OH-}$$

Loosely speaking, the pH of a solution is defined to be

$$\mathrm{pH} = -\log_{10}[\ce{H+}]$$

and from the ionic product of water, we know that $[\ce{H+}][\ce{OH-}]$ is always a constant (equal to $10^{-14}\ \mathrm{M^2}$ at $25\ ^\circ\mathrm{C}$).

Therefore, a higher pH implies a lower concentration of $\ce{H+}$, and therefore a higher concentration of $\ce{OH-}$. This means that the truth of option 1 implies the truth of option 2, and vice versa, i.e. option 1 is logically equivalent to option 2.

My first suspicion would therefore be a typo in your book. In any case, your reasoning is correct.


The compounds you are referring to are called catenanes (see also rotaxanes, as they are a close relative).

(image source)

  one molecule chain of one benzene molecule, go through the center of
  another, before closing back up as a ring

While catenanes involving larger rings can be prepared, having one benzene ring threaded through a second benzene ring is not likely.  If you build a model you can see that there would be severe steric interactions, I've tried to depict this in the following drawing.

The carbon-carbon bond length in benzene is 139 pm.  Therefore, the approximate side to side distance in benzene is $\mathrm{2 \cdot 139~ pm \cdot cos(30°)= 241~pm}$.  The van der Waals radius is defined as "half of the internuclear separation of two non-bonded atoms of the same element on their closest possible approach" (source).  The van der Waals radius for carbon is 170 pm, meaning that the closest approach for two non-bonded carbon atoms is around 340 pm.  This value is significantly larger than the 241 pm side-to-side distance we calculated above for benzene.  This means that a carbon atom cannot pass through the interior of a benzene ring, at least not for carbon atoms with energies typically found in an organic laboratory.


What is the most acidic neutral hydrocarbon?

In this answer a hydrocarbon anion called Kuhn's anion has the formula $\ce{C_{67}H_{39}^-}$ has reported $pK_b=8.1$ which would correspond to the neutral hydrocarbon having $pK_a=5.9$.  The anion, together with several hydrocarbon cations with which it forms stable salts, is shown below (taken from the answer referenced above; primary reference J. Org. Chem. 1990, 55 (3), 996–1002):


Small scale gas-to-liquids via Fischer–Tropsch — is it even possible?

It probably isn't practical to do Fischer-Tropsch on a small scale
Gasification of biomass is easy. And produces syngas, a mixture of carbon dioxide, carbon monoxide, hydrogen and other compounds.
But the Fischer-Tropsch process to convert that mixture into useful liquids (like diesel) is not easy. The process requires carefully designed reactors, controlled heat and specific catalysts which are poisoned by many of the typical minor constituents of biomass originated syngas (especially any containing sulfur). So industrial reactors have to clean the syngas so the catalysts are not poisoned.
It is unlikely that you will be able to build a carefully controlled reactor that guarantees the right conditions (moderate but controlled temperature and pressure, complex geometry, specific catalysts, clean syngas). Even less likely that the result will be remotely economical on a small scale given the required cost and energy consumption. Heck, major chemical and oil firms have been trying for decades and few of their efforts have been  economically viable.
And there is another reason why this is futile on a small scale. Syngas itself is a usable fuel without further processing. It has been used to power engines (when diesel or petrol/gasoline are scarce as they were during WW2). And variants of syngas were once common as a piped domestic fuel for heating and cooking (before natural gas which is mainly methane became far cheaper and less toxic for most customers). So why not just burn the syngas?