For a system to undergo a spontaneous change at constant temperature and pressure the Gibbs free energy must decrease, but how?

If you want to track where the energy is going, do a first-law analysis (compute heat and work for a given change in internal energy or enthalpy).

If you want to predict whether a process occurs spontaneously or not, or you want to study an equilibrium, the second law is the tool you need. It's convenient to apply the second law by looking at free energy changes, since they let you focus on the system rather than the entire universe.

When we're interested in experimental internal energies or enthalpies we're tracking energy transfers. When we're interested in the Gibbs free energy we're usually answering questions about process spontaneity, or equilibrium conditions, or maybe we want to compute non-PV work. Different tools for different jobs.

Let me edit some of my comments below into my answer, because the comments are scrolling off the page and they make it clearer why you can't really track energy transfers just by looking at the Gibbs free energy:

The Gibbs free energy is an energy in the sense that it is the part of the energy that is "free" to do useful work. It does indeed have units of energy, and $\Delta G$ can be used to compute the maximum  possible nonexpansion work.

It is not an energy in the sense that it is not conserved in a process, the way internal energy and enthalpy are. $G$ is derived partly from entropy. It's quite possible that $G_\text{system} + G_\text{surroundings}$ isn't constant in a process.

If you have an isothermal process, $\Delta G= \Delta H−T\Delta S$. You can see from that equation that it's possible to decrease the Gibbs free energy simply by increasing the entropy, without any energy transfer at all. DavePhD's answer gives a concrete example of that.


In a mass spectrum, how can the sum of relative abundances be greater than 100?

2.3:10, 23:100 and (23/123):(100/123) are all correct values for the ratio of the relative abundance of B-10 to B-11.  However, only the last value, (23/123):(100/123) or 18.7:81.3 shows the normalized ratio of relative abundance of these isotopes.  It is this normalized value that adds up to 100.


Your question is how to do the conversion of the units of mole/liter to gram/liter.
Mole/liter, written as mol/L, is an amount (in units of mole) per liter.

To do this conversion, we need to know the convert mole to gram or vice versa.

To this end, we need to use the physical property of molar mass in units of gram/mole. This is the mass per mole of substance, in your case the substance is Vitamin B-12.
According to Wikipedia, the molar mass of Vitamin B-12 is 1355.37 g/mol.

$$\frac{200\times 10^{-12} \frac{g}{mL}}{1355.37 \frac{g}{mol}}=0.146\times10^{-12}\frac{mol}{mL}=146\times10^{-12}\frac{mol}{L}=146\frac{pmol}{L}$$


Deviation from Raoult's law because of adhesive and cohesive forces

Foreword
This topic is treated in most books and sites so I've tried to explain it to you in a more unusual and friendly way to grasp the concept but maybe the canonicals explanations are more suited for exams and serious conversation.
First step idealize and simplify the Raoult law
You can see Raoult law as the most basic attempt to predict the partial vapour pressure of a multi-components solution. If $A$ is your first component with higher vapour pressure an $B$ is the second component with lower vapour pressure, it seems quite obvious that if you mix an equal amount of $A$ and $B$ you have a vapour pressure equal to the mean of the two.
Back to reality
In fact, the reality is more complex and you have to take into account the force between the molecules of the two components. If you have strong adhesive forces the molecules of the first component near the molecules of the second keep themself more stuck together. So this creates a sort of "net" that doesn't allow the molecules to "escape" from the solution and so the vapour pressure (that in fact quantify the ratio of molecules that "escape" from the bulk) is lower and finally  you have a negative deviation.
But what if the interaction between the molecules of the same component is greater than the interaction between the two components? As you state, this is the case of strong cohesive forces. In this case you have to imagine that the components act as if they were separate.  Molecules of the component A lower the positive interaction between the molecules of the component B interposing them self and avoiding the formation of bonds between molecules of component B and finally increasing the vapour pressure, a positive deviation!
Deep into fantasy
So if you think the molecules as humans exiting from an elevator, humans A are more claustrophobic and so tend to exit more rapidly than humans B when the doors open. If they interact positively with humans B they will keep calm when exiting. If humans B don't find funny humans A they preserve their panic attitude and at the same time separate B humans from each other and rushing out from the elevator pull humans B out more rapidly too!


Why can't this molecule avoid being antiaromatic by becoming nonplanar?

The central atom is approximated to be tetrahedral and thus approximately $\mathrm{sp^3}$ hybridised. The three atoms of the outer ring it is bonded to are approximately $\mathrm{sp^2}$ hybridised as they are all part of double bonds.

If you build this, you will notice that while these three atoms are necessarily in a plane they have different orientations. It is simply not possible to make a planar ring without some significant distortions of the geometry. I thus challenge your professor’s assumption.

I would expect this system to have strongly localised double bonds, i.e. significant bond length alteration, in such a way that the double bond are twisted with respect to each other to minimise strain.


How can I measure the pH of high-viscosity fluids like dough?

In addition to buckminst’s answer, for weak acids, which are not fully ionized, the pH of the solution depends on the ratio of the concentrations of the conjugate base $\ce{A-}$ to the neutral acid $\ce{HA}$, especially in the pH regime close to the $\mathrm{p}K_\text{a}$ of the acid (the buffer region). Diluting the solution will change the concentrations, but in the same way. The ratio will not change. The Henderson-Hasselbalch equation can be derived from the mass action expression for $K_\text{a}$ of a weak acid:

$$\ce{HA <=>H+ + A-}$$
$$K_\text{a}=\dfrac{\ce{[H+][A^{-}]}}{\ce{[HA]}}$$
$$\text{p}K_\text{a}=-\log{K_\text{a}}=-\log \left(\dfrac{\ce{[H+][A^{-}]}}{\ce{[HA]}}\right)=-\log[\ce{H+}]-\log\left(\dfrac{\ce{[A^{-}]}}{\ce{[HA]}}\right)$$
$$\text{p}K_\text{a}=\text{pH}-\log\left(\dfrac{\ce{[A^{-}]}}{\ce{[HA]}}\right)$$
$$\text{pH}=\text{p}K_\text{a}+\log\left(\dfrac{\ce{[A^{-}]}}{\ce{[HA]}}\right)$$

If the volume increases by a factor of 10, then $[\ce{A-}]$ and $[\ce{HA}]$ both decrease by a factor of 10, and $\dfrac{\ce{[A^{-}]}}{\ce{[HA]}}$ remains constant.

Since the acids in dough are likely acetic or lactic from fermentation; carbonic or tartaric from leavening; or citric or ascorbic from preservatives – all of which are weak acids – diluting the dough might not change the pH.


What is the difference between polyatomic ions and compounds?

The ions, by themselves, are charged and thus cannot be complete compounds which are electrically neutral.

We sometimes do have names for polyatomic ions as If they are compounds, like cyanidin (which is a cation containing a pyrylium ion function).  But we understand that a "generic" counterion is needed to complete the compound.  For example, "cyanidin" (a polyatomic cation) can't be completely isolated by itself, but it could be incorporated into cyanidin chloride (a compound), which in principle could be isolated.


Most of the things to be told about magnetic stirring are already in Rons comment. A rotating magnetic field is induced by a rotating permanent magnet inside the hot plate magnetic stirrer. This field drives the magnet inside your reaction flask. To ensure chemical inertness in most situations the magnet is coated with PTFE (Teflon) or protected by a thin layer of glass. Magnetic stirring can replace traditional mechanical stirring is most situations and has many benefits over it: it does not require an additional inlet to the reaction flask, which is very handy for reaction under reduced pressure and is also more easily adjustable to different scales due to the possibility to replace the magnet in the reaction flask. Two exceptions are 1) very large/very viscous reactions or reactions in which excessive amounts of solid are present 2) reactions that involve ferromagnetic compounds that interfere with the rotational movement of the magnet inside the flask (as in Urushibara nickle catalyst preparation).

Vogel has a section on reaction heating and stirring in which this and more is discussed extensively. See pages 71-80.


What is the "rate factor" or "second-step rate constant" in the reaction rate equation?

The implication is that the reaction mechanism for any particular reaction, if written like this:

$$A + B → C$$

might not proceed as one step. Having a $k_2$ indicates that there is a second step to the mechanism, and potentially even several more. $k_{obs}$ is probably a better name for it, because sometimes you see it written as a simplified "one-step" process, such as in Michaelis-Menten Kinetics, when really, it might be $1$ more step, or it might even be $20$, depending on the reaction (I'm exaggerating a bit, but you get the point).

If we take Michaelis-Menten enzyme kinetics as an example:

$$E+S⇌ES→E+P$$

The first step is an equilibrium in which the concentration of the intermediate is assumed to not change because the intermediate is being produced about as fast as it is being consumed. This would be the Steady State Approximation. In other words:

$\frac{d[ES]}{dt} = 0 = k_1[E][S] - k_{-1}[ES] - k_2[ES]$

Eventually...

$[ES] = \frac{k_1}{k_{-1} + k_2}[E][S]$

The second step is where the $ES$ complex does something in such a way that the enzyme unbinds somehow, and the product is made... somehow. Depending on the reactants and products, the mechanism is different.

Anyways, the point is that $k_2$ describes this second step in this type of reaction (which is what you were wondering---why the "second step"?), hence the $2$.

In any case, $k_2$ corresponds to the reaction step(s) that proceed after an intermediate product is formed on the first step for a reaction mechanism.


List item

All the reactions eliminate chloride (a) by the obvious direct attack, the other products b and c arise from methoxide attack on the 3 and 5 positions of the furan and chloride loss via an exomethylene intermediate. The system is somewhat similar to a polyallyl choride where attack can be direct or at the 3 or 5 position.


Why is 1,3-pentadiene favored in the reaction of pent-3-en-2-ol with sulfuric acid?

Hydrogenation of 1,trans-3-pentadiene (your top molecule) and the allene, penta-2,3-diene (your bottom molecule), both produce pentane plus heat.  Heat is given off because both of the dienes plus hydrogen is less stable than pentane.  With 1,trans-3-pentadiene, 53.1 kcal/m of heat is given off; with penta-2,3-diene 66.8 kcal/m of heat is given off (data from Frances Maron Fraser and Edward J. Prosen, Journal of Research of the National Bureau of Standards Vol. 54, No. 3, March 1955 Research Paper 2575, p. 143-148). Clearly the allene is of higher energy than the 1,3-diene by about 15.7 kcal/m.

The reaction conditions (strong acid and heat) suggest that an equilibrium mixture of possible products will result.  Around room temperature, each 1.4 kcal/m difference in energy between two isomers will shift the equilibrium between them by a power of 10 ($\ce{\Delta G=-RTlnK}$).  A 15 kcal/m energy difference between your two dienes suggests that the equilibrium will be ~ $\mathrm{10^{10}:1}$ favoring the 1,3-pentadiene. If any of the allene forms, it will rapidly convert to the 1,3-pentadiene.

Side Note: Why are cumulenes generally less stable than 1,3-dienes?

Conjugation which results in resonance stabilization is present in the 1,3-diene but not in the 2,3-diene.
Allenes (in fact, all cumulenes) have considerable strain built into them. If you view a double bond as a two-membered ring, it is easy to see this.  The bond angle in the two-membered ring is severely pinched down.  In turn, the bond angle external to the double bond tries to open up and relieve strain.  This strain relief, can't happen with the central carbon in an allene because it has a double bond on both sides.


In the elementary steps of a reaction, why do we assume that the coefficients of a reactant correspond to its order?

If a reaction has just one step, it's named an elementary reaction. In this reaction all of reactant directly participate on reaction and effectively collide with each other. So, the order of reaction corresponds to its coefficient, but if a reaction is not a elementary reaction, there isn't any reason for colliding all reactants with each other. For example, consider the following overall reaction:

$$\ce{A + B + D -> F}$$

Assume the corresponding mechanism is constituted of two elementary reactions:

$$\ce{A + B -> C}$$
$$\ce{C + D -> F}$$

As shown above, there is no need to collide A with D or A with C ... Then, in these type of reactions coefficients are not corresponded to order of reaction .


Which solution has the highest total molarity of ions and why?

You are right in thinking that a strong electrolyte is something that dissociates completely in solution.$\ce{Ba(OH)2}$ and $\ce{KI}$ then, being a strong electrolytes, will effectively dissociate into $\ce{Ba^2+}$+ $\ce{2OH-}$ and $\ce{K+}$+ $\ce{I-}$, respectively.

Going into your example, it's a just a matter of multiplication. We can assume that the volume of the solution is 1L for simplicity. Since you begin with 0.008 M $\ce{Ba(OH)2}$, we can say that you begin with 0.008 mol of $\ce{Ba(OH)2}$. In solution this becomes 0.008 mol of $\ce{Ba+}$ and 2(0.008) mol of $\ce{OH-}$, or 0.024 mol of ions in total. Likewise for KI, you are left with 0.010 mol of $\ce{K+}$ and 0.010 mol of $\ce{I-}$, or 0.020 mol of ions in total.

Converting these values back to molarity (by dividing by 1), we get that the concetration of ions in the $\ce{Ba(OH)2}$ solution is 0.024 M, whereas it is only 0.020 M in the $\ce{KI}$ solution.


Why catalyst does not alter the entropy of a reaction?

TL;DR Catalysts by definition do not alter the magnitude of changes in thermodynamic properties that accompany conversion of reactants to products, they only alter the conversion (reaction) rates.
The reason a catalyst does not alter the entropy of a reaction is that this is one part of the definition of a catalyst! Ideally none of the changes in thermodynamic functions accompanying a reaction are altered by the catalyst. However, that is a circular and not very helpful argument. The fundamental reason is that catalysts are not consumed in chemical reactions. Since the products and reagents remain the same, it follows from Hess's law that catalysts do not contribute to thermodynamic changes associated with the reaction.  Only the rate with which reactants and products interconvert (kinetics) changes.
Catalysts alter the reaction rate by altering the rate of encounter of the reagents or products or by assisting in formation of new lower energy (more stable) reaction intermediates. A catalyst may be as simple as a material that increases the proximity of reagents by increasing their local concentration, without otherwise altering their reactivity (bond enthalpies). That can be the case in either enzymes or solid supports. However catalysts often also alter reactivity by weakening specific bonds. An example of a catalyst that provides both functions is platinum.


Is it possible to calculate energy of electron in any orbital and atom in Schrödinger wave model theory?

It is surely possible. Note though, that for many-electron systems orbital picture is an approximate description, but that is an approximation on which the whole general chemistry is based on, so that is fine from that perspective. The approximation is known as the Hartree-Fock (HF) model, and thus, to get the energies of electrons occupying some orbitals all you need to do is to solve the Schrödinger equation in this approximation.

As a quick exercise one can do HF calculation in, say, Gaussian, with the following input file:

#P HF/aug-cc-pVTZ Pop=Full

Na-

-1 1
 Na                 0.00000000    0.00000000    0.00000000

to get this picture of occupied orbitals:

The HOMO with the energy of $-0.01288 \, \mathrm{Hartree} = -0.3505 \, \mathrm{eV}$ is basically the $\mathrm{3s}$ orbital you're looking for (trust me). Note, however, that this energy is approximate since the basis set (aug-cc-pVTZ) is finite. We could do better than that, but that is a different story.

In response to permeakra (since he does not trust his colleagues) I quote an authoritative reference in the field which explicitly talks about individual electrons occupying individual spin-orbitals and orbital energies. As a reference I choose the infamous book entitled "Modern Quantum Chemistry" written by Attila Szabo and Neil S. Ostlund.

Quote #1 (Szabo & Ostlund, p. 50)

  This Slater determinant has $N$ electrons occupying $N$ spin orbitals
  $(\chi_i, \chi_j, \dotsc, \chi_k)$ without specifying which electron
  is in which orbital.

Quote #2 (Szabo & Ostlund, p. 54)

  $$   f(i) \chi(x_i) = \varepsilon \chi(x_i) \tag{2.52} $$ [...]

  The
  solution of the Hartree-Fock eigenvalue problem (2.52) yields a set
  $\{\chi_k\}$ of orthonormal Hartree-Fock spin orbitals with orbital
  energies $\{\varepsilon_k\}$.

Below I also quote what Szabo & Ostlund have to say about Koopmans' theorem, because I already know that won't trust me.

  Quote #3 (Szabo & Ostlund, p. 110)

  The first theorem [Koopmans' theorem] constitutes an interpretation of
  the Hartree-Fock orbital energies as ionization potentials and
  electron affinities.

See, orbitals with their energies & occupancies do exist in the Hartree-Fock theory. It is the interpretation of orbital energies as ionization potentials which requires an additional theorem.


Under acidic conditions, a part of the divinylether will probably just undergo cationic polymerization:


What is the difference between a reaction of Lewis acids with Lewis bases and a redox reaction?

It's indeed redox - Lewis theory doesn't account for redox, but on grounds of Usanovich's theory ...

A Lewis acid is an electron pair acceptor; a Lewis base is an electron pair donor. The simplest Lewis acid is the hydrogen proton. H$^{+}$ has no electrons by itself. It accepts electron pairs from bases in Lewis acid-base reactions. Here's an example of a Lewis acid-base reaction; note how a pair of electrons from the Lewis base attacks the Lewis acid.

Redox - a portmanteau of reduction and oxidation - involves the transfer of electrons. Hey, there are electrons being "transferred" above, right?

At first glance, it seems that all Lewis acid-base reactions would be redox reactions. However, there is an artificial line in the sand that one would do well to heed.

Redox reactions, as taught at the general chemistry level, involves changes in formal oxidation state. What's an oxidation state? It's the hypothetical charge that an atom would have if all bonding were ionic (as opposed to covalent, or electron-sharing). This means that the more electronegative atom in a bond would have complete possession of the bonding electrons.

If we go back to the reaction pictured above, then we see that electrons are not being "transferred" in the formal sense. Hydrogen is of much lower electronegativity than oxygen; hence, in a formal sense, it has gained no electrons - remember, when assigning formal oxidation states, we take all bonding as ionic.

Obviously, in reality, no bond is 100% ionic, but this is exactly the artificial line in the sand I was talking about.

  For example I came across this equation:
  $\ce{2Al_{(s)} + 6HCl_{(aq)}} \longrightarrow \ce{2Al^{3+}_{(aq)} + 6Cl^{-}_{(aq)} + 3H_2_{(g)}}$
  And thought this was a redox reaction but my book says it shows basic nature of aluminium?

This indeed is a redox reaction, as aluminum is oxidized and hydrogen is reduced as we go from left to right.

It also shows that aluminum can react with hydrochloric acid to make hydrogen gas; this removes hydrogen protons from the solution and thereby makes the system more basic.

This is not an acid-base reaction in the Lewis sense, as electron pairs aren't being transferred. It's not an acid-base reaction in the Bronsted sense either, as hydrogen protons are not being moved around.

  On grounds of Usanovich's theory Al as reductor would indeed be basic.

Usanovich defines an acid that accepts negative charge; a base as something that accepts positive charge. Therefore, Al is a base here; it becomes positive from left to right. This theory is an even more generalized view of acids and bases than Lewis acid-base theory.


Commercial drying agents, such as Drierite (anhydrous calcium sulfate), or other desiccant, should do the job for small quantities of oxygen.

Most inorganic desiccants should not react with diatomic oxygen; as long as the oxygen is produced separately from the desiccant, there should be no issue. (Dropping 90% or stronger $\ce{H2O2}$ on many substances can cause dangerously rapid decomposition, but with 30% or lower concentration, that danger is minimal.)

As far as preventing atmospheric gases from entering, first bleed the apparatus of residual air by discarding the initial output. Seal the later containers underwater, e.g. by sliding a glass plate across the mouth. See Janice Crowleys' demo on YouTube.


Is hydrolysis of polypeptides and polysaccharides "anabolic" or "catabolic"

1) Is the process of hydrolysis that breaks up polypeptides and polysaccharides a net endothermic or exothermic process?

Under physiological conditions, it is a process that goes forward, i.e. the Gibbs energy is negative. As a consequence, it can happen outside of cells in the absence of ATP. When we eat, the hydrolysis of polysaccharides starts in our mouths, while the hydrolysis of proteins occurs under harsher (acidic) conditions in the stomach and continues in the intestine (slightly basic conditions).

For warmblooded animals like us, exothermic or endothermic is less important, but you could look it up.

  2) Do the free amino acids and monosaccharides have more or less stored energy than the polypeptide or polysaccharide that they were broken down from?

Less Gibbs energy, as the hydrolysis is spontaneous. However, this Gibbs energy is not captured (different from downstream oxidation of building blocks to $\ce{CO2}$, which is coupled eventually to synthesis of ATP).

  3) Is it proper to call the isolated process of "hydrolysis" of proteins and polysaccharides "catabolic"?

If your definition is just breaking down complicated molecules into simple ones, then you would call it catabolic. If you also require that Gibbs energy is captured in a coupled reaction, then you would classify these reactions as digestion, and the downstream oxidation to carbon dioxide as catabolism.


How to explain Gibbs free energy is a pressure-dependent state function?

$G$ is defined as $H - TS$, and $H$ is $U + PV$.

Since in closed system you have: $\mathrm{d}U = T \mathrm{d}S - P \mathrm{d}V$, it follows that:

$\mathrm{d}H = T \mathrm{d}S + V \mathrm{d}P$
$\mathrm{d}G = -S \mathrm{d}T + V \mathrm{d}P$


Rate of dehydration of alcohols having keto group

A carbonyl group normally destabilizes carbocations but stabilizes carbanions.
In this particular case, the rate of dehydration isn't guarded by carbocation stability, but by CH- acidity. In acidic conditions, OH group in alcohols is relatively easily protonated. Then, if there is a relatively acidic hydrogen in $\beta$-position it immediately dissociates, together with a water molecule.
Excercise: Try to draw a mechanism for dehydration of (a) in basic conditions. Hint: it does occur and produces the same product. It does require more heating, though.


Why HBr has stronger attraction between molecules than CH3NH2?

EDIT (after title edited): $\ce{HBr}$ doesn't have stronger interaction than $\ce{CH2NH2}$, but it has dipole-dipole interaction as the strongest forces between it's molecules, which is obviously weaker than H-bonding.

Hydrogen bonding is the strongest intermolecular attraction. It is a type of dipole-dipole interaction1, but it is specific to Hydrogen.
In general, dipole-dipole interactions are considered weaker than H-bonding. The relative strength of forces are as follows,

Hydrogen bonding >  Van der Waals dipole-dipole interactions > Van der Waals dispersion forces

As per your question, (a), (c) and (d) have dispersion forces (as they're non-polar), while (b) has H-bonding as it's strongest interaction (not dipole-dipole).
Finally, as only (e) has dipole-dipole as it's strongest attraction between molecules, hence it's the answer.

Notes:
1: Dipoles are created due to difference in electronegativity. Below image shows that there is sufficient electronegativity difference to create a dipole. H-bonding and dipole-dipole interactions have same origins.

(Source: Master Organic Chemistry)


Transesterification of propylene carbonate and methanol

I know it is used to obtain  dimethyl carbonate (DMC) I think the others products should vary from which catalyst you choose. Have a look to Transesterification of propylene carbonate with methanol using $\ce{MgAlCO3}$
hydrotalcite as solid base catalyst: Indian Journal of Chemistry
Vol. 49A, September 2010, pp. 1182-1188 to choose the catalyst. I think there are two little typos in Scheme 2 where the mechanism is shown: miss $\ce{^-}$ from $\ce{CH_3O^-}$ (first step) and miss $\ce{C}$ from one part ($\ce{CH_3CO}$) of DMC formula (second step) but is a good article.


1H-borepine is isoelectronic to the cycloheptatrienyl cation. In both cases, you have a planar ring of seven atoms, whose hybridisation is close enough to $\mathrm{sp^2}$ to allow for a third p-orbital to be perpendicular to the ring plane. And in both, you are then filling these seven π orbitals with 6 π electrons, resulting in a Hückel-aromatic system ($4n+2 = 6$ for $n=1$).

There is no requirement for each atom along the chain of conjugation to have a lone pair; for example, in benzene no carbon atom has a valence lone pair. The only requirement for Hückel aromaticity is through conjugation, i.e. each atom must have an accessable p-type orbital that can overlap with the neighbouring atoms’ p-orbitals in a favourable π manner. However, neutral aromatic systems with empty p-orbitals are rare if boron is not used as most non-metals do not have favourable electronic configurations.


What is the relationship between the molecular weight of a polymer and that of polystyrene?

You have to have something to calibrate the column with. Polystyrene is typically used since it can be made cheaply (well, relatively cheaply) with a pretty small molecular weight distribution.

Yes, there will be differences between the true molecular weight of a polymer and what the comparison to polystyrene gives you. But does that really matter? Why do you need to know the absolute molecular weight, when a relative comparison works just fine?

It's similar to how we determine the mass of something. At present, there is that infamous block of Pt/Ir in Paris that is the standard kilogram. All mass measurements are made through a series of comparison to that mass.


How is kinetic energy accounted for when describing diatomic bonds with a Morse potential?

The kinetic energy is not ignored. Each quantized state (labeled by the vibrational quantum number) has a total energy given by the expectation value of the Hamiltonian operator for the wavefunction. The contribution of the kinetic energy can be computed as the expectation value of the QM kinetic energy operator, or as the difference between the expectation values of the total energy and potential energy.
Because the math is simpler I will limit the discussion to the harmonic oscillator (ie quadratic potential). For the QM harmonic oscillator the expectation value of the potential energy turns out to be
$$\left< E_\text{pot} \right> =\frac12 \left(v+\frac12\right)\hbar \omega=\frac12 E_\text{tot}$$
The expectation value of the kinetic energy is then
$$\left< E_\text{KE}\right>=E_\text{tot}-\left< E_\text{pot} \right>= \left< E_\text{pot} \right>$$
So it turns out that the kinetic energy and potential values have identical expectation values, consistent with the prediction of the virial theorem for a quadratic potential. This  is all outlined very nicely for instance in Atkins' physical chemistry textbook.
It is difficult to tell what energy levels are displayed on the diagram but they might very well correspond to the total energy given by $ E_\text{tot}  = \left(v+\frac12\right)\hbar \omega$.


To start, the figure and explanation in the link are almost comically misleading. Potassium chloride solution does not "attract" oxygen. The oxygen flowing through the wire is pretty silly and

  Where the probe joins the wire, oxygen mingles with the electricity.
  Oxygen is not very ionized, meaning that it does not have a negative
  charge as electricity does, so the oxygen dilutes the current at the
  electrode beyond the probe.

is just plain ridiculous. This one is better.

The way this type of probe works is by amperometric reduction of $\ce{O2}$, the first example being the Clark electrode. Essentially, an electrolyte is separated from the measurement solution by a porous film of poly(tetrafluoroethylene) (Teflon). These pores are small enough that only small substances like oxygen can diffuse through. On the electrolyte side, there are two electrodes in one of two possible configurations:

In the original Clark-type electrode, the working electrode is platinum, which is chemically-inert but catalyzes the reduction of oxygen. The counter electrode is a AgCl-coated Ag electrode that acts as a reference for the measurement. (Its potential is supposed to remain constant) To reduce the oxygen, an external potential must be applied across the two electrodes (normally using a potentiostat).
The galvanic cell–type electrode has one electrode made of silver (where the oxygen reduction occurs) and one made of an easily oxidized metal like lead. It works the same way as the Clark electrode, but instead of applying an external potential, the current required to reduce the oxygen is supplied by the oxidation of the lead electrode.

Any oxygen that reaches the working electrode is reduced and produces a current that is measured to produce the analytical signal. The amount of current produced depends on how much oxygen diffuses through the membrane, which depends on the concentration of dissolved oxygen in the test solution.

Now, for your questions:

  how would you know that this sensor is working - meaning that only
  dissolved oxygen triggers the electrical change in the sensor, and not
  other atoms?

This one is a bit tricky. You're right that oxygen isn't the only thing that could produce a signal—amperometry isn't a very selective technique on its own. You can control the redox reactions that occur at the working electrode, to a limited degree, by changing the potential you apply, i.e. different substances require different potentials to be reduced and won't be reduced if an insufficient potential is applied. In the case of this specific type of electrode, selectivity comes mainly from the PTFE membrane. Since it limits diffusion to only a few very small molecules, that greatly increases the selectivity of the electrode, though depending on the solution you're trying to measure, there still may be interferences.

  and how would you calibrate the sensor to translate changes in
  electric signal to the change in number of dissolved oxygen atoms?

In principle, you could calculate the number of reduction reactions happening at the surface from Faraday's laws, but to turn this into a measure of the exact concentration in solution requires a lot of thought of how the transport of oxygen to the electrode occurs, from diffusion from the electrolyte to the electrode surface and transport from the test solution into the electrolyte. In practice, the assumption that the measured current is directly proportional to the concentration in solution is made (and is valid under many conditions), and a calibration is made using a standard solution.

  also, how would one get pure dissolved oxygen in varying amounts as 'positive control' for the sensor?

Calibration solutions seem to be commonly made using tables of oxygen saturation at different temperatures. It's also possible to measure dissolved oxygen through other means, such as titration. Solutions can be made up, titrated, then the value used to calibrate an electrode.

Now, for the measurement specifically in cell cultures, a membrane electrode may not be the best bet, at least not without sample pre-treatment. The PTFE membrane is quite susceptible to adsorption of proteins and the like, which will probably affect the electrode's response. It may be necessary to somehow remove these contaminants (without affecting the dissolved oxygen concentration). Other options that might be viable are colourimetric measurements or titration, though there are problems with these, as well.


Calculate partial pressure of gases at equilibrium

The equations you need are
$$K_p=\frac{p_{\ce{COCl2}}}{p_{\ce{CO}} p_{\ce{Cl2}}}=1.47\cdot10^8$$
and
$$p_\text{tot}=p_{\ce{COCl2}}+p_{\ce{CO}}+p_{\ce{Cl2}}=5.0~\text{atm}$$
Additionally, $n_{\ce{CO}}=n_{\ce{Cl2}}\implies p_{\ce{CO}}=p_{\ce{Cl2}}$, making the system solvable.


Does enthalpy of dissolution change with temperature?

I want to know whether an increase in temperature affects the enthalphy of dissolution?

It certainly can. You can treat the temperature dependence of dissolution with a formalism analogous to that applied to chemical reactions. The key parameter is the difference in heat capacity of the solid and dissolved solute, since the enthalpy of solubilization can be written as

$$\Delta H(T) = \Delta H(T_\mathrm{ref}) + \int_{T_\mathrm{ref}}^{T} \Delta C_p\,\mathrm{d}T$$

where $\Delta C_p = C_{p,\mathrm{solute}} - C_{p,\mathrm{s}}$. The $\Delta C_p$ term accounts for the different temperature sensitivities of the enthalpy (which for condensed phases approximates the energy) of solid and solute. If the heat capacity of the solute is greater than that of the solid, the enthalpy of dissolution will increase (become more positive) with an increase in $T.$

Note that if we associate a solubility constant $K_\mathrm{sp}$ with the dissolution process

$$\ce{A(solid) <=> A(solution)}$$

then we can write

$$\frac{\mathrm{d}\log K_\mathrm{sp}}{\mathrm{d}T}=\frac{\Delta H^\circ}{RT^2}$$

as suggested in a comment. Here both $K_\mathrm{sp}$ and $\Delta H^\circ$ are functions of $T.$


How to identify whether a molecule is planar

The concept of a plane of symmetry is easier to understand with molecules which are planar by itself.  Benzene, for example, has such a plane which may accommodate simultaneously all carbon and hydrogen atoms:

However, planarity of molecules is not a requirement to identify such planes.
Even if a construction of the cyclobutane molecule with model kit illustrates you the non-planarity of the molecule

(source)
you drawing depicts it like it were planar; because you defined drawing plane where carbon atoms alternate to be either closely above, or below this level in common.  For compound 1 illustrating 1,3-dichloro cyclobutane, however, the symmetry analysis is much affected by the configuration of the two special carbon atoms.  Either you find the two chlorine atoms on the same side of this plane (syn)

or you find them directing to different sides (anti):

The interactive training set by Symmetry@Otterbein may be a helpful resource for you.  Note that the representation of compound 2 is a simplification, and the symmetry attributed strongly depends on the underlying geometry.


Why does delocalization (only) occur in molecules represented by resonance hybrids?

A short answer is that we mostly use resonance structures to represent delocalization when the bond orders or formal charges predicted by VBT are not approximately integers, simply because non-integer bond orders and charges are more cumbersome to draw. A typical example is a carboxylate, which we could draw with 1.5 bonds between both C-O pairs and -0.5 charge on each O, but it is simpler to just draw a resonance structure with one C=O double bond and one C-O single bond and a full -1 charge on one oxygen.
As others have noted, electrons are still delocalized across the entire molecule in molecules for which we do not use resonance structures. The difference is that we can approximate the bond orders in those cases as integers, so only one structure is needed to represent that.


Mechanism for acid-catalyzed dehydration of 2,3-dimethylcyclopentanol


How does existence of alpha and beta form of glucose prove that it exists as a cyclic structure

According to Wikipedia:

  Glucose is usually present in solid form as a monohydrate with a closed pyran ring (dextrose hydrate). In aqueous solution, on the other hand, it is an open-chain to a small extent and is present predominantly as α- or β-pyranose, which partially mutually merge by mutarotation.

Glucose predominantly occurs in nature in the form of the D‐enantiomer, which is generally believed to exist in three crystalline forms: $\alpha$‐D‐glucose monohydrate (Figure 1A)(Ref.2), anhydrous $\alpha$‐D‐glucose (Figure 1B)(Ref.2), and anhydrous $\beta$‐D‐glucose (Figure 1C)(Ref.1,3). Both anhydrous $\alpha$‐D‐ and $\beta$‐D‐glucose crystals are orthorhombic while $\alpha$‐D‐glucose monohydrate crystals are monoclinic (see Fig. 1A-C). However, a fourth form, which is metastable in solution phase at $\pu{38\!-\! 50 ^{\circ}C}$ and thought to be a hydrated form of $\beta$‐D‐glucose, has been reported as well (Ref.1,4).

The crystal structure of $\beta$-D-glucose published in 1960 (Ref.5) clearly showed the exsistence of pyranose ring system. As in the inserted box in Figure 1 state that, in aqueous solutions, 99% of D‐glucose exists as a mixture of the $\alpha$- and $\beta$-forms (approximately 62%  $\beta$ and 38% $\alpha$ when equilibrated at $\pu{31 ^{\circ}C}$ (Ref.1). Recent NMR study using fully $\ce{^13C}$ labelled glucose (Ref.6) clearly showed $\alpha/\beta$ ratio of $37/63$, which is almost identical to this literature value (Figure 2):

References:

F. W. Schenck, "Glucose and Glucose-Containing Syrups," In Ullmann's Encyclopedia of Industrial Chemistry: Ullmann's Food and Feed, Vol. 2; B; Elvers, Ed.; Wiley‐VCH Verlag GmbH & Co. KGaA: Weinheim, Germany, 2017, pp. 781-802 (https://doi.org/10.1002/14356007.a12_457.pub2).
W. B. Newkirk, “Manufacture and Uses of Refined Dextrose,” Industrial & Engineering Chemistry 1924, 16(11), 1173-1175 (DOI: 10.1021/ie50179a028).
G. R. Dean, “Optical-Crystallographic Properties of $\beta$-D-Glucose,” Anal. Chem. 1973, 45(14), 2440–2441 (DOI: 10.1021/ac60336a005).
G. R. Dean, “An unstable crystalline phase in the D-glucose-water system,” Carbohydrate Research 1974, 34(2), 315–322 (https://doi.org/10.1016/S0008-6215(00)82906-7).
W. G. Ferrier, “The crystal structure of $\beta$-D-glucose,” Acta Cryst. 1960, 13, 678-679 (doi: 10.1107/S0365110X60001588).
T. Richter, S. Berger, “A NMR method to determine the anomeric specificity of glucose phosphorylation,” Bioorganic & Medicinal Chemistry 2013, 21(10), 2710–2714 (https://doi.org/10.1016/j.bmc.2013.03.008).


carbon electrodes have potential difference even with no electrical input?

Some possibilities:

Porous carbon electrodes do act as a capacitor, the basis for supercapacitors (aka ultracapacitors).
You are electrolytically depositing copper on the negative electrode when you run current through the solution, even though it might not be enough to see. This creates an electrochemical cell.
Simply disconnecting the power supply leads will leave a high-impedance voltmeter floating, so it might be detecting random static charges. You might need to bypass the voltmeter with a high resistance (10 or 20 megohms) to reduce static sensitivity.


Isn't the reaction of chloral with sodium hydroxide a disproportionation reaction?

You should not call the haloform reaction of trichloroacetaldehyde with sodium hydroxide a comproportionation. (And even less a disproportionation.)

Disproportionations are defined such that you start at two species of the same type and oxidation state (e.g. two formal $\ce{Cr^4+}$ species) and end up with two different species (e.g. a formal $\ce{Cr^3+}$ and a formal $\ce{Cr^6+}$ species). This is not the case here, the oxidation states start out differently.

Comproportionations are defined oppositely: You start at two different species and create two of the same. But in no way can chloroform and sodium formiate be called ‘the same’, even though their oxidation states are. Carbon monoxide (also $\mathrm{+II}$) is not the same as either, either.


Enthalpy change vs. change in internal energy in a system

Both $U$ and $H$ are not directly related to any particular process.  They are physical properties of the material being processed, and are unique functions of the thermodynamic state of the material, and not the process.  They are directly proportional to the amount of material, and its thermodynamic state is characterized by its temperature and pressure (and composition for a multicomponent system).  The internal energy is a fundamental property, but the enthalpy is a defined property given by $H = U + PV$, and is just a convenient parameter to work with is solving many kinds of thermodynamics problems.

For a process carried out at constant pressure in a closed system, the amount of heat $Q$ added to the system turns out to be equal to the change in enthalpy between the initial state of the material and the final state of the material.  For a chemical reaction carried out at constant pressure, the amount of heat $Q$ that must be added to the system in order for the final temperature of the products to equal the initial temperature of the reactants turns out to be equal to the enthalpy of the pure products minus the enthalpy of the pure reactants at that temperature.  Both these results follow from the first law of thermdynamics.

For a chemical reaction, generally both the internal energy and the enthalpy change between the initial and final equilibrium states.

  And what about cases where $P$ is not constant? What happens then? Can we calculate the value of $ΔH$? And does $ΔH$ even have a meaning in this case?

Since $H$ is just defined as $U + PV$, we can always calculate the change in enthalpy between two thermodynamic equilibrium states of a system.  $\Delta H$ certainly has a value in this case, even if it is not equal to the amount of heat added for a process between the two states.

The critical thing to remember is that $U$ and $H$ are physical properties of the materials comprising a system, and are independent of any specific process.


How is it possible to account for the fact that NH3 is a weak base in water but a strong base in HCl?

The difference between a weak base and a strong base is that a strong base is completely converted to ions in a given solvent, whereas a weak base is not. Although this depends on the actual basicity of the solute, it is also very dependent on the solvent involved.

In water, $\ce{NH3}$ can react to form $\ce{NH4+}$ and $\ce{OH-}$. However, most of the ammonia remains as $\ce{NH3}$. This is due to the difference in pKas:1

$\ce{NH4+}$ = 9.2

$\ce{H2O}$ = 15.7

Because the ammonium ion is more acidic than neutral water, very little $\ce{NH4+}$ will be present at equilibrium.

In contrast, the relevant pKas for the $\ce{HCl}$ solution are:

$\ce{NH4+}$ = 9.2

$\ce{HCl}$ = -8

This tells us that $\ce{HCl}$ is much more acidic than the ammonium ion, and therefore essentially all of the ammonia will be protonated when dissolved in $\ce{HCl}$. As a result, it is considered a strong base in this case.

1Evans' pKa tables


One possibility would be to check the IR spectrum.  If there's any azide in there the azide band should show up as a strong band at around 2150 cm-1.  The trouble is of course that this isn't quantitative, you can only prove the absence of azide, not how much is present.

The Staudinger reaction might be a viable option.  PPh3 won't go up in aqueous solution, but sulfonated PPh3 will.  One should be able to distinguish starting material and product by 31P NMR.  With a known quantity of TPPTS you can even determine % azide in the starting material.


Why are van der Waals' forces between gas molecules strong near its critical temperature?

Critical temperature is  kind of measure of the strength of intermolecular van der Waals force.

Comparison of actual and critical temperature is kind of comparison of average kinetic energy of molecules with energy needed to break the intermolecular bonds.

So temperature itself does not say anything about the bonding strength. It only indirectly determine probability these bonds get broken by thermal molecular motion.


This Wikipedia article on "Insect repellent" pretty much answers your question.

An insect repellent isn't necessarily the same as an insecticide.

The repellent, well, "repels" insects; i.e- it discourages them from approaching you (or whatever you sprayed the repellent on) for as long as the chemical(s) linger. These (particularly those based on essential oils) aren't much of a health hazard to other animals.

An insecticide is actually designed to kill insects ("No sh!t Sherlock") rather than simply keeping them at bay. The term "insecticide" refers to substances that kill insects/arthropods at very low concentrations ( "very low" is subjective... I don't think there's an authoritative, precise definition for "insecticides" anyway).

FYI, any substance (including common stuff like air and  water), supplied in excess (again, "excess" is also highly subjective) can kill. Ever heard of "Too much of anything can kill"?

The opposite effect is something called "Hormesis"; where just the right amount of anything (including radiation and toxins) can have a positive effect on you.

Back to question.

The insect repellent you're using contains (according to the sticker on the cap) lemon-eucalyptus oil, whose active ingredient is p-menthane-3,8-diol (and its geometrical isomers). Also, it's supposed to be used against mosquitoes (Mücken) and ticks (Zecken).

It serves as a deterrent, on account of how strong it smells.

Ectoparasitic insects, such as the said mosquitoes and ticks, rely on (to a large extent) their sense of "smell" to locate hosts. So spraying around strong odorants will do a good job of interfering with the bug's ability to track you down.

Remember my "anything can kill" rant from earlier? Well, the same goes for your repellent. If I were a mosquito or a tick (or whatever insect for that matter), I doubt I'd make it out alive if you decided to smother me with some godforsaken liquid/aerosol. For all I know, your fly could've died because it suffocated, rather than being poisoned.

Heck, you could achieve the same effect if you decided to drown the fly in deodorant or even perfume.

So there isn't really much behind the mysterious death of your fly; in all likelihood it suffocated (You murderer!). Take @Mithoron's advice; a fly swatter works better.

If you want to know what biochemical processes the repellent would tamper with in a bug, then that's a question for the Biology.SE. ;)


How does dissolving a hydrate affect concentration?

There are two methods to dissolve the calcium chloride salt.

For the first method you'd dissolve the salt in a "small" amount of water and dilute to some particular volume. So if the solution was 0.100 molar, the amount of water would be the same regardless if you started with $\ce{CaCl2}$ or $\ce{CaCl2\cdot2H2O}$.

The other method would be to make say a 1.00 % solution by weight of the calcium chloride salt. Here it does matter. Do you want 1.00 % solution of $\ce{CaCl2}$ or a 1.00 % solution of $\ce{CaCl2\cdot2H2O}$?


To me this looks like an apparatus to measure water in fat, butter or oil. A weighed sample of butter is placed in the conical flask. The tube A goes below the level of the melted butter or oil. It's purpose is to maintain atmospheric pressure or in case of a bump in the comical flask or back pressure or suction from the boiling tube D or the condensation flask C. Flask C should have a hole open to the atmosphere through the bung. The hole at the bottom of C is to remove water from the condensed ice. The conical flask is heated. The butter (oil) has a much higher boiling point than the water in it. So the water boils out. The tube from the conical flask will go well down into the boiling tube D. More than you have indicated in your diagram. Flask C should be full of an ice water salt mixture at −4 °C surrounding the boiling tube up to below the ventilation hole in D. It must not cover it or spill over into D as this would give you a wrong result. The thermometer measures the temperature in the region of D where the water condenses. If the thermometer indicates a rise in temperature above a set limit (I do not now have access to data where I could look up this temperature. However, I think it should be in the region of 5–10 °C) then the rate of heating show be reduced as you might overload the condenser and loose moisture. If the test has been running for some time it may be necessary to add ice as the ice melts to water the cooling efficiency of your condenser would be compromised. To do this some water should be run off from the tube at the bottom of flask C. This would be done Using the glass spout which you have shown. What you have not shown is the piece of rubber tubing (now long perished) and the clip attached to the spout at the bottom of C which would allow you to remove excess water from C. If you do not remove the excess water from C you could cause water to spill over into flask D as the ice is added from the top. If water were to get into D from C then your result would be destroyed and after cleaning everything out you would have to start over again. Remember it is necessary to have some water in C so that the ice can make better contact with D and act as a much more efficient condenser. The volume of water in the butter can be read on the side of flask D. It works well on low fat spreads which contain lots of water. If you want to use do not send for me. Good luck.


Is it possible to identify the source of a protein hydrolysate

DNA is not a protein, as @MARamezani states. See http://www.answers.com/Q/Is_DNA_a_protein.
There is a prevalence (or lack) of some amino acids in certain sources. For example, maize ("corn" in America) is deficient in tryptophan and lysine, so if your sample is short on those two, it is more likely that it is from corn. A list of some foods and amino acid content is at http://www.ljhs.sandi.net/faculty/aquesnell/Biology%20Notes/Chapter%204/4.8-4.13/Essential%20Amino%20Acids.pdf [there appear to be some errors, though]. According to that source, yeast may be deficient in phenylalanine, so that is another indicator.

You would need an analysis of the relative abundance of each amino acid, some reference showing their abundance in various sources, and additional analysis to calculate relative abundances in a mixture, e.g. corn supplemented with yeast hydrolyzate.


Is the flint used in disposable lighters toxic?

The "flint" in disposable lighters is made of pyrophoric alloy called "ferrocerium", which consists of approx. 50% cerium, 25% lanthanum and 20% iron; the remaining 5% is magnesium, neodymium and praseodymium. Exact composition may vary depending on the manufacturer.

If you rub such an alloy on a rough surface, tiny metal particles are formed, which are instantly ignited and produce metal oxides. Neither of the expected products are particularly toxic: lanthanoid oxides, particularly $\ce{CeO2}$ and $\ce{La2O3}$, are not toxic and can only cause mild irritation when swallowed, and iron oxide $\ce{Fe3O4}$ is basically rust.

To sum it up, unless you are allergic to transition metals (which people usually discover very early), I cannot think of any health hazards coming from the "flint" or the sparks.


Elevated boiling points in alkanes with even number of carbons

There is not really any difference between the boiling points of even and odd carbon alkanes.  For single-chain alkanes the boiling point just goes up as a smooth curve versus chain length.

The melting points do show a small oscillating component with the even chains being higher.  If we imagine the chains in their idealized zig-zag configuration, which would be relatively favorable for the solid state, we find that the even-length chains have a center of inversion which makes them fit better than odd-length ones.  This effect fades, however, with increasing chain length.

Reference:

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


Why can copper have both four and six bonds?

$\ce{Copper(II)}$ complexes are $\ce{d^9}$ type complexes. This means that there are 9 d-electrons for the metal centre. When a complex forms, a number of ligands coordinate to the metal centre and the species adopts certain geometry. The five d-orbitals are no longer the same energy. They split into sets of different energies depending on the geometry and how ligands interact with them.

For an octahedral complex, orbitals split into two sets of degenerate orbitals - a lower $\ce{t_{2g}}$ ($\ce{d_{xy}},\ce{d_{yz}},\ce{d_{xz}}$) which is bonding to non-bonding and a higher $\ce{e_g}$ ($\ce{d_{z^2}}, \ce{d_{x^2-y^2}}$) which is slightly anti-bonding.

Copper is a bit funny as it experiences what is called a Jahn-Teller distortion where there is an elongation of the metal-ligand bonds along the z-axis. This is due to uneven occupancy of the $\ce{e_g}$ orbitals set. Then, the orbitals further split to remove the degeneracy and gain more stability. For the square planar complex of $\ce{tetraaquacopper(II)}$ to form you would need to go even further and remove those two ligands along the z-axis. The stabilization energy gained overall by going from octahedral to square planar configuration is what allows this switch to occur; it wouldn't happen if it were unfavorable. The gain in stability is borderline and that is why I assume that both configurations are possible.

Also, square planar fields tend to prefer stronger field ligands and water is not such ligand and this might also be considered as a reason why you get a mixture of the two types of geometry.


Why are there only 7 types of unit cells and 14 types of Bravais lattices?

All quotes will be from Solid State Physics by Ashcroft and Mermin.

Bravais Lattice:

  A fundamental concept in the description of any crystalline solid is that of the Bravais lattice, which specifies the periodic array in which the repeated units of the crystal are arranged.  The units themselves may be single atoms, groups of atoms, molecules, ions, etc., but the Bravais lattice summarizes only the geometry of the underlying periodic structure, regardless of what the actual units may be."

Primitive Unit Cell:

  A volume of space that, when translated through all the vectors in a Bravais lattice, just fills all of space without either overlapping itself or leaving voids is called a primitive cell or primitive unit cell of the lattice.

Unit Cell; Conventional Unit Cell:

  One can fill space up with nonprimitive unit cells (known simply as unit cells or conventional unit cells). A unit cell is a region that just fills space without any overlapping when translated through some subset of the vectors of a Bravais lattice. The conventional unit cell is generally chosen to be bigger than the primitive cell and to have the required symmetry.

Crystal Structure:

  A physical crystal can be described by giving its underlying Bravais lattice, together with a description of the arrangement of atoms, molecules, ions, etc. within a particular primitive cell.

So, one comes up with 14 Bravais lattices from symmetry considerations, divided into 7 crystal systems (cubic, tetragonal, orthorhombic,monoclinic, triclinic, trigonal, and hexagonal). This comes solely by enumerating the ways in which a periodic array of points can exist in 3 dimensions.

Now, what is on those points is a unit cell, which will itself have some symmetry. Thus, the combination of Bravais lattice and unit cell symmetry can again be enumerated and one comes up with 230 space groups.

Now for some of your related questions:

All cubic-related Bravais lattices will have 90 degree angles because they are based on cubic symmetry.  The trigonal Bravais lattice has no 90 degree angles, but isn't talked about much in more basic textbooks because, well, it looks weird.

Why no pentagonal unit cells? Well, because you can't fill space with a 5-fold symmetric Bravais lattice. Quasicrystals, while they have 5-fold symmetry, are a tiling through space that does not obey the rules for a Bravais lattice.


Software/server for calculation of protonation states of small molecules depending on pH

This is a challenging problem and there are many programs that do this.

One review was published a few years ago: "Comparison of Nine Programs Predicting pKa Values of Pharmaceutical Substances", J. Chem. Inf. Model., 2009, 49 (12), pp 2801–2812

  Knowledge of the possible ionization states of a pharmaceutical substance, embodied in the pKa values [...] is vital [...] We compare nine commercially available or free programs for predicting ionization constants. [...] We compared their performances by applying them to 197 pharmaceutical substances with 261 carefully determined and highly reliable experimental pKa values from a literature source. The programs ADME Boxes 4.9, ACD/pKa DB 12.0, and SPARC 4.2 ranked as the top three with mean absolute deviations of 0.389, 0.478, and 0.651 and $r^2$ values of 0.944, 0.908, and 0.894, respectively. ACD/pKa DB 12.0 predicted all sites, whereas ADME Boxes 4.9 and SPARC 4.2 failed to predict 5 and 18 sites, respectively.

Surprisingly, the quantum program did poorly.

There's even a full book on the subject: Computational Approaches for the Prediction of pKa Values by George Shields and Paul Seybold. It covers both quantum and QSAR approaches.

Some of the best seem to be:

ACD/pKa
ADMET Predictor
Propka Also available as open source code: https://github.com/jensengroup/propka-3.1
Marvin / ChemAxon

But there are dozens out there. I'd check the review articles and try a few.


How does a copper piece added in nitric acid change after nitrogen dioxide starts appearing?

Metals often have corrosion on their surface, such as oxides.  By dipping in nitric acid and then rinsing with water you would achieve a true copper metal surface free of oxide or other corrosion products.


Looking for a toxic substance that cannot exist at room temperature

As Ivan Neretin already explained in his comment, you can't have it all.

Anyway, cyanogen, $\ce{N#C-C#N}$ would be a nice candidate. It is colourless, but highly toxic (H330). If it's cold enough, that is below -21 °C, you can even have a pond of it ;-)

In addition, it adds a historical space touch to the story. Back in 1910, when the return of Halley's comet was expected, cyanogen was spectroscopically identified in the comet's tail and the New York Times titled COMET'S POISONEOUS TAIL ;-).


How would I calculate the Normality of a 0.00167 molar solution of KIO₃?

Hmm.. I think that your data is wrong, I get that the normality:molarity ratio must be 5 in this case.

This ratio (I call it the $\eta$-factor) for a compound participating in a reaction is the net change of its oxidation state.

In this case, one iodine in $\ce{IO3-}$ (+5) becomes one iodine in $\ce{I2}$ (0). The net change in oxidation number is 5, so $\rm normality=5\times molarity$. Normality is like the reducing/oxidising "power" of a solution--or the concentration of electrons that it can furbish/absorb.

Your density, your $\mathrm{g~L^{-1}}$, or your value for normality is wrong here.

Note that one needs to be careful with this reaction. It is comproportionation, where the same element is oxidising itself (Basically, two forms of the element in different oxidation states are reacting with each other). We can have some trouble when dealing with the products of such a reaction, though dealing with the reactants is fairly simple, as shown above. I'm writing a detailed self-answered post on this later.


Formation of peracetic acid from acetic acid and hydrogen peroxide and its stability in their presence

The formation of peracetic acid from acetic acid and hydrogen peroxide is an equilibrium reaction and so in order for the peracetic acid to remain at a constant concentration as desired, acetic acid and hydrogen peroxide must also be present so that the reaction can be at equilibrium. $$\ce{CH3COOH + H2O2 <=> CH3COOOH + H2O}$$

  what exactly keeps the excess hydrogen peroxide from reacting with the excess acetic acid to simply form more peracetic acid?

Nothing. The hydrogen peroxide reacts with the acetic acid to form peracetic acid and water but the peracetic acid and water also react to reform hydrogen peroxide and acetic acid. At equilibrium the rates of these reactions are equal and so the concentrations of the species do not change. At room temperature the equilibrium constant is somewhere around 2.5 (see references) which indicates that significant amounts of reactants and products are present, supporting the reasoning above.

The paper Chin J Proc Eng 2008 February, 8 (1), 35–41 has some data on the values of the equilibrium constant and the variation in concentration of peracetic acid formed.

Another value for the equilibrium constant on page vii of Unis, Melod (2010) Peroxide reactions of environmental relevance in aqueous
solution. Doctoral thesis, Northumbria University.


How much of Sodium dichloro-s-triazinetrione dihydrate should I use to replace bleach?

According to a relevant reference (Ref.1) Following reactions happen when sodium hypochlorite and sodium dichloro-s-triazinetrione dissolve in water, respectively:

$$\ce{NaOCl + H2O -> HOCl + NaOH} \tag {1}$$

$$\ce{NaCl2(NCO)3 + 2 H2O <=> 2 HOCl + NaH2(NCO)3} \tag {1}$$

Hence it is safe to assume that you need only $\pu{0.5 mol}$ of $\ce{NaCl2(NCO)3}$ (sodium dichloro-s-triazinetrione) or in your case $\ce{NaCl2(NCO)3.2H2O}$ (sodium dichloro-s-triazinetrione dihydrate), in place of each $\pu{1.0 mol}$ of $\ce{NaOCl}$ used. Let's see how much of $\ce{NaOCl}$ in $\pu{1.0 US gal}$ of $5.25\%$ bleach solution:

$$5.25\%(w/v) = \\
 \frac{\pu{5.25 g} \;\ce{NaOCl}}{\pu{100 mL} \;\text{solution}}\times \frac{\pu{1.0 mol} \;\ce{NaOCl}}{\pu{74.44 g} \;\ce{NaOCl}}\times \frac{\pu{1000 mL} \;\text{solution}}{\pu{1.0 L} \;\text{solution}}\times \frac{\pu{3.79 L} \;\text{solution}}{\pu{1.00 US gal} \;\text{solution}}\\
 = \pu{2.673 mol/US gal}$$

As states above, you need only half of this amounts in solid $\ce{NaCl2(NCO)3.2H2O}$. You can calculate the amount of $\ce{NaCl2(NCO)3.2H2O}$ in grams as follows:

$$\pu{1 US gal} \: \text{of } 5.25\%(w/v)  = \\
\pu{2.673 mol}  \; \text{of } \ce{NaOCl} \times \frac{\pu{0.5 mol} \; \ce{NaCl2(NCO)3.2H2O}}{\pu{1.0 mol} \; \ce{NaOCl}} \times \frac{\pu{255.98 g} \;\ce{NaCl2(NCO)3.2H2O}}{\pu{1.0 mol} \;\ce{NaCl2(NCO)3.2H2O}} \\
= \pu{342.1 g}  \;\ce{NaCl2(NCO)3.2H2O}$$

Thus, you can dissolve $\pu{342.1 g}$ of $\ce{NaCl2(NCO)3.2H2O}$ in your 30-gal (US) water barrel to get the same effect of 1-US gal of 5.25% bleach solution. Keep in mind that rate of chlorine release from $\ce{NaCl2(NCO)3.2H2O}$  is slower than that of bleach solution (Ref.1).

Late Edit:

This is about comment made by OP on my post. In that comment, OP states that the cost for dichloro-s-triazinetrione dihydrate (Di-Chlor) and beach are almost same. But I do not agree with that. Just for curiosity, I checked the prices for two chemicals: You can buy $\pu{50 lbs}$ bucket of Di-Chlor granules [online] from Pool Supply World for $\pu{132.99 USD}$; and You can buy $\pu{121 fl \: OZ}$ container of 6% bleach solution from Target for $\pu{4.49 USD}$ (let's assume its about $\pu{1 US\:Gal}$ of 5.25% bleach solution). According to my calculations above, $\pu{50 lbs}$ bucket of Di-Chlor granules equal to:

$$\pu{50.0 lbs} \; \ce{NaCl2(NCO)3.2H2O} \times \frac {\pu{454 g}}{\pu{1.0 lbs}}  \times \frac{\pu{1.0 US\:Gal}  \;\text{5.25% }\ce{NaClO}}{\pu{342.1 g}  \;\ce{NaCl2(NCO)3.2H2O}}\\
 \approx  \pu{66.5 US\:Gal} \text{ of 5.25% }\ce{NaClO} \text{ solution} $$

As a consequence, the cost for $\pu{50 lbs}$ bucket of Di-Chlor granules is $\pu{132.99 USD}$ while the cost for $\pu{66 US\:Gal}$ of 5.25% bleach solution  containers is $\pu{4.49 USD} \times 66  \approx \pu{296.34 USD}$ (more than twice the cost for Di-Chlor granules needed).

References:

Thomas Clasen, Paul Edmondson, "Sodium dichloroisocyanurate (NaDCC) tablets as an alternative to sodium hypochlorite for the routine treatment of drinking water at the household level," International Journal of Hygiene and Environmental Health 2006, 209(2), 173-181 (https://doi.org/10.1016/j.ijheh.2005.11.004).


Synthesis for turning synephrenine into N-methylphenethylamine

Suggested route - step 1 acetylate the amine with 1 eq of acetyl chloride in pyridine.

step 2 hydrogenation of the benzylic alcohol over Pd/C in the presence of acid or use TFA/triethylsilane

step 3 form the triflate of the phenol

step 4 catalytic transfer hydrogenation of the triflate, conditions here

Step 5 acid hydrolysis of the N-acetyl group with cHCl to give product

Thanks to @PCK for the suggestion of forming and reducing the triflate


What is the most common oxidation state of gold?

It's not obvious, but common oxidation state for gold is +3. It caused by destabilization of the $5d^{10}$ orbital. Detailed explanation you can find in The Chemistry of Gold, in Chapter 1.1.3.


Are van der Waals coefficients independent of temperature?

The van der waals constant can be directly obtained from the critical temperature and pressure. We have,

${a = \frac{27 R^2 T_c^2}  {64 P_c}}$
${b =  \frac{R T_c }{ 8 P_c}}$

Since, $T_c$ and $P_c$ are constants, we can conclude that the van der waals constants are really "constants."


How to synthesize 3‐cyclopentylpropanal from (chloromethyl)cyclopentane?

3 Steps:
Step 1 - react the starting material with the preformed anion of ethyl cyanoacetate (NaOEt/EtOH should do this comfortably)
Step 2 - mild base hydrolysis of the product (NaOH/EtOH) followed by acidification to remove the ethyl ester group.
Step 3 - Dibal-H reduction of the nitrile to aldehyde 1.


How will liquid nitrogen affect a steel instrument's ability to function?

Metals become brittle as low temperature (and that's why you cool your Parr bomb in a dry ice bath, never in liquid nitrogen when you condense gases in).  But the worst thing that can happen is that a claw might break off, which shouldn't cause any damage to equipment.  Why not try in a small dewar first.  Liquid nitrogen is cheap.

A good alternative to racks of storage boxes are storage canes, such as these.


Aren't unit cell and the relevant crystal similar?

The statement from your book is correct, though the devil lies in the details. The point here is that the book is treating a theoretical crystal (i.e. one that is infinitely large). Let me elaborate and then answer your question.

A unit cell is simply a representation of a periodic structure. It is a means of conveying translational symmetry (and perhaps other types of symmetry as well, but let us focus on translational symmetry). If you have for instance a cubic unit cell, than a 3x3x3 grid of these unit cells is also a unit cell of that same lattice. If you extrapolate this idea, then the reasoning you pose in your question makes a lot of sense!

Let us now discuss the details. Assume that we have a theoretical crystal of infinite dimensions. What is then the unit cell? Can we define a unit cell which is also the infinite crystal? Every finite unit cell would obviously not be that crystal. This explains why we even use a unit cell in the first place: By defining the translational symmetry by means of the unit cell, we can define an infinite crystal.

In the field of solid state physics, it is attempted to keep things simple and well-defined and therefore a so-called primitive unit cell is introduced. This unit cell only contains one lattice point and you could see this as the smallest unit cell possible that is still able to convey the translational symmetry of the infinite crystal.

Your last question relates to why we want to use, for instance, the primitive unit cell or some other unit cell rather than the complete crystal? Let me answer this question by giving an example: If you cut a crystal, you will create a surface facet. Such surfaces also have translational symmetry and you can define these using Miller indices. Your textbook will probably treat these as well. Those Miller indices are defined by the reciprocal lattice vectors of your unit cell. You already touched upon on the fact that there exists an infinite set of unit cells that define that infinite crystal, yet the question arises which unit cell to use if we, for instance, want to use these Miller indices. In principle, any unit cell in the set would be valid. To avoid any kind of ambiguity, we tend to use the smallest unit cell possible.

A nice book treating these concepts is "Introduction to Solid State Physics" from Charles Kittel.


acyclically stereocontrolled reaction: organolithium and 2-chiral compound

I have just had a look at the paper and actually they reported that their main product was S,R:

"...treated with excess 2-lithio-1,3-dithiane in THF at -50°C; this treatment generated the tertiary alcohols 22a and 22b in 72% combined yield with 6.2:1 diastereoselectivity as predicted by Cram's model based on participation of the piperidine nitrogen atom, giving a rigid cyclic complex with the lithium compound".

Compound 22a: (2S)-2-[(R)-l-(1,3-Dithian-2-yl)-l-hydroxyethyl]piperidine

Compound 22b: (2S)-2-[(S)-1-(1,3-Dithian-2-yl)-l-hydroxyethyl]piperidine

I do not find any mistake.


How many electrons with l = 1 does Si in its ground state have?

You're on the right track, but those superscripts in the electronic configuration don't refer to orbitals. They refer to the number of electrons in that type of orbitals.

For more information, you might want to look at this question: What are the maximum number of electrons in each shell?


Why is sodium carbonate less soluble in water than sodium bicarbonate?

Oops! The question contains an incorrect assumption. But chemists are amazing - they can still answer such a question:

Sodium carbonate is way more soluble than sodium bicarbonate at all temperatures above about freezing. $Na2CO3$ has several crystalline hydrates that are temperature sensitive and make the curve less smooth than most other salts. (I suppose the caffeine is in there to wake us up.)
So, I woke up to the fact that when we calculate the weight of "sodium carbonate" which dissolves at 0 C, it's the decahydrate, whereas sodium bicarbonate is anhydrous, so at 0 C, the solubility of sodium bicarbonate is 0.833 moles per liter, while the same weight of sodium carbonate decahydrate dissolved in a liter is only 0.245 moles.
Now I'm more puzzled by why sodium carbonate has such a complex hydrate system, and sodium bicarbonate doesn't.


Confidence in conformation search results from molecular mechanics

We have published² a probabilistic treatment of this question, but for the case of MD simulations (and not MC). The principal idea is to use Good-Turing statistics to estimate the probability of unobserved species (ie conformations that have not -as yet- been observed). I have no idea whether this treatment can be transferred to MC, but there is an open source program³ -based on R- that you could try to use.

² https://pubs.acs.org/doi/10.1021/ci4005817

³ https://github.com/pkoukos/GoodTuringMD


For this example, the volume of the solution doesn't matter. It's a thought experiment, doesn't say what sovent is used, and you could replace ethane by helium or miraculane.

As long as you talk about the same compound, mass is proportional to molar quantity. Plus the units cancel out in the calculation anyway.

Think of the hypothetic experimentator as a guy who only has one glass beaker, without graduation. ;)


How to designate the position of an amine substituent in an alkyl-chain?

In general usage, the practice of omitting locants when there is no ambiguity is widespread. In particular, the locant ‘1’ is often omitted when there is no ambiguity.

Clearly, the locant ‘1’ for the amine group in unsubstituted ethanamine can be omitted since the isomer ‘ethan-2-amine’ does not exist.

Furthermore, the locant ‘1’ for the amine group can be omitted in substituted ethanamines provided that the amine group is the principal characteristic group (i.e. the characteristic group chosen for citation at the end of the name by means of a suffix or a class name), for example 2-chloroethanamine.

Hence, the omission of the locant ‘1’ for the amine group in 1-phenylethanamine is permissible in general usage. Since locants are placed immediately before that part of the name to which they relate, it is clear that that the remaining locant ‘1’ in 1-phenylethanamine refers to the phenyl group.

However, for absolute clarity in preferred IUPAC names, the current IUPAC recommendations are prescriptive about when omission of locants is permissible. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), locants are omitted in preferred IUPAC names in the cases that are mentioned in Subsection P-14.3.4, in particular:

  P-14.3.4.2 The locant ‘1’ is omitted:

  (a) in substituted mononuclear parent hydrides;

  (b) in monosubstituted homogeneous chains consisting of only two identical atoms;

  (…)

Thus, for example, methanamine (a) and ethanamine (b) are preferred IUPAC names.

Generally, however, no locants can be omitted if any locants are required to identify the specific isomer.

  P-14.3.3 Citation of locants

  In preferred IUPAC names, if any locants are essential for defining the structure of the parent structure or of a unit of structure as defined by its appropriate enclosing marks, then all locants must be cited for the parent structure or that structural unit. For example, the omission of the locant ‘1’ in 2-chloroethanol, while permissible in general usage, is not allowed in preferred IUPAC names, thus the name 2-chloroethan-1-ol is the PIN. (…)

Therefore, the omission of the locant ‘1’ for the amine group in 1-phenylethanamine, while permissible in general usage, is not allowed in preferred IUPAC names. The preferred IUPAC name is 1-phenylethan-1-amine.


Electrolytes in electrochemical half cells

For example, a silver electrode is immersed in a silver solution and so on, but is why is this necessary? Surely any positive metal ion will do?

There is a fundamental reason, which is the Nernst equation. It requires that the metal be in contact with its own ions. What is the functional meaning of the electrode potential? It is the tendency of an ion to reduce itself. Yes one can make salt water battery by dipping an iron nail (or copper) in one half cell and a piece of aluminum in another half cell but you cannot predict the cell potential beforehand.


How to write an entire trajectory into a single PDB file?

#!/usr/bin/env python2

import MDAnalysis

# file formats are automatically deduced; this is a standard topology and binary
# (NetCDF) trajectory from AMBER
topology = 'peptide_1.prmtop'
trajectory = '02_peptide_1_equil_NVT.nc'
u = MDAnalysis.Universe(topology, trajectory)

with MDAnalysis.Writer("all.pdb", multiframe=True) as pdb:
    for ts in u.trajectory:
        pdb.write(u)

One thing that may be important: I see lots of warnings when trying to use Python 3, so 2.7 is probably better at this point.


Can one predict the entropy change for aqueous phase reactions without calculation?

Some ions when dissolved in water may form a solvation shell that orders the solvent molecules in a way that the entropy decrease through this ordering outweighs the entropy increase through the ions leaving the crystal structure. Then you would have an overall decrease in entropy when dissolving the salt.

Small ions with high charge have a tendency for this.


What is the specific functional group in MDMA that caused addictive effect on user?

Firstly, MDMA is widely regarded not to be physically addictive, which simply means that the compound (actually the absence thereof) will not cause potentially life-threatening symptoms on discontinuation, as is the case with alcohol or morphine. Note however that this does not make any statement on the addictive potential of the substance, as e.g. cocaine also falls into this category of substances, which are called psychologically addictive.

In general psychological addiction (which is not necissarily associated to substances, but can also involve habits such as gambling or eating) is widely believed to be strongly associated with the dopamine system and the overstimulation that dopamine facilitates during the addictive actions.

Which brings me to the second point:
The active site of the molecule. In a way, looking at both structures you will notice a set of similarity between dopamine and MDMA. It is not that simple however. The effects of MDMA are quite complicated and due to its status as illegal substance research is kind of limited.

Known effects are:

Known and experimentally confirmed increase in dopamine levels, although the mechanism is not known.
Serotonine release from vesicles to synaptic cleft (has effects on dopamine levels through mediation/coupling)
MAO inhibition (decreases dopamine and serotonine degradiation - among others)

Other effects like oxytocin response are also proposed and may also have effect on dopamine levels. But in this case the effect could as well be a down-regulation.

Which structural component is responsible for all of this? All of them, as you can tell from the complexity of the physiological reaction on the substance, especially if compared to similar compounds with in some regards drastically different physiological effects. Think of (meth-)amphetamine(very high dopamine response, next to no serotonin effects) or mescaline(direct agonist on specific serotonin and dopamine receptors, as well as dopaminergic effects). (It would really blow the reach of this question, but taking a look at other compounds, such as 2-CB or 3,4-CTMP will make this picture even more complicated, but also more complete...)

These substances can give you at least some hints at two honorable mentions in terms of 'addiction-causing functional group':

(N-Me-)Phenethylamine
3,4-(+M)-substitution pattern on the aromatic ring

A small sidenote on the difference between physical and psychological addiction: I am not 100% sure about the current research in that topic, but afaik in both cases there is in overexpression of receptors that causes the addiction, only that in the case of a psychological addiction the effects of the receptor-agonist is not live-threatening. In that sense the terminology which kind of suggests that psychological addiction is not associated with any changes in brain structure is somewhat misleading.


Is it possible to calculated the concentration of hydroxide ions from the known concentration of hydronium ions?

Yes, that works. In fact, some textbooks and teachers ask this type of question (asking for the lesser ion of the autodeprotonation of water) to test which students actually read the question and which gloss over details. I fell into the trap of calculating the wrong one and stopping too early, too.

So if your question had asked for the hydroxide concentration, you would indeed calculate the hydronium concentration first (because that is actually accessible) and then invert and multiply $10^{-14}$ to get the desired one as you suggest.


How many electrons can an orbital of type f hold?

You're correct that there are seven possible spatial orientations for an f-type orbital, and hence seven possible orbitals in one f-type sub-shell. However, the question specifically asks for the maximum number of electrons in one such orbital, and any single atomic orbital, regardless of the sub-shell type specified by $l$, can only hold two electrons. This is by virtue of the Pauli exclusion principle. Fourteen would be the maximum number of electrons across an entire f-type sub-shell, but the question only asks about one orbital.


What is the meaning of writing a catalyst name with two compounds split by a hyphen?

If we consider palladium on carbon or alumina then it means that the palladium particles are dispersed onto the carbon or alumina. The carbon or alumina acts as a support for the catayltic particles of the palladium.

If we consider Lindlar's cataylst this can be written as Pd/CaCO3/Pb

The Pd (palladium) is the hydrogenation cataylist
The CaCO3 is the support
The Pb is a poison which is used to modify the chemistry of the palladium

Sometimes one of the things separated by a / line is either a promotor or a poison.


Gibbs energy derivative vs equilibrium for aqueous reaction

To derive the Gibbs' free energy of an ideal solution you would invoke Raoult's law, which says that
$$\mu_i(l) = \mu_i^*(l)+RT\ln\left(\frac{p_i}{p*}\right)= \mu_i^*(l)+RT\ln\chi_i$$
Here the asterisk refers to the pure substance. Using the notation in the OP, the total number of moles of reagent and product in solution is

$$n_{tot}=2(n_{ini}-\xi)+\xi+n_{solv}=2n_{ini}-\xi+n_{solv}$$

where $n_{ini}$ is the initial amount of A or B (assumed equal), so that one can write for reagents A and B:
$$\mu_i(l) = \mu_i^*(l)+RT\ln\left(\frac{n_{ini}-\xi}{2n_{ini}-\xi+n_{solv}}\right)$$
and for product C:
$$\mu_C(l) = \mu_C^*(l)+RT\ln\left(\frac{\xi}{2n_{ini}-\xi+n_{solv}}\right)$$
There is also a term for the solvent:
$$\mu_{solv}(l) = \mu_{solv}^*(l)+RT\ln\left(\frac{n_{solv}}{2n_{ini}-\xi+n_{solv}}\right)$$
which I ignored in earlier edits.
Adding terms to compute the total Gibbs energy and taking the derivative wrt the extent of reaction $\xi$ then results in

$$\begin{align} \Delta G' &= \Delta \mu^* + RT \ln\left(\frac{\xi(2n_{ini}-\xi+n_{solv})}{(n_{ini}-\xi)^2}\right)  \\  &= \Delta \mu^* + RT \ln\left(\frac{n_Cn_{tot}}{n_A n_B}\right)  \\&= \Delta \mu^* + RT \ln \left(\frac{\chi_C}{\chi_A \chi_B}\right) \end{align} $$

($^\dagger$)

Note that here the standard state for each substance refers to the pure liquid in equilibrium with its vapor at the same temperature. To obtain an expression relating a change in standard free energy to the concentration equilibrium constant requires expressing mole fractions in terms of concentrations. This is generally done for dilute solutions where $n_{solv} \approx n_{tot}$ by writing $\chi_i\approx c_iM_{solv}/\rho_{solv}$, which leads to

$$\Delta G' = \Delta \mu^* - RT \ln \left( \frac{M_{solv}}{\rho_{solv}} \right) + RT \ln \left(\frac{c_C}{c_A c_B}\right)  $$
Then by redefining the standard states as
$$\mu_i^\circ = \mu_i^* + RT \ln \left( \frac{M_{solv}}{\rho_{solv}} \right) $$ leads to $$\Delta G' = \Delta \mu^\circ + RT \ln \left(\frac{c_C}{c_A c_B}\right)  $$
If you think the standard state looks strange, consider that if $\chi_i = c_iM_{solv}/\rho_{solv}$ then $$\chi_i(\pu{c^\circ_i= 1 M}) = c_i^\circ \frac{M_{solv}} {\rho_{solv}}$$ and so we can write

$$\mu_i^\circ = \mu_i^* + RT \ln \left( \chi_i(\pu{c^\circ_i= 1 M}) \right) $$ which is more in line with canonical definitions of standard states. The new standard state is defined as a 1 M ideal solution with chemical potential related to that of the pure solute by the above equation.

$^\dagger$/TL;DR :  The mysterious extra $RT$ term in the OP is gone. It's important to keep a term for the solvent. And a lesson from these repeated edits: careful doing math when tired!


When is donation into an anti-bonding MO stabilising?

The difference between the "stabilising" and "destabilising" examples that you have chosen are that the "stabilising" examples are static molecules, whereas the "destabilising" examples are chemical reactions, in which a lot is happening. We'll look at the "stabilising" cases first, as these are easier to understand.

As with any question involving "stability", it is important to consider what we are measuring stabilisation with respect to. In the example of the anomeric effect (e.g. in esters), we are comparing the true electronic state of the ester against a hypothetical scenario, where the lone pair and σ* do not overlap, but everything else is kept the same. In such cases, it is nearly always true that allowing overlap, or delocalisation, leads to a lower energy. This is captured quite well in the last picture in your question body. (See also: Physical reason for why delocalisation leads to stability?)

In a chemical reaction, as we move along the reaction coordinate (RC), we are creating overlap between two different orbitals, just as in the previous case. Naively we might think that this necessarily leads to stabilisation. But this is not physically sensible: if the reaction proceeds via a transition state (TS), this TS must be higher in energy than the reactants (at infinite separation).[1] The problem is that when we bring two molecules together in space, it is not just overlap that is affected: the energies of every orbital will change as nuclear positions are moved, electron-electron repulsions are different, etc. This means that all the orbital energies change in a way which depends not only on overlap with other orbitals (as was the case before), but also a number of other factors.

Instead of going from reactants at infinite separation to the TS directly along the RC, as would be physically realistic, it may be instructive to consider hypothetically doing it in two stages. In the first step, we bring the reactants together from infinite separation to the TS geometry while forbidding any orbital overlap; secondly, we stay at the TS geometry, but turn on the overlap. In real life, you cannot separate these two things: changing the geometry comes with formation of overlap. But thinking about it this way allows us to link it to the "stabilising" cases talked about before.

Let's use the addition of a nucleophile to a carbonyl group as an example to make this more concrete. In the reactants, there are three key frontier orbitals we need to consider: the nucleophile lone pair ($n$), and the π-type orbitals of the carbonyl group ($\pi$ and $\pi^*$).

In general, as we bring the nucleophile and carbonyl together, mixing of all three orbitals will occur continuously. As the reaction proceeds, we will therefore have three MOs $\{\psi^{(i)}\}$ $(i = 1,2,3)$ which are linear combinations of the three frontier orbitals listed earlier:

$$\psi^{(i)} = c^{(i)}_n\phi_n + c^{(i)}_\pi \phi_\pi + c^{(i)}_{\pi^*}\phi_{\pi^*}$$

For well-separated reactants, the relevant MOs will (for the most part) be the original three frontier orbitals. So, for example, we could have $\psi^{(1)} = \phi_\pi$, $\psi^{(2)} = \phi_n$, $\psi^{(3)} = \phi_{\pi^*}$. But at the end of the reaction, the MOs of the product will be a complicated mixture of the three orbitals. For example, the $\ce{C-Nu}$ $\sigma$ orbital might be made up of something that resembles this:

$$\psi^{(1,\text{new})} \approx \frac{1}{\sqrt{3}}(\phi_n + \phi_\pi + \phi_{\pi^*})$$

The reason why these linear combinations (or mixtures) can form is because there is non-zero overlap between all of these orbitals as the reaction proceeds. If there is no overlap between two orbitals, then they cannot mix. So, the way to forbid any overlap from occurring in the first step is to simply not change the coefficients of the MOs as we bring the reactants closer to each other (towards the TS geometry). Of course, this is not physically possible, which is why it is purely a hypothetical scenario.

Even though the MOs themselves do not change in this first step, their energies do change, because the Hamiltonian is changing (by virtue of the nuclear positions changing). So, the energies of the MOs will change. To throw an additional wrench into the works, it is not just the electronic energy that we must consider, but also the nuclear energy (when we bring nuclei closer to each other, they will repel each other, and this is not captured in MO diagrams which only illustrate electronic energy). As you can imagine, it is very hard to make any real analysis as to what happens in this step.

In any case, once we reach the TS geometry, the energy of the system will definitely have changed in some way. We can now carry out the second step by "turning on" the overlap, i.e. finding the appropriate linear combinations of ${n, \pi, \pi^*}$. Since nothing else here is changing (in particular, no nuclear movement), except for us "allowing" overlap to occur, it is exactly analogous to the "stabilising" case which was discussed earlier. So the second step always leads to a decrease in energy.

The problem is that, from the infinitely-separated starting materials, we have carried out two steps to get to the TS. In the second step, the energy has decreased – a "stabilisation" of sorts. But in the first step, we don't know what has happened. If a TS is to be higher in energy than the reactants, then it follows that there must have been an increase in energy in the first step.[2]

Simple LCAO-MO diagrams don't quite capture the complexity involved in the first step – they can only really properly describe the second step. A better way of drawing it might be to do something like this:

I haven't drawn the orbitals in the TS, but you can imagine that they are starting to morph into what they will be in the products. So $\psi^{(1)}$ will be somewhat intermediate between $\pi_\ce{C=O}$ and $\sigma_\ce{C-Nu}$. Of course, this is purely hypothetical - especially in the middle, the orbitals and their energies have no physical meaning at all. In a real reaction you would go straight from SM to TS, and not via an unphysical intermediate stage as is depicted here.

So, therein lies the difference between the two cases you have described. In the "stabilising" case, the only thing that is happening is a magical "switching on" of overlap. In the "destabilising" case of a chemical reaction, in addition to the overlap (which is stabilising), you also have a glut of other factors which can lower or (more frequently) raise the energy of the system as it move towards the TS. The problem is that we cannot separate the two from each other. Creating overlap comes with all the other strings attached, and so overall as we move along the RC towards the TS, there is (typically) an increase in energy.

Finally, how do we relate this back to the description of nucleophilic addition in Clayden? The wording is definitely somewhat loose (organic chemists frequently do this for ease of use), but it does have a grounding in quantum theory. I'll reproduce the image here and try to relate each phrase:

  Electrons in HOMO begin to interact with LUMO

This refers to the developing overlap that we looked at in step 2, i.e. the formation of MOs which are linear combinations of the HOMO ($n$) and LUMO ($\pi^*$). On its own this seems to be a stabilising effect, but recall that this does not come on its own: we also have to consider the different nuclear positions and its implications on the energy.

  Filling of $\pi^*$

I think the way to interpret it is that, at the start of the reaction the two filled orbitals $\psi^{(1)}$ and $\psi^{(2)}$ are simply $n$ and $\pi$, so the coefficients $c_{\pi^*}^{(1)}$ and $c_{\pi^*}^{(2)}$ are zero. As the reaction progresses, these coefficients steadily increase, which in a sense corresponds to a population of the $\pi^*$-orbital, even though the $\pi^*$-orbital ceases to exist in its own right once we bring the starting materials together.

  causes $\pi$ bond to break

In a sense we could define a "$\pi$ bond order" to be some kind of difference between the coefficients of the $\pi$ orbital in the filled MOs, and the coefficients of the $\pi^*$ orbital in the filled MOs. At the start these are $1$ and $0$ respectively, for a bond order of $1$. As the reaction proceeds and the orbitals mix, we find that $c_\pi$ decreases in the filled MOs, $c_{\pi^*}$ increases, and at the end of the reaction we have a bond order of $0$. So loosely speaking, it is true that "population of the $\pi^*$ orbital" (i.e. increasing $c_{\pi^*}$ in the filled MOs) leads to the cleavage of the $\pi$-bond.

  Electrons from $\pi$ bond end up as negative charge on oxygen

This is somewhat true again since the filled MOs morph from $\pi_\ce{C=O}$ plus a lone pair to $\sigma_\ce{C-Nu}$ and a lone pair, but it's important to remember that it is not a one-to-one correlation, i.e. the $\pi_\ce{C=O}$ MO does not become the oxygen lone pair all by itself.[3] More properly, the oxygen lone pair is a linear combination of multiple MOs in the starting material.

It's also important to not ascribe too much importance to exactly which electrons go where, as this violates quantum indistinguishability.

Notes

[1] There are barrierless reactions which do not proceed via a transition state, such as $\ce{H + H -> H2}$, and in such cases it is quite true that we always get stabilisation as we move along the RC towards products. However, these reactions don't fully capture the complexity involved in other reactions with TSes and it's better to think of these as being special cases of the other reactions.

[2] Barrierless reactions don't obey this line of thinking, of course. In fact, they don't even have a TS, so there isn't a well-defined geometry for us to distort towards in the first step. You can pick any geometry along the reaction coordinate between reactants and products, and it would probably work.

[3] Under some special circumstances it is possible to have a one-to-one relationship between reactant MOs and product MOs. See e.g. the first correlation diagram in Theoretical basis behind orbital correlation diagram for pericyclic reactions. But note that here there is only one orbital of each symmetry species, so there is no mixing going on and no linear combinations being formed. See also the last section in the answer on "Real pericyclic reactions".


G and H are like altitude.  They are not absolute; rather their values are always relative to whatever you set as a reference. So they can be either positive or negative.  There is no general constraint on the sign of their values.  For this reason, we usually talk about $\Delta G$ or $\Delta H$, since it's the changes that are meaningful; giving them arbitrary values isn't typically useful.
By contrast, there is a non-arbitrary zero for entropy.  In classical thermodynamics, it's the entropy of a pure substance at absolute zero. Since the temperature can never be lower than absolute zero in a system at thermodynamic equilibrium, $S$ (as opposed to $\Delta S$) can never be negative.
In statistical mechanics, there are various equivalent ways of defining the entropy, depending on the ensemble you are using.  The simplest definition is that for the microcanonical ensemble:
$$S = k_B \ln \Omega,$$
where $\Omega$ is the number of possible microstates of the system, and $k_B$ is the Boltzmann constant.  Since the minimum number of microstates is 1 (which happens at absolute zero), the minimum value for $S$ is zero.


Can we determine apriori the volume of solvent in an electrolyte

You can calculate it using partial molar volumes.

The partial molar volume is defined as the differential change in volume with respect to change in mole fraction, and is determined experimentally. From the definition of partial molar properties, you can calculate the amount of one component needed to achieve a certain volume, if you know all of the partial molar volumes and pure component molar volumes.

There may be some complications to that for this system because you are talking about mixing a solid-phase ionic compound with water, instead of two liquids directly.

In practice, it is much easier to just use a volumetric flask so that you know the total volume is 1 L. If you measure the amount of water you actually used you will find that it is very close to 1 L.


Temperature changes in exo/endothermic reactions

Your confusion is understandable.  I looked at the answers linked in the comment on your OP, and I'd like to explain it differently from how they do.  Hopefully this explanation will get to the heart of your confusion.
You are thinking (incorrectly) that an exothermic reaction works something like this:
The system has a certain amount of thermal energy.  In an exothermic reaction, some of the system's thermal energy is relased to the environment, and thus the system must lose thermal energy and thus decrease in temperature.  I.e., you're imagining something like what happens when a warm block of metal is placed in a cool cup of water.  The metal releases some of its thermal energy to the water, and thus cools.
But that's not what happens in an exothermic reaction.  In an exothermic reaction, the system internally generates extra thermal energy. This is generated through a chemical reaction in which chemical energy (essentially, the energy in chemical bonds) is converted to thermal energy.  Now, if the system is internally generating extra thermal energy, any of a number of things can be done with this energy.  The two most common pictures are as follows:

The system is in a large (relative to the system) constant-temperature heat bath, and is surrounded by diathermal (heat-conducting) walls.  In this case, because the temperature of the system is kept constant, the thermal energy flows out (i.e., is released) into the bath.

The system is adiabatic, i.e., is surrounded by a perfect thermal insulator.  In this case, the extra thermal energy stays within the system, causing its temperature to increase.

The reverse of what I wrote above applies to an endothermic reaction.