Does exchange energy affect completely filled orbitals?

It seems like you are just being introduced to quantum mechanics and is confused with the concept of exchange energy. The concept can be confusing at first, since it neither involves exchanging electrons as classical particles nor has a classical analogue. I will try explaining the problem with a simple two electron system.

Before we discuss, here is a short introduction. In quantum mechanics, one solves the Schrodinger's equation and tries to estimate the wave function of the system. We are talking about time-independent case here, so we solve time independent Schrodinger's equation to obtain the wavefunction.
$$\hat{H}\psi=E\psi$$
Here $\hat{H}$ is the Hamiltonian (energy operator), $\psi$ is the wavefunction and $E$ is the energy. This equation is an eigenvalue equation, which essentially means any valid wavefunction should have a fixed energy irrespective of coordinates. In the equation, the $\hat{H}$ is the only known quantity (operator), $\psi$ and $E$ has to be estimated based on boundary conditions.

Let us consider a system with two electrons, with wavefunction $\psi_1(r)$ and $\psi_2 (r)$, where $r$ denotes the four dimensional space spin coordinates. Now, we know that electrons are indistinguishable and two electrons (fermions) cannot have same space-spin coordinates (Pauli's exclusion principle). Two electrons not having same space-spin coordinates is same as saying that the electrons should not have all same quantum numbers.
So, in order to consider the indistinguishability  of the electrons, we can have two cases of the total wave function: (1) $\psi_1(r_1)\psi_2(r_2)$ or (2) $\psi_1(r_2)\psi_2(r_1)$. On a side note, these are known as Hartree products and are valid for bosons. However, in case we wish to include the Pauli's exclusion principle, we write the system as a Slater determinant, i.e.
$$\psi(r_1,r_2)=\frac{1}{\sqrt{2}}\begin{vmatrix} \psi_1(r_1) & \psi_2(r_1)\\\psi_1(r_2) & \psi_2(r_2)\end{vmatrix}=\frac{1}{\sqrt{2}}\left[\psi_1(r_1)\psi_2(r_2)-\psi_1(r_2)\psi_2(r_1)\right]$$

Now let us talk about the Hamiltonian (energy operator). The Hamiltonian for an atom/molecule can be divided into three parts:

Constants: This include any constant terms, for example, nuclear-nuclear repulsion for poly-atomic molecules.
One electron operator: This part contains terms dependent on single electrons, for example, kinetic energy and nuclear-electron attraction.
Two electron operator: This part contains terms dependent on two electrons, and usually contains only electron-electron repulsion terms. This part gives rise to the exchange energy. We shall expand this part.

Before we go into details, if one is provided with a operator $\hat{A}$ and a wavefunction $\psi$, the expectation (average) value of the observable corresponding to the operator is given as $\langle\psi|\hat{A}|\psi\rangle$, where $\langle\cdot\rangle$ implies integral over all coordinates.

In atomic coordinates, the electron-electron repulsion is given as $1/r_{12}$, where $r_{12}$ is the distance between the electronic coordinates. Now, if we use the wavefunction in the Slater determinant, we get the total two electron interactions as:
$$\left\langle\psi(r_1,r_2)\left|\frac{1}{r_{12}}\right|\psi(r_1,r_2)\right\rangle \\= \frac{1}{2} \left[ \left\langle\psi_1(r_1)\psi_2(r_2)\left|\frac{1}{r_{12}}\right|\psi_1(r_1)\psi_2(r_2)\right\rangle + \left\langle\psi_2(r_1)\psi_1(r_2)\left|\frac{1}{r_{12}}\right|\psi_2(r_1)\psi_1(r_2)\right\rangle - \left\langle\psi_1(r_1)\psi_2(r_2)\left|\frac{1}{r_{12}}\right|\psi_2(r_1)\psi_1(r_2)\right\rangle - \left\langle\psi_2(r_1)\psi_1(r_2)\left|\frac{1}{r_{12}}\right|\psi_1(r_1)\psi_2(r_2)\right\rangle \right]$$
Now, in this cases the first two terms are called Coulomb terms and are easy to imagine classically, i.e. what is the repulsion of electron 1 if present in coordinate $r_1$ with electron 2 if present in coordinate $r_2$ for the first term and vice-versa for the second term. These are repulsive in nature.
The third and fourth terms are attractive terms, and are known as exchange terms. There is no classical analogue of these, and appears purely from the indistinguishability of electrons and Pauli's exclusion principle.

Now remember we discussed that $r_1$ and $r_2$ are space-spin operators, and $r_{12}$ depends only on space? This essentially means that we can separate the space and spin parts while calculating the terms. So the exchange terms vanish while the two electrons have different spins, and remains only when electrons have same spins. This is the origin of exchange energy and IT DOES NOT INVOLVE PHYSICALLY EXCHANGING THE ELECTRONS.

I hope I could explain you in brief the origin of exchange energy. In reality, it is very difficult to calculate the exact energy of a system with N-electrons and say half filled/full filled systems are more stable due to "exchange interactions" simply because:

We do not exactly know how the wavefunctions look. We assuming hydrogenic wave functions but that may not work out always.
Optimising the wavefunctions based on Hamiltonian is carried out self consistently and is costly.
Let us say one resolves the first two issues somehow, still total energy has other terms like kinetic energy, electron-nuclear attraction, Coulomb terms, and even nuclear-nuclear repulsion for polyatomic molecules. So it is very difficult to assert only "exchange energy" is responsible for the extra stability.

Whatever I have tried to explain in short takes up at least two quantum chemistry courses in Bachelor's degree and Master's degree. In case you have some doubts, please feel free to ask in comments. In case you wish to study more, I will suggest the following books (in the order given):

D. J. Griffiths, Introduction to Quantum Mechanics
A. Szabo and N. Ostlund, Modern Quantum Mechanics
I. Levine, Quantum Chemistry


How can benzaldehyde have a pKa of 14.9?

Significant amount of geminal diol of benzaldehyde exists in an aqueous solution of benzaldehyde at 25 °C because $\mathrm{p}K_{\text{hyd}} = 2$ (Ref. 1)

The $\mathrm{p}K_{\mathrm a}$ of benzyl alcohol is listed as 15.40 (Wikipedia). Thus, one can reasonably assume that the given value of $\mathrm{p}K_{\mathrm a}$ 14.9 represents a composite equilibrium constant for the hydration of benzaldehyde and dissociation of the geminal diol thus formed.
Late addition to the answer:
In his paper, "Acidity constants of benzimidazolium ketone and pyridinium aldehyde hydrates" (Ref.2), Terence C. Owen states that:

It is known that the acidity constants of gem-diols typically are about 2.5 units lower than those of the corresponding monohydric alcohols.

When do some extensive literature survey, one can find few examples to backup that statement. For example, $\mathrm{p}K_{\mathrm a}$ of methanol is reported as 15.7 while that of formaldehyde hydrate is 13.3, between which $\Delta\mathrm{p}K_{\mathrm a} = 2.5$ (Ref.2). Interestingly, $\mathrm{p}K_{\mathrm a}$ of 1,1,1,3,3,3-hexafluoropropan-2-ol is reported as 9.22 while that of hexafluoroacetone hydrate is 6.45 where $\Delta\mathrm{p}K_{\mathrm a} = 2.77$ (Ref.3). However, $\mathrm{p}K_{\mathrm a}$ of 2,2,2-trifluoroethanol is reported as 12.37 (Ref.4) while that of 2,2,2-trifluoroethanal hydrate is 10.05 (Ref.3) where the difference is < 2.5 ($\Delta\mathrm{p}K_{\mathrm a} = 2.33$).
Thus, we can conclude that the $\mathrm{p}K_{\mathrm a}$ of benzaldehyde is derived from its hydrate (gem-diol).
Also see: Yoshiro Ogata and Atsushi Kawasaki, In The Chemistry of Carbonyl
Group, Volume 2; Jacob Zabicky, Ed.; John Wiley & Sons Ltd.: New York, NY, 1970, Chapter 1: Equilibrium additions to carbonyl compounds, pp 1–69 (https://doi.org/10.1002/9780470771228.ch1).

Late Edit:
It is also noteworthy that $\mathrm{p}K_{\mathrm a1}$ of the cyclic hydrate of hthalaldehyde is reported to be $11.50 \pm 0.10$ at $\pu{25 ^\circ C}$ (Ref. 5):


As orthocresol also mentioned, it's probably copper sulfate. As you said, the blue color essentially gives it away.

(picture taken from Science Madness Wiki)


Baking soda is sodium bicarbonate NaHCO$_3$. It is stable in air indefinitely. Baking powder is a mix of sodium bicarbonate and a powdered acid, typically an aluminum or calcium hydrogen phosphate, which, in the absence of moisture, is stable for a long time (maybe a year? maybe more). After exposure to moisture, it will release CO$_2$ and become inactivated.
The third item, "chalk", has many definitions. Blackboard chalk is not just a stick of CaCO$_3$. A stick of CaCO$_3$ could be marble, or limestone - oh - no, you would like it to be nice and powdery but sticking together. Yep, it can be done: glue. Nah, too scratchy. Plaster of Paris! Mix with water, add some CaCO$_3$ chalk, maybe some white clay... You get the idea. Blackboard chalk is not just CaCO$_3$. Not even that, but CaCO$_3$ reacts very fast with hydrochloric acid (but that's too hazardous for grade school experiments), and its reaction with vinegar (5% CH$_3$CO$_2$H) is slower initially, and slows down markedly as the reaction progresses because the acetate ion produced inhibits the ionization of the remaining (weak acid) CH$_3$CO$_2$H.
The key lesson is that demonstrations are very temperamental, and depend on doing the same thing with the same materials many times, to see how the "same" experiment can vary so much. You may even begin to doubt that any experiment can ever be duplicated - but it can, if you keep everything constant.


1) How come I've never seen my professor's method mentioned anywhere? I've Googled, looked at scholarly papers; looked in advanced analytical textbooks, etc.

It is entirely possible that no one has published that approach! It is certainly not common in introductory chemistry textbooks.

  2) How were you taught?

I was taught (and currently teach) the "standard" algorithm you found on the Chem Wiki.

  3) Which way do you think is more advantageous?

It depends on your goal.

If your goal is to make sure students understand all of the steps and have a good intuition for what is physically happening, then your professor's method makes a lot of sense. This is because you need to already have a clear understanding of what things like "thermodynamically favorable" and "relative acid strength" really mean in order for that approach to work.

On the other hand, if your goal is to make sure students have the functional ability to work through a particular type of problem, regardless of how well they grasp the underlying concepts, then the algorithmic method makes more sense.

And finally, if your goal is to solve a problem quickly without having to think too hard about details, the algorithmic method also makes more sense.


What function is used to graph the potential curve of two hydrogens with parallel spins?

I believe that you wont be able to find a closed analytic expression for $\mathrm{H_2}$, since no closed analytic expresion is possible for a many-body problem (more than 2 bodies). That said you can calculate as many points on the triplet curve as you like by using a QM program able of doing unrestricted Hartree-Fock (UHF). Here you should be aware that UHF does not give very good potential curves, because the Hartree-Fock approximation lacks the electron correleation energy. $\mathrm{H_2}$ is a quite small system, so you will be able to make a potential curve using unrestricted MP2 or unrectriced CCSD. I do believe that PySCF is capable of doing such calculations. Regarding the experimental data, I am not aware of any databases for such kind of data. I would suggest finding a paper where a curve is presented and then use engauge digitizer to extract the points you need.


Why does covalent nature in ionic solids increase the lattice energy?

(First of all I have to note that I'm just a student, not a specialist in solid body physics or chemistry)

Let's consider chemical bond in general, starting with diatomic molecule (our findings for this case we will generalize to crystalline structure later):

When we say that chemical bond is formed, we mean that for electrons of each atom by atomic interaction (interaction of atomic orbitals of each atom to each other) become available to be situated on molecular orbitals(MO) (defining as superposition with different contributions of atomic orbitals(AO)). Now we must say that depending from atomic orbital energy and its symmetry, result of interaction with another AO may be different.
If we have two interacting AOs with big (bigger than stated value) difference in energies it leads to formation of two MOs: one of them, bonding MO, must have energy closer to such AO that is lower by energy, another relatively must have energy closer to another AO (see picture). In this case we have low energy of stabilization (system weakly feel energy changing before and after interaction).
If difference as big as whole system looks like we give both electrons to one of AOs it means that electron density of bond is shifted to one of atoms. In this case we say that we deal with IONIC bond.
Now COVALENT bond: energies of interacting orbitals are close. This way we have big stabilization energy (see picture). Because of it we can say that covalent bond is stronger then ionic.

Simplifying: ionic bond is provided by Coulomb interactions (in approximation), but covalent bond is provided not only by them, and this is the reason why it's stronger.

Generalization to crystalline lattice: now we must consider group of interacting orbitals, but main findings remain the same: bond forming by AOs with same energy is stronger.

And, as lattice energy is energy of interacting particles forming lattice, it's clearly determined by MO's energy.

That's why I think yes, we may generalize your finding as a fact.


In Bohr's model what is angular momentum in the lowest orbital?

You are trying to mix the Bohr model with quantum mechanics. The Bohr model is a semi-classical treatment of the hydrogen atom in which an electron is circling around a proton. Classically, this electron has angular momentum, even in the lowest orbital because it has a well defined position vector $\vec{r}_n$ and velocity $\vec{v}_n$. The classical angular momentum is thus given by

$$
l_n=m_e|\vec{r}_n\times\vec{v}_n|,
$$
where $m_e$ the electron mass. Bohr postulated that the angular momentum must be quantized (this is the semi-classical part) resulting in a set of orbitals.

Though Bohr's model works remarkably well it is not perfect because electrons are not really particles in the classical sense, but also behave like waves. A better treatment (but not the best) of the hydrogen atom is therefore to use quantum mechanics. In quantum mechanics the electron is described by a wavefunction that is spread out over all space and has a certain shape. For $s$ orbitals, such as the lowest orbital in the hydrogen atom, the wavefunction is spherical symmetric and has no angular momentum. The second expression you mention is the quantum mechanical operator to obtain the angular momentum.

To conclude, you cannot compare the angular momentum of the Bohr atom to the Schrodinger atom because they are based on different assumptions. Note further that for principal quantum number $n$, $\ell=0, 1, \ldots, n-1$.


Applying Henderson–Hasselbalch equation to amino acids: which pKa to use to calculate Z-/HZ ratio?

There are two approaches.

The two groups are treated as independent. You would calculate the ratio of protonated and deprotonated group one using the pKa for that group. Then, you would repeat that for the second group. Independent treatment would be a good approach, for example, to estimate the average charge of the hundreds of side chains in a typical protein that have a pKa near physiological pH.
The two pKa values are obtained experimentally from a titration.$$\ce{AH2 <=> AH- + H+ <=> A^2- + 2 H+}$$ Here, the $\ce{AH-}$ species could be ambiguous if the pKa values are close to each other (group 1 deprotonated and group 2 protonated or vice versa). Or, for a case like hydronium, water and hydroxide where the same atom can accept two protons, $\ce{AH-}$ would represent a single defined species. You can get the ratios of the concentrations from the pH and the respective pKa ($\ce{AH2}$ vs $\ce{AH-}$ via $K_1$, and $\ce{AH2}$ vs. $\ce{AH-}$ via $K_2$), and turn that into fractions of the total for each species. You can use that for a weighted average to calculate the average charge, for example.

  Let's say I want to know the ratio of HZ to Z− for a pH of 1.00 and 11.00, which pKa should be used in the Henderson–Hasselbalch equation and why?

This depends what HZ and Z- refer to. With two ionizable groups, there are really four species, $\ce{HZYH, HZY^-, ^-ZYH and ^-ZY-}$. If the pKa values are far apart, one of the mixed species would be very unlikely in aqueous solution (in your example, protonated carboxylate with deprotonated amino group). At pH 1.00, the amine would be 100% protonated, and the carboxylic acid would be mostly protonated (comparing its pKa of 2.33 to a pH of 1.00). At pH 11, the carboxylic acid would be 100% deprotonated, and the amine would be mostly deprotonated (comparing its pKa of 9.74 to a pH of 11.00).


How do you find the pH of lime juice through titration?

As said, titration and the concept of $\mathrm{pH}$ is very different. The process of titration is used to figure out an estimate of how much of acid $\ce{A}$ (of come concentration) will neutralize base $\ce{B}$ in an acid-base reaction (of come concentration) (also the reverse). The acid and base reaction is solely dependent on the coefficient of a chemical equation regardless of the $\mathrm{pH}$.

e.g., One mole of aqueous $\ce{HCl}$ will react always with one mole aqueous of $\ce{NaOH}$.

The $\mathrm{pH}$ in this case only indicates the "acidity" of either of the reactants based on the amount of $\ce{H3O+/OH-}$ ions that dissociated form them:
$$\mathrm{pH} = - \log[\ce{H3O+}]$$
You can of course figure out the concentration of a base/acid in titration if you know the basicity of one of the solution and the mole of acid/base in the other. You can do this by:

Let's say you want to know the concentration of a solution of ethanoic acid ($\ce{CH3COOH}$), we will have to prepare a suitable base to use as a titrant. We mix a universal indicator with the solution of $\ce{CH3COOH}$ and add it to a flask. Dissolving one mole of $\ce{NaOH}$ into x volume of water and put it in the buret. Run the buret until the resultant solution is neutral.

If the original volume of the $\ce{CH3COOH}$ solution is y and the amount of $\ce{NaOH}$ solution deposited is k.
Concentration of original acid is given by:
$$[\text{acid}] = {\frac{k}{x*\text{(original basicity of acid)}*y} }$$
as in
\begin{align}
 \\ [\text{acid}] &= {\frac{(\text{mole of}\, \ce{NaOH}\,\text{used})\cdot(\text{volume of}\, \ce{NaOH}\,\text{used})}{(\text{volume of}\, \ce{NaOH})\cdot(\text{original basicity of acid})\cdot(\text{volume of acid})} }
\end{align}
However, we still wouldn't be able to determine the $\mathrm{pH}$ of the original acid as we do not know how many of the acid in the original solution actually ionized:
$$[\text{acid}]*(\text{basicity}) \neq [\ce{H3O+}]$$
Thus we do not know the $[\ce{H3O+}]$ and the $\mathrm{pH}$ cannot be determined.


Some background, since most readers would probably need it.

A thermoplastic is a polymer that becomes pliable/moldable upon heating, and then returns to a solid (keeping its new shape) when cooled. Thermoplastics are generally high molecular weight polymers with very few or no crosslinks. Polymer chains are held together by intermolecular forces (most often van der waals) and physical entanglement, both of which can be overcome with addition of heat.

A thermoset is a polymer that is irreversibly cured, usually through covalent crosslinking (during or after polymerization). The covalent crosslinks hold the chains together in such a way that the chains cannot be separated by the addition of (small amounts of) kinetic energy. Thermosets tend to dramatically fail after the addition of lots of heat, following which they do not return to their original state. It is possible for the cumulative forces from many strong non-covalent interactions (ionic, hydrogen-bonding, and even pi stacking) to make a polymer behave like a thermoset. For example, proteins are thermosets.

  Can you tell if a monomer will produce a thermoplastic or themoset based on its structure?

Yes. All polymers produced from only mono- (addition polymerizations) or bi- (condensation polymerizations) functional monomers will always produce thermoplastics. All of the hydroxybutanoic acids (e.g. $\gamma$-hydroxybutanoic acid) are difunctional monomers for condensation polymerizations. Their polymers will be thermoplastics.

Unless...

The addition of tri- or tetra- functional monomers produces crosslinks, which always produces thermosets. For example, glycerol, malic acid, or lysine could be trifunctional monomers in condensation polymerizations. 2-Amino-3-hydroxybutanoic acid (threonine) is a trifunctional monomer. It has three functional groups (alcohol, amine, carboxylic acid), that can all react in a condensation polymerization. Any synthetic polymer containing threonine is likely to contain crosslinks and be a thermoset.

Or...

Many thermoplastic polymers can be converted into thermosets by curing them afterwards. Some curing is thermal, some is irradiative (espccially using UV), and some is chemical. A prime example is polyisoprene (natural latex rubber), which is a thermoplastic (with a very low melting point). It is also an elastomer - it will stretch and return to its original shape (although melting it erases the "memory"). Rubber is converted into a thermoset by vulcanization to make it more useful while still having the elastomeric properties.


The carbonyl oxygen can act as a nucleophile but it is strongly dependent on the conditions of the reaction.

Enolates can react as a nucleophile through either the $\alpha$-carbon or the oxygen.

(source)

The ratio of carbon to oxygen substituted products depends on the balance of orbital interactions and electrostatic interactions between the reactants. As you can see from the diagram above, the HOMO of the enolate has the greatest contribution from the $\alpha$-carbon and so reactions which are controlled by orbital energy factors take place through there. Conversely, the greatest negative charge is on the oxygen due to its greater electronegativity and so if the reaction is controlled by electrostatic factors then it will take place at the oxygen.

Changing the electrophile:
Alkylation is a common reaction of enolates and provides a good insight into the factors affecting the selectivity. If methyl iodide is used as the alkylating agent in a polar aprotic solvent then the reaction produces mostly the carbon alkylated product. Switching to methyl bromide or methyl chloride increases the amount of oxygen alkylated product and if methyl tosylate is used then the reaction produces mainly the oxygen alkylated product. This is because the increasing electronegativity of the substituent increases the positive charge on the methyl carbon, meaning that electrostatic interactions are more important. Additionally, the $\ce{C-X}~\sigma^*$ orbital (the LUMO of the electrophile) will be lower in energy in methyl iodide than in methyl tosylate due to the poorer energy match between the $\ce{C}$ and $\ce{I}$ atomic orbitals compared to $\ce{C}$ and $\ce{O}$. This brings the electrophile LUMO closer in orbital to the enolate HOMO, making the orbital interactions more important.

This thinking can be extended to other electrophiles. Many hard Lewis acids such as $\ce{R2BCl}$ and $\ce{(CH3)3SiCl}$ (credit to @Jan for pointing this out) will give almost exclusively oxygen substituted products. This is because the central atoms are small and highly charged, and so have strong electrostatic interactions with the oxygen end of the enolate. This can be viewed as an application of 'hard soft acid base theory', where the silicon and boron based electrophiles are considered hard acids and the carbonyl oxygen is a hard base. By comparison, neutral carbon electrophiles are soft acids and so exhibit less affinity for the oxygen.

Changing the solvent:
Changing the solvent is one of the easiest ways to influence the outcome of the reaction, particularly in the case of alkylation. In polar aprotic solvents, the metal cation is strongly solvated but the enolate is weakly solvated due to the lack of acidic hydrogens on the solvent to form hydrogen bonds. In polar protic solvents, the oxygen end of the enolate is strongly solvated by hydrogen bonding which hinders reaction at the oxygen, strongly favouring C-alkylation, even with an alkyl tosylate.

Some more reading can be found here, here, here and here.


Propanol can refer to both propan-1-ol and propan-2-ol.

Earlier we used notations like 1-propanol or 2-propanol. But according to latest IUPAC recommendations, we prefer names like propan-1-ol and propan-2-ol.

For compounds like aldehydes, carboxylic acids and for other compounds involving carbon centered functional groups, we need not mention position because we always consider the carbon of these functional groups as the 1 carbon.

But, for alcohols, as in this case, position should be mentioned for being more specific.


How does hydrogen gas build up in hot water lines?

I think that it is probably due to a reaction between an acid and a metal. These reactions can lead to the formation of hydrogen following this generic forumula:
$$\ce{M + 2H+ -> M^{2+} + H2}$$

For example iron could be a source of electrons. If you look at the corrosion of iron: $$\ce{Fe->Fe^{2+} + 2e^{-}}$$

At low pH (notice that there is a correlation between pH and temperature) hydrogen gas is formed:
$$\ce{2H^{+} +2e- -> 2H2}$$


Isomerization reaction involving carboxylic acid derivatives

tl;dr A slightly anthropomorphic approach with some basis in general organic chemistry:

Step 1: The protonation of the carbonyl (An acid-base reaction)

Here, we have three options – we can protonate the nitrogen (deactivated by the resonance with oxygen), the alcohol group ($\mathrm{sp^3}$) or the carbonyl group ($\mathrm{sp^2}$). As you can see this is basically an acid-base reaction and therefore the strongest base (having highest electron density) would react the fastest and this means since the carbonyl has the highest electron density here the carbonyl is protonated.
Step 2: The intermolecular attack ($\ce{O}$ is very sneaky)

Well, we began with $\ce{HCl}$. Time for the $\ce{Cl-}$ to make an appearance? Not really. As you can see, we have an alcohol group in the compound which has electrons to donate. The carbon on the (protonated) carbonyl group is highly electropositive (It was positive enough when there was no positive charge on the oxygen and now it just got more electropositive). Now, the oxygen with the lone pair makes an appearance. It attacks the carbonyl carbon and at the same time the π electrons move to the previously protonated oxygen making it neutral.
Step 3: An intermolecular proton transfer (Another acid-base reaction?)

Now, as you may have learnt, oxygen is not the most electropositive element. If possible, it doesn't stay positive. Nitrogen comes to the rescue. Another acid base reaction takes place where the newly active amine group deprotonates the oxygen forming $\ce{>NH2+}$.
Step 4: The final rearrangement (We're done here(?), you can figure it out)

We still need the final isomer and we also need to remove the positive charges on the compound before we end. Therefore, the ring breaks and the electrons move to the nitrogen removing the positive charge on it. This also means a carbocation is formed in the carbon where the alcoholic group is attached which stabilises it using resonance to form a $\ce{C=OH+}$ bond. (try drawing this step).
But we aren't done.
Step 5: The true final step (A small deprotonation)

Nobody needs any positive oxygens, do we? Therefore the oxygen is deprotonated. (Note: this is in equilibrium since medium is acidic). All $\ce{H+}$ used has been released which also makes this a catalytic reaction.
There might be one final question in the back of your mind (for the curious).
Why not go back to the starting compound?
This is because as @user55119 very well pointed out in a comment to the other answer. My additions in italic.

In the absence of acid, heating amino ester (a) would give the starting material, the hydroxyamide because esters are more labile (easily broken down or displaced) than amides. The reason this reaction can occur is that the amino group is protonated in acid which shifts the equilibrium.

Welcome to the realms of organic chemistry.


The expression
$$v/(\mathrm{mol\ s^{-1}\ kg^{-1}})=30$$
means
$$v=30\ \mathrm{mol\ s^{-1}\ kg^{-1}}$$

This notation is used in the table, so that the entries are all just numbers without any unit symbols.

For an explanation, I copy the following section from my meta answer, which is mainly a collection of rules and examples taken from various standards and guides (see the meta answer for references).

  Unit symbols are mathematical entities and not abbreviations. In expressing the value of a quantity as the product of a numerical value and a unit, both the numerical value and the unit may be treated by the ordinary rules of algebra.

  (…)

  In expressing the value of a quantity as the product of a numerical value and a unit, both the numerical value and the unit may be treated by the ordinary rules of algebra.

    the equation $T=293\ \mathrm K$ may equally be written $T/\mathrm K=293$

  It is often convenient to write the quotient of a quantity and a unit in this way for the heading of a column in a table, so that the entries in the table are all simply numbers.

    $$\begin{array}{ll}
\hline
T/\mathrm K & p/\mathrm{kPa} \\
\hline
300.00&3.5368\\
310.00&6.2311\\
320.00&10.546\\
330.00&17.213\\
\hline
\end{array}$$

  The axes of a graph may also be labelled in this way, so that the tick marks are labelled only with numbers.

    do not write “mass [kg]” for “mass in kilogram”

Also note that expressions for units shall contain nothing else than unit symbols and mathematical symbols. Any attachment to a unit symbol as a means of giving information about the special nature of the quantity or context of measurement under consideration is not permitted. Therefore the use of the unit “$\mathrm{mol\ s^{-1}\ kg^{-1}\ \text{protein}}$” as written in the question is not permissible. The correct unit is $\mathrm{mol\ s^{-1}\ kg^{-1}}$.


Is there any relationship between ultrasonication and temperature

The temperature is highly nonlinear with sound pressure and dependent on external pressure, tensile strength of liquid etc. Cavitation, http://en.wikipedia.org/wiki/Cavitation, can produce local shock waves and temperatures of thousands of K, causing sonoluminescence. See also http://www.scs.illinois.edu/suslick/sonochembrittanica.html.

Because of that nonlinearity and all the dependencies, I do not know if any equation for calculating the peak temperatures exists, and there is still research going on to see how much of sonoluminescence is due to black-body radiation, the recombination of radicals etc.

BTW, you can view a mantis shrimp with a sonoluminescent punch at https://www.youtube.com/watch?v=sNzI8K-lCqo and https://www.youtube.com/watch?v=ti2Uoc1RXuQ, and another species at https://www.youtube.com/watch?v=QXK2G2AzMTU.


Mechanism for interconversion of spin isomers of hydrogen

The first question is:

  What is the mechanism by which spin isomers of hydrogen switch between the ortho and para forms?

There is some explanation in ChemPhysChem 2006, 7 (3), 551–554 (non-paywall version here):

  One can define three situations. In the first, a magnetic conversion occurs without bond elongation or breaking. For example, in solid dihydrogen dipolar magnetic nuclear interactions are responsible for an extremely slow spin conversion.[4] In the presence of unpaired electrons the conversion is accelerated by the magnetic hyperfine interaction.[5] The second situation also involves a magnetic conversion mechanism, but is assisted by an intermediate H–H bond elongation, for example, by binding to a transition metal center. Finally, in the third situation, a H–H bond splitting and re-formation with other hydrogen atoms occurs, which corresponds to a chemical spin conversion.

The second question is:

  If such a mechanism exists, why does ortho-hydrogen not increase in concentration on increasing temperature?

The equilibrium fraction of ortho-hydrogen does increase from 0% at absolute zero to 75% at high temperature.

para-Hydrogen is the lowest energy, most stable form, since it is capable of accessing the $J=0$ rotational level (whereas ortho-hydrogen cannot). There are three microstates corresponding to ortho-hydrogen:

$$|\!\uparrow\uparrow\rangle, |\!\downarrow\downarrow\rangle, |\!\uparrow\downarrow\rangle + |\!\downarrow\uparrow\rangle,$$

but only one microstate for para-hydrogen:

$$|\!\uparrow\downarrow\rangle - |\!\downarrow\uparrow\rangle.$$

So, according to the Boltzmann distribution, the equilibrium fraction of ortho-hydrogen goes from 0% at 0 K to 75% at infinite temperature (in practice, a temperature at which the effects of rotational quantisation are not seen).


Converting to answer: While the energy to introduce a grain boundary into a single crystal depends on the relative orientation of the two resulting grains, there will be some atomic rearrangement (strains or reconstruction), and you will end up with a number of bonds across the grain boundary. The boundaries are usually weaker than the bulk crystal, which is why they preferentially fracture along the grain boundary.


Do dishwashing detergent and stain remover powder/stick have the similar ability to remove collar rings?

Detergents seem to consist of both polar and non-polar compounds. Probably because they need the non-polar part of the mixture to dissolve the grease, and the polar compound part of the mixture to get washed off with the water so there is not much left over on your shirt. The ideal solution would probably depend on just how bad the stain is (how much grease needs to be dissolved). If your home solutions did not work the first time, try to look for signs if it did anything at all, and it might not be something that will remove all the grease the first go around.

I have no experience using detergents or removing stains, but I am going off a lot of speculation here as to how the process might work.

"Like dissolves like"

According to this article if there are lots of Mg, Ca, or Iron ions in your water (likely) then it could reduce the effectiveness of soaps. A good question to ask yourself is whether or not your water contains many ions such as these.


Placing the atomic mass and number of an element...

In the periodic table, the largest number is indeed the atomic number. This is the number of protons. Since this defines the element, it is considered the most important number.

What you probably mean is
$^A_Z\mathrm{Element}$. Here, by convention the lower $Z$ (from the German word Zahl is the atomic number, while $A$ is the sum of protons and neutrons. This defines a particular isotope. This notation is often used in the context of NMR spectroscopy ($^{13}\ce{C}$ NMR), or when radioactive decays are described:

\[\ce{^{14}_6C -> ^{14}_7N} + e^- + \bar\nu_e\]

In summary, I think this is just a set of different conventions which show that in differents contexts either the element itself or a particular isotope and its properties are considered more important.


Density and concentration of hydroiodic acid

In 1981, Nishikata et al. published Viscosities of Aqueous Hydrochloric Acid Solutions, and Densities
and Viscosities of Aqueous Hydroiodic Acid Solutions in the Journal of Chemical Engineering Data.

Table III from this paper shows what you are after.

I don't know what temperature you used to make your density measurement, but assuming it was 20 °C, then your $\ce{HI}$ looks to have a concentration of about 39 weight percent.


Maurice notes that alkali metal diatomic molecules are at best a very minor constituent.  But alkali metals do fill their $s$ valence shells in other ways.

Most alkali metals are known to form alkalides, in which an electron is transferred from one alkali metal to another (or from barium to an alkali metal, in a more recent example).  To effect this electron transfer, a complexing agent is used to stabilize the cation, such as 2.2.2 cryptand in the compound $\ce{[Na}\text{(2.2.2-crpytand)}\ce{]^+Na^-}$.  A combination of low ionization energy, positive electron affinity and the stabilization offered by complexing the cation enable this outcome.

Lithium apparently fails to form such alkalides, but not because it refuses to adopt an $s^2$ valence electron configuration.  Quite the opposite: it tends to form polar covalent bonds with that configuration.  Organolithium compounds are a widely used example of this electron structure, and the covalent bonding component in lithium salts may contribute to their solubility in some organic solvents.


Why does increasing the available surface area of a pure solid not effect equilibrium?

You can change the amount of $\ce{B}$ available and the forward rate, but that doesn't change the rate constant. And the equilibrium constant is directly related to the rate constants of the forward and reverse reactions:

$$K = \frac{k}{k'}, \text{where }k\text{ is forward rate cosntant, }k'\text{ is reverse rate constant}$$

As @MaxW points out, you'll get to equilibrium faster, but it's the same equilibrium.

Edit:

Here's another way to think about it. Consider the reaction:
$$\ce{H2O_{(l)}<=>H2O_{(g)}}$$

Does the equilibrium between liquid water and water vapor change if you increase the surface area of a pool of water? No, because even though evaporation is faster, there's also more area for condensation to take place.

In your specific case, the reverse reaction from $\ce{C}$ needs to interact with some amount of solid to redeposit $\ce{B}$. Otherwise, you're making $\ce{B_{(g)}}$ and that's not the same thing...

Edit 2 (for OP's comments):
Notice that your reaction is actually extremely unrealistic. Gases don't just form solid or liquids on their own; they need some nucleation seed. Otherwise, there would be no point in seeding rain clouds.

As @IvanNeretin points out, invoking microscopic reversibility, you'd need to invoke these nucleation points in your reaction, forward and backward. These points are of course whatever surface is available on $\ce{B}$.

This makes the reaction:
$$\ce{A_{(g)} + B_{(g)} + n B_{*} \overset{k}{\underset{k'}{<=>}} n B_{*} + C_{(g)}}$$

$\ce{B_{*}}$ refers to the surface. Notice that it cancels out so it doesn't affect the equilibrium constant, but having more surface area will increase the forward rate if you increase the surface. It will also increase the reverse rate, which is why the equilibrium constant can stay the same.


Why does a real gas with attractive forces between particles experience an increase in internal energy when expanding isothermally?

The internal energy change with temperature can be found from one of the Maxwell equations and is

$$ \left (\frac{\partial U}{\partial V}   \right)_T = T\left (\frac{\partial p}{\partial T}   \right)_V -p$$

and  a real gas can be represented  with the empirical van-der-Waals equation $\displaystyle (p+a/V^2)(V-b)+RT$ where a represents the attractive potential and b the molecular volume.

Performing the calculation gives

$$ \left (\frac{\partial U}{\partial V}   \right)_T = \frac{a}{V^2}$$

which shows what you note in your question.

As the gas expands work has to be done to separate the molecules because they have an attractive  potential between them, thus the internal energy has to increase  because the temperature is held constant. If the gas was thermally isolated during expansion it would cool and this is observed in the Joule-Thomson effect.


Rietveld (FullProf) general background function

If I recall correctly, Chebyshev polynomials are preferred over simple polynomials because of their termination behavior (i.e. they tend not to diverge wildly), although I don't have a reference for this handy.

Dinnebier's admonition to use higher-order polynomials with care is effectively a warning not to over parameterize your fit. If you inspect the two fits with 6- and 24-term backgrounds, my guess is that you will find the over parameterized version nudged your Bragg peaks up or down as necessary to better match your observed data. Hence, your residual has diminished but your fit is not more meaningful.

An overfitted background precludes accurate extraction of the observed Bragg intensities, leading to correlated errors in related factors (e.g. occupancy, position, etc.).


Does a Na2SO3 solution oxidize to Na2SO4 in an oxygen rich atmosphere?

I'm not sure what is the stable form at room temperature, since we have both sodium sulfite ($\ce{Na2SO3}$) and sodium sulfate ($\ce{Na2SO4}$) in our lab and they're both stable, in air. I do know that the sulfite will oxidise to sulfate with the addition of bleach.

That said, your redox reaction is not quite accurate. In sulfite, sulfur is in the $\ce{S^4+}$ oxidation state. In sulfate, it is $\ce{S^6+}$. What happens in the reaction is that the sulfur gives away two electrons, which then go to an oxygen that changes from $\ce{O^0}$ to $\ce{O^2-}$.

One way of writing it would be $\ce{S^4+ + 1/2O2^0 -> S^6+O2^2-}$. Total charge is preserved and there are no free electrons.


Which white powders form acids when mixed with water?

From the pH and solubility information alone, I would propose sodium bisulfate: cheap, readily available, safe in the powerfully oxidizing solution. (Possibly also potassium bisulfate.)

One way of manufacturing $\ce{HCl}$ produces sodium bisulfate as a byproduct:
$$\ce{NaCl + H2SO4 -> HCl + NaHSO4}$$ The odor of $\ce{ HCl}$ could be a minor impurity in this mystery acid. The other impurities also suggest it is not a reagent grade product.


What happens when you mix milk and coke?

The phosphoric acid causes denaturation of milk proteins which precipitate out of the solution (milk is an emulsion, to be exact). The flock that is formed also adsorbs dye molecules (caramel) which is why the solution become clear.

There is nothing coke-specific about it. Any acid would denature milk just like that. And nearly any huge organic molecule would get absorbed, just like caramel dye.


Why is there a subtractive volume correction in the van der Waals' equation for real gases?

The van der Waals equation is an attempt to explain the pressure-volume relationship for real gases. For example, what will the pressure be of a given amount of gas in a vessel of a fixed volume.

In ideal gases the size and interactions between the atoms or molecules of the gas don't matter at all. We think of them as point-like entities that don't interact except to bounce elastically off each other and off the walls of the vessel. Their behaviour is described by the ideal gas equation and doesn't depend at all on the nature of the particles making up the gas.

Real molecules and atoms deviate from this ideal behaviour because the particles of the gas both have finite volume and attractive interactions which mean they don't behave exactly as the ideal gas equation describes. The van der Waals equation is a simple attempt to explain those deviations from ideality. The volume correction is one of those factors. It takes into account the fact that real gases are made from particles that actually have a finite size and are not mathematical points. The factor is about the volume of the particles of the gas not about the vessel the gas is in. The result of the correction is that the pressure of a given amount of gas in a vessel of fixed size is slightly different than would be expected from the ideal gas equation.


Should I expect any surprise when trying to shoot green fireballs like this?

I would not expect boric acid to cause ignition of nitrocellulose. However, boric acid may react with un-nitrated hydroxyl groups on the cellulose backbone--likely a benign reaction. You might need more than a few drops of boric acid solution to see a green color - perhaps immersion in boric acid solution will be needed. Copper salts might give a good green color, so that is also worth trying. Either way, you will want to let the nitrocellulose dry thoroughly before trying to ignite it.

Also, be sure to carefully practice the words: "I AM OZ, THE GREAT AND POWERFUL" before you go public.


What are the possible changes that might occur if I add an inert gas to a chamber containing another few different gases?

Quoting from the Wikipedia page on Le Chatelier's principle, we have to distinguish two cases:

The total volume is constant:

Adding an inert gas into a gas-phase equilibrium  does not result in a shift of the equilibrium in this case. This is due to the fact that adding an inert gas does not change the partial pressures of the other gases in the container. While it is true that the total pressure of the system increases, the total pressure does not have any effect on the equilibrium constant; rather, it is a change in partial pressures that will cause a shift in the equilibrium.

The total volume is not constant:

If, however, the volume is increased, the partial pressures of all gases will decrease which results in a shift of the equilibrium towards the side with the greater number of moles of gas.


Why does sulfonation of alcohols even work?

There are always two sides of a reaction. In the case of an alcohol attacking a sulphonic acid chloride, we can look closely at the alcohol and say: ‘Oh my, this is a bad nucleophile and we cannot deprotonate it with the conditions we have. Probably, this reaction won’t work’. But we can also look at the sulphonic acid chloride and say: ‘Oh my, this is a very strong electrophile and it has a very good leaving group. This reaction will work like a charm!’

So it is important, that you compare the reactants extensively. In this case, the electrophilicity of the sulphur(VI) is high enough to even allow rather bad nucleophiles to work. On the other hand, oxygen atoms aren’t even that bad a nucleophile. Give them enough time and alcohols will attack via $\mathrm{S_N2}$ — it might just take forever.

We can also take electrostatics into consideration. Oxygen is rather negatively polarised since it has a very high electronegativity. The sulphur atom, on the other hand, is strongly positively polarisied, since it is connected to two formal oxide anions and a chlorine atom — all three are more electronegative than sulphur by a noticeable margin. A rather negative particle such as the alcohol oxygen, will happily attack a rather positive particle such as the sulphonic acid chloride sulphur.

Mechanistically, this attack forms an intermediate pentavalent sulphur which will then displace the best leaving group: The chloride atom. Therefore, even the hydrochloric acid that is produced isn’t as bad as it may look. We loose the chloride first and the proton much later. We aren’t immediately removing a strong acid but rather a good leaving group and the proton just has to deal with it.

However, I will allow the objection that a significant concentration of $\ce{HCl}$ would favour the reverse reaction. Thus, when employed synthetically, a base is usually added that serves no purpose other than to capture the proton and keep it somewhere where everybody can be happy.

Too-long-didn’t-read-version:

Alcohol might be a bad nucleophile, but oxygen is negative.
Sulphonic acid chloride is a very good electrophile and the sulphur atom is positive.
Positive and negative like each other, thus alcohol will attack sulphur.
Chloride is displaced because it is a pretty good leaving group.
The proton just has to do what the others force it to do.
Driving force is the liberation of the chloride and turning a weak $\ce{S-Cl}$ bond into a strong $\ce{S-O}$ one.
Add a base to capture the proton for further enhancement.


Boiling point of a liquid increasing with temperature?

It sounds that they actually meant:

  Boiling point of a liquid increases with increase in temperature pressure.


An important case of acidity constants changing with solvent, with consequences in organic chemistry, can be found for compounds with hydroxyls, especially water. It is possible to deduce that hydroxyl compounds are more acidic in protic solvents compared to aprotic solvents, which is attributed to an abnormally high stabilization of the resulting conjugate base via strong $\ce{O^{-}\ ^...\  H-O}$ hydrogen bonds.

The difference is most clear for water. Water molecules solvated by water itself have a $\mathrm{p}K_\mathrm{a}$ of 15.7 in ambient conditions, but if the unusual stabilization from hydrogen bonding were removed, they would have a $\mathrm{p}K_\mathrm{a}$ of around 28. In other words, $\ce{H2O}$ is a weaker acid by some 10 orders of magnitude in non-protic solvents compared to water, and correspondingly, its conjugate base $\ce{OH^{-}}$ is a stronger base by the same factor. This means hydroxide ions can effect quantitative deprotonation of even rather weak acids, including the hydrocarbons cyclopentadiene, indene and fluorene, in solvents such as 1,2-dimethoxyethane.

Unfortunately, compounds containing hydroxide ions have somewhat limited use as superbases in non-protic solvents because the solubility of common hydroxide bases such as $\ce{NaOH}$ or $\ce{KOH}$ is rather small. Other hydroxides such as quaternary ammonium hydroxides have appreciable solubility in non-protic solvents, but they are always found in the form of hydrates. Attempts to prepare anhydrous $\ce{R_4N^+OH^{-}}$ inevitably result in decomposition, because the crystallized water solvates the hydroxide ion and tames its reactivity; once the water is removed, the bare hydroxide ion immediately attacks the quaternary ammonium cation creating an alcohol and a tertiary amine.

As mentioned previously, other compounds containing hydroxyls also show changes in $\mathrm{p}K_\mathrm{a}$ comparing water and non-protic solvents, though the difference tends to be smaller. For ethanol, the difference is at most about 4 units of $\mathrm{p}K$.


What chemical file formats are compatible with 3D printing?

None of the chemical file formats are typically accepted by 3D printing services. The reason is that chemical file formats are fundamentally different from those describing 3D objects.

In the simplest case, a chemical file format describes a list of atoms and their positions (plus possible metadata). An example is the XYZ format, which stores exactly that. Other “chemical objects” which can also be described in some chemical file formats include bonds, protein structure, crystalline structure information (unit cell, symmetry, occupancy of each atomic position, …), etc.

3D file formats, on the other hand, describe a series of objects. In the simplest case (e.g., the STL format or the Wavefront .obj file format), the object itself is not stored, but an unstructured triangulated surface delimiting the object. In those, you can't describe a sphere, but only a tesselation of the sphere.

So, you need a conversion step to create a 3D representation of your chemical structure. Chemical file formats don't typically contain representation information: you need to pick the one you want (CPK, liccorice, protein cartoon, …) and use software to create a 3D scene. VMD, for example, allows export to STL format, which is suitable for 3D printing.


How many resonating structures of Carbon Monoxide are there?

You have said it yourself. There are many possible resonance structures but most of them have a negligible contribution to the overall structure. Your proposed structure is one of them.

The major resonance contributors of carbon monoxide are: $$\ce{\!\overset{\ominus}{C}#\overset{\oplus}{O}\! ~<->~ \!C=O\! ~<->~\!\overset{\oplus}{C}-\overset{\ominus}{O}\!}$$


You need to understand some details of using point group tables. The pyrazine  belongs to  $\ce{D_{2h}}$ point group and pyridazine belongs to $\ce{C_{2v}}$
In the third column of the point group are found x, y, z which are used to identify dipole transitions, e.g. vibrational symmetries that can exhibit  IR transitions and in the fourth $x^2, z^2, xz$ etc. which are used to identify Raman transitions. If there is a squared term in column 4 and in the top line of the table (totally symmetric representation) labelled $\ce{A_g, A_1}$ then a Raman transition is totally polarised. Raman transitions from other symmetry species vibrations are not fully polarised. (Ignore the $\ce{R_x, R_y,R_z}$ they do not refer to IR or Raman transitions)
The point groups are shown below

In $\ce{C_{2v}}$ all vibrational species $\ce{A_1}$ etc. can have Raman transitions but in the IR the $A_2$ does not have an IR transition, no x, x, z in column 3. The other three types of vibrations could have both IR and Raman transitions. In the $\ce{D_{2h}}$ you can see that, because of the centre of inversion, Raman and IR vibrational transitions are mutually exclusive (compare symmetry species of $x^2, xy$ etc. with x, y, z ) because there is no vibration with a symmetry species that can generate both IR and Raman transitions.
Thus, with your compounds, the one with $\ce{C_{2v}}$ symmetry will have IR and Raman transitions with the same frequency, but for the $\ce{D_{2h}}$ compound there should be no common IR and Raman frequencies.
(See the answer to this question Understanding group theory easily and quickly more details on using point groups)


A basic question about tetramethylammonium hydroxide --strong or weak?

TETRAMETHYLAMMONIUM HYDROXIDE Journal of the Chemical Society 87 (1905): 955-961:

  Practically the preparation of tetramethylammonium hydroxide
  from its salts resolves itself into a question of solubility as follows.
  In general, the equation

  $$\ce{NMe4X + MOH = NMe4OH + MX}$$

  will represent a real action proceeding nearly to completion if M, X,
  and the solvent are so chosen that all the substances except MX shall
  be soluble, or at least that MX shall be much less soluble than either
  of the original reacting substances. This principle was applied by
  Hofmann in his preparations with water as solvent. For M and X
  he chose either the pair Ag,I or Ba,SO4, in both of which cases MX is
  practically insoluble in water.

  It is clear that if the general application of the above principle is
  justifiable, tetramethylammonium hydroxide may be prepared from a
  tetramethylammonium salt by means of potassium hydroxide if we so choose X and the solvent that of the substances represented in the equation

  $$\ce{NMe4X + KOH = NMe4OH + KX}$$

  all shall be soluble except KX.

...

  An estimation of the strength of the base by means of the velocity
  of saponification of methyl acetate in N/80 solution showed that it was
  somewhat weaker than sodium hydroxide. The velocity constants obtained at 25 [degrees] were 0.0106 and 0.0115 respectively, so that if the
  strength of sodium hydroxide is represented as 100, that of tetramethylammonium
  hydroxide, under the above conditions, will be represented by 92.

So by two independent methods it is shown that tetramethylammonium hydroxide  is much stronger than pKb=4.2.  Firstly, the halide cannot not be converted to the hydroxide by adding metal hydroxide, unless an insoluble halide is produced to force the equilibrium forward.  Seconding, it is shown by velocity of saponification to be only slightly weaker then NaOH.

See also Essential Principles of Organic Chemistry (1956)

  It follows that a quaternary ammonium hydroxide is completely dissociated and should be a strong base, as is sodium hydroxide. The experimental facts are entirely consistent with this view. Although trimethylamine in water solution gives only a small concentration of hydroxide ion, tetramethylammonium hydroxide gives a full molar equivalent of hydroxide ion in solution.


Sounds like a chemical mixture of sucrose and glucose.  In which case, you only talk about the molar mass of the two sugars separately.  The mass or number of moles of the mixture can be determined by knowing the ratio of sucrose and glucose in the mixture.  FYI... the molar mass of glucose is 180 g/mol (not 170).


When would the presence of lone pair result in a polar molecule?

A polar molecule must have a finite net dipole moment.

A dipole moment is a result of separation of charges.

The below explanation requires you to have a basic understanding of dipole moment, vectors and resonating structures.

Ozone is not polar because of the presence of lone pair on oxygen atoms. In ozone, all the oxygen atoms are not same. Some oxygen atom wants to hold on the electrons more than the others which leads to separation of charges. This separation of charges results in a dipole which makes a molecule polar.

Electronegativity difference in a bond causes one atom to have more affinity for the electrons. This makes that atom attract the pair of electrons more, hence developing a slight negative charge on itself(because of more electron density) and a slight positive charge on the other atom(because of less electron density). This leads to separation of charges and creates a dipole. In the figure, a dipole is shown from carbon atom to chlorine atom. A dipole is represented as a vector with its tail pointing towards the atom having more electron density.

It is not the dipole but the net dipole moment that makes a molecule polar. The net dipole moment is a vector sum of all the dipole vectors. Now, why need vectors for finding the net dipole moment?
Suppose you are being surrounded by a bunch of strong guys who are pulling you through a rope from all sides. They are applying heavy forces to pull you towards them but you won't move because the net force on you is zero. Similarly, in the second figure, the four chlorine atoms are pulling the bond pair of electrons toward itself from 4 different orientations. To account for those different orientations which we know will affect the net pull of electrons by chlorine atoms, we need vectors.

Consider this case of carbon tetrachloride. The carbon can be thought of being placed at the centre of a regular tetrahedron and the chlorine atoms are placed at the corner of the tetrahedron. All the chlorine atoms are at equal distance from carbon. Even though you might say there is electronegativity difference between carbon and chlorine, the net pull on the electrons of carbon is zero. Hence, the dipole moment is zero despite the electronegativity difference.

The non-polarity of oxygen atom has got to do nothing with the presence of lone pair on oxygen atom. The oxygen molecule is not completely non-polar. There is instantaneous dipole in oxygen molecule.

No molecule is completely non-polar at every instant of time. H-At is also polar, although its dipole moment is very less. H and At are different atoms. One atom must have a little more affinity toward the electron and hence there will be separation of charges. The electronegavity difference between H and At is very less, therefore the separation of charges is very less which makes its dipole moment near to zero, but still finite.


tldr: Manually or script dihedral angle changes, then use a GFN or another DFTB method to find any/all conformers

Without seeing the compound, it's really hard to know the problem. Most programs, including those you mentioned, have a set of rules to identify "rotatable bonds" and suggest likely dihedral angles.

As you might expect, such rules are derived for neutral, ground-state organic molecules.

There are a few options:

As suggested by Jan Jensen's comments above, you can use RDKit, Open Babel or another tool to drive your own dihedral angle search.
As suggested by Martin's comments above, there are some tools in Grimme's xtb  program to sample conformations. These will use the GFN1 or GFN2 approximate density functional (pretty much DFTB + dispersion correction) to evaluate energies.

I'd probably suggest a combination of both. From what I understand of Grimme's conformer search scripts, it calculates the vibrational modes of the molecule and samples geometries along vibrational modes. This is a good idea, although in some molecules it may not sample all minima (e.g., if you have many conformations in a large molecule, you might never "reach" the whole potential energy surface from any given initial geometry).

So you can change the dihedral angles using your program of choice, generate a few initial structures, then use xtb to optimize the geometries and confscript to sample a few more geometries.

I feel pretty confident that in small to medium geometry molecules, this will do a good job.

A few notes:

If you're sampling a radical, you definitely want to use a good DFT or quantum method to evaluate the energies. I doubt any current force field will do what you want.
We haven't finished our evaluation, but right now, GFN2 is performing better than any current method on conformer energetics for neutral organic molecules. We haven't checked radicals, but I would trust it more than any other current "fast" method.
If you don't have the xtb package, the method and conformer sampling scripts are now available as part of ORCA 4.1 too. There's a description of this kind of conformer sampling in the manual when discussing prediction of NMR spectra.


Why is the melting of ice a spontaneous reaction?

First of all,

  what exactly is spontaneous reaction?

In very easy language, reaction that occurs in a given set of conditions without intervention is called spontaneous reaction.

Now let us consider melting of ice example,

Take a ice on a plate and leave it for half an hour(conditions). After half an hour we will notice water instead of ice.

This process took place without any intervention from your side. Heat absorbed by the ice from surrounding is due to set of conditions not due to your intervention.

Thus we can consider melting of ice as a spontaneous process(or reaction).


Uncle Al is correct.  Colligative properties are properties of solution in which only the ratio between number of particles of solute and solvent are considered (often expressed in different forms, such as moles of solute per kilograms of solvent for molality); the identities of either are irrelevant.  So we could dissolve sodium chloride in water or magnesium bromide, but the general properties will be the same (though note that the magnitude of freezing point depression and similar properties will be slightly different depending on the substance).

In particular for salt water, the point at which the solution will freeze is depressed.  To understand this, we must understand what occurs when pure water freezes.  In general terms, when a liquid freezes, its normally hectic and randomly moving particles form a stable, packed structure that we call a solid; a hexagonal structure in particular for water at standard conditions.  The specifics are more complicated, but this is the main idea.

However, if you introduce a solute into the pure water, such as salt, the solute particles are getting in the way of the formation of the packed structure.  As a result, the temperature must be lowered slightly further in order for the solvent (i.e. water) molecules to properly overcome the solute molecules acting as obstacles for the formation of the solid structure.

In fact, when the solvent freezes, more often than not, the solute particles are often excluded, either pushed outside or trapped inside small pockets.  In the case of salt water, a eutectic system is created, if you're interested in learning of them.  Salt water itself cannot freeze into salty-ice or otherwise.


Why do sulfur and oxygen form sulfur dioxide?

One atom of oxygen is a fictional species outside of a plasma.  Oxygen is naturally diatomic, and the process by which it oxidizes sulfur is radical-based, leading to both oxygen atoms reacting.  The fact that SO2 forms instead of SO is a result of the [redundant answer] thermodynamics favor SO2 over SO.

Given that simple fact here is a quantum mechanical rationalization: Sulfur can expand its octet, which means in SO2 it has 12 electrons 'around' it.  This allows resonances forms where O- has a significant contribution to resonance, while S still has at least 8 electrons around it.  In SO no such resonance forms exist, meaning the highly electronegative oxygen must share its electrons with with sulfuer to avoid exposing a six-eletron sulfur species.  Hence SO2 should be a better structure than SO overall.


Is there ANY chemical that can destroy PTFE, or Teflon?

Corrosion Resistant Products, Ltd., with the help of Dupont, has established this source of information on what can and cannot eat teflon.

Here's a list:

Sodium and potassium metal - these reduce and defluorinate PTFE, which finds use in etching PTFE
Finely divided metal powders, like aluminum and and magnesium, cause PTFE to combust at high temperatures

These reactions probably reduce PTFE in a manner that starts:

$$\ce{(CF2CF2)_{n} + 2Na -> (CF=CF)_{n} +2NaF}$$

The world's most powerful oxidizers like $\ce{F2}$, $\ce{OF2}$, and $\ce{ClF3}$ can oxidize PTFE at elevated temperatures, probably by:

$$\ce{(CF2CF2)_{n} + 2nF2 -> 2nCF4}$$

Similar things can occur under extreme conditions (temperature and pressure) with:

Boranes
Nitric acid
80% NaOH or KOH
Aluminum chloride
Ammonia, some amines, and some imines


In the absence of a physical model, how can I use free software to visualize stereochemistry?

I used Jmol to make the following animated GIF:

First I used NCBI's Pubchem Sketcher to get the SMILES code for the molecule: CC(C(C(C)F)(Br)F)O
Then, I loaded it into JSmol running here using the load molecule by name button.
Internally, this triggers the command
load /*file*/"https://cactus.nci.nih.gov/chemical/structure/CC(C(C(C)F)(Br)F)O/file?format=sdf&get3d=true
I fixed the stereochemistry by clicking on the carbons with incorrect chirality after issuing the command "set picking INVERTSTEREO". To rotate around the bonds, I used "rotate branch @3 @2 180 180" and "rotate branch @3 @4 180 180". Finally, I chose the viewing orientations, displayed the orientationwith "show moveto" and issued the resulting moveto commands before and after rotating around the single bonds.
A more elaborate example with glucose (4 chiral centers) can be found on Proteopedia.


What ratio of 73% calcium hypochlorite powder to water mimics the strength of sodium hypochlorite (bleach)?

To start, we must first understand that we can ignore the counterions to the $\ce{Ca(ClO)2}$ and the $\ce{NaClO}$. Typical household bleach claims to have a concentration of 6% (w/v). Now we can do some basic arithmetic to convert the percent weight by volume of sodium hypochlorite in water to a molar concentration:

$\mathrm{\frac{0.06\,g}{1\,mL}}$$\mathrm{\frac{1\,mol}{74.44\,g}}$$\mathrm{\frac{1000\,mL}{1\,L}}$$\mathrm{\,= 0.8060\,M \,OCl^-}$

The first fraction in the above equation is based on the definition of w/v, which is the number of grams of a substance dissolved in one milliliter of a solution times 100%. In the next fraction, we convert the grams $\ce{NaClO}$ into moles of $\ce{NaClO}$. Finally, in the last fraction, we convert milliliters into liters to give the number of moles per liter of $\ce{NaClO}$, or simply $\ce{ClO^-}$ without the counter ion.

Now that we know the molar concentration of $\ce{ClO^-}$ we are aiming for, we can convert this into the number of grams of pure $\ce{Ca(ClO)2}$ we would need to to achieve this concentration in one gallon of water

$\mathrm{\frac{0.8060\,mol}{1\,L}}$$\mathrm{\frac{1\,mol}{2\,mol}}$$\mathrm{\frac{142.98\,g}{1\,mol}}$$\mathrm{\frac{3.78541\,L}{1\,gal}}$$\mathrm{\,=\frac{218.1\,g\,OCl^-}{gal}}$

In the second fraction, we converted the moles of $\ce{ClO^-}$ into the number of moles of $\ce{Ca(ClO)2}$ (as there is only one unit of $\ce{Ca(ClO)2}$ per two units of $\ce{ClO^-}$). In the third fraction, we converted the number of moles of $\ce{Ca(ClO)2}$ into grams. In the last fraction, we converted the number of liters into the number of gallons.

As you stated, the powder only contains 73% $\ce{Ca(ClO)2}$ by mass. We know we need a total of 218.1 g of $\ce{Ca(ClO)2}$, so we can use the following conversion factor:

$\mathrm{\frac{218.1\,g\,OCl^-}{1\,gal}}$$\mathrm{\frac{1\,g}{0.73\,g}}$$\mathrm{\,=298.8\,g\,powder}$ (Rounded to 0.3 kg since we only have one significant figure)

In the second fraction, we know that per every gram of your powder, 73 percent (0.73 g) of the $\ce{Ca(ClO)2}$ will be present. Thus, the conversion factor is valid.

Final Answer: 0.3 kg of Powder


The reason why statement #2 doesn't make sense to you is because it is not generally true!
For a closed system of constant composition, with P-V work only,
$$dU = \left(\frac{\partial U}{\partial T}\right)_V dT +  \left(\frac{\partial U}{\partial V}\right)_T dV = C_v dT + \left(\frac{\partial U}{\partial V}\right)_T dV$$
The difference between that statement and your statement is the extra term at the end, which is the amount by which U changes with V at constant T, times the change in V.
[Minor note: I've dropped the "n" you use, as I'm using total heat capacities, not molar heat capacities.]
However, if we have an ideal gas, U is independent of V, which means:
$$\left(\frac{\partial U}{\partial V}\right)_T=0,$$
Hence, with an ideal gas, the term on the right goes away, giving us your statement #2:
$$dU = C_v dT$$
I.e., your statement #2 doesn't follow directly from #1.  Instead, it's specific to ideal gases.
So:  If a process invoving an ideal gas is carried out at constant pressure instead of constant volume, it is true that $dV$ may not be zero.  But since that non-zero $dV$ doesn't matter when it comes to calculating $dU$ for ideal gases, we can use $C_V dT$ even when $dV \neq 0$.


Does the addition of base to 3-pentanone give the cis or trans enolate?

As a prenote: My answer is somehow speculative because I have no experimental data or reference values of similar reactions used. In other words: The argumentation pathway below might be wrong.

Addition of LDA mostly gives the trans-enolate if a primary alkyl substituent next to the carbonyl group has been used.

Carbonyl(O) is smaller than the methyl group [ref.], therefore the allylic A13 interaction of Me//Me or even Me//i-Pr is bigger than the allylic A12 interaction.

An allylic A13 Me//Me interaction is about 3,7 kcal/mol, an allylic A12 Me//Me is may be around 3,2 kcal/mol (this is just a guess).

If HMPA is added Li might coordinate to four HMPA molecules forming a cluster and inhibiting the cyclic transition state of LDA and hence no longer providing the trans-enolate, meaning an open transition state must be lower in activation energy for the cis-enolate formation.

So I tried to draw the conformations of 3-pentanone and then treated the base as "point bulk", meaning the base is like a sphere with different sizes.

Hard bases have a bigger repulsion to the carbonyl oxygen.

I have used arbitrarily numbers for the stereoselectivity.

Here is my answer which base gives more cis-enolate than the other:


Are there amide versions of lactides and lactones?

Amide analogue of lactons are lactams, their tautomeric forms are lactims. Citing from the IUPAC Nomenclature of Organic Chemistry (Preferred names 2013):

  P-66.1.5.1 Lactams and lactims

  Intramolecular amides of amino carboxylic acids, $\ce{-CO-NH\bond{-}}$, are
  called ‘lactams’ and their tautomers, $\ce{-C(OH)=N\bond{-}}$, are ‘lactims’. Lactams named in two ways:

    (1) as heterocyclic pseudoketones;

    (2) by substituting ‘lactam’ for the ‘ic acid’ ending of the systematic ‘oic acid’ name for the parent acid without the amino substituent, and inserting a locant designating the position of the amino group between the ‘o’ and the ‘lactam’. Method ‘lactams’. Lactims are named in the same way, using ‘lactim’ in place of ‘lactam’.

  (1) generates preferred IUPAC names.

  Examples:

  pyrrolidin-2-one (PIN)
  butano-4-lactam

  (…)

  3,4,5,6-tetrahydropyridin-2-ol (PIN)
  pentano-5-lactim

(Note that the numberings depicted are for the preferred names (PINs) based on nitrogen heterocyclics)

There's no (at least IUPAC) term for lactides analogue (‘lactide’ is not used in IUPAC names themselves anyway), but they exist.

In traditional or general names, the Greek letter numbering is used, e.g. ε-caprolactam, or β-lactam four-membered ring part in bicyclic penicillin skeleton.


Time and entropy has not much to do with each other.

From practical point of view: Equilibrium systems can hang around for infinite time, and their entropy does not change at all - so time would be an impractical measure.

From conceptual point of view: Entropy comes from statistical behavior of the system of many bodies. Time is unrelated to statistics; it exist even in single body systems.


What do the quantum numbers actually signify?

Quantum numbers give information about the location of an electron or set of electrons.  A full set of quantum numbers describes a unique electron for a particular atom.

Think about it as the mailing address to your house.  It allows one to pinpoint your exact location out of a set of $n$ locations you could possibly be in.  We can narrow the scope of this analogy even further.  Consider your daily routine.  You may begin your day at your home address but if you have an office job, you can be found at a different address during the work week.  Therefore we could say that you can be found in either of these locations depending on the time of day.  The same goes for electrons.  Electrons reside in atomic orbitals (which are very well defined 'locations').  When an atom is in the ground state, these electrons will reside in the lowest energy orbitals possible (e.g. 1$s^2$ 2$s^2$ and 2$p^2$ for carbon).  We can write out the physical 'address' of these electrons in a ground-state configuration using quantum numbers as well as the location(s) of these electrons when in some non-ground (i.e. excited) state.

You could describe your home location any number of ways (GPS coordinates, qualitatively describing your surroundings, etc.) but we've adapted to a particular formalism in how we describe it (at least in the case of mailing addresses).  The quantum numbers have been laid out in the same way.  We could communicate with each other that an electron is "located in the lowest energy, spherical atomic orbital" but it is much easier to say a spin-up electron in the  1$s$ orbital instead.  The four quantum numbers allows us to communicate this information numerically without any need for a wordy description.

Of course carbon is not always going to be in the ground state.  Given a  wavelength of light for example, one can excite carbon in any number of ways.  Where will the electron(s) go?  Regardless of what wavelength of light we use, we know that we can describe the final location(s) using the four quantum numbers.  You can do this by writing out all the possible permutations of the four quantum numbers.  Of course, with a little more effort, you could predict the exact location where the electron goes but in my example above, you know for a fact you could describe it using the quantum number formalism.

The quantum numbers also come with a set of restrictions which inherently gives you useful information about where electrons will NOT be. For instance, you could never have the following possible quantum numbers for an atom:

$n$=1; $l$=0; $m_l$=0; $m_s$=1/2

$n$=1; $l$=0; $m_l$=0; $m_s$=-1/2

$n$=1; $l$=0; $m_l$=0; $m_s$=1/2

This set of quantum numbers indicates that three electrons reside in the 1$s$ orbital which is impossible!

As Jan stated in his post, these quantum numbers are derived from the solutions to the Schrodinger equation for the hydrogen atom (or a 1-e$^-$ system).  There are any number of solutions to this equation that relate to the possible energy levels of they hydrogen atom.  Remember, energy is QUANTIZED (as postulated by Max Planck).  That means that an energy level may exist (arbitrarily) at 0 and 1 but NEVER in between.  There is a discrete 'jump' in energy levels and not some gradient between them.  From these solutions a formalism was constructed to communicate the solutions in a very easy, numerical way just as mailing addresses are purposefully formatted in such a way that is easy that anyone can understand with minimal effort.

In summary, the quantum numbers not only tell you where electrons will be (ground state) and can be (excited state), but also will tell you where electrons cannot be in an atom (due to the restrictions for each quantum number).

Principle quantum number ($n$) - indicates the orbital size.  Electrons in atoms reside in atomic orbitals.  These are referred to as $s,p,d,f...$ type orbitals.  A $1s$ orbital is smaller than a $2s$ orbital.  A $2p$ orbital is smaller than a $3p$ orbital.  This is because orbitals with a larger $n$ value are getting larger due to the fact that they are further away from the nucleus.  The principle quantum number is an integer value where $n$ = 1,2,3... .

Angular quantum number ($l$) - indicates the shape of the orbital. Each type of orbital ($s,p,d,f..$) has a characteristic shape associated with it.  $s$-type orbitals are spherical while $p$-type orbitals have 'dumbbell' orientations.  The orbitals described by $l$=0,1,2,3... are $s,p,d,f...$ orbitals, respectively.  The angular quantum number ranges from 0 to $n$-1.  Therefore, if $n$ = 3, then the possible values of $l$ are 0, 1, 2.

Magnetic quantum number ($m_l$) - indicates the orientation of a particular orbital in space.  Consider the $p$ orbitals.  This is a set of orbitals consisting of three $p$-orbitals that have a unique orientation in space.  In Cartesian space, each orbital would like along an axis (x, y, or z) and would be centered around the origin at 0,0.  While each orbital is indeed a $p$-orbital, we can describe each orbital uniquely by assigning this third quantum number to indicate its position in space.  Therefore, for a set of $p$-orbitals, there would be three $m_l$, each uniquely describing one of these orbitals.  The magnetic quantum number can have values of $-l$ to $l$.  Therefore, in our example above (where $l$ = 0,1,2) then $m_l$ would be -2, -1, 0, 1, 2.

Spin quantum number ($m_s$) - indicates the 'spin' of the electron residing in some atomic orbital.  Thus far we have introduced three quantum numbers that localize a position to an orbital of a particular size, shape and orientation.  We now introduce the fourth quantum number that describes the type of electron that can be in that orbital.  Recall that two electrons can reside inside one atomic orbital.  We can define each one uniquely by indicating the electron's spin.  According to the Pauli-exclusion principle, no two electrons can have the exact same four quantum numbers.  This means that two electrons in one atomic orbital cannot have the same 'spin'. We generally denote 'spin-up' as $m_s$ =1/2 and spin-down as $m_s$=-1/2.


Why is entropy change in an irreversible adiabatic process positive?

The friction they are talking about is viscous friction of the gas itself.  This comes into play when the gas is deformed rapidly, and is roughly proportional to the square of the rate of gas deformation.
Regarding entropy, the change is entropy is equal to the integral of dq/T only for a reversible process.  In an irreversible process, in addition to this exchange of entropy with the surroundings, there is also entropy generated within the system itself.


Why can’t I use solubility product for normally soluble compounds?

In addition to Mithoron's and Poutnik's comments, the solubility product may not be relevant to your solution chemistry anyway.  Solubility products enter into solution equilibria when the precipitates are present, as in this answer where relative solubility products predict whether mostly magnesium or mostly iron will be dissolved in water when their hydroxides are present.  With readily soluble materials, you likely do not get the precipitate so solubility products, even if you can compute then on an activity basis, have no bearing on the equilibrium.


How does work from other sources affect Gibbs free energy of the system?

I will treat the first and second question as if they were part one and part two of a single question. I think the OP wants to know how non-PV work plays a role when looking at the Gibbs energy of a reaction, and its interpretation.
To start off, the Gibbs energy for a given reaction is a state function, so it does not depend on how much work or how much heat are exchanged between system and surrounding. The total change in entropy, however, is affected by what is going on in the system and in the surrounding, so work vs heat is relevant for the second law of thermodynamics and the direction of the reaction.

First Question: In the case of an electrolytic cell, in the system, there is also a power source. How would this external voltage source affect the Gibbs free energy equation? Would it just be an extra $-VIt$ term or something else which would cause $\Delta_\mathrm rG < 0?$

The easiest way to deal with the power source is to place the system/surrounding boundary such that the power source is in the surrounding. This shows it does not affect the Gibbs energy of the reaction at all. The Gibbs energy is positive, but the reaction can go forward because it is offset by the work done on the system. As a result, the combined entropy increases, and we are not breaking the second law.
In general, the criterion is not that the Gibbs energy of reaction has to be negative, but that
$$\Delta_r G < w$$
When there is no work done on the system or by the system, this simplifies to the requirement that the Gibbs energy be negative.
As an aside, when the Gibbs energy of reaction is negative, we can not have the reaction do an arbitrary amount of work on the surrounding, but that amount is limited to the Gibbs energy. That's why the Gibbs energy is sometimes called the maximal work (this is a good term for cases when it is negative; when it is positive, it is the minimal work that needs to be done on the system for it to be possible for the reaction to go forward).

Second Question: In the case of a biological cell, the work usually comes from the hydrolysis of ATP. How would this alter the equation and cause $\Delta_\mathrm rG < 0?$

The hydrolysis of ATP usually does not do work on the surroundings. Instead, the uncatalyzed reaction has a positive Gibbs energy of reaction. To use up the reactants and make the products, the cell runs a different reaction that includes the hydrolysis of ATP. It is not sufficient that ATP be hydrolized somewhere while the desired reaction goes on somewhere else. Instead, the two reactions are coupled so that ATP hydrolysis can not occur without the other part of the reaction happening. Here is an example how that might look for the reaction $\ce{A + B -> C}$:
$$\ce{A + ATP -> A-P + ADP}\tag{1}$$
$$\ce{A-P + B + H2O -> C + P_i}\tag{2}$$
In this scheme, A-P refers to a phosphate group bound to A, and $\ce{P_i}$ refers to phosphate. Both steps can have a negative Gibbs energy of reaction, even thought the Gibbs free energy of reaction for $\ce{A + B -> C}$ is positive. As net result, A and B turn in to C, without work done on the system.
There are other cases where there is work done on the system or by the system. These include proton pumps, and myosin hydrolyzing ATP to contract muscle fibers. You could also discuss photosynthesis in terms of work done to drive the synthesis of ATP.


Similarity between Ecell and equilibrium constant Kc

$E^\circ_{cell}$ and $K_c$ are related. Both indicate which direction of the reaction is spontaneous. Both tell you something about the position of equilibrium.

However, $E^\circ_{cell}$ is a directly measurable value. If you construct an electrochemical cell as using a zinc electrode immersed in a 1.0 M solution of say zinc chloride and a copper electrode immersed in a 1.0 M solution of copper (II) chloride with a salt bride of sodium chloride connecting the solutions, a voltmeter connected between the two electrodes will read the value of $E^\circ_{cell}$, at least initially. The voltage will change as the reaction progresses, but the initial voltageis $E^\circ_{cell}$ which is defined for standard conditions of 1.0 M concentration, 1 bar of pressure, and 25 $^\circ$C.

$$\ce{Zn|ZnCl2}(1.0 \text{ M})\parallel\ce{CuCl2}(1.0 \text{ M})|\ce{Cu} \ \ \ \ E^\circ_{cell}=1.1\text{ V}$$

The relationship between $E^\circ_{cell}$ and $K_c$ is the Nernst Equation. Note the use of both $E^\circ_{cell}$, the standard cell potential and $E_{cell}$, the observed cell potential under other conditions. The remainder of the alphabet soup is defined below.
$$E_{cell}=E^\circ_{cell}-\frac{RT}{zF}\ln{Q}$$

$E^\circ_{cell}$ - Standard cell potential
$E_{cell}$ - nonstandard cell potential at other conditions
$R$ - the ideal gas constant (which appears here because it is equal to Avogadro's number times Boltzmann's constant
$T$ - the temperature in kelvins
$z$ - the number of moles of electrons being transferred in this redox process. If we wanted to use the number of individual electrons, then we would need to replace $R$ with $k_B$
$F$ - Faraday's constant
$Q$ - the reaction quotient

We need to use $Q$ in the Nernst equation, because this equation can relate the cell potential to any range of conditions, not just equilibrium.

  At equilibrium $Q = K_c$ and $E_{cell} = 0 \text{ V}$ (there is no net change in the reaction composition, so there is no potential difference across the half-cells.

Thus, the Nernst Equation becomes:

$$0=E^\circ_{cell}-\frac{RT}{zF}\ln{K_c}$$
$$E^\circ_{cell} = \frac{RT}{zF}\ln{K_c}$$

  To determine the values of $E^\circ_{cell}$ and $K_c$ for the reverse reaction...

For $E^\circ_{cell}$:

$$E^\circ_{cell} = -E^\circ_{rev}$$

For $K_c$

$$K_{rev} = \frac{1}{K_c}$$


Can I synthesize an Amide simply by heating an amine and a carboxylic acid (with a vacuum pump)?

It can be done but it is not a good reaction. From Lundberg et al. [1]:

Formally, the amide bond is formed through the condensation of a carboxylic acid and an amine with the release of one equivalent of water. This reaction has been considered challenging due to the competing acid–base reaction, which occurs when the amine and the carboxylic acid are mixed. Although the amide bond can be formed from the corresponding ammonium carboxylate salt upon heating, this reaction has generally been considered to be of limited preparative value. Furthermore, the high activation barrier for the direct coupling of a carboxylic acid and an amine can only be overcome using forcing reaction conditions.

Use of borate ester catalysis under Dean–Stark conditions has recently been reported to give good yields without resorting to acyl activation groups [2].
References

Lundberg, H.; Tinnis, F.; Selander, N.; Adolfsson, H. Catalytic Amide Formation from Non-Activated Carboxylic Acids and Amines. Chem. Soc. Rev. 2014, 43 (8), 2714–2742. DOI: 10.1039/C3CS60345H. (PDF)
Sabatini, M. T.; Boulton, L. T.; Sheppard, T. D. Borate Esters: Simple Catalysts for the Sustainable Synthesis of Complex Amides. Sci. Adv. 2017, 3 (9), e1701028. DOI: 10.1126/sciadv.1701028.


How can melting point equal freezing point?

Because melting point and freezing point describe the same transition of matter, in this case from liquid to solid (freezing) or equivalently, from solid to liquid (melting).

What you may not realize is that while water is freezing or melting, its temperature is not changing!  It is stuck on $0\ \mathrm{^\circ C}$ during the entire melting or freezing process.  It is easier to see this for boiling points.  if you put a thermometer in water and heat it, the temperature will rise until it reaches $100\ \mathrm{^\circ C}$, and then it starts boiling.  And while it boils, it will stay at $100\ \mathrm{^\circ C}$!  All the way until the water has all boiled away.  Now if you could somehow trap the steam (gaseous water) and keep heating it, the steam could have a temperature higher than $100\ \mathrm{^\circ C}$.

So to sum this all up, when matter is transitioning from solid to liquid (melting) or liquid to solid (freezing), its temperature is fixed at the melting/freezing point, which is the same temperature.


Numerous chemicals are linked with decreasing male fertility.
In fact, studies suggest that unhealthy lifestyles and environmental factors (e.g. plasticisers in the water system) may be causing this decrease in fertility.
See: http://www.telegraph.co.uk/men/active/mens-health/10695991/Why-are-mens-sperm-rates-falling.html etc
Obviously conclusive studies are limited at this stage.

But if you want to be more specific, I recall that chloroform is known to cause miscarriages and sterility in mice. Googling that to find a citation led me to this article:
http://www.ncbi.nlm.nih.gov/pubmed/10969727
Which found evidence of an otherwise non-toxic antifertility agent for male rats.

To be honest, I would be surprised to find studies which have tested a anti-fertility drug on men. Such a thing should not make it past an ethics board (given the ability to prevent impregnation by far less drastic measures, e.g. condoms).

If however you were talking about the sort of chemical castration they did to someone like Alan Turing, they were not designed to sterilise him (heck, his 'crime' was homosexuality and children were not a likely outcome of that), but to alter his hormones and personality. Anyway, modern drugs might be something like Cyproterone acetate or Depot medroxyprogesterone acetate - though that is also used as a birth control drug for women.


Why does a mixture get warm in an exothermic reaction?

In thermodynamics, we like to divide any situation into two parts, the system and the surroundings. In general the system is the part of the universe that is being studied while the surroundings is everything else that interacts with the system. Between the boundary of the system and the surroundings, energy and heat can be transferred (in an open system).

Now in fermentation, the yeast and glucose molecules are said to be the system while the surround water is the surroundings. Since the reaction is exothermic, this means that energy is moving from the system to the surroundings. Therefore the water should gain more heat energy which will increase its temperature.

However I doubt that the heat generated from the reaction is enough for the enzymes in the yeast to stop working. When I did this experiment we used a Bunsen burner to heat up the solution to approximately 35 degrees. So really what is going to cause the solution to heat up to above 45 degrees (this approximately where yeast begin to stop working) is that you keep the Bunsen burner on and keep heating it.


As has already been mentioned in comments, those letters and numbers for each state correspond to their symmetry, specifically the symmetry of the (purely electronic) wavefunction. For more information regarding symmetry notation, see here, and see here for more discussion about how symmetry can be applied to molecular orbitals.

It might not be clear from the second question, but because the symmetry of the overall N-electron wavefunction results from the direct product of all occupied 1-electron wavefunctions (at least in Hartree-Fock), exciting electrons into molecular orbitals with different symmetries will lead to excited N-electron wavefunctions (excited states) with different symmetries.

To consider the specific case of biphenyl,

the $\tau = 0^{\circ}$ structure, corresponding to the leftmost values on the x-axis, will look like the 2D structure just above where both rings lie in the same plane. As an exercise left to the reader, you can figure out that the point group of the planar form of biphenyl is $D_{\mathrm{2h}}$. The point group $D_{\mathrm{2h}}$ has the following character table:

$$\begin{array}{c|cccccccc|cc} \hline
D_\mathrm{2h} & E & C_2(z) & C_2(y) & C_2(x) & i & \sigma(xy) & \sigma(xz) & \sigma(yz) & & \\ \hline
\mathrm{A_g} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2,y^2,z^2 \\
\mathrm{B_{1g}} & 1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 & R_z & xy \\
\mathrm{B_{2g}} & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & R_y & xz \\
\mathrm{B_{3g}} & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 & R_x & yz \\
\mathrm{A_u} & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & & \\
\mathrm{B_{1u}} & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 & z & \\
\mathrm{B_{2u}} & 1 & -1 & 1 & -1 & -1 & 1 & -1 & 1 & y & \\
\mathrm{B_{3u}} & 1 & -1 & -1 & 1 & -1 & 1 & 1 & -1 & x & \\ \hline
\end{array}$$

The three-dimensional symmetry of any object can always be broken down into one or more irreducible representations (irreps) belonging to the object's point group; each irrep corresponds to a row in the point group's character table. The symmetry table is the leftmost column in each irrep row. To enumerate them,

The labels from the figure are $1^1A, 1^1B_{1}, 1^1B_{2}, 1^1B_{3}, 2^1B_{3}, 2^1B_{2}, 2^1A, 2^1B_{1}$.
The labels from the character table are $\mathrm{A_g}, \mathrm{B_{1g}}, \mathrm{B_{2g}}, \mathrm{B_{3g}}, \mathrm{A_u}, \mathrm{B_{1u}}, \mathrm{B_{2u}}, \mathrm{B_{3u}}$.

You can see that these labels are almost but not quite the same as those in the figure. A minor difference is italicized versus regular/upright text; this is a matter of typographic choice and there is no physical or chemical difference between them. There are two significant differences:

The labels in the figure are prefixed with a regular number and a superscript number. The regular number helps uniquely defines the state ordering: $1^1B_{1}$ is the first state with $^1B_{1}$ symmetry, and $7^1B_{1}$ is (hypothetically) the seventh state with $^1B_{1}$ symmetry. Without the superscript number, each symmetry label only specifies the spatial symmetry, and with the superscript defined as $2S + 1$, the spin multiplicity, each label can also specify the spin symmetry. $1^1B_{1}$ is the first singlet ($S = 0$) state with $B_{1}$ symmetry, and $1^3B_{1}$ is (hypothetically) first triplet ($S = 1$) state with $B_{1}$ symmetry. See this excellent answer for more information, in particular the figure.
The labels in the figure do not have u/g in the subscript, because for most of the biphenyl structures, where $\tau \neq 0^{\circ}$, the molecular point group is no longer $D_{\mathrm{2h}}$, but one of lower symmetry. When the rings are perfectly perpendicular to each other ($\tau = 90^{\circ}$ and $\tau = 270^{\circ}$), the molecular point group will be $D_{\mathrm{2d}}$, like hydrogen peroxide, and every other intermediate value will be even lower in symmetry: $D_{\mathrm{2}}$, similar to skewed ferrocene (rather than staggered or eclipsed). To see how the irrep and its symmetry operations from a point group of higher symmetry map to those in a point group of lower symmetry, one needs a descent table. Here, most notably, all skewed structures are missing the inversion operation that would necessitate the gerade/ungerade subscript. Because most of the structures are of lower $D_{2}$ symmetry, it is easiest to represent all structures with $D_{2}$ point group labels.

For completeness,

$$\begin{array}{c|ccccc|cc} \hline
D_\mathrm{2d} & E & 2S_4 & C_2 & 2C_2' & 2\sigma_\mathrm{d} & & \\ \hline
\mathrm{A_1} & 1 & 1 & 1 & 1 & 1 & & x^2+y^2,z^2 \\
\mathrm{A_2} & 1 & 1 & 1 & -1 & -1 & R_z & \\
\mathrm{B_1} & 1 & -1 & 1 & 1 & -1 & & x^2-y^2 \\
\mathrm{B_2} & 1 & -1 & 1 & -1 & 1 & z & xy & \\
\mathrm{E} & 2 & 0 & -2 & 0 & 0 & (x,y),(R_x,R_y) & (xz,yz) \\ \hline
\end{array}$$

$$\begin{array}{c|cccc|cc} \hline
D_2 & E & C_2(z) & C_2(y) & C_2(x) & & \\ \hline
\mathrm{A} & 1 & 1 & 1 & 1 & & x^2, y^2, z^2 \\
\mathrm{B_1} & 1 & 1 & -1 & -1 & z, R_z & xy \\
\mathrm{B_2} & 1 & -1 & 1 & -1 & y, R_y & xz \\
\mathrm{B_3} & 1 & -1 & -1 & 1 & x, R_x & yz \\ \hline
\end{array}$$

See how there are fewer symmetry operations (middle set of columns) in $D_{2}$ than in $D_{\mathrm{2d}}$ than in $D_{\mathrm{2h}}$.


Is Cracking of Every HydroCarbon Family possible? (Alkane ,Alkenes,Arenes)

Simple arenes (benzene, toluene, xylenes) are a typical product of a steamcracker. https://en.wikipedia.org/wiki/Cracking_(chemistry)#Thermal_methods

Catalytic hydrocracking otoh. (with added hydrogen at high pressure) leads to an increase in the aliphatic fraction and lowers the aromatic content. https://en.wikipedia.org/wiki/Cracking_(chemistry)#Hydrocracking

in itself, benzene is quite stable. It's a questions if it has sth. to react with.


When you say "alcohol", I am not sure if you are referring to ethanol (drinking alcohol) or alcohols in general so in my example I will use ethanol. When an alcohol is combined with a carbocylic acid and a catalyst, they will produce esters (fragrances). For example, ethanol and ethanoic acid would react to produce ethyl ethanoate (a fruity smelling substance) and water.

There are a plethora of different reactions from all different types of alcohols and carbocylic acids that produce countless different smells. The fragrances also come in a variety of colors as well. If you want to find out more, I would suggest looking up ester synthesis. Hope my answer helps.


How do they ensure there is one electron on an oil drop in Millikan's oil drop experiment?

There does not have to be just one electron per drop.  Say you have a drop which, in reality, picked up four electrons, another with five and a third drop with seven.  None has one electronic charge, but when you measure charges you find they have a common factor; the first drop has four times that factor, the second has five times that factor and the third drop shows a multiplier of seven.  It was this common factor, not necessarily the charge on any specific drop, that was recognized as one electronic charge.