Who coined the term "gas evolution reaction"?

There is no specific term as "gas evolution reaction" in modern chemistry but the word evolve itself conveys this meaning of emission. I am curious as to why you are interested in this terminology? Just for the sake of historical interest, the unabridged Oxford Dictionary (subscription required) shows that the term dates back to 1770s in relation to gas, heat, light etc. However there is no guarantee that the following references are the earliest. If you have more energy or you are a serious scholar of history, you might need to see German and French literature before the 1770s and what words Lavoisier used when he heated the oxide of mercury and discovered oxygen.

IV. The process of emission.

Chiefly Chemistry. The emission or release of gas, heat, light, etc.; the disengagement of a substance from combination; an instance
of such a process. Cf. evolve v. 5.

1783   Philos. Trans. (Royal Soc.) 73 350   The method put in practice
by Mr. Hutchins to settle the freezing point of quicksilver depends
entirely upon this generation, re-appearance or evolution of heat.
1800   H. Davy Res. Nitrous Oxide 425   This is on account of the
increased evolution of nitrogene from the blood.

Similarly, evolve is even older

1772   [implied in: J. MacLurg Exper. Human Bile 38   The colouring
matter of the Bile discovers also a much stronger attraction to the
mineral acids, and gives indubitable marks of a more evolved
phlogiston. (at evolved adj. 2)].
1786   Edinb. New Dispensatory i. 7   In the incipient state a
quantity of gas is still evolved, and along with it a quantity of
alkaline air.


Why do double bonds cause kinks in fatty acid chains?

Instead of zig-zag, zig-zag, zig-zag etc. it now goes zig-zag, zig, zig-zag, zig- etc (the bold zig is due to the cis double bond). So the direction of the alkyl chain on one side of the cis double bond is different from that of the alkyl chain on the other. You could partially correct for that "wrong turn" by rotating a single bond by $180^\circ$, but that changes a staggered conformation into an eclipsed one.

Here is a link to a 3D model: http://www.biotopics.co.uk/jsmol/fattyacids.html and a snapshot in case the link goes away:


The standard set of orbitals (s, p, d, f, ...) form an orthogonal basis set, such that you can describe any continuous function as a linear combination of them.  Now, this also means that you can take some subset of them, and make a new set of orthogonal functions, by appropriate normalized linear combinations, that will contain the full symmetry of the subset chosen. The $sp^{3}$ hybridization is just such a set of combinations of $s$ and $p$ orbitals, chosen to make a set of equivalent orbitals emphasizing a mixed symmetry.

Note that the standard orbitals are not the only orthogonal basis set that exists - it just happens to be the one that nicely gives solutions to the Schrodinger equation separating out $n$ and $l$. One could use spherical Bessel functions, or any number of other orthogonal basis sets, but it would not be as clean of a solution just because of the form of the differential equation under consideration.


Why is HCl a stronger acid than Acetic Acid?

A simpler question, would be why is water a weaker acid than $\ce{HCl}$?  If we were to look at that issue, we might clear some of the mess up.

Strongly reactive compounds produce weakly reactive products.

In the case of acid base theory, chloride ion is a much weaker base than hydroxide.  Why is this so?  The hydroxide is much smaller, and what is referred to as 'harder'.  The guts of the issue is that in order for a base to deprotonate anything it has to donate two electrons to a bonding region near a proton (and in doing such render the orbital manifest, and the bond existent).  In oxygen based anions, those two electrons are more likely to be in a confined region, and hence more likely to donate to the putative $\ce{HO-H}$ bond that is forming, while $\ce{Cl^-}$, being larger has a harder time getting two electrons into the bonding space required to form a $\ce{Cl-H}$ bond.  So hydroxide is a stronger base.  This means that when donating a proton, it is likely to just yank it back again from whatever it just gave it up to, as compared to $\ce{Cl^-}$.  $\ce{HCl}$ gives up the proton and is more likely to not recapture it.

Groovy?  Back to acetate.  The resonance in acetate creates some diffusion of the negative charge and makes it harder to form said $\ce{AcO-H}$ bond, and as such it is a weaker base than hydroxide.  However, the smaller oxygen atoms are still better and forming bonds than $\ce{Cl^-}$ and as such it is a stronger base than chloride, and therefore its conjugate acid is less acidic.

Strong acids: weak conjugate bases.
Strong bases: weak conjugate acids.


Reaction mechanism for 7-hydroxy-4-methylcoumarin synthesis with an iodine catalyst

There is a paper about this exact reaction, but in addition to the use of $\ce{I2}$ it mentions the use of microwaves. In the reaction, $\ce{I2}$ acts as a Lewis acid, catalyzing both the transesterificaiton and electrophilic aromatic substitution reactions. It can be seen acting as such in the first step of the mechanism:

$\hspace{2cm}$

Next, a proton transfer occurs:

Next, the resorcinol derivative can act as a nucleophile, and ethanol is ejected, completing the transesterification reaction:

Next, $\ce{I2}$ coordinates to the keto carbonyl, activating the carbon towards electrophilic addition. Because this resorcinol derivative is so activated by the $\ce{-OH}$ and $\ce{-OOCR}$ groups, a weak Lewis acid like $\ce{I2}$ is sufficient to catalyze the reaction:

$\hspace{1.5cm}$

$\hspace{1.8cm}$

$\hspace{1.6cm}$


During phase change in matter, why doesn't the temperature change?

From Changes of Phase (or State):

...

So, how could there be a change in heat during a state change without a change in temperature?

"During a change in state the heat energy is used to change the bonding between the molecules. In the case of melting, added energy is used to break the bonds between the molecules. In the case of freezing, energy is subtracted as the molecules bond to one another. These energy exchanges are not changes in kinetic energy. They are changes in bonding energy between the molecules.

"If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance. The example we will use here is ice melting into water. Immediately after the molecular bonds in the ice are broken the molecules are moving (vibrating) at the same average speed as before, so their average kinetic energy remains the same, and, thus, their Kelvin temperature remains the same."


How to calculate the rotational constants for different states from a stick spectrum?

In your question, it's not clear whether the quantum number assignments for the transitions are known in advance, so for generality I'll assume that you don't know what they are, and help you find them.

Combination-differences is the most common paper-based method to determine line assignments (for more detailed analysis, a computer-aided fitting program such as PGopher is best). From the line assignments, it's easy to find the $B$-values.

Rotational energy levels are given by
$$F_\nu(J)\approx B_\nu J(J+1)+D_\nu [J(J+1)]^2$$
where $B_\nu$ and $D_\nu$ are the rotational constant and centrifugal distortion constant associated with the manifold $\nu$.

Denoting the initial and final states by $1$ and $2$ respectively, the $P,R$ transition energies are given by
$$P(J)=F_2(J-1)-F_1(J)
\\R(J)=F_2(J+1)-F_1(J).$$
Combination-differences is can be done in two ways:

If your rovibronic transition originates from an initial manifold with well-documented spectral characteristics (including $B_1$), note that
$$R(J-1)-P(J+1)=2 (2 J+1) \left(B_1-2 D_1 \left(J^2+J+1\right)\right)\approx2B_1(2J+1)$$
when $D_1$ is small, and so then you can just try to find pairs of $P,R$ peaks with spacing $2B_1(2J+1)$ for various small values of $J$. This allows you to number the peaks as to their identity.
If instead the upper state has well-known $B_2$, then note that
$$R(J)-P(J)=2 (2 J+1) \left(B_2-2 D_2 \left(J^2+J+1\right)\right)\approx2 B_2 (2 J+1)$$
and so then you can just try to find pairs of $P,R$ peaks with spacing $2B_1(2J+1)$ for various small values of $J$. Again, this allows you to number the peaks as to their identity.

Once the quantum numbers are assigned, finding both $B_1$ and $B_2$ is easy. For example, $R(J)-R(J-1)=2 \left(B_2 (J+1)-B_1 J\right)$, so if you measure it for two values of $J$, then you have two unknowns and two datapoints, so just solve for $B_1$ and $B_2$.

Fun Pictures

Just to give a simple idea of how spectra are simulated, the following Mathematica code does a basic rotational spectrum assuming Boltzmann statistics at a reasonable temperature (hit Cell>Convert To>StandardForm to get it to render the subscripts correctly):

Subscript[F, k_][J_] :=
  Subscript[B, k] J (J + 1) - Subscript[D, k] J^2 (J + 1)^2;
R[J_] := Subscript[F, 2][J + 1] - Subscript[F, 1][J];
P[J_] := Subscript[F, 2][J - 1] - Subscript[F, 1][J];
ListPlot[Transpose[
  Table[{{R[J], (2 J + 1) Exp[-(1/30) Subscript[F, 1][J]]}, {P[
       J], (2 J + 1) Exp[-(1/30) Subscript[F, 1][J]]}} /. {Subscript[
      D, k_] :> 0, Subscript[B, 1] -> 1, Subscript[B, 2] -> 0.9}, {J,
    0, 15}], {2, 1, 3}], Filling -> Axis]


Is p-bromophenol more acidic than p-chlorophenol?

Disclaimer: This is not necessarily the "correct" answer as many factors play roles and it would be difficult to describe the observation so simply.

I'm not sure I can provide an easily "correct" answer, but I can try to explain some of the many factors that influence the acidity.

Most notably, the factors we should consider are inductive effects and delocalization of charge.

The inductive effects, based on electronegativity, would dismiss the experimental observation, so I'll look at delocalization of charge.

The delocalization of charge, which introduces a negative charge by the halogen is greatest by F>Cl>Br>I. This would reduce acidity, as a more negative benzene ring would not desire to gain another negative charge. Thus, this is one factor that supports the experimental observation.

Thus, we may be able to explain the observation, if only partially, by the delocalization of charge.

I adapted the answer to a similar question on http://www.quora.com/Organic-Chemistry/Which-is-more-acidic-ortho-fluorophenol-ortho-chlorophenol-ortho-bromophenol-or-ortho-iodophenol-Why . I would have preferred more sources but I could not find a very good source.


Why is the oxidation of sulfur dioxide not termed as heat of formation of sulfur trioxide?

Heat of formation is the enthalpy difference to the elements in standard state as defined by IUPAC, which is in most cases their most thermodynamically stable form at STP.

It is completely irrelevant if the respective chemical equation makes any sense, i.e. it would work out in a laboratory. However high the number of intermediate steps to reach the desired product is, energy is always conserved.


What type of bonding occurs in isolated silicon?

In addition to @Mithoron's comments and the Wikipedia he linked:

This is an important consideration in photovoltaics, as silicon forms the basis of the internal structure of solar technology (and image sensors). According to the PVEducation.org page Semiconductor Structure,

  Each silicon atom has four valence electrons which are shared, forming covalent bonds with the four surrounding Si atoms.

Diagrammatically, the website includes the following schematic:

The caption reads:

  Schematic representation of covalent bonds in a silicon crystal lattice. Each line connecting the atoms represents an electron being shared between the two. Two electrons being shared are what form the covalent bond.

In terms of germanium, according to the WebElements page Group 14 Elements,

  Tin and lead are very metallic although one modification of tin known as grey tin has the same diamond structure [as carbon] as does germanium and silicon.


How can two electrons lie together in an orbital?

Preludium

In chemistry we often use some techniques to factorize the (quite complicated) many-particle ground state of atoms and molecules into a series of one-particle states that have to fullfil certain requirements; examples are Hartree-Fock-Theory or Kohn-Sham-Density-Functional-Theory. These one-particle states are what we usually call orbitals or, when you differentiate them according to their spin, spin-orbitals. In the theories that I quoted you then have to solve a series of one-particle Schroedinger-equations which will give you certain discrete energy levels that correspond to bound states of the system. Those energy levels are a fundamental property of the system, i.e. the eigenvalues of its Hamilton operator. You can't have energy levels in between or below. Now, the problem is that the orbitals and their energy levels depend on the number of electrons you have in the system, i.e. considering some species $\ce{A}$ you will find different orbitals at different energies for its anion $\ce{A-}$, neutral species $\ce{A}$ or cation $\ce{A+}$. That is so, because the electrons interact with one another and populating one orbital changes the potential which the other electrons feel.

Main answer

But let's assume for a moment that the orbitals for $\ce{A-}$, $\ce{A}$ and $\ce{A+}$ are the same and only taking into account the electron-electron-repulsion of electrons occupying the same orbital: Starting from $\ce{A+}$ where we have all lower-lying orbitals doubly occupied and one orbital $\psi_n$ with energy $E_n$ singly occupied. Now, if you want add one more electron to this system to get to $\ce{A}$ you have to choose where to put this electron: You can either put it into $\psi_n$ or into some higher-lying orbital, like $\psi_{n+1}$ or $\psi_{n+2}$. If you put it into $\psi_n$, thus pairing the electrons up, you have to pay for that with the electron-electron-repulsion energy between the paired electrons (also known as spin pairing energy) $\Delta E_{\mathrm{pair}}$. So, compaired to $\ce{A+}$ the energy of your new system $\ce{A}_{\text{paired}}$ will be higher by an amount of $E_n + \Delta E_{\mathrm{pair}}$. If you instead put the new electron into $\psi_{n+1}$ you don't have to pair up any electrons, so you don't have to pay $\Delta E_{\mathrm{pair}}$. But the energy $E_{n+1}$ of orbital $\psi_{n+1}$ is higher than $E_n$. So, compaired to $\ce{A+}$ the energy of your new system $\ce{A}_{\text{unpaired}}$ will be higher by an amount of $E_{n+1}$. Now, the question: Which situation is better? For that you have to look at the energy difference between $\ce{A}_{\text{paired}}$ and $\ce{A}_{\text{unpaired}}$. This will be

\begin{equation}
  \Delta E = E(\ce{A}_{\text{paired}}) - E(\ce{A}_{\text{unpaired}}) = (E_n - E_{n+1}) + \Delta E_{\mathrm{pair}} \ .
\end{equation}

If $\Delta E$ is negative then $\ce{A}_{\text{paired}}$ is lower in energy and thus energetically favored, i.e. pairing electrons is favored when $|(E_n - E_{n+1})| > \Delta E_{\mathrm{pair}}$. So, coming back to your original question: Typical values of the pairing energy are between 200 and 300 kJ/mol (or 2-3 eV) - for an example have a look here: there you can see that the pairing energy for carbon and nitrogen lies at about $20000 \, \mathrm{cm}^{-1}$ which is ca. 240 kJ/mol. The seperation between energy levels in atoms is usually much greater as can be seen in the table of ionization potentials given here (the differences between the ionization potentials of $\ce{s}$ and $\ce{p}$ valence orbitals are usually greater than 6 eV). Only in molecules and transition-metal complexes with frontier-orbitals that lie close in energy (or are degenerate) you have so called high-spin configurations where it is favorable not to pair up the electrons in a lower-lying orbital but instead to occupy the higher-lying orbital. So in conclusion, it really does "cost" energy to pair up electrons in a single orbital but it usually would cost a lot more energy to force an electron into a higher-lying orbital instead.

Finally, a word of caution: My answer oversimplifies things as I tried to point out in my preludium. There are other effects at work that might have a big influence in some cases.


I really like the concept but the problem is to find a membrane that is permeable to water and H3O+ ions but impermeable in respect to the indicator.
I understand you need something as in the picture below.
Dialysis tubing are commonly used to prevent bigger molecules from crossing the membrane, but unfortunately the typical lower limit is for molecules of around 1 KDa so much more than a pH indicator molecule.
 The only option available seems reverse osmosis membranes, for example used in water purification filters. However, the resulted filtered water has an altered pH (ref.1) therefore not suitable.
Finding a membrane that is permeable to water and H3O+ ions but not to small polar molecules (i.e.Phenolphthalein, blue bromothymol, Anthocyanins etc.) is something difficult to achieve especially if you have also the requirement of not altering the pH.
I think that is the reason why we still don't have a pH meter that works as you described.

References:
1)Nir, Oded & Bishop, Noga & Lahav, Ori & Freger, Viatcheslav. (2015). Modeling pH Variation in Reverse Osmosis. Water Research. 87. 328–335. 10.1016/j.watres.2015.09.038.


Why does Area of Interface depends on size of particles?

Area of interface is the area of contact between two phases. If one phase is dispersed in another one, it then consists in the sum of the surface for each particle/droplet.

If you consider a spherical particle, its surface is $S=4\pi r^2$ and its volume $V=\frac{4}{3}\pi r^3$. Now consider the ratio $\frac{S}{V}=\frac{3}{r}$, so when the size of a sphere (but it is true for any shape) diminishes, the ratio $\frac{S}{V}$ increases. In other words, if you take an initial volume that you divide in small spheres, the global volume does not change (the sum of all spheres) but the surface increases.


You'd need to have some sort of ligand dictionary. For many PDB ligands, this exists, and you can look up the atom naming.

For example, the PDB ligand expo allow you to search for ligands by 3-letter code (e.g., HEM).

The PDBeChem also lets you find atom names based on 3-letter codes (the link is HEM).

However, as far as I know, there isn't (yet) any sort of pattern matching that would allow you to go through PDB files and automatically perceive your ligands, sort the atoms, and name them systematically.

If you find one, please let me know - it would be a great addition to programs like Open Babel


Definition of internal energy:  U is a function of state, representing the total kinetic and potential energy of the molecules.  U = U(T,V)

First Law of Thermodynamics:$$\Delta U=\delta Q+\delta W$$where the symbol $\Delta$ is used to represent the change in a (path-independent) function of state (like U) between an initial and final thermodynamic equilibrium state of a closed system and the symbol $\delta$ is used to represent the change in a parameter that depends on the process path between and initial and final thermodynamic equilibrium state of a closed system. $\delta Q$ is the heat added to the system over the path, and $\delta W$ is the work done on the system over the path.

Relationship for the Work:$$\delta W=-\int{P_{ext}dV}$$where, for both for reversible and irreversible process paths, $P_{ext}$ is the force per unit area exerted by the gas on the piston face, and, by Newton's 3rd law, the force exerted by the piston face on the gas.  For an irreversible process path, the pressure typically varies with spatial location within the cylinder, so that the average gas pressure does not match $P_{ext}$ at the piston face.  For a reversible process, the gas pressure is uniform within the cylinder, so $P_{ext}=P$ where P is the gas pressure calculated from the equation of state of the gas (such as the ideal gas law), based on the number of moles in the cylinder, the gas pressure in the cylinder, and the gas temperature in the cylinder.

Combining the First Law with the Relationship for Work:$$\Delta U=\delta Q-\int{P_{ext}dV}\tag{rev and irrev processes}$$
$$\Delta U=\delta Q-\int{PdV}\tag{rev processes}$$

Constant Volume ($V_i=V_f)$:$$\delta W=0$$
$$\Delta U=\delta Q$$

Constant Pressure ($P_{ext}=P_i=P_f=P$):$$\delta W=-P_{ext}\Delta V=-P\Delta V$$
$$\Delta U=Q-P\Delta V$$
So,$$\Delta U+P\Delta V=\delta Q$$But, since P is constant,
$$\Delta U+\Delta (PV)=\delta Q$$
The definition of enthalpy is $$H\equiv U+PV$$
Therefore, for a constant pressure process (a process in which $P_{ext}$ over the entire process path is equal to the equilibrium pressures in both the initial and final equilibrium states of the system) $$\Delta H=\delta Q$$


Why is vicinal diiodopropane unstable, leading to deiodination?

That 1,2-diodopropane is unstable does not indicate whether it is as the neat halide or in solution. Breaking (BDE) two $\ce{C-I}$ bonds costs $\pu{+114 kcal/mol}$ while forming the $\pi$-bond of propene is worth approximately $\pu{-62 kcal/mol}$ and the formation of iodine, $\pu{-36 kcal/mol}$. The net reaction is endothermic by $\pu{+16 kcal/mol}$. The reaction is entropically favored and the products are volatile. Therefore, the unfavorable equilibrium can be shifted to the right.


These orbitals represent the angular part of the wavefunction. The solution obtained directly from solving the Schrödinger equation produces equations containing complex numbers so cannot be drawn on normal $xyz$ axes and are hard to visualise.
The angular part of the wavefunction is given by functions called Spherical Harmonics these usually are given the symbol $Y$ and require two quantum numbers $\ell$ and $m_z$ making $Y_\ell^{m_z}$.
What is done to make the orbitals real functions, as opposed to complex ones, is to take linear combinations of the spherical harmonics, thus choosing the $z$-axis then the $\mathrm{d}_{z^2} = Y_2^0 = N(3\cos^2(\theta)-1)$) where $N$ is the normalisation constant. In cartesian coordinates ($xyz$) rather than polar ones (which are in $r,\;\theta,\;\phi$), $\displaystyle Y^0_2= N\frac{3z^2-r^2}{r^2}$ so you can see where the $z^2$ comes from and $r$ is the radial distance.
The other orbitals are also linear combinations e.g. $\mathrm{d}_{xz}=(Y_2^{1}+Y_2^{-1})/\sqrt{2}$ and in cartesian coordinates $\displaystyle Y_2^1 \sim \frac{z(x+iy)}{r^2},\; Y_2^{-1}\sim \frac{z(x-iy)}{r^2}$ so you can see where the notation $zx$ comes from and by making a linear combination the sum is no longer a complex number ($i=\sqrt{-1}$).
For completeness $\mathrm{d}_{yz}=(Y_2^{1}-Y_2^{-1})/\sqrt{2},\; \mathrm{d}_{x^2-y^2}=(Y_2^{2}+Y_2^{-2})/\sqrt{2},\; \mathrm{d}_{xy}=(Y_2^{2}-Y_2^{-2})/\sqrt{2}$  and in cartesians $\displaystyle Y_2^{\pm2}\sim \frac{(x\pm iy)^2}{r^2}$.


How to determine the number of stereoisomers of a di-substituted allene, exhibiting both axial and central chirality?

These are the three diastereomers with their enantiomers on the right side. The top and middle diastereomers are connected via axial chirality (R and S are used to mark the configuration of A groups).


Factors affecting strengths of carboxylic acids

1)"Why does withdrawing of electron density from an acidic centre result in the stabilization of the anion formed from dissociation? Or is this a misconception?"

Imagine you're an acid molecule. You can only show your "acid" characteristics if you're able to dissociate your proton (H+) when you're in water. If you lose your proton easily, you're a strong acid.If you can't lose your proton, you are not a good acid(weak acid). This is the background you need to know about acids first.

Next to the question about the importance of stability: After you lose your proton (suppose from your hand), you are now a negatively charged ion floating in water (with a negative charge at your hand). Because there is now a -1 charge in just your hand, there is a dipole between the rest of your body and your hand. This is considered not as "stable" as let's say if the -1 is dispersed throughout your body. The presence of the dipole would "encourage" the proton you lost to return to you. And if it does, you're no longer an acid.

If it requires more work for you to remain safely as a negatively charged ion in water (because of an uneven charge spread), it would be less probable that you exist in that state. Hence, it is said that you are less "stable".I've tried to simplify it.

"Also, what determines whether a group is electron withdrawing or electron donating?"

This is easier to answer. One part, Electronegativity. It is a measure of how strongly an atom pulls electron (cloud) towards itself. Highly electronegative atoms are Fluorine (F), Oxygen (O) and Nitrogen (N) in that order too. F > O > N. you can google "electronegativity chart" and look at images of the values. You can also see how the pKa changes with the atoms near the charge in your picture. (Lower the pKa the stronger the acid, btw).

Carbon and Hydrogen are much less electronegative and they both have similar electronegativity. Hence, compared to F, O, N, there can be considered to be electron donating as they are at the other end of the spectrum.


Reaction of primary nitroalkane with nitrous acid

As the comments say, it's both.  The two products can tautomerize in the same way as a ketone and it's tautomeric enol.  In fact, this is basically a keto-enol tautomerization, disguised by occurring around a nitrogen atom instead of carbon.


How does an electrolyte conduct electricity at low potential?

Under low potential differences, electrolytes conducts electricity in a similar way that they does at high potentials differences. I think, after reading the question and the comments, that the key point to understand this is: electric potential difference between which spatial points?

The conductivity of a solution is measure of its ability to conduct electricity when the solution is under a difference of electric potential. That is: the solution must be under the influence of an electric field (remember that $\boldsymbol E = - \nabla \boldsymbol \phi$) for having a non zero flux of charges.

You have to look the potential differences inside the solution and not at the electrodes.

You are right in that in order to maintain for long time a continuous current  electrolysis is required (at least for simple experimental devices). If there is not electrolysis you won't see a current. However, this is not because the solution (lets say the bulk, that is where the solution as such is (homogeneous)) does not have the ability of conduct electricity, but because there would not be a net external electric field at all in that region.

In short, the conductivity of the solution is not dependent of your ability for creating an external electric field on it.

EDIT

  But then again, what if the potential difference passes a treshold and
  at a certain point allows electrolysis, wouldn't the conductivity
  increase very steeply at that point?

If the potential difference between electrodes is low enough to ignore electrolysis, then there will be a charge accumulation in their surroundings that generates a contrary effect in the potential of the solution "far" from electrodes. So, the mean electric field in the bulk will be null.

According to Ohm's law:

$$ \boldsymbol J = \sigma \, \boldsymbol E$$

For the reasons stated above: $\boldsymbol E = \boldsymbol 0$, and so $\boldsymbol J = \boldsymbol 0$. What is the value of $\sigma$ for the solution? As any value satisfies the equation: We can say nothing from Ohm's law
! (technically, it can be computed even if no electric field is present, but it is not important here).

If the potential allows electrolysis, then a current appears so in principle you can measure the conductivity of the solution using Ohm's law. Notice that it is not an step function the current vs. potential. Due to this, there is not a truly step, but it seems because the growing is almost exponential, so we do not notice the increment till we notice a big increment. You can check out the Butler-Volmer equation for more details.

Once the current due to migration (allowed by electrolysis) is measurable, you could measure the solution conductivity. BUT it is a very bad idea, as normally the solutes concentrations are very small and the process of electrolysis change them! (and of course, many times (but not always) we want to know the conductivity to estimate concentrations)


Why does iodine solution turn transparent on adding base?

Iodine disproportionates, forming iodide and iodate ions, in the presence of base.  But it can be restored by neutralizing the base.  Neutralization will drop the hydroxide concentration low enough so that the disproportionation does reverse.

Let's punch in some numbers using the equilibrium constants given here.  We have

$$
\begin{array}{ccc}
\ce{I2 + 2OH- &<=>& I- + IO- + \ce{H_2O}}&K=30\\
\ce{3IO- &<=>& 2I- + IO3-}&K=10^{20}
\end{array}
$$

To superpose the reactions we take three units of the first one plus one unit of the second, canceling out the unstable hypoiodite ions.  So, rendering that three units of the first reaction mean three factors of that value of $K$, we end with:

$$\ce{3I2 + 6OH- <=> 5I- + IO3- + 3H2O}, K=2.7×10^{24}$$

If we consider a hypothetical case where all the iodine species are in their standard states, including iodine as the element, then the equilibrium hydroxide ion concentration is  $\pu{8.5×10^{-5} M}$.  Taking the base 10 logarithm and adding $\mathrm{p}K_\mathrm{w}=14$ then gives a $\mathrm{pH}$ just below $10$, so a neutral solution gets your elemental iodine back.


How do the molecules know the future? How do they proceed to form "Most Stable" compound?

Chemical reactions occur because the thermal energy in a substance makes molecules bump into each other, and each time they bump into each other there is some chance for a reaction to take place.  This is an oversimplified model, but it suffices to get some intuition for what's going on.

The molecules of course do not "know" that some configuration is more stable than others.  What happens is that the molecules react randomly, again and again; they take a sort of random walk through the available configurations.  Every now and then, a molecule will purely randomly find its way into the stable state.  Once it's there, it's "stuck" - it has a much smaller chance of getting out of the stable state.  That's more or less what it means for a state to be stable.

Over time, more and more molecules will find their way into the stable state and get stuck, so the result looks as if the molecules "knew" which state to go to and just went straight there.  But actually they only got there through the process of randomly bouncing around the configuration space, plus the stable state acting as a trap that molecules can enter but not (as easily) leave.


The concept of the half reaction has nothing to do with balancing redox reactions; how one arrives at a balanced reaction is not really important. The interesting thing about redox reactions is that electrons do not necessarily have to be transferred directly between the oxidized and reduced species. In many types of electrochemical cells, the anode and cathode are isolated from one another by a salt bridge and the species in each half-cell never approach one another. In techniques such as potentiometry, often one cares only about one half of the reaction, e.g. in a combination pH electrode, the response from the half reaction that contains hydronium ions produces the analytical signal and the the other half is a reference electrode reaction whose job is to simply produce a constant potential. In things like voltammetry, one is often only concerned about the half reaction occurring at the working electrode and the half reaction that balances it at the counter electrode isn't analytically significant (and may in fact be more than one different reaction).

Because the two half-reactions need not necessarily occur at the same place and one half reaction can generally be substituted without affecting the other, it makes sense to think of them separately. This lets us do things like think about different lithium batteries. There are many types and they all use the $\ce{Li -> Li+ + e-}$ oxidation half-reaction, but it's coupled with different reduction half-reactions based on cobalt, iron, etc.


Disproportionation of 2 KO₂ + 2 H₂O → 2 KOH + H₂O₂ + O₂

The disproportionation happens to elements, not to compounds.

In this particular case, to the part of oxygen,

$$  \overset{\ce{2 KO2}}{ 4 \times -0.5}=  \overset{\ce{H2O2}} {2\times -1} +\overset{\ce{O2}}{ 2\times 0}$$

while the other part keeps the oxidation state constant:

$$  \overset{\ce{2 H2O}} {2 \times -2}=  \overset{\ce{2 KOH}} {2\times -2}$$

$\ce{H2O}$ does not get directly involved in disproportionation.

$\ce{KO2}$ oxidizes and at the same time reduces  itself.

This is the nature of isproportionation, which happens , if the particular compound of the element in the intermediate oxidation state has higher Gibbs energy then particular compounds of this element in higher and lower oxidation states. Therefore redistribution of oxidation states is thermodynamically preferred.

Typical cases are $\ce{2 Cu^{I}->Cu^{II} + Cu^{0}}$ and $\ce{2 Hg^{I}->Hg^{II} + Hg^{0}}$


How many hydrogen bonds in these compounds?

Hydrogen bonds require a $\ce{\delta{+}}$ hydrogen atom capable of forming a hydrogen-bond, so $\ce{H2O}$ makes 2 H-bonds, and $\ce{NH3}$ has 3 H$\ce{\delta{+}}$ atoms capable of forming these bonds.

However hydrogen bonds also require an "active" lone pair capable of accepting a hydrogen-bond, where $\ce{H2O}$ can do 2 of these again, but $\ce{NH3}$ can only accept 1.

So, does water make 4 bonds or 2? Does ammonia make 1 bond or 2?  The answer is both, but it depends on how you count it.  If you look at an individual molecule in a cluster of water molecules it may appear to be surrounded and held in place by four H-bonds; it makes two with the hydrogen and accepts two with the lone pairs.  That molecule is involved in 4 hydrogen bonds.

But if you take 100 water molecules and count how many hydrogen bonds there are between them, the answer will be about 200 because each molecules makes 2 bonds.  If you consider each molecule making 4 bonds then you are double counting each bond being made and accepted.  The same is true for ammonia, each molecule makes one hydrogen-bond, it also accepts one.  You could say that an individual molecule is involved in two bonds, but if you count them all up each molecule only makes one each.

Because of these two different viewpoints, some texts will say water makes 4 and alcohols make 2 bonds, and some will say water makes 2 and alcohols make 1 bond.  They are both right, but right in different ways. Whichever way you count them, you must be consistent. If this is related to an exam syllabus, the correct way is defined by that.


Deriving heat capacity in terms of internal energy U and natural variables S & V

Usually this part of thermodynamics is not presented in the most efficient way. To derive the expression for, e.g., $C_P - C_V$, one goes through a series of steps involving Maxwell's relations and the triple product rule. Figuring out what to do in each step is done mostly by guesswork.

However, all relations of these sorts really come from just a single mathematical relation that more generally holds. That is to say,

  Consider a twice-differentiable,  two-variable function $f(x,y)$ and its Legendre transform with respect to $y$, defined by $g(x,f_y) := f - f_y y$. 1  Here, $f_y$ denotes the derivative of $f$ with respect of $y$ with other independent variables held fixed. Then, the following identity is satisfied:
  \begin{equation}
\boxed{g_{xx}  - f_{xx}  = - \frac{f_{xy}^2}{f_{yy}} = \frac{g_{x f_y}^2}{g_{f_y f_y}}}.
\end{equation}
  Proof. We have
  \begin{equation}
\bigg(\frac{\partial f}{\partial x} \bigg)_{f_y} = f_x + f_y \bigg(\frac{\partial y}{\partial x}\bigg)_{f_y} = f_x - f_y \frac{f_{xy}}{f_{yy}}.
\end{equation}
  On the above, the first equality holds because of the chain rule [for the change of variables $(x,y)$ $\to$ $(x, f_y)$], and the second equality because of the triple product rule. Then,
  \begin{equation}
g_x = \bigg(\frac{\partial f}{\partial x} \bigg)_{f_y} - f_y\bigg(\frac{\partial y}{\partial x}\bigg)_{f_y}   =f_x,
\end{equation}
  and
  \begin{equation}
g_{xx} = \bigg(\frac{\partial f_x}{\partial x} \bigg)_{f_y}= f_{xx} + f_{xy}\bigg(\frac{\partial y}{\partial x}\bigg)_{f_y} = f_{xx} - \frac{f_{xy}^2}{f_{yy}}.
\end{equation}
  Also, using the chain rule for the change of variables $(x,y)$ $\to$ $(x, f_y)$, we get
  \begin{equation}
g_{f_y} = \bigg(\frac{\partial f}{\partial f_y} \bigg)_{x} -\bigg(\frac{\partial y}{\partial f_y} \bigg)_{x}f_y - y = -y.
\end{equation}
  Therefore, Legendre transform of $g$ with respect to $f_y$ is simply $g - g_{f_y} f_y = f$. Then, performing a similar analysis as above for $g$ gives
  \begin{equation}
f_{xx} = g_{xx} - \frac{g_{x f_y}^2}{g_{f_y f_y}}.
\end{equation}

Now, all that remains to be done is to choose the variables and functions $x$, $y$, $f$ and $g$. Below are some examples.

(1) $x = S$, $y = V$, $f = U$, $g = H$, and $f_y = -P$:
\begin{equation}
H_{SS} - U_{SS} = -\frac{U_{SV}^2}{U_{VV}}.
\end{equation}
Note that
\begin{equation}
U_{SS} = \bigg(\frac{\partial^2 U}{\partial S^2}\bigg)_V = \bigg(\frac{\partial T}{\partial S}\bigg)_V = \frac{T}{T(\partial S/\partial T)_V} = \frac{T}{C_V},
\end{equation}
and that, similarly,
\begin{equation}
H_{SS} = \bigg(\frac{\partial^2 H}{\partial S^2}\bigg)_P = \bigg(\frac{\partial T}{\partial S}\bigg)_P = \frac{T}{T(\partial S/\partial T)_P} = \frac{T}{C_P}.
\end{equation}
Hence, we have
\begin{equation}
\boxed{\frac{1}{C_P} - \frac{1}{C_V} = -\frac{U_{SV}^2}{T U_{VV}} = -\frac{\Big(\frac{\partial^2 U}{\partial S \partial V}\Big)^2}{\Big(\frac{\partial U}{\partial S}\Big)_V \Big(\frac{\partial^2 U}{\partial V^2}\Big)_S}}
\end{equation}
which is equivalent to the relation OP wanted to derive.

(2) $x = T$, $y = V$, $f = A$, $g = G$, $f_y = -P$, and noting that $f_{-y} = -f_y$,
\begin{equation}
G_{TT} - A_{TT} = \frac{G_{T,-P}^2}{G_{-P,-P}} = \frac{G_{TP}^2}{G_{PP}}.
\end{equation}
Here,
\begin{equation}
A_{TT} = \bigg(\frac{\partial^2 A}{\partial T^2}\bigg)_V = -\bigg(\frac{\partial S}{\partial T}\bigg)_V = - \frac{1}{T}C_V,
\end{equation}
\begin{equation}
G_{TT} = \bigg(\frac{\partial^2 G}{\partial T^2}\bigg)_P = -\bigg(\frac{\partial S}{\partial T}\bigg)_P = - \frac{1}{T}C_P,
\end{equation}
\begin{equation}
G_{TP} = \frac{\partial^2 G}{\partial T \partial P} = \bigg(\frac{\partial V}{\partial T}\bigg)_P = V \alpha,
\end{equation}
where $\alpha := \frac{1}{V}\Big(\frac{\partial V}{\partial T}\Big)_P$ is the volume expansion coefficient,
\begin{equation}
G_{PP} = \bigg(\frac{\partial^2 G}{\partial P^2}\bigg)_T = \bigg(\frac{\partial V}{\partial P}\bigg)_T = - V\kappa_T,
\end{equation}
where $\kappa_T := -\frac{1}{V}\Big(\frac{\partial V}{\partial P}\Big)_T$ is the isothermal compressibility. Then, it follows that
\begin{equation}
\boxed{C_p - C_V = \frac{TV\alpha^2}{\kappa_T}}.
\end{equation}

(3) $x = V$, $y = S$, $f = H$, $g = G$, and $f_y = T$:
\begin{equation}
G_{PP} - H_{PP} = \frac{G_{TP}^2}{G_{TT}}.
\end{equation}
We have
\begin{equation}
H_{PP} = \bigg(\frac{\partial^2 H}{\partial P^2}\bigg)_S = \bigg(\frac{\partial V}{\partial P}\bigg)_S = - V\kappa_S,
\end{equation}
where $\kappa_S := -\frac{1}{V}\Big(\frac{\partial V}{\partial P}\Big)_S$ is the adiabatic compressibility. We have already shown that
\begin{equation}
G_{PP} = -V\kappa_T, \quad G_{TP} = V \alpha, \quad G_{TT} = - \frac{C_P}{T},
\end{equation}
whence we obtain
\begin{equation}
\boxed{\kappa_T - \kappa_S = \frac{TV\alpha^2}{C_P}}.
\end{equation}

1 To be more precise, one should first write down $f(x,y) - y f_y$ and then express $y$ as a function of $x$ and $f_y$. Also, for the Legendre transform to exist, $f(x,y)$ should be a convex or concave function of $y$.


As OP suggested there are three resonance structures of diazomethane contributing to the structure:

If you want to neutralize the formal charges of last resonance structure, you can put a double bond on negatively charged carbon and adjacent positively charged nitrogen to get the fourt resonance structure. However, traditionally we don't do that because it make the middle nitrogen hypervalent (or pentavalent nitrogen). On other word, it gave that nitrogen 10 valance electrons in its outer cell, which can't have in a 2nd row element.

It's just like resonance structures of nitrate ion, in which central nitrogen contains a full positive charge in all three resonance structures and two of three oxygen atoms contain full negative charges. So far, every one accept those resonance structures without questioning hypervalency of the central nitrogen atom until recently.

In 2015, Durrant has published his work on quantum mechanical calculations, suggesting possibility of hyper valency in diazomethane and some other possible molecules (Ref.1). According to its abstract:

  From the inception of Lewis' theory of chemical bonding, hypervalency has remained a point of difficulty that has not been fully resolved by the currently accepted qualitative definition of this term. Therefore, in this work, a quantitative measure of hypervalency has been developed. The only required input is the atomic charge map, which can be obtained from either quantum calculations or from experiment. Using this definition, it is found that well-known species such as $\ce{O3, \color{blue}{CH2N2}}$ and $\ce{ClO4−}$, are indeed hypervalent, whilst others such as $\ce{XeF4, PCl5}$ and $\ce{SO4^2−}$, are not. Quantitative analysis of known species of general formulae $\ce{XF_n^m−, XCl_n^m−}$, and $\ce{XO_n^m−}$ shows that there are no fundamental differences in chemical bonding for hypervalent and non-hypervalent species. Nevertheless, hypervalency is associated with chemical instability, as well as a high degree of covalent rather than ionic bonding. The implications for accepted Lewis structure conventions are discussed.

The experimentally determined atomic charges from electron densities obtained by X-ray crystallography have been compared with theoretical values obtained by QTAIM analysis of the output from quantum calculations. In order to calculate the overall electron count at the $\ce{X}$ atom, the author has defined a parameter called the valence electron equivalent, $\gamma$, as ‘the formal shared electron count at a given atom, obtained by any combination of valid ionic and covalent resonance forms that reproduces the observed charge distribution’. Mathematically:

$$Q(\ce{X}) = \displaystyle \sum_{i=1}^n{a_iq_i(\ce{X})}; \ \text{and then, } \ \gamma(\ce{X}) = \displaystyle \sum_{i=1}^n{a_i\gamma_i(\ce{X})}$$

where $Q(\ce{X})$ and $\gamma(\ce{X})$ are the overall charge and valence electron equivalent of atom $\ce{X}$, $q_i(\ce{X})$ and $\gamma_i(\ce{X})$ are the formal charge and electron count of each contributing resonance form, and $a_i$ are the coefficients required to reproduce $Q(\ce{X})$.

According to the paper:

  It follows that for any given atom $\ce{X}$, if $\gamma_i(\ce{X}) = 8$, the atom obeys the original Lewis octet rule. If $\gamma_i(\ce{X}) \lt 8$, the atom obeys the ‘modified octet rule’, but if $\gamma_i(\ce{X}) \gt 8$, neither form of the octet rule is obeyed and the atom is hypervalent. In such a case, no combination of formally charged Lewis octet-compliant structures can reproduce the observed charge distribution and it is necessary to include a contribution from an ‘expanded octet’ structure.

In the case of central nitrogen atom of diazomethane, $\gamma_i(\ce{N}) = 10$ ($ \gt 8$), thus it is considered to be hypervalent (Note: for nitrate ion: $\gamma_i(\ce{N}) = 8.30$ and hypervalent as well). The author has compared the results of $\gamma$ calculations on a variety of species including diazomethane, using quantum method 5 plus the QTAIM, Natural Bond Orbital (NBO), Hirshfeld, and Mulliken charge models. All of these methods agree with the findings, giving value of $\gamma_i(\ce{N})$ of diazomethane $10, 9.91, 9.78,$ and $9.56$, respectively.

References:

M. C. Durrant, "A quantitative definition of hypervalency," Chemical Science 2015, 6, 6614-6623 (https://doi.org/10.1039/C5SC02076J).


Thermodynamic vs Kinetic Sulphonation of Naphthalene

In contrast to many other electrophilic aromatic substitution reactions, aromatic sulfonation is reversible, in other words it is an equilibrium. If you use a large excess of $\ce{SO3}$ you push the reaction to the sulfonation side; if you heat the sulfonated product in the absence of $\ce{SO3}$ it will de-sulfonate.  This makes sulfonation a nice method to 1) control where a second substituent will be introduced or 2) serve as a protecting group (block second substituent introduction) - and then when you are done, you simply de-sulfonate.

Now to your question

If we look at the intermediates leading to the 2 isomers we see that we can draw 2 resonance structures that preserve aromaticity in one ring for the transition state leading to naphthalene-1-sulfonic acid, but only one resonance structure where the other aromatic ring remains intact for the 2-isomer (draw the resonance structures to convince yourself).

(image source)

Therefore the transition state leading to the 1-isomer will be lower in energy and naphthalene-1-sulfonic acid will be formed fastest - it is the kinetically favored product.

However, naphthalene-2-sulfonic acid is a few kcal/mol more stable than its 1-isomer. This is due to an adverse 1,8 steric interaction present in the 1-isomer.

(image source)

Because the sulfonation reaction is reversible, if we run it long enough at high temperature (favoring reversibility), eventually we will wind up with the thermodynamically preferred (more stable) product, the 2-isomer.


What causes "stringy" melted cheese (and similar behavior in other materials)?

When cheese is heated to 140 degrees fahrenheit, the proteins line up in rows.  Soaking the cheese in a salt solution before pulling it helps to stiffen it and keep it pliable, as the salt neutralizes acids which could cause the long lines of protein to break.

Polymers, similar to cheese, can be stringy because their molecules are very large (macromolecules), and tend to entangle with one another.  @IanF1 is correct in saying this may be a chemistry question.  Here is a good explanation of polymerization:

http://courses.chem.psu.edu/chem112/materials/polymers.html


Would adding water to a fish tank with sand lower the level of sand?

If your sand is packed in a optimal sphere packing, like a hexagonal packing for example,

nothing would happen to the sand level, because it can't be packed more dense.

If the packing of the dry sand wasn't optimal (which it probably isn't 100%), then the water may help settle it through motion and friction and it would lower a bit. In any case it wouldn't lower so much as that you could easily see it with your eyes.


Help with determining concentration of magnesium hydroxide for a specific pH level

If $c$ is the concentration of $\ce{Mg^2+}$ in the $\ce{Mg(OH)2}$ solution, the concentration $[\ce{OH-}] = 2c.$
At $\mathrm{pH}~9,$ $[\ce{OH-}] = \pu{1E-5 M},$ then $c = \pu{5E-6 M}.$ So you have to dissolve $\pu{1.25 μmol}$ $\ce{Mg(OH)2}$ in $\pu{250 mL}$ water. This is $\pu{71.3 μg}$ of $\ce{Mg(OH)2}.$ This is difficult to do in practice, as such a small amount is difficult to weigh. It may be done in two steps: first prepare a moderately concentrated solution, then dilute it to $\mathrm{pH}~9.$


Oxidation: losing electron(s), Oxidizer / oxidizing agent: a chemical that can oxidize another reagent. Reduction: gaining electron(s) - think of it as reduction of charge! Reducing agent: A chemical that can reduce another reagent.

And oxidation can never occur without a reduction, meaning if you use an oxidizing agent to oxidize a substance, when the oxidizing agent is in turn being reduced itself. It gains the electrons of the oxidized substance.

To describe how well a substance functions as an oxidizer, we use reduction potentials, measured in volts! A substance with a large, positive reduction potential is easy to reduce, which in turn means it is a good oxidizing agent. Similarly, if a substance has a large, negative reduction potential, it means that it is hard to reduce this substance. This is because it's good at reducing other substances itself - it's a reducing agent.

To figure out if a redox reaction is spontaneous (if it can potentially proceed on its own), we use standard reduction potentials, $E_0$. This is for concentrations of 1 M and a temperature of 25 degrees celcius. You look these up in your textbook or online.

An example: Is the following reaction spontaneous?
$$\ce{Cu^{2+}(aq) + 2Ag(s) -> Cu(s) + 2Ag+(aq)}$$

To find out we first find the standard reduction potentials for each participating species:

$\ce{Cu^{2+}(aq) + 2e- -> Cu(s)}, E_0=+0.34V$

$\ce{Ag^{+}(aq) + e- -> Ag(s)}, E_0=+0.80V$

This describes how much they want to be reduced (GAIN electrons). The silver ion wants it the most! However, notice that in our reaction we're asking for it to LOSE electrons (be oxidized). THEREFORE: No, the reaction is not spontaneous under the aforementioned standard conditions. However, the REVERSE reaction is spontaneous. To calculate the potential for our reaction, we simply say:

$$E_0 = (+0.34V) - (+0.80V) = -0.46V$$

The reduction potential for the substance that we want to gain electrons (reduced), minus the potential for the substance we want to lose electrons (be oxidized). The fact that the result is negative, is what tells us the reaction will not proceed - and that it will in fact proceed in the OPPOSITE direction.

Take notice that these reduction potentials are not given the unit Volt for no reason! They are true electric potentials. You can think of electric potentials as "electron pressures". The pole with the highest electron pressure will be (-), because electrons are negatively charged, and the other will be (+) relative to it.

Remember that the reduction potential (the willingness to take electrons) for the silver ion is higher than that of the copper ion. Silver wants to take up electrons more than copper, meaning that electrons will travel from the copper to the silver. you could just as well say that the copper wants to get rid of its electrons more than the silver - copper has the higher electron pressure.

What about electronegativity?
Electronegativity does indeed correlate somewhat with reduction potentials, at least for simple reactions involving pure elements. And you can see that Ag does indeed have a higher electronegativity than Cu, which makes sense. Electronegativity doesn't take into account the oxidation states of different species etc., and quickly it becomes a better idea to just leave electronegativity out of the picture when dealing with redox reactions.


What are known examples of drugs that racemize/stereoconvert in vivo, and how are they converted?

Although I can't think of any drug examples other than thalidomide, here's information on thalidomide's mechanism:

The chiral carbon of thalidomide can tautomerize in basic conditions into an enol, which is achiral.  A reversal back to the ketone results in a mix of (R) and (S) enantiomers.

In the body, this tautomerization is generally catalyzed by basic amino acids.  Specifically, albumin is the main catalyst in humans.

While this is beyond the scope of your question, the reason that only (S)-thalidomide causes birth defects is that it can insert itself into DNA and suppresses certain genes necessary for embryonic development.

Reddy - Chirality in Drug Design and Development
http://www.ncbi.nlm.nih.gov/pubmed/9860497
http://www.ncbi.nlm.nih.gov/pubmed/9499573


What does addition of iodine and base do this case?

$\ce{I2}/\ce{OH-}$ will perform the iodoform reaction on the the product of previous reaction. The base deprotonates an alpha hydrogen, and the resulting enolate nucleophile attacks iodine. After three substitutions, the $\ce{CI3}$ group is a passable leaving group, and a nucleophilic acyl substitution occurs. The reaction is done in basic conditions, so the carboxylic acid is deprotonated. The leaving group is a strong base, so it is protonated. $\ce{CHI3}$ will be formed along with the conjugate base of m-bromo-benzoic acid. Here is a generic version:

$$\ce{RC(O)CH3 + 3I2 + 4OH- -> RCO2- + 3I- + 3H2O + CHI3}$$


Degassing HI from Reaction (Via Ultrasound Maybe?)

In short, to answer your question "Why side products are being produced",  you appear to have ignored all possible radical activity!
Per this source: Free radical formation from sonolysis of water in the presence of different gases, for example, to quote:

$\ce{H2O + Sonolysis -> .H + .OH}$

Here, replacing H2O with HI likely results similarly in:
$\ce{HI + Sonolysis -> .H + .I}$
Some possible further reactions:
$\ce{.H + .H -> H2 (g)}$
$\ce{.I + .I -> I2 (s)}$
And, in the presence of an organic (RH):
$\ce{.H + RH -> .R + H2 (g)}$
$\ce{.OH + RH -> .R + H2O (l)}$
$\ce{.R + .I -> RI }$
$\ce{.R + .R -> R2 (g)}$
Also, with any dissolved oxygen (or even nitrogen, see cited reference), expect even more, as for example:
$\ce{.H + O2 -> .OOH}$
$\ce{.HO2 + .HO2 -> H2O2 + O2}$
$\ce{.R + .OOH -> ROOH}$
....
As such, expect a range of products, as would occur, by the way, if you performed photolysis in lieu of sonolysis. However, there are related reactions, like in the presence of ammonia, where the final photolysis products can display a high degree of product selectivity.


How would you separate sugar from water?

If you start with a known 'pure' mixture of sugar and water, then what you propose is fine, to an extent, if you are just after separating the two ingredients.

Crystallisation is a purification method for extracting a solid from solution; it will allow you to separate pure solute from the solvent. The remaining solution would still contain a small amount of dissolved material, and hopefully all of your impurities.

If you wanted to isolate pure solvent from the solution, you would need to use distillation. After collecting the distillate, you would be left with a gummy residue of semi-decomposed sugar (and your impurities).

Sugar is hygroscopic, and starts decomposing at just over 180°C (~365°F). The problem that you would have with your proposed method is that in order to remove the final remnants of the water, you will need to heat well above the boiling point of water. Google 'making caramel' if you can't think of what will happen. Even a very small amount of decomposition of your sugar will mean you are left with an impure solid. Back to square one. The reason that you can simply boil the water off salt is that it is far more stable and can be heated safely to much higher temperatures.

So, there are a few ways to separate these two ingredients, depending on your end goals:

To collect just solute (sugar): heat over very low heat to evaporate as much water as possible to give yourself a saturated solution. Then crystallise sugar and filter (in the lab you'd wash with a suitable solvent and dry under vacuum)
To collect just the solvent (water): distill over moderate heat to collect water. To collect as much as possible, the heat used to boil off the solvent needs to be slowly increased, especially as you approach the final portion.
To collect both solute (sugar) and solvent (water): You would remove solid first, then distil solvent. Unless you are starting with a saturated solution, you would need to do a partial distillation to remove some solvent, then crystallise your sugar, then complete the distillation. You'd still be left with a horrible brown gummy residue.

Sounds like a great home project, though. Just be sure to have done some suitable research to allow you to make use of your materials if it doesn't quite go to plan :-)


Actually it is not necessary to dig deep into quantum mechanics. There are several reasons why noble gasses are stable (as gasses at room temperature).

First of all, there is the obvious full valence shell.
Trend in the periodic table make clear that the charge of the nucleus grows from left to right in every period. The attractive force towards the electrons therefore increases. (This holds also from top to bottom.) Electrons in lower orbitals ($n<\text{period}$) now shield the charge of the nucleus. This will be somewhat the same throughout the period. (The orbitals also shrink due to higher charge of the nucleus.) In noble gasses the valence shell is completely filled, providing quite a good shield for the next shell. Also the increase in the main quantum number means a significant increase in the energy level of the next orbital. This makes it unlikely, that a noble gas will accept another electron.

Due to the high charge of the nucleus it is also not easy to remove an electron from an orbital. However, this is possible from (at least) Argon on downwards. For example: $\ce{HArF}$ is stable in a matrix at room temperature. It has a strong covalent $\sigma$ bond ($\ce{H\bond{-}Ar+}$) and a not quite as strong ionic bond ($\ce{[HAr]+\cdots F-}$). Krypton already does some fancy chemistry, that is stable at room temperature. As Uncle Al has pointed out Xenon is well known for its reactivity.

But why the sudden change? It is fairly intuitive, that the maximum electron density for each orbital with increasing main quantum number is also further away from the nucleus. That makes the valence shell very well polarizeable. Also the electrons are better shielded by the previous shells. However, the natural occurrence of these elements are as (homonuclear) gasses.

However, if you bring these elements in close contact with each other it was found out, that they have a very small dissociation energy ($D_e<1~\text{meV}$). This is due to dispersion, and van-der-Waals Forces, which is the main interaction between these elements. However, in $\ce{He2}$ no binding mode was observed (instant dissociation).

But that still does not answer, why those elements are stable as gasses, as pointed out that there are in fact attractive forces. The reason is as simple as obvious: Entropy. If two noble gasses would form a molecule/adduct, the bond/association energy of this molecule would have to compensate for the loss of entropy (Two volume elements would become one, hence the gas would have to expand to cover the room previously occupied, which requires work energy).

The explanation by tschoppi also covers, why from a MO point of view there could not be a bond in $\ce{He2}$. Go ahead and ask yourself the question if this would be true for $\ce{He3}$. We also know that orbital overlap is only one component to the truth. There are many other more. If it was not for such a nice thing like dispersion and other weak chemical interactions we would not exist.


Which of the free access molecular modelling program apply the B3LYP/G-31G(d,p) level of theory?

There are several free packages, see here.

Like orthocresol mentioned, you might want to be tutored by someone of experience at first. Quantum calculations are very easy to start (too easy, some might say), but harder to actually interpret correctly. The actual method can also greatly influence the results and needs to be chosen carefully (for example, B3LYP has fallen out of favor after some weird behaviours have been observed (Why the Standard B3LYP/6-31G* Model Chemistry Should Not Be Used in DFT Calculations of Molecular Thermochemistry: Understanding and Correcting the Problem); consider M06 or M062X as good alternatives)


Determine surface area of heat exchange tube

The outlet temperature  should be 53.3.  The log-mean temperaturr difference should be positive.  Calculate the heat load and divide by the log-mean temperature difference and by the heat transfer coefficient to get the area.

ADDENDUM

This addendum is for the benefit of @MaxW (the downvoter).

The heat load $\dot{Q}$ of the heat exchanger is given by:  $$\dot{Q}=\dot{m}_HC(T_{H,in}-T_{H,out})=20(4.183)(80-60)=1673\ kW$$This heat load represents the rate of heat transfer from the hot fluid to the cold fluid.
The exit temperature of the cold stream is determined from $$\dot{m}_CC(T_{C,out}-T_{C,in})=12(4.183)(T_{C,out}-20)=\dot{Q}=1673$$

Solving for the exit temperature of the cold stream gives  $$T_{C,out}=53.33\ C$$
3.  The log-mean temperature difference is given by:
$$(\Delta T)_{LM}=\frac{(T_{H,in}-T_{C,out})-(T_{H,out}-T_{C,in})}{\ln{[(T_{H,in}-T_{C,out})/(T_{H,out}-T_{C,in})]}}=\frac{26.67-40}{\ln{[26.67/40]}}=32.88\ C$$

The heat transfer area A is determined from the equation:  $$U_oA(\Delta T)_{LM}=2.0A(32.88)=\dot{Q}=1673$$Solving this equation for the heat transfer area A gives:
$$A=25.4\ m^2$$

Questions?


The Born-Haber Cycle might help you visualizing how those values are used. Notice the values used bellow are a bit different from the ones you got.

Electron affinity

  The amount of energy released when an electron is added to a neutral
  atom or molecule to form a negative ion.

Ionization energy

  The amount of energy required to remove an electron from the atom.

Enthalpy of formation (ΔH)

  The change of enthalpy that accompanies the formation of 1 mole of the
  compound from its elements.


Static electricity with calcium carbonate particles

Some links of a commercial equipment to avoid problems of static electricity in powders during weighing:

https://www.mt.com/us/en/home/products/Laboratory_Weighing_Solutions/Accessories/anti_static_devices.html

http://lab.mt.com/antistatic/

https://www.mt.com/dam/mt_ext_files/Editorial/Generic/6/Anti-static_0x000246700004f0770006d93b_files/ion_e.pdf


If you search online for Uraninite then you find various websites saying that uraninite is called pitchblende.

On wikipedia you find :

  Uraninite is a radioactive, uranium-rich mineral and ore with a
  chemical composition that is largely UO2, but also contains UO3 and
  oxides of lead, thorium, and rare earth elements. It is most commonly
  known as pitchblende

even the word pitchblende will automatically redirect you to the uraninite wikipedia page .

Another trustworthy website is mindat.org : uraninite information can be found on this link:http://www.mindat.org/min-4102.html (It seems that there are various names associated to the mineral):

  Re-named in 1845 by Wilhelm Karl von Haidinger for its composition
  containing uranium. Originally recognized before the element uranium
  was known and called "schwarz beck-erz" by Franz Ernst Brückmann in
  1727. Subsequently described with several names including "pseudogalena" and "pitch-like zinc-blende" by Johan Gottschalk
  Wallerius (Vallerius) in 1747. Named pechblende by Axel Cronstedt in
  1758. Named "uranerz" by Martin Klaproth after he discovered the element uranium in this mineral. Many spelling variations of the above
  names known.


What gas is released during dissolving NaOH

I don't think any gas is released.  You are probably experiencing dust or mist that contains NaOH itself, or tiny amounts of gas phase NaOH.

There is related discussion at a soapmaking website

http://www.tikvah.com/medical/lye.html

See also "Pneumothorax following inhalation of caustic soda fumes" Archives of Emergency Medicine, 1988, 5, 45-47


A photon has no mass and moves at relativistic speeds, therefore you can't use the usual expression for the kinetic energy.

The energy of a photon is given by (as you already stated)
$$
E = \frac{hc}{\lambda}
$$
where $h$ is Planck's constant, $c$ the speed of light and $\lambda$ the wavelength of the photon.

You can derive it from the relativistic energy
$$
E^2 = p^2c^2 + m^2c^4
$$
with $m =0$ and using the fact that for a photon
$$
p = \hbar k = \frac{h\nu}{c} = \frac{h}{\lambda}.
$$


What volume does one mole of an ideal gas occupy?

The common saying is a hold over from when STP was defined to be $\pu{273.15 K}$ and $\pu{1 atm}$. However, IUPAC changed the definition in 1982 so that $\pu{1 atm}$ became $\pu{1 bar}$. I think the main issue is a lot of educators didn't get the memo and went right along either teaching STP as $\pu{1 atm}$ or continuing with the line they were taught ("$\pu{1  mol}$ of any gas under STP occupies $\pu{22.4 L}$") without realizing it didn't hold under the new conditions.

Just as a "proof" of this working for the old definition.
\begin{align}
V &=\frac{nRT}{P}\\
  &=\frac{\pu{1  mol} \times \pu{8.2057338 \times 10^-2 L * atm//K * mol}
    \times \pu{273.15 K}}{\pu{1 atm}}\\
  &=\pu{22.41396 L}\\
  &\approx \pu{22.4 L}
\end{align}


What is the formally correct notation of t-Bu, i-Pr, p-NO2, n-Bu, a-HCl, etc.?

In the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), single italic lower-case Latin letters are used in some stereodescriptors (such as ‘c’, ‘m’, ‘p’, ‘r’, ‘s’, and ‘t’), in descriptors for fusion sites (e.g. dibenzo[c,g]phenanthrene), in locants (‘o’, ‘m’, and ‘p’ in place of ortho, meta, and para, respectively; however, this usage is strongly discouraged), and in other structural descriptors (e.g. s-indacene).

However, single lower-case Latin letters that are used for numbering are not italicized (e.g. naphthalene-4a,8a-diol).

All these symbols are typically used in names (which are usually not typeset using MathJax/mhchem) and do not necessarily appear in formulas.

Furthermore, common abbreviations for organic structure fragments may be used in formulas, but not in text. According to the Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), the permissible abbreviations include:

isopropyl: iPr
isobutyl: iBu
sec-butyl: s-Bu
tert-butyl: t-Bu

However, the ACS Style Guide recommends different abbreviations:

isopropyl: i-Pr
isobutyl: i-Bu
sec-butyl: sec-Bu
tert-butyl: t-Bu

Furthermore, authors may create their own abbreviations, but any abbreviations not included in the recommendations shall be defined clearly when they are used.

Also note (since the question mentions letters followed by an element) that several structural abbreviations in current usage are identical to an element symbol:

Ac (acetyl, actinium)
Ar (aryl, argon)
Cm (carboxymethyl, curium)
Nb (p-nitrobenzyl, niobium)
Np (p-nitrophenyl, neptunium)
Pr (propyl, praseodymium)


Can aluminium form blue-colored solution in liquid ammonia?

The SE answers quoted in the comments provide a good broad answer but with respect to the question of aluminium – aluminium will form a blue solution in liquid ammonia following electrolytic reduction of $\ce{AlI3}$ in liquid ammonia according to J. Am. Chem. Soc. 1950, 72 (11), 5178–5180.
Europium and ytterbium have also been reported to dissolve in liquid ammonia in J. Phys. Chem. 1956, 60 (11), 1590–1591.


Coordination number of carbon in methyllithium tetramer

TL; DR: coordination number of carbon atom in methyllithium tetramer is indeed 7, arising from 6 intramolecular interactions, as you suggested, and an extra bond with the lithium atom of the next tetramer.

The key point is that you need to look beyond the tetramer. I'm sorry to inform you that the image from Wikipedia of the standalone cubane $\ce{Li4(CH3)4}$ is useless for the determination of coordination number (C.N.) of carbon. In fact, there is a 3D network of interlinked clusters, and the description given in J. D. Lee's Concise Inorganic Chemistry is in fact very precise [1, pp 304]:

  The structure of the $\ce{(LiCH3)4}$ cluster is unusual. The four $\ce{Li}$ atoms occupy the corners of a tetrahedron. Each methyl $\ce{C}$ atom is above a face of the tetrahedron, and forms a triple bridge to the three $\ce{Li}$ atoms that make up the face of the tetrahedron. The intramolecular $\ce{Li-C}$ distance is
  $2.31~\mathrm{Å}$. The $\ce{C}$ is bonded to the three $\ce{H}$ atoms in the methyl group. The $\ce{C}$ is also bonded to a $\ce{Li}$ atom in another tetrahedron (with an intermolecular $\ce{Li-C}$ distance of $2.36~\mathrm{Å}$). The coordination number for the $\ce{C}$ atom is therefore $7$.

The problem I stumbled upon is that I didn't find any ready-to-use CIFs for methyl lithium with 3D coordinates (there are plenty for the derivatives and some powder XRD of questionable quality). I ended up creating my own CIF based on data published by Weiss et al. [2] with VESTA program and visualized the content of the unit cell with OLEX2:

Figure 1. Unit cell of the methyl lithium tetramer with the central cubane core $\ce{Li4(CH3)4}$ linked via four crystallographically equivalent $\ce{C}$ atoms with 4 neighboring cubanes displayed as tetrahedra $\ce{Li4}$ for clarity. Color code: $\color{#EEEEEE}{\Large\bullet}~\ce{H}$; $\color{#CC80FF}{\Large\bullet}~\ce{Li}$; $\color{#909090}{\Large\bullet}~\ce{C}$.

Figure 2. Coordination environment of the carbon atom and interatomic distances (Å).

In addition, I run the test for C.N. with Dirichlet subroutine of the ToposPro, which also suggests the C.N. 7:

Central atom:C1 0.320 0.320 0.320 Rsd:0.900
 D(CP):0.624  ( 0.3698 0.3698 0.3698 )
 D(VDP):0.364  ( 0.3490 0.3490 0.3490 )
Atom:0.980 < r < 2.370  <r>=1.747   Top: 1.229 < R < 1.556  <R>=1.357
CN=7:0:0 NV=10 V=3.052/6.126 S=13.376 Cpac=0.161 Ccov=5.175
G3=0.114818789
Face distribution: {3/1 4/3 5/3 }
Vertex distribution: {3/10 }

  N  Atom   x      y      z     Dist.   D1      D2     SAng.
  1  H_1  0.3510  0.1920  0.3510  0.97957  0.89982  0.07975  27.31224
  2  H_1  0.1920  0.3510  0.3510  0.97957  0.89982  0.07975  27.31224
  3  H_1  0.3510  0.3510  0.1920  0.97957  0.89982  0.07975  27.31224
  4  Li1  0.3690  0.3690  0.6310  2.30686  0.89982  1.40704   5.76860
  5  Li1  0.3690  0.6310  0.3690  2.30686  0.89982  1.40704   5.76860
  6  Li1  0.6310  0.3690  0.3690  2.30686  0.89982  1.40704   5.76860
  7  Li1  0.1310  0.1310  0.1310  2.37007  0.89982  1.47025   0.75748

Figure 3. Voronoi-Dirichlet coordination polyhedron for the carbon atom.

Content of the CIF file:

#======================================================================

# CRYSTAL DATA

#----------------------------------------------------------------------

data_VESTA_phase_1

_chemical_name_common                  'Methyl lithium tetramer'
_cell_length_a                         7.24000
_cell_length_b                         7.24000
_cell_length_c                         7.24000
_cell_angle_alpha                      90
_cell_angle_beta                       90
_cell_angle_gamma                      90
_space_group_name_H-M_alt              'I -4 3 m'
_space_group_IT_number                 217

loop_
_space_group_symop_operation_xyz
   'x, y, z'
   '-x, -y, z'
   '-x, y, -z'
   'x, -y, -z'
   'z, x, y'
   'z, -x, -y'
   '-z, -x, y'
   '-z, x, -y'
   'y, z, x'
   '-y, z, -x'
   'y, -z, -x'
   '-y, -z, x'
   'y, x, z'
   '-y, -x, z'
   'y, -x, -z'
   '-y, x, -z'
   'x, z, y'
   '-x, z, -y'
   '-x, -z, y'
   'x, -z, -y'
   'z, y, x'
   'z, -y, -x'
   '-z, y, -x'
   '-z, -y, x'
   'x+1/2, y+1/2, z+1/2'
   '-x+1/2, -y+1/2, z+1/2'
   '-x+1/2, y+1/2, -z+1/2'
   'x+1/2, -y+1/2, -z+1/2'
   'z+1/2, x+1/2, y+1/2'
   'z+1/2, -x+1/2, -y+1/2'
   '-z+1/2, -x+1/2, y+1/2'
   '-z+1/2, x+1/2, -y+1/2'
   'y+1/2, z+1/2, x+1/2'
   '-y+1/2, z+1/2, -x+1/2'
   'y+1/2, -z+1/2, -x+1/2'
   '-y+1/2, -z+1/2, x+1/2'
   'y+1/2, x+1/2, z+1/2'
   '-y+1/2, -x+1/2, z+1/2'
   'y+1/2, -x+1/2, -z+1/2'
   '-y+1/2, x+1/2, -z+1/2'
   'x+1/2, z+1/2, y+1/2'
   '-x+1/2, z+1/2, -y+1/2'
   '-x+1/2, -z+1/2, y+1/2'
   'x+1/2, -z+1/2, -y+1/2'
   'z+1/2, y+1/2, x+1/2'
   'z+1/2, -y+1/2, -x+1/2'
   '-z+1/2, y+1/2, -x+1/2'
   '-z+1/2, -y+1/2, x+1/2'

loop_
   _atom_site_label
   _atom_site_occupancy
   _atom_site_fract_x
   _atom_site_fract_y
   _atom_site_fract_z
   _atom_site_adp_type
   _atom_site_B_iso_or_equiv
   _atom_site_type_symbol
   Li1        1.0     0.131000      0.131000      0.131000     Biso  1.000000 Li
   C1         1.0     0.320000      0.320000      0.320000     Biso  1.000000 C
   H1         1.0     0.351000      0.351000      0.192000     Biso  1.000000 H

References

Lee, J. D. Concise Inorganic Chemistry, 4th ed.; Chapman & Hall: London ; New York, 1991. ISBN 978-0-412-40290-6.
Weiss, E.; Hencken, G. Über Metall-Alkyl-Verbindungen: XII. Verfeinerung der Kristallstruktur des Methyllithiums. Journal of Organometallic Chemistry 1970, 21 (2), 265–268. DOI: 10.1016/S0022-328X(00)83621-9. (In German)


How did early chemists measure mass of atoms?

Careful measurements were the foundation of chemical theory

Let's take the questions out of order to simplify the answers.

It doesn't matter what yours you use to measure volume as long as you stick to the same units in all your calculations. Cubic widgets is fine as long as you do all your calculations using them.

The key to understanding how chemists came to know about atomic mass is careful measurements of the widths and volumes involved in reactions. Chemists observed that certain reactions always involved the same mass ratios. More interestingly, when gases are reacted they also involve the same volume ratios. When hydrogen and oxygen react to for  water you start with two volumes (2 cubic widgets if you want) of hydrogen and one of oxygen react together to give water. How do we know it is just that ratio? Do the experiment with 3 volumes of hydrogen and one of oxygen and observe that there is still one volume of hydrogen left after the reaction is done.

Those experiments showed that the mass ratios were constant but different to the volume ratios if the reactants were gases. Oh, and we know the weight of the gases because we can measure the weight of a given volume of gas using a vessel of known volume containing a gas at a specific pressure using a carefully constructed vessel. A litre of oxygen weighs about 1.5g which is easy to measure using a good balance and good experimental conditions even in the 19th century.

Chemists hypothesised (and later proved) that the volume of gases were proportional to the number of molecules they contained so the ratio of gas volumes told them something about the number of units of substance (or molecules) involved in reactions. Comparing many reactions led to an understanding of more complicated reactions than hydrogen burning in oxygen. Comparing the known results from gas volume ratios to the known weights of substances involved in liquid and solid reactions extended this idea (with an early glitch caused because chemists didn't realise that many gases consist of diatomic molecules).

The actual number of molecules or atoms in a give volume took a lot longer to determine than the basic principles involved in chemistry (that reactions always involved specific ratios of inputs and outputs). Initial methods involved atomic theory applied to gasses or measuring the charge on a single electron. Later methods involved measuring the mass of a very pure crystal and comparing it with density derived from x-ray measurements to give more accurate values. (see this).

The basic experiments about atomic ratios can be repeated with good experimental technique and apparatus. But you need a very good way to measure weights of gases and volume. Repeatable, but not easy. You won't be easily able to measure the number of atoms in anything with household apparatus, though. That requires fairly sophisticated equipment.


Can we have Aromatic rings in a non-aromatic structure

Sounds like a contradiction in terms to me. If a compound has aromatic ring(s), then it is aromatic - that's the definition.
Though aromatic compounds with aliphatic side chains are possible, of course.

Your compound is azulene; look it up on Wikipedia or elsewhere. Indeed, it is aromatic, and so is each of its rings. You may be confused by the fact that cyclopentadiene (which looks just like one of these rings) as an individual compound is not aromatic, but that's another story. See, a 5-membered ring needs an extra electron to be truly aromatic. At the same time, the 7-membered ring has one electron more than it needs. So they exchange that electron and live happily ever after.


what's the mass of the original sample (g)?

It is first of all important to recognise that all the chloride in the original sample will precipitate upon treatment with excess AgNO3. This will give you a starting point, which is a measurable amount of AgCl.

The molecular mass of AgCl = (107.87+35.45)=143.32g

Therefore, the proportion of chloride in AgCl = (35.45/143.32) = 0.2473

Therefore in 2.013g of AgCl, there is (0.2473 x 2.013) = 0.4978g of chloride.

If 0.4978g of chloride constitutes 75% of the mass of the original sample, then the amount of original sample is (0.4978 / 0.75) = 0.6637g


How to obtain curve energy and non-bonded interaction parameters (e.g. Lennard Jones parameters) with Gaussian?

To start off, you definitely need to use dispersion-corrected density functional methods if you want to obtain reasonable non-bonded interaction parameters. I've seen many cases where B3LYP (with no dispersion) yields a fully-repulsive potential energy scan.

I also agree with your assertion about doing a rigid scan. In principal, a relaxed scan is a good idea, but consider that most force fields separate out the non-bonded interactions. Thus, for parameterizing, you don't really want to change the bond lengths, angles, etc. from a relaxed scan.

There are a few ways to get the parameters. As mentioned above, there's already a similar question with tips on curve fitting.

Multiple papers have also been published on fitting force field parameters from quantum chemical results, e.g:

QubeKit J. Chem. Inf. Model. 2019, 59, 4, 1366-1381 - code at GitHub
QuickFF J. Comput. Chem. 2015, 36, 1015– 1027 - code at GitHub
ForceBalance J. Chem. Theory Compute. 2013, 9, 1, 452-460 and J. Phys. Chem. Lett. 2014, 5, 11, 1885-1891 - code at GitHub
ForceFit J. Comput. Chem., 2010 31: 2307-2316
Parfit J. Chem. Inf. Model. 2017, 57, 3, 391-396 - code at GitHub

Update there's a paper on this very specific problem:

J. Chem. Theory Comput. 2020, 16, 2, 1115-1127

  Here, we utilize a previously described Minimal Basis Iterative Stockholder (MBIS) method to carry out an atoms-in-molecules partitioning of electron densities. Information from these atomic densities is then mapped to Lennard-Jones parameters using a set of mapping parameters much smaller than the typical number of atom types in a force field. This approach is advantageous in two ways: it eliminates atom types by allowing each atom to have unique Lennard-Jones parameters, and it greatly reduces the number of parameters to be optimized.


How many grams of each substance will be present at equilibrium in the reaction of ethanol and ethanoic acid?

Knowing that $K_c=4.0$, equilibrium is achieved when $Q_c = K_c$.

Let's set up an ICE table and write down our equilibrium expression like you did:

Reactants:

$$\ce{C2H5OH + CH3COOH <=> CH3COOC2H5 + H2O}$$

\begin{array} {|c|c|c|c|c|}
\hline
\text{Initial conc.} & \pu{0.373 mol} & \pu{0.396 mol} & \pu{0.552 mol} & \pu{3.96 mol}\\
\hline
\text{Change conc.} & -x & -x & +x & +x\\
\hline
\text{End conc.} & \pu{0.373 mol} - x & \pu{0.396 mol} - x & \pu{0.552 mol} + x & \pu{3.96 mol} + x \\
\hline
\end{array}

Assuming that our reaction is at equilibrium:

$$ K_c = \ce{\frac{[CH3COOC2H5][H2O]}{[C2H5OH][CH3COOH]}}$$

We plug in our values for our change in equilibrium:

$$ K_c = 4.0 = {\frac{[0.552 + x][3.96 + x]}{[0.373 - x][0.396 - x]}}$$

Solving for $x$, which is the change in concentration:

\begin{align}
4 &= \frac{x^2+4.512 x+2.18592}{x^2-0.769 x+0.147708}\\
0 &= 3x^2-7.588 x-1.59509
\end{align}

Using the quadratic equation, or thankfully polysolv on modern calculators, we get two values for $x$:

\begin{align}
x_1 &= 2.72449 \\
x_2 &= -0.195154
\end{align}

The negative zero is chosen, since the reaction will evolve to the left (being $Q_c>K_c$), thus increasing the concentration of the reactants. This will only be accounted for, if one takes the negative zero. If one chooses the other zero, one will go to the wrong opposite direction (towards a diminution of reactants concentration). In the particular case of the present exercise, one would even result in a negative concentration. Therefore:

\begin{align}
\ce{[CH3COOH]}
  &= \pu{0.373 mol} - (\pu{-0.195154 mol}) &
  &= 0.568154\\
\ce{[C2H5OH]}
  &= \pu{0.396 mol} - (\pu{-0.195154 mol}) &
  &= 0.591154\\
\ce{[CH3COOC2H5]}
  &= \pu{0.552 mol} + (\pu{-0.195154 mol}) &
  &= 0.356846\\
\ce{[H2O]}
  &= \pu{3.96 mol} + (\pu{-0.195154 mol}) &
  &= 3.764846
\end{align}

Therefore:

\begin{align}
\pu{0.568154 mol} \ \ce{CH3COOH} \times \frac{\pu{60.1 g}}{\pu{1 mol} \ \ce{CH3OOH}} &= \pu{34.1 g}\\
\pu{0.591154 mol} \ \ce{C2H5OH} \times \frac{\pu{46.1 g}}{\pu{1 mol} \ \ce{C2H5OH}} &= \pu{27.3 g}\\
\pu{0.356846 mol} \ \ce{CH3COOC2H5} \times \frac{\pu{88.1 g}}{\pu{1 mol} \ \ce{CH3COOC2H5}} &= \pu{31.4 g}\\
\pu{3.764846 mol} \ \ce{H2O} \times \frac{\pu{18.0 g}}{\pu{1 mol} \ \ce{H2O}} &= \pu{67.8 g}
\end{align}

Initial product sum is $\pu{160.8 g}$.

Equilibrium product sum is $\pu{160.6 g}$. $\pu{0.2 g}$ is likely due to truncation error in calculations.

This problem is really cruel... whoever assigned it... :/


Effect of redefining Avogadro's constant and kilogram on molar mass

Will these changes result in the molar mass of a substance no longer being exactly equal to the molecular mass expressed in grams?

Yes. The ratio between the molar mass and the relative molecular mass is called the molar mass constant, and it used to be 1 g/mol exactly. Now that Avogadro's constant is set to a fixed value, it's value has to be determined experimentally.

NIST quotes its value as 0.999 999 999 65(30) x 10-3 kg mol-1 (CODATA 2018).

The relative molecular mass of the carbon-12 isotope is still exactly 12 and the atomic mass unit u (or dalton Da) is still defined as 1/12 of the mass of a carbon-12 atom, so the relationship between relative molecular mass and molecular mass still holds. However, its molar mass is no longer exactly 12 g/mol.

As @MaxW says in a comment, these changes are so minute that they have no practical implications outside of metrology (which was one of the goals of the International Bureau of Weights and Measures when they suggested the change).


How does a catalytic poison reduce a catalyst's activity?

Basically the catalytic poison reduces the activity by slowing or inhibiting the reaction by occupying the surface of the catalyst and therefore the surface area available for the reactants reduces. You can get a general idea from the following text.

Substances which themselves are not catalysts, but when mixed in small quantities with the catalysts increase their efficiency are called as promoters or activators.
Catalyst poisoning refers to the partial or total deactivation of a catalyst caused by exposure to a range of chemical compounds. Poisoning may be desirable when it results in improved selectivity (e.g.Lindlar's catalyst) but may be undesirable when the catalyst is rendered ineffective (e.g. in catalytic converters). Poisoning refers specifically to chemical deactivation, rather than other mechanism of catalyst degradation such as thermal decomposition or physical damage.
Poisoning involves compounds which bonds chemically to the active surface sites of a catalyst. This may have two effects: the total number of catalytic sites or the fraction of the total surface area that has the capability of promoting reaction always decreases, and the average distance that a reactant molecule must diffuse through the pore structure before undergoing reaction may increase. Poisoned sites can no longer accelerate the reaction with which the catalyst was supposed to catalyze. Large scale production of substances such as ammonia in the Haber–Bosch process include steps to remove potential poisons from the product stream.

For more info, read the Wikipedia article on Catalytic poisoning.


Is there a mnemonic to study the spectrochemical series?

I would not recommend a mnemonic for the spectrochemical series for a number of reasons:

Often, it depends on both metal and ligand and coordination geometry how big a certain splitting is.
The splitting of some ligands ($\ce{NH3, H2O}$) is even dependent on the pH — even though that could sort-of be overcome by including their corresponding anions in the list.
Any mnemonic is bound to be incomplete. N-heterocyclic carbenes (NHCs) are very important ligands especially in organometallic chemistry, but they aren’t on your list.
Some ligands (e.g. $\ce{NO}$) can display very complex redox chemistry that defies them being ordered in a simple list.
Rather than learning some list off by heart and rote memorising it later, it is much better to know why a ligand is strong- or weak-field and to deduce that property from the ligand’s structure.

The last point is most important. Rather than learning or attempting to remember that chloride is a weak-field ligand, ask yourself why. It is a π donor ligand that will split the $\mathrm{t_{2g}}$ orbitals of the central metal in a bonding–antibonding way, lifting the metal’s $\mathrm{t_{2g}}$ orbitals and lowering the energy difference. That way, I immediately know what to expect of (say) selenide $\ce{Se^2-}$ even if it is not present in the list. The strength of a certain ligand in this discussion is equivalent to the usual trends in the periodic table. Quickly memorised, no long mnemonic and applicable even to new ligands.


You might want to look at this link which has a nice general discussion about cathodic protection, with an abbreviated activity series for the metals of interest in cathodic protection.  Typically, aluminum, magnesium and zinc are used as sacrificial anodes for cathodic protection.

Strictly speaking from a thermodynamic standpoint, one would suspect that magnesium would be needed to protect aluminum.  However aluminum boat manufacturers state that zinc can be used for protecting aluminum boats in salt water.

I'm completely unfamiliar with the die casting industry, but they use Al/Zn alloys which, if affordable, may provide an interesting pedagogical component to your corrosion activity (namely, what do scientists do when the properties of the elements on the periodic table don't quite fit what we have in mind).


For an electron arrangement diagram, is it necessary to show the inner electron shells or is it customary to only show the valency electrons?

Both are correct: for $\ce{Cl-}$, the inner set of electrons would be implicit, as $\ce{Cl}$ has 17 electrons. The point of drawing these orbital diagrams, especially in a molecular context, is to show the nature of bonds and to show how the octet/duplet rules are being followed.

If you have difficulties in using the second diagram (finding it harder to read or are confused by the similarities in $\ce{H}$ and $\ce{Cl}$), you should use the first diagram.


There are a number of rumors on the internet about different ways to treat your battery. Some of these involve putting batteries in the refrigerator or freezer for varying lengths of time to improve performance in some way. Snopes has concluded that there is no basis for this. However, batteries and battery aging mechanisms are very complex and so I think it's difficult to completely rule out the possibility that under some situations there could be some beneficial effect from freezing.

Nevertheless, I would recommend that you avoid using your battery in any way that it was not designed for. The reason is that internal damage to the battery (such as from extreme temperatures, moisture, or mechanical stress) could create a safety hazard. Even if cooking, freezing, or bashing does have some positive effect, it is probably not worth putting yourself at risk. The best thing to do is to recycle your old battery and buy a replacement.


Rearrangement considerations in feasibility of SN1 in given reactants

The proposed rearrangement would produce a primary carbocation which, relatively lacking in hyperconjugation, is itself unstable.  The proposed rearrangement therefore is not really plausible.
The "instruction" that the highly strained 1-methyl-cyclopropyl cation cannot be formed may be wrong anyway.  This paper reports a substituted cyclopropyl cation as an intermediate in a reaction between a more complex chlorocyclopropane and an alcohol.


Is there a systematic naming system to name carbonyl derivatives?

The names (2 of 6) are rather correct (but note that acetyl and  acetic names are preferred over ethanoyl and ethanoic), except following:

$\ce{Ac-SH}$: ethanethioic acid, more specifically ethanethioc S-acid (since there are two possibilities of “monothiation”)

  P-15.5.3.4.1

  (…)

  $\ce{CH3-CO-SH}$
  thioacetic S-acid
  ethanethioic S-acid (PIN)

  $\ce{CH3-CH2-CS-OH}$
  thiopropionic O-acid
  propanethioic O-acid (PIN)

$\ce{Ac-O-NH2}$: Anhydride name cannot be used as hydroxylamine is not a name of acid. It's also not a recognized acid functional group, so the name probably should be regarded a substituted ketone, with preselected prefix aminooxy (P-68.3.1.1.1.2), 1-(aminooxy)ethan-1-one. (Although pubchem entry lists wrong name: amino acetate; O-acetyl hydroxylamine is acceptable I think.)
$\ce{Ac-SO3H}$: “α-oxosulfonic acid“, 1-oxoethane-1-sulfonic acid
$\ce{Ac-PH2}$: phosphide as alleged analogy to amide (with different meanings in inorg. and org. chem.) is not used, the name should be 1-phosphanylethan-1-one (or acetylphosphane).

  P-64.1.2.1 Pseudoketones

  (…)

  $\ce{H2P-\overset{1}{C}O-\overset{2}{C}H2-\overset{1}{C}H3}$
  1-phosphanylpropan-1-one (PIN)
  propanoylphosphane


Where do we report yields in a research report?

Your professor is right! Giving the yield in the method section, that is, together with detailed information on synthesis, workup and characterization of the products obtained, is standard in scientific journal I know or I have published in.


What is the liquid used in instant shoe shiner?

It's silicone oil.

You can use a mineral oil. You can find it as baby oil and if you don't like the odor, also can be found as fleet enema products (eg. http://fleetlabs.com/fleet_enema_products.php).
For larger and cheaper amounts, you can buy it in computer hardware store, farm supply store, or even in vet clinic.


Why does a molecule with a Thiol group form Hydrogen bonds?

Actually it's all four.  See here for I.  The article describes a hydrogen bonding interaction between methane and water.  It's weaker than the "usual" hydrogen bonds but still contributes to forming methane-water clathrates.

The larger lesson here is that hydrogen bonding is not just the electrostatic interaction often described in textbooks.  It has a molecular orbital component that works even with primary bonds of low polarity.


What's the difference between aqueous NaOH and dilute NaOH?

Dilute simply means that the concentration of the solute of interest, in this case $\ce{NaOH}$, is low relative to some reference.  One could imagine a non-aqueous, dilute solution of $\ce{NaOH}$ in, say, methanol.

An aqueous solution is one where the solvent is water, regardless of the concentration of the dissolved solute(s).

So, for example, the following combinations of the terms are possible:

Concentrated aqueous solution - High $\ce{NaOH}$ concentration in a water solvent
Dilute aqueous solution - Low $\ce{NaOH}$ concentration in a water solvent
Concentrated methanolic solution - High $\ce{NaOH}$ concentration in a methanol solvent
Dilute methanolic solution - Low $\ce{NaOH}$ concentration in a methanol solvent


Does boiling water really treat it to make it safer than filtered/chlorinated water?

What do you mean by "safer"? If the water is pure to start, there's no point in boiling it. For mildly contaminated water, chlorination is inexpensive, but can form some toxic halogenated carbon compounds. Boiling is very energy intensive, and may require use of fossil fuels (unless you use a solar water heater). Both eliminate most microorganisms, though there are a few, mostly nonpathogenic ones, such as Deinococcus radioduran and others found at undersea volcanic vents, that can survive radiation, heat and other chemicals. Bacteria have even been found living in the common biocide, povidone-iodine! See this source for more information on biocides.

Better alternatives might be ultraviolet sterilization or ozonation.


New relationship between exponential factors and entropy of a reaction?

Yes, your expression is correct. It's nothing new though. Entropy of activation can be interpreted as part of the pre-exponential factor in the Eyring equation:
$$k=\bigg(\frac{k_B T}{h}e^{\frac{\Delta^\ddagger S^\circ}{R}}\bigg)\cdot e^{-\frac{\Delta^\ddagger H^\circ}{RT}}$$


Please allow me to answer the question in two parts.

Stability of carbonates

We observe this order when we compare the thermal stability of group I-A carbonates: $$\ce{Li2CO3 \lt Na2CO3 \lt K2CO3 \lt Rb2CO3 \lt Cs2CO3}$$

On similar grounds, the thermal stability of group II-A carbonates can be predicted.

This order is in accordance with Fajans' Rules. Here's the table we need to glance at from the Wikipedia page.

  $$\begin{array}{|c|c|} \hline
\mathrm{Ionic} & \text{Covalent}\  \\ \hline
\text{Low positive charge} & \text{High positive charge} \\
\text{Large cation} & \text{Small cation} \\
\text{Small anion} & \text{Large anion} \\ \hline \end{array}$$
  Polarization will be increased by high charge and small size of the cation.

Thus, larger the ratio of charge and radius, higher is the polarization power of the cation, more is the covalent character in the compound and lesser is the ionic character; Ionic compounds are more stable when exposed to heat. A higher ratio will therefore mean lesser stability.

Stability of complex compounds

Thermodynamic stability of complexes, on the other hand, is determined by the Irving–Williams series, according to which general stability sequence of high spin octahedral metal complexes for the replacement of water by other ligands is:
$$\ce{Mn^2+ < Fe^2+ < Co^2+ < Ni^2+ < Cu^2+ > Zn^2+}$$
Owing to this observation, a lot of theories were formulated to explain this phenomenon. One such theory was that since Crystal Field Theory (CFT) was based on the idea that a purely electrostatic interaction existed between the central metal ion and the ligands, the stability of complexes should be somehow related to the size by charge ratio (otherwise known as the ionic potential) of the central metal atom. [Source]

An interesting thing to note here is that Wikipedia has stated in its article that:

  None of the explanations can satisfactorily explain the success of the Irving–Williams series in predicting the relative stabilities of transition metal complexes.

Thus from manganese till copper, it can be said that higher the ionic potential, more is the stability.


I do not know whether the following is a real answer or not but Wikipedia 3-methylpyridine article shows

3-Methylpyridine is produced industrially by the reaction of acrolein with ammonia:
:$$\ce{2 C3H3O + NH3 -> 3-CH3C5H4N + 2 H2O}$$ This reaction also affords substantial amounts of pyridine.

Wikipedia Nicotinonitrile article

A colorless solid, it is produced by ammoxidation of 3-methylpyridine:[2]. Nicotinonitrile is a precursor to the vitamin niacin.

In other words, the question could become one of separating pyridine from 3-Methylpyridine.


What alcohol concentration makes bathroom cleaner "disinfecting"?

According to this Center for Disease Control document:

  Ethyl alcohol (70%) is a powerful broad-spectrum germicide and is considered generally superior to isopropyl alcohol. Alcohol is often used to disinfect small surfaces (e.g. rubber stoppers of multiple-dosemedication vials, and thermometers) and occasionally external surfaces of equipment (e.g. stethoscopes and ventilators).

Although 70% alcohol is ideal, another CDC article states that it can be effective at 50%.  According to this article:

  In the healthcare setting, "alcohol" refers to two water-soluble chemical compounds—ethyl alcohol and isopropyl alcohol—that have generally underrated germicidal characteristics.  FDA has not cleared any liquid chemical sterilant or high-level disinfectant with alcohol as the main active ingredient. These alcohols are rapidly bactericidal rather than bacteriostatic against vegetative forms of bacteria; they also are tuberculocidal, fungicidal, and virucidal but do not destroy bacterial spores. Their cidal activity drops sharply when diluted below 50% concentration, and the optimum bactericidal concentration is 60%–90% solutions in water (volume/volume).

So, alcohol has certainly been shown to be a powerful, broad-spectrum antimicrobial agent, though not at the dillution you proposed. I would just use the straight vodka if it were greater than 50% alcohol, or probably beter (cheaper) yet would be denatured alcohol if that's available where you live.


Relative O-O bond lengths in O2X2 molecules (X = H, F, Cl)

First things first: the shorter a bond, the higher the multiple bond character. I.e. a double bond is shorter than a single bond, and a bond with a bond order of $1.5$ is in the middle of the two. Thus, the shorter the bonds in $\ce{X2O2}$, the higher the corresponding double bond character.

The key to understanding why the bond in $\ce{O2F2}$ has a very high double bond character is the gauche effect. The effect has its name because of the observation that $\ce{O2F2}$ prefers a gauche configuration rather than the expected (due to sterics) anti configuration. The reason is electronic in nature: a lone pair of oxygen A can interact with the $\sigma^*(\ce{O_B-F})$ orbital. This lowers the energy of the populated orbital at the expense of the unpopulated one. Since we have an interaction along $\ce{O_A-O_B}$, this increases the double bond character of the $\ce{O-O}$ bond. At the same time, since we are using the $\sigma^*$ for stabilisation, the $\ce{O-F}$ bond decreases in bond order — it becomes less than a single bond. In Lewis resonance depictions, this can be explained by the following mesomeric structures:

$$\ce{F-O-O-F <-> F-O^+=O\bond{...}F-}$$

This effect is strongest in $\ce{O2F2}$ since fluorine is more electronegative than oxygen and therefore the $\sigma^*$ orbital has a much higher oxygen contribution.

In hydrogen peroxide, a similar mechanism is at play that also contributes to reducing the bond length and increasing the bond order. In $\ce{Cl2O2}$, the effect is weakest.


What is the appropriate sizing for fiberglass compatible with polyamide?

I think epoxy resin is compatible with polyamide because its chemical structure. Epoxy is compatible with silica (glass) and poyamide functional groups then it could be a good sizing .