Dissolving of non polar gases in water (liquid)

Certainly there are attractive interactions between water molecules and nonpolar solutes, there are always dispersion attractions.  Additionally, water will "layer" against a solute  particle, producing solvation shells, even in the complete absence of attractions, just because of the excluded volume forces packing  everything together.   However, for nonpolar gases the new gas-water interactions are significantly weaker than the water-water interactions they replace, so the enthalpic change on dissolve is strongly positive (unfavorable).  Hence the only reason gases dissolve is because the entropy increases, and that is simply because the volume available to the gas molecules is larger if it includes the volume of the water as well as the volume of the gas above it.

Another way to put that is that molecules of nonpolar gas are found in the water simply because they can be.  Id est, those molecules will zoom around in the gas phase, occasionally strike the surface of the water, plunge in, and wander around randomly for a while before exiting.  Net result: a low concentration of nonpolar gas dissolved in the water.  The concentration will increase with pressure simply because a higher pressure means a higher density of the gas, which means more gas molecules  will strike each cm2 of the water surface per second, and at equilibrium that must be matched by a higher rate of exit from the surface, which requires a higher equilibrium concentration in the bulk water.  For small pressures, the effect is approximately linear in pressure, and the coefficient is called the Henry's Law coefficient.

The dissolved gas particles do not occupy spaces between the  water molecules, if by that you mean preexisting spaces.  They shove the water molecules out of the way, create a hole, and live in that.  The volume of the solution may not increase, however, or may not increase as much as one might think, because as I mentioned above, the water will form solvation shells around the dissolved particle, and will moreover disrupt the hydrogen-bonding network in the water.  Both effects could result in an increase in density in the regions near the hole created by the solute particle.  So predicting whether and by how much the volume of the water expands (or contracts) is in general difficult.


How can sodium be reduced in the Castner–Kellner process?

I am not sure if you are familiar with the term "overpotential." With mercury electrode, as a cathode, it takes more negative potential to reduce hydrogen ions to hydrogen gas, rather it is easier for sodium ions to get reduced on a mercury surface and form an amalgam (an alloy of Hg and Na). This is a peculiarity of the mercury electrode.

Mercury cell in chlor-alkali industry


Why is the Maxwell–Boltzmann formula inapplicable at low temperatures?

Since the explanation was a little more complicated than I initially thought, I figured it would be worth it to combine my comments (and info from Physics SE) into an answer.
Quantum particles satisfy Fermi–Dirac or Bose–Einstein statistics depending on whether they are fermions or bosons. These distributions have the form $$\langle n_i\rangle=\frac{1}{\exp[(\epsilon_i-\mu)/k_bT]\pm1}$$ where the plus/minus is for fermions/bosons.
To consider the high temperature limit, we need to note that not only is there a direct temperature dependence, but $\mu$ is also dependent on temperature. Specifically, in the high temperature limit, $\mu<0$ and $|\mu|>k_bT$. Combining this information, we can make the (very accurate) approximation $$\exp[(\epsilon_i-\mu)/k_bT]\pm1\approx\exp[(\epsilon_i-\mu)/k_bT]$$ as the exponential function will be much larger than $1$. So at high temperatures $$\langle n_i\rangle\approx\frac{1}{\exp[(\epsilon_i-\mu)/k_bT]}$$ which matches the form of the Maxwell–Boltzmann distribution. We can see that at low temperatures these two distributions would not agree, as the additional $\pm1$ term in the denominator would become more significant as the exponential got smaller.
It's important to remember that the particles are always indistinguishable; all we have done in the high temperature limit is made an approximation that simplified the functional form. We should not take this coincidental agreement with the MB distribution (for which the particles are assumed to be distinguishable) to imply that the quantum particles have become distinguishable at high temperatures.


Can ring inversion in Cis-Decalin allow the E2 reaction?

The issue in this Grob fragmentation is whether or not the $\ce{C-C}$ and $\ce{C-OTs}$ bonds are anti-periplanar. The $\ce{C-OH}$ bond is free to rotate and the unpaired electrons of the negative oxygen can align with the other two bonds. In the first example a), the red bonds are gauche to one another. Standard E2 elimination is likely the source of the mixture. Examples b), c) and d) all have red bonds anti-periplanar. The blue bonds predict the geometry of the double bonds in the product. Examples b) and c) give E-alkenes while d) affords a Z-alkene.


Detecting trash (plastic) burning smoke with sensors

The typical gas sensor marketed to hobbyists and enthusiasts (except for $\ce{CO2}$) are of the metal oxide semiconductor type, not to be confused with MOS as in MOSFET.
How do these work?
In these detectors, a semiconductor material made of the oxide of a metal, usually tin.
I'll quote from this question:

...sintered composite based on the semiconductor material $\ce{SnO2}$, which is the metal oxide. The different sensors have different admixtures and operating conditions to achieve different sensitivities to different gasses.
The principle of operation of MOS-type gas sensors is explained in Figaro sensors' Operating principle; MOS type.

The Figaro link shows the operating principle in the GIF below:
 GIF (Source: Figaro)
When the $\ce{SnO2}$ is heated by the (usually 5V but not always) heater supply, oxygen in the air will dissociate at the surface and oxygen ions will stick to the surface, depleting the semiconductor and cutting off the flow of current. When a carbon-containing gas (e.g. $\ce{CO}$ or $\ce{C_xH_y}$ (but not $\ce{CO2}$)) comes along, it will combine with the adsorbed oxygen and remove it from the surface. The more gas, the lower the equilibrium concentration of adsorbed oxygen and the higher resulting current flow through the semiconductor.
Not as gas-specific as you might think
By adjusting the amount and types of additives to the sintered $\ce{SnO2}$ and adding catalyzing metal clusters to its surface, the manufacturer can enhance sensitivity to some gases and reduce it to others, but there is no single-gas specificity with these sensors.
Since the carbon in $\ce{CO2}$ is already fully oxidized, it won't react with the adsorbed oxygen. Those sensors (e.g. MG-811) use a chemical cell sensor based on a solid electrolyte. They need very high impedance amplifiers that measure millivolts, a bit like pH sensors, in fact, a lot like pO2 sensors!
CO and VOCs
I would recommend you use a $\ce{CO}$ sensor (e.g. MQ-7, MQ-9, MQ-307A or MQ-309A) since burning garbage will likely be incomplete (not in an industrial incinerator) and a "Air Quality" VOC sensor like MQ-135.
You can experiment with others, some people like to build an "electronic nose" which is an array of a bunch of these, then they plot the data for different kinds of smoke and environments to see what happens, and try to use the pattern to identify the likely problem.
PM2.5
Inefficient combustion can also produce particles as soot, and there are more expensive "dust sensors" or "PM2.5 sensors" which measure small particles (2.5 microns or larger) in the air. Soot is a hallmark of incomplete combustion (think of the smoke from a badly tuned engine) and a huge problem from incomplete or low-temperature combustion from a garbage fire.
None of these will be conclusive, you'll have to look at your data and look at when there is a fire, and learn how the signals correlate to the events.
Ultraviolet light
There are UV light based fire detectors as well, but you'd need direct line-of sight access to the flames. If the sensor is designed to be solar-blind by adding a filter that removes all wavelengths from the Sun that make it to the ground, then it might be helpful here. However, they are usually designed to work indoors where indoor lighting may have less UV than outdoor sunlight. This would take some more investigation.
read the data sheets carefully
Finally, read each sensor's data sheet carefully and note carefully the specific voltages that each sensor needs. The "Air quality" sensors need you to run for a certain amount of time (about a minute) at one voltage, then a different amount of time at another voltage because the surface chemistry is a bit more complex.
Sensors can also be found here:

http://wiki.seeed.cc/search.html?q=gas+sensor
https://www.pololu.com/category/83/gas-sensors


Why does solvent particles move towards solution when semi permeable membrane is added?

Here what happens is that the fresh water is a pure solvent and a pure solvent always has a greater vapor pressure than that of a solution. So due to the greater vapor pressure on water's side, the molecules of water will experience a force in salt water's direction as the pressure is less in salt water side. Therefore the water will flow towards the salt water.


What is this fluorinated organic substance?

The name of the compound is 1-chloromethyl-4-fluoro-1,4-diazoniabicyclo[2.2.2]octane bis(tetrafluoroborate) (CAS #: 140681-55-6), which is commonly known as Selectfluor, a trademark of Air Products and Chemicals (see Waylander's comment elsewhere). Different view of the compound is given below (to you to understand the zig-zag feature):

Introduced in 1992, this organic salt is used as a fluorine donor in organic synthesis (Ref.1). For example of using Selectfluor as a source of fluorine, see Ref.2:

References:

R. Eric Banks, Suad N. Mohialdin-Khaffaf, G. Sankar Lal, Iqbal Sharif, Robert G. Syvret, "1-Alkyl-4-fluoro-1,4-diazoniabicyclo[2.2.2]octane salts: a novel family of electrophilic fluorinating agents," J. the Chem. Soc., Chem. Commun. 1992, (8), 595-596 (DOI: 10.1039/C39920000595).
Timothy J. Barker, Dale L. Boger, "$\ce{Fe(III)}$/$\ce{NaBH4}$-Mediated Free Radical Hydrofluorination of Unactivated Alkenes," J. Am. Chem. Soc. 2012, 134(33), 13588-13591 (https://doi.org/10.1021/ja3063716).


The diazo dye (1), known as Solvent Yellow 124 or Somalia Yellow is soluble in aliphatic hydrocarbons. In the European Community, it is used to mark low-taxed diesel fuel (for heating purposes). In order to prevent its use in cars, samples are acidified. This results in acetal cleavage and protonation of the diazo-fragment, yielding the red, water-soluble iminoquinone species (2)

Since this is probably not exactly what you wanted, what about dyes that exist in a neutral and a zwitterionic form, depending on the polarity of the solvent?

Brooker's merocyanine is one of the most prominent examples and show strong solvatochromism, i.e. the colour changes with the polarity of the solvent


Historically, how was the connection between NMR peaks and hydrogen atoms made?

TL;DR: Scientists knew that electrons had a spin, and thought, "wouldn't it be nice if atomic nuclei did too?" When they passed molecular hydrogen through a homogenous magnetic field (in a modified version of the Stern–Gerlach experiment) and subjected it radio waves, they found energy was absorbed at precise frequencies. This set the stage for the usefulness of $^1\ce{H}$ NMR in the distant future.
Taken from A Brief History of Nuclear Magnetic Resonance:

In the mid-1920s it became apparent that many features in atomic spectra could be accounted for only if certain atomic nuclei likewise possessed spin and a magnetic moment. Refinements of the Stern-Gerlach experiment verified this concept by 1933.
In 1939 Rabi et al. made a major improvement in beam techniques by sending a stream of hydrogen molecules through not only the inhomogeneous magnetic field required for deflection, but also through a homogeneous magnetic field, where they were subjected to radio frequency electromagnetic energy. Energy was absorbed by the molecules at a sharply defined frequency, and the absorption caused a small but measurable deflection of the beam. This was the first observation of NMR, and Rabi received the Nobel Prize in 1944. However, such studies were limited to nuclei in small molecules under very high vacuum in a molecular beam, the deflection of which served to detect the resonance.


Place cooling bath apparatus in freezer to speed up the process?

The freezing point remains the same, but its measured value may depend on the way of measurement.
In case the solution freezing point is too close to the ice/salt 3:1 bath temperatures, it may be the reason for too long freezing time. The bath ice/$\ce{CaCl2 . 6 H2O}$ bath 1:2 could help, if applicable.
Note that fast freezing may lead to biased value, especially, if supercooling occurs. E.g. a PET bottle with water can get supercooled in a freezer deep below the freezing point.


Are these 3 molecular structures of ammonium bicarbonate the same?

Like Punarbasu Roy already stated, the depicted structures are certainly not the same. To be frank, only one of them (A) is correct, one of them is not completely wrong (B - but only because it is incomplete), and one is completely, utterly wrong (C).

In lieu of an actual crystal structure of ammonium hydrogencarbonate,[1] I'll refer to the structure of the parent compound ammonium carbonate monohydrate to make my point.[2] It is actually quite difficult to assign one correct Lewis structure. In its solid form ammonium (bi)carbonate is a salt, in which ammonium and (bi)carbonate ions are interacting via ionic bonds in a lattice. In solution this certainly is broken up and we will find separate ions, according to the following equilibria:
\begin{align}
\ce{(H4N)HCO3~(s) &~->C[(H2O)] NH4+ ~(aq) + HCO3- ~(aq)}\\
\ce{NH4+ ~(aq) + H2O &~<=> H3+O~(aq) + NH3~(aq)}\\
\ce{HCO3- ~(aq) + H2O &~<=> H2CO3 ~(aq) + {}^{-}OH~(aq)}\\
\ce{HCO3- ~(aq) + H2O &~<=> CO3^2- ~(aq) + H3+O~(aq)}\\
\end{align}
The same equilibria will establish for ammonium carbonate with different concentrations.

In the crystal structure of ammonium carbonate monohydrate you can see the different molecular entities (ions) held together via hydrogen bonds, reproduced from [2]:

Wikipedia[3] therefore offers the best possible Lewis structure:

Strictly speaking a Lewis structure may only contain covalent bonds, i.e. two-centre-two-electron bonds, while ionic bonds are implied by placing the (molecular) ions in close proximity to each other. However, the hydrogen bond is a well known concept in chemistry and it was already extensively discussed by Linus Pauling. In his book "The Nature of the Chemical Bond" he uses dashed lines to represent the hydrogen bond.[4] Certainly another interesting read is the approach of a more unified theory about the hydrogen bond.[5]
Another possibility is to introduce the donor-acceptor notation with arrows, to indicate a bonding interaction. Therefore one of the following structures may also be considered correct in an extended Lewis concept.

The following article reads like this is still a work in progress: R. Brooks, and T. C. Alcock, Nature, 1950, 166, 435-436. (It does also not come with any supporting information.
A. D. Fortes, I. G. Wood, D. Alfè, E. R. Hernández, M. J. Gutmann, and H. A. Sparkes, Acta Cryst. Sec. B, 2014, 70, 948-962. This quite interesting article is open access.
Ammonium bicarbonate, Ammonium carbonate
Linus Pauling: The Nature of the Chemical Bond. Cornell University Press, 1960.
G. Gilli, P. Gilli, J. Mol. Struc., 2000, 552 (1-3), 1-15.


Why does the concentration of the acid equals that of the base at the mid-point of titration?

At the midpoint of titration, the concentration of acid is just approximately equal to its conjugated base, as there must be considered the acid dissociation and ( in less extend ) salt hydrolysis.

If we take the definition(simplified using concentrations):

$$K_\mathrm{a}=\frac {\ce{[H+][A-]}}{\ce{[HA]}}$$

we can see that by definition:

$$\mathrm{pH}=\mathrm{p}K_\mathrm{a} + \log { \frac {\ce{[A-]}}{\ce{[HA]}}}$$

$\mathrm{p}K_\mathrm{a}$ is $\mathrm{pH}$ when both items of the conjugate pair have the same concentration.

$\mathrm{pH}$ near titration midpoint is approximately equal

$$\mathrm{pH}=\mathrm{p}K_\mathrm{a} + \log { \frac {n_\mathrm{base}}{
n_\mathrm{acid, init} - n_\mathrm{base}
}}$$

This function has minimal slope = maximal buffer capacity at $$n_\mathrm{acid, init} = 2 \cdot n_\mathrm{base}$$


According to this article a mixture of products is formed, with alkylchlorsulfonates $\ce{Cl-SO2-O-R}$ being dominant products. These molecules are resistant to abstraction of sulfur trioxide, probably because trioxide is a very strong Lewis acid, unlike dioxide.


How are there two C3 rotation axes in ammonia?

In a character table for $C_{3v}$, $ 2C_3$, means that the element is actually 2 $C_3$ elements in the same conjugacy class.  That means the two $C_3$ operations are related in some way.
In fact, it means they share the same single axis.  One $C_3$ operation is a $120^\circ$ turn around the axis, and the other $C_3$ operation is a $240^\circ$ turn around the same axis in the same direction - or equivalently a $120^\circ$ turn around the same axis in the opposite direction.
In the image below the two elements $C_3$ and $C_3^{-1}$ are the two symmetry elements of the $2C_3$ class, which are rotations about the same axis out of the plain of the image.

Symmetry elements of the $C_{3v}$ point group
Image taken from Fig S5 in supplementary information of T. Karin, et al, Phys. Rev. B 94, 041201


Does chemistry change under high voltage?

Long story short, no, they don't.

Charging a vessel to 1 MV is not a big deal, if you look at it from the inside. To put things into perspective, imagine a vessel of about 10 cm across, which is kinda OK for a flask. Imagine it spherical to make the calculations easier.

Now, the capacitance of a sphere is $C=4\pi\varepsilon_0R$, which evaluates to 5.56 pF (yes, "p" stands for "pico").
So charging your vessel to 1 MV implies putting $5.56\;\mu\rm C$ of electric charge on it.
That's $3.48\cdot10^{13}$ extra electrons (or some negative or positive ions, that doesn't matter) ...
... which may seem big until you compare it to the amount of water that would fit into the vessel: $1.75\cdot10^{25}$ molecules.

So it is like 1 electron per quite a few billions of molecules. That would not make much of an influence. The molecules will just sit there as usual, feeling nothing; most of them have never seen an electron in their entire lives, and have no idea that their vessel is being charged.

Things will change a great deal if you touch the vessel from the outside, but that's just ordinary electrochemistry which you seem to put out of scope of your question.


Ok your question is a little ambiguous but I'll try.

a. Why should it happen? - If you look at Na, it has a lone electron in its outer shell and that essentially unbalances the atom, so it prefers to lose it and become an ion, this is almost spontaneous and hence requires less energy. When $\ce{Na+}$ joins up with $\ce{Cl-}$ they neutralise the electrostatic charge introduced by them becoming ions, releasing more energy. More energy because they have gone from an energy state A (medium energy) to B (higher energy) to become an ion. Then when they come together they both, each atom, falls to a lower energy state C than state A even and hence the change (NB) from B to C is greater than the change from A to B. In your book the change from my A to B is the change of an electron ( first two equations in your book) and then my change from B to C is the Na + Cl part.

b. How do attractive forces tend to reduce energy? - This is an important one to note the point of the text book, which is pretty badly written if you ask me, is referring to the energy of the system and not the energy of attractive forces. When it says system it means the newly bonded together $\ce{NaCl}$. The energy is reduced because of the neutralisation mentioned above. Positive and negative prefer to be neutral and hence, are attracted to one another. But before that, you must realise that being charged or an 'ion' is less stable and therefore in a state of higher energy than a neutral system. So it is a matter of stability as well.

If you want quantum explanations please say but I get the feeling you'd prefer the simpler option.

I hope this helps, even if only a bit.


How can I safely carry out an electrolytic etch on stainless steel without producing relatively harmful by-products?

Sand or shot blast it instead.  Rub it with crushed aluminum foil and saltwater.

Everything is harmful.  Voltage determines products.  Use the minimum voltage (slow way to travel).  Don't use a chloride electrolyte.  (TEST FIRST!) Use hydroxide, carbonate, (nitrate - Homeland Severity and OSHA will beat on you), sulfate, electrolyte (note solubility of metallic byproducts after reaction with electrolyte).  The larger the charge on the ion, the slower the kinetics.  Vent the chlorine plus inert air through lye solution to make bleach, or through a solution of reducing agent.

Cr(III) is OK, Cr(VI) is a carcinogen. Reduce spent electrolyte with sulfite, Vitamin C, zinc granules, FeCl2, shredded scrap iron, steel wool, etc - and have a way to test for Cr(VI) residual.

Before you vent or sewer anything, talk with your Safety Officer.  The EPA and OSHA are carnivorous.  it's always a good idea to be a good upstream neighbor.


What happens when you put a battery in saltwater?

Salt dissolves in water to left behind positively charged sodium ion and negatively charged chloride ion. once you put battery in salt water, the sodium ion migrate towards the "negative tank" and chloride ion migrate towards the "positive tank". Salt water is much more conducive than regular water, so the battery would discharge more quickly and the current will break down the salt by electrolysis, producing hydrogen and chlorine gas.


I searched CCDC for the infinite structures with the following building block:

where where each vertex contains any non-metal atom linked by any type of bond. The only known crystal structure that would meet these criteria is $\ce{[B12]2[CBC][C2]Mg_{1.42}}$ where $\ce{B12}$ icosahedra are linked via $\ce{B–C}$ and $\ce{B–B}$ bonds forming the moieties of the previously shown topology:

  Figure 2. Projection of the $\ce{B25C4Mg_{1.42}}$ structure onto the $0ac$ plane (the block of icosahedra of boron carbide structure is marked by shading).

Unfortunately, no similar structures are found for nitrogen yet. Also, the structural motif breaks on boron clusters, but this is the closest as I was able to get.

Reference

Ponomarev, V. I.; Konovalikhin, S. V.; Kovalev, I. D.; Vershinnikov, V. I.; Borovinskaya, I. P. Mendeleev Communications 2014, 24 (1), 15–16. DOI: 10.1016/j.mencom.2013.12.004.


What is this brown liquid that I got when I boiled glycerol and water solution?

If your mixture reached 160 degrees C then you must have boiled off all the water. If there was some water left, it would stay at around 100 degrees C and all (most of) the heat would be used to evaporate water.

The boiling point of glycerol is about 290 degrees C and so you didn't boil that off. I disagree with @Skeptical Chymist about you making the caramel (for the sake of god, don't eat it to test it out). Not all that gets brown is caramel and there were no sugars there I presume. If i had to guess I'd say that at 160 C you've managed to completely obliterate any complicated organic structures and what's left is glycerol, charcoaled bits of any organic impurities that you happened to have in the starting material and possibly some acrolein which is a common product of dehydration of glycerol (you can recognize it by it's smell).

Overall; it's hard to tell what one gets, when they heat the hell out of something.


Find the final temperature and volume of the gas

The process should indeed be considered irreversible, as the problem statement gives a rather explicit hint stating that the pressure was released suddenly, so that your 3rd suggestion is correct. With an irreversible expansion against a constant external pressure, the adiabatic equation $pV^\gamma = \text{const}$ is inapplicable, so one must use the first law of thermodynamics. Work is done at the expense of the loss of internal energy:

$$W = - \Delta U = nC_\mathrm{v}(T_1 - T_2) \tag{1}$$

For ideal monoatomic gas the following is applied:

$$n = \frac{p_1V_1}{RT_1} \tag{1.1}$$
$$C_\mathrm{v} = \frac{3R}{2} \tag{1.2}$$

so that (1) can be rewritten as such:

$$W = nC_\mathrm{v}(T_1 - T_2) = \frac{3p_1V_1R}{2RT_1}(T_1 - T_2) = \frac{3p_1V_1}{2} - \frac{3p_1V_1T_2}{2T_1} \tag{1.3}$$

At the same time work by gas expansion is

$$W = p_2(V_2 - V_1) = p_2V_2 - p_2V_1 \tag{2}$$

and from the ideal gas law the reduced product is

$$p_2V_2 = nRT_2 \tag{2.1}$$

so that (2) can also be rewritten:

$$W = p_2(V_2 - V_1) = nRT_2 - p_2V_1 \tag{2.2}$$

I would leave for OP the task to to equate (1.3) and (2.2) and express the final temperature $T_2$ from there:

$$\frac{3p_1V_1}{2} - \frac{3p_1V_1T_2}{2T_1} = nRT_2 - p_2V_1$$

Here I bring up the final result:

$$T_2 = \frac{3p_1V_1 + 2p_2V_1}{5p_1V_1} T_1 = \frac{3 \cdot \pu{10 atm} \cdot \pu{10 L} + 2 \cdot \pu{1 atm} \cdot \pu{10 L}}{5 \cdot \pu{10 atm} \cdot \pu{10 L}} \cdot \pu{273.15 K} = \pu{174.7 K} \tag{3}$$

Volume after expansion is also to be found using ideal gas law and (1.1):

$$V_2 = \frac{nRT_2}{p_2} = \frac{p_1V_1RT_2}{RT_1p_2} = \frac{p_1T_2}{p_2T_1}V_1 = \frac{\pu{10 atm} \cdot \pu{174.7 K}}{\pu{1 atm} \cdot \pu{273.15 K}} \cdot \pu{10 L} = \pu{64.0 L} \tag{4}$$


Elucidating reaction mechanism and purpose

I have no idea what the addition of benzyl bromide accomplishes. There doesn't seem to be any need for a protecting group in any of the subsequent reactions. The rest of the reactions are relatively easy to explain.

This paper reports essentially 100% yield for bromination of 4-hydroxyacetophenone with $\ce{CuBr2}$, refluxed in chloroform-ethyl acetate for 15-20 mins. There is no mention of any need for a protecting group for the phenol. No mechanism is given but you are correct in your assumption that radicals will be involved.

The second reaction is a regular $S_N2$ reaction. The transition state of the reaction is highly stabilized by the adjacent carbonyl group and so the reactivity is dramatically increased compared to a regular primary halide. 18-crown-6 is used to solvate the $\ce{K+}$ cations, for which it has a very high affinity. This leaves the fluoride free to react, making it far more nucleophilic than it would otherwise be. The reaction will go to completion easily because bromide is a far more stable leaving group than fluoride.

The final reaction is a standard ether cleavage, catalysed by a strong acid. The oxygen is protonated by the acid and leaves to reform the phenol and the stabilized benzyl cation.


Calculating the electronegativity of an element from bond dissociation energies

Considering the data you have been given, the equation you need is the one that defines the Pauling electronegativities:

$$\left|\chi_\ce{A} - \chi_\ce{B}\right|=\left(\text{eV}\right)^{-1/2}\sqrt{E_\mathrm d(\ce{A-B})-\frac{E_\mathrm d(\ce{A-A})+E_\mathrm d{\ce{(B-B)}}}{2}}$$

In the equation, $\left(\text{eV}\right)^{-1/2}$ is in the equation to negate the units of the bond dissociation energies. This means that you need to convert all your bond dissociation energies from $\pu{kcal/mol}$ into $\pu{eV}$. You can use the BDEs in units of $\pu{kcal/mol}$, but then the formula will require a constant in front to adjust for the difference between units.

$\pu{1 eV}$ is equal to $\pu{1.602 \times 10^-19 J}$, so you need to be able to convert from $\pu{kcal/mol}$ to $\pu{J/mol}$, then divide through by Avogadro's constant $N_\mathrm A = \pu{6.022 \times 10^19 mol-1}$. There is a handy website which can help with this.

$$\begin{align}
E_\mathrm d(\ce{H-H}) &= \pu{4.519 eV} \\
E_\mathrm d(\ce{F-F}) &= \pu{1.59 eV} \\
E_\mathrm d(\ce{H-F}) &= \pu{5.837 eV}
\end{align}$$

Plugging the information you have been given, we find that

$$\begin{align}
\left|2.1 - \chi_\ce{F}\right| &= \sqrt{5.837 - \frac{4.519+1.59}{2}} \\
&= 1.668
\end{align}$$

This gives you two possibilities for $\chi_\ce{F}$, $0.4$ or $3.8$, depending on whether the quantity $(2.1 - \chi_\ce{F})$ is positive or negative. This is where we need to use some knowledge of chemistry: fluorine is more electronegative than hydrogen, so we should have $\chi_\ce{F} > \chi_\ce{H}$, and consequently we take the solution $\chi_\ce{F} = 3.8$.

Why should bond energies be related to electronegativity? The idea that Pauling had when he devised this equation was that in a covalent bond between two of the same atom, there is no electronegativity difference and hence no ionic contribution to the bond strength. Hence, the "average covalent contribution" of each atom to a bond could be measured by the mean of the two bond strengths, i.e. $[E_\mathrm d(\ce{A-A}) + E_\mathrm d(\ce{B-B}) ]/2$.

However, when there is a difference in electronegativity, the bond strength has an ionic contribution. The larger the difference in electronegativity $\left|\chi_\ce{A} - \chi_\ce{B}\right|$, the larger the ionic contribution. This ionic contribution could be roughly measured by subtracting the covalent contribution, $[E_\mathrm d(\ce{A-A}) + E_\mathrm d(\ce{B-B}) ]/2$, from the $\ce{A-B}$ bond strength.


Heating p-Bromobenzenesulfonic acid in the presence of dilute HCl

I think @ron has hit the bull's eye in the comments! I would like to elaborate upon that:

Since each step in the sulfonation of benzene is an equilbirum, sulfonation is a reversible reaction. According to Le-Chatlier's principle addition of $\ce{H+}$ ions in the solution would shift the equilibrium to the left in the last step of sulfonation where arenium is deprotonated, hence addition of $\ce{HCl(aq)}$ accompanied with heating for kinetic requirements results in the production of 4-bromobenzenesulfonic acid in the given reaction.  This process is called protodesulphonylation.


Relationship between surface area of electrode and reaction rate of hydrogen in fuel cells

Any galvanic cell produces its equilibrium voltage at open-circuit.  Let's say your cell is at +3 V before you allow any current to flow through it.  As soon as you throw the switch, you allow that 3 V to push a current through the cell and some external load that you connect to it. There are several resistances within the cell that act to reduce the cell voltage from the initial 3 V.  The electrolyte resistance, concentration dependent resistances at the electrodes and electron transfer resistances all contribute to the total voltage drop in the cell, and hence, effectively limit the amount of current it can generate.  Typically, one of these resistances is much larger than any of the others depending on the exact configuration and operating conditions of the cell.  It is the electron transfer resistance that you are concerned with in your question.  The traditional way of expressing the voltage drop due to electron transfer resistance is the Tafel equation:

$$\eta _{a}=a+b\, log\, i$$

where $a$ and $b$ are empirical constants, $i$ is the current density and $\eta _{a}$ is what is called the activation overpotential.  We see from this expression that the amount of voltage drop in the cell due to electron transfer increases linearly with the logarithm of the current density.  It is desirable then to minimize $\eta _{a}$ for a battery or flow cell application so that we have the maximum amount of voltage available for pushing current through the load.  A very good way to do this, as you pointed out, is to simply increase the electrode area.  By doing so, we decrease the current density which decreases $\eta _{a}$.

So, the exact mathematical relationship you seek would be the empirical Tafel expression given above, or even better, the Butler-Volmer equation which is really just a more fundamental way of expressing the Tafel relationship.  All you should need to do is locate some references on platinum polarization in aqueous solutions.  There just happen to be tons of papers on this subject.  Here is one to get you started:

  Sheng W. et al, Journal of The Electrochemical Society, 157 11
  B1529-B1536 2010


Azathioprine CIF data file rendering problem in Accelrys

The coordinates of the hydrogen atoms of the methyl group in question are there (H16a, H16b, H16c).

The problem is a rendering problem of the Accelrys visualizer.

Other tools don't have a problem with this CIF file. The image below is a medium-quality PovRay rendering generated through PLATON.


The reason is indeed $\text d$-to-$\text d$ transitions. The d orbitals of both $\ce{CuCl}$ and $\ce{CuCl2}$ have two possible energy levels (based on crystal field theory).

$\ce{Cu(II)}$ is a $\text{d}^{9}$ metal, and $\ce{CuCl2}$ exists as a (distorted) octahedral around the copper atom. Based on crystal field theory, the d orbitals of an octahedral metal center can be represented as shown below. In this case, because $\ce{Cu(II)}$ is $\text{d}^{9}$, the three lower energy suborbitals are fully filled in the ground state, and but the top two suborbitals only contain a combined three electrons. Therefore, the energy required to excite that electron is $\Delta _o$. When that electron returns to the ground state, it gives off a different wavelength of light than it absorbed (in this case, the light emitted is apparently not visible light).

$\ce{Cu(I)}$, on the other hand, is a $\text{d}^{10}$ metal ion, so each of the d suborbitals is fully filled. In order for an electron to be excited in this situation, it would have to leave the 3d orbital (and the $n=3$ shell altogether) and enter the $n=4$ shell. This is rather large  energy jump compared to the intraorbital jump an electron in $\ce{CuCl2}$ has to make. Therefore, $\ce{CuCl}$ does not readily absorb visible light as $\ce{CuCl2}$ does, so it appears colorless. That is, until it see moist air.


Unknown anion (one carbon and two nitrogen atoms, charge 2−)

The parent molecule is cyanamide.

The ion is usually referred to as either the cyanamide dianion or the carbimide dianion


Baking soda bath to deacidify newsprint?

It seems that adding bicarbonate is how acid-free paper is made, though more modern paper is neutralized by the use of chalk as filler. The wiki page says that magnesium or calcium bicarbonate are commonly used, but that may be just due to cost rather than any specific issues with using sodium bicarbonate. Source


Why does the concentration of hydrogen ions increase in the electrolysis of dilute sulphuric acid?

Replace the anode equation with your new equation. Then find the overall equation for the reaction. You are correct that the net number of $\ce{H+}$ is not changing as a result of this process.  However, something is being consumed. What is it? How does that change the concentrations of the other species present?


How to figure out which substance has the highest osmotic pressure?

$\ce{MgCl2}$ has the highest osmotic pressure because it breaks up into three ions in water, as opposed to all the others, which only break up into two.


The reaction is used industrially to make hydroquinone (source: wikipedia), but often using hydrogen peroxide as the oxidation agent in the presence of a catalyst.
A study on catalytic hydroxylation of benzene comes to the following conclusion:

experimental evidences have allowed first discarding Fenton-like mechanisms and later proposing the plausible competition between an electrophilic aromatic substitution pathway with the alternative rebound (hydrogen abstraction) route.

The Fenton-like mechanism refers to addition of a free oxygen radical species to benzene, which they reject based on experiments with radical scavengers. Instead, they hypothesize formation of a radical oxygen:catalyst complex. This either reacts with benzene in an electrophilic substitution, or abstracts a hydrogen from benzene, which then rebounds and reacts with the oxygen still attached to the catalyst.
Although the study I am citing is for hydroxylation of benzene instead of phenol, the paper also talks about substituted benzenes.

[OP:] My guess is that the double bond of phenol will react with the oxidizing agent, but not sure of how this will work.

You can't pin down a single double-bond in an aromatic system. What you can say is that the existing hydroxyl group will steer the reaction to make an ortho- or para- product. Once hydroquinone is formed, it oxidizes quickly to benzoquinone under the oxidizing conditions.

[OP:] What is the mechanism of oxidation of phenol to benzoquinone?

For the reactions conditions given by the OP, I did not manage to find a mechanism. However, the mechanisms I discussed probably are sufficiently general that they apply in this special case as well. That is just a guess, though.


Which Lewis Structure for BeCl2 is more commonly seen?

Beryllium chloride has a significant ionic contribution to its electronic structure. A quick calculation, DF-BP86/def2-SVP, reveals that the charge of beryllium is $\mathbf{q}(\ce{Be})=0.8$ and of chlorine $\mathbf{q}(\ce{Cl})=-0.4$, based on natural bond orbital analysis.

The Lewis structures best describing this are depicted below. The percentage indicates how much Lewis character this model actually describes based on the before mentioned calculation.

Of course this is just an approximation and the truth lies somewhere in between, like Eljee already mentioned. For these bonding situation, resonance is unavoidable. In quantum chemistry these phenomena are usually very well described by valence bond theory.

Another possibility, which is quite complementary, is looking at the structure with molecular orbital theory. Since the molecule has $D_\mathrm{\infty{h}}$ symmetry, you can observe $\pi$ bonds, that are delocalised over the whole molecule. And these are only depicted in the right hand side structure above. Below are the valence orbitals of $\ce{BeCl2}$, DF-BP86/def2-SVP.

So you can see, that one Lewis structure is not enough to actually describe the bonding in that molecule well, but as a first approximation, the left hand structure above should be sufficient.


Why do elements change their colour in reactions?

Elements don't have inherent colors at all; substances do. Furthermore, while there are situations when some compounds (and even copper sulphate) change their colors, this is most definitely not one of them. Here copper sulphate is blue and never changes to pale green. It's just the reaction... wait, didn't you write the equation yourself? See, in the end there is no more copper sulphate. It is gone, and so is its color. What you see now is the color of $\ce{FeSO_4}$, or more precisely, of $\ce{Fe}^{2+} (aq)$.

As to why compounds have certain colors... well, this is a really huge body of knowledge. Long story short, color is caused by absorption of light with some specific wavelength. What wavelength is it, is determined by possible electron transitions within the substance in question. The transition may be between some molecular orbitals, if the compound is molecular. Or it may be between the orbitals of some metal ion, as is the case in your example. In many cases you may be able to predict the color of a compound you have never seen, using simple techniques.

Come to think of it, this question must have been asked before, and surely there is already a nice answer with lots of pics and stuff.


Is 2-oxobicyclo[2.2.1]heptan-1-ide resonance stabilized?

I think, in contrast to cyclohexanone, a resonance stabilization by

is less likely to happen as you already start with a bi-cyclic compound where the cycles involved are small to moderate. The then additional strain introduced by a (partial) double bond at the bridge-head carbon atom would destabilize the molecule too much.  I extrapolate here from Bredt's rule.


solve steady state of simple product reaction

There are $3$ equations for $3$ unknowns $[E],[S],[ES]$ given by

$$\left\{\begin{array}{l}
k_f[E][S]=k_r[ES],\\
[E]+[ES]=E_0,\\
[S]+[ES]=S_0.
\end{array}\right.$$

We can use the last two equations to cancel $[E]=E_0-[ES]$ and $[S]=S_0-[ES]$ and plug them into the first equation to get a quadratic equation for $[ES]$ given by

$$k_f(E_0-[ES])(S_0-[ES])=k_r[ES].$$

Then we plug in numbers $k_r/k_f=1$, $E_0=10$, $S_0=5$ to obtain $[ES]=8\pm\sqrt{14}$. Since we have $[ES]\leq\min\{E_0,S_0\}=5$, the solution is $[ES]=8-\sqrt{14}\approx 4.26$.


Finding molar mass of uranium fluoride using Graham’s law of diffusion

Graham's law of diffusion deals with the molar effusion rate rather than mass effusion rate.
This can be confirmed by the description of Graham's Law on Wikipedia.
An excerpt from the aforementioned page,

$$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$$
where:
$r_1$ is the rate of effusion for the first gas (volume or amount per unit time)
$r_2$ is the rate of effusion for the second gas
$M_1$ is the molar mass of gas 1
$M_2$ is the molar mass of gas 2

Now the mathematical part,
$$\begin{align}
r_{\ce{UF6}} &= \pu{17.7 mg/h} = \frac{\pu{17.7 mmol/h}}{M_{\ce{UF6}}} \\
r_{\ce{I2}} &= \pu{15 mg/h} = \frac{\pu{15 mmol/h}}{M_{\ce{I2}}}
\end{align}$$
By Graham's Law,
$$\begin{align}
\frac{r_{\ce{UF6}}}{r_{\ce{I2}}} &= \sqrt{\frac{M_{\ce{I2}}}{M_{\ce{UF6}}}} \\
\frac{17.7}{M_{\ce{UF6}}} \cdot \frac{M_{\ce{I2}}}{15} &= \sqrt{\frac{M_{\ce{I2}}}{M_{\ce{UF6}}}} \\
\frac{17.7}{15} &= \sqrt{\frac{M_{\ce{I2}}}{M_{\ce{UF6}}}} \cdot \frac{M_{\ce{UF6}}}{M_{\ce{I2}}} \\
\frac{17.7}{15} &= \sqrt{\frac{M_{\ce{UF6}}}{M_{\ce{I2}}}}\\
M_{\ce{UF6}} &= \left(\frac{17.7}{15}\right)^2 M_{\ce{I2}}\\
  &= \pu{353.39 g mol^-1}
\end{align}$$


Why is HCl not considered to have hydrogen bonding?

Very, very related. Check it for a more in depth answer.

Well, first-of-all you need to understand that the $\ce{H}$ bonding isn't actually bonding. It is just a covalent attraction. Also since $\ce{Cl}$ is larger than $\ce{N}$, $\ce{F}$ and $\ce{O}$ it does not make a strong $\ce{H}$ bond. The size of the $\ce{Cl}$ makes the dipole-dipole attraction weaker. However $\ce{N}$, $\ce{F}$ and $\ce{O}$ are smaller and thus have an $\ce{H}$ bond.

Although in reality, compared to other covalently bonded structures, $\ce{HCl}$ has a very strong covalent bond.


How can electron move between two metals without the presence of electric field?

The wording of the question hides the answer.

Electrons do not move from Zn to Cu. They move from neutral zinc to cupric ion Cu++. The field arises from the mix of H2O, H+, OH-, SO4--, and Cu++ all around the Zn rod. Could electrons be forced onto Zn from OH- or SO4-- ? No.There is no gain in stability; energy would be needed. Or could electrons be attracted to H+; yes, and even more so to Cu++. And the final products (Zn++ and neutral Cu) would be more stable.


As for acidity of boric acid you can find on https://en.wikipedia.org/wiki/Boric_acid that it dissociates in both ways, but is very weak. However it's acidic enough that it's only a matter of time until it would become almost completely deuterised if dissolved in D2O. It happens because in mixture there would be much more deuterium then normal hydrogen. The acid is too weak to capture or detect its anions. Maybe kinetics of deuterisation would give us some info about properties of its anions.


Why is weight of 1 mole of substance equal to atomic/molecular mass in grams?
According to me, it happens because mole has been defined in such a way.

Yes! That is correct.

It is defined as the numbers of particles in 12 g of C12. If it were 24 g, instead of 12 g, then the weight of 1 mole of substance would equal 2 times the atomic/molecular mass in grams.

Also correct, assuming that the definition of unified atomic mass units (amu) remained the same. @Martin's answer is correct, but we can also arrive at the same conclusion using a simple dimensional analysis approach.
First we need the definition of an amu:
$$\pu{1 atom}~ \ce{^12C} = \pu{12 amu}$$
Now take the real definition of a mole:
$$\pu{1 mol}~ \ce{^12C} = \pu{12 g}$$
Now, divide the first equation by the second:
$$\frac{\pu{1 atom}~ \ce{^12C}}{\pu{1 mol}~ \ce{^12C}} = \frac{\pu{12 amu}}{\pu{12 g}}$$
Cross-multiply and reduce:
$$1~\frac{\pu{g}}{\pu{mol}~ \ce{^12C}} = 1~\frac{\pu{amu}}{\pu{atom}~ \ce{^12C}}$$
What this tells us is that the ratio of g/mol to amu/atom is exactly one - and we made sure it would work out that way by carefully choosing how we defined moles and amus.
Since the masses of all elements are defined in terms of amu, which are ultimately based on the mass relative to carbon-12, this ratio holds for all any atom or molecule.
Let's take it further and put in your hypothetical doubling of the number of grams of carbon-12 per mole:
$$\frac{\pu{1 atom}~ \ce{^12C}}{\pu{1 mol}~ \ce{^12C}} = \frac{\pu{12 amu}}{\pu{24 g}}$$
Now when we cross-multiply and reduce, we get:
$$2~\frac{\pu{g}}{\pu{mol}~ \ce{^12C}} = 1~\frac{\pu{amu}}{\pu{atom}~ \ce{^12C}}$$
Since we kept the definition of an amu the same, but changed that of the mole, what we see is that 1 mole of carbon-12 using these units would have to weigh 24 g (the molar mass), even though its atomic mass would still be 12.
That's a lot harder to keep track of mentally, and so it is a good thing for us that the definitions of amus and moles were chosen so carefully!


One half of a nail is exposed to vinegar. Un-exposed half develops a brown film. What is it?

The nail is in two solutions. The lower solution is 5% acetic acid, saturated with air (20% $O_2$, 80% $N_2$). At 0.2 atmospheres, $O_2$ solubility in water is 40 mg/L x 0.2 = 8 mg/L, or about 8 parts per million.
The upper solution is a thin film on the nail (perhaps 10 microns thick - or less!) of $H_2O$ with some acetic acid (probably about 3-4%) and 8 ppm $O_2$. This film may have deposited from the lower solution by vapor deposition due to temperature differences, or by capillary action, as the meniscus crept up in a thin layer.
In the lower solution, $Fe^0$ goes into solution as $Fe^{++}$ and electrons go to $O_2$.
4$e^-$ + $O_2$ + 2 $H_2O$ --> 4 $OH^-$
The 8 ppm $O_2$ in the lower solution doesn't go very far, but the meniscus, and any film going higher up, will be continually refreshed with $O_2$, and will develop a higher pH and will precipitate iron hydroxide. Any iron dissolving in the lower solution will stay in solution because of the acidity, and the nail may not even show obvious signs of corrosion damage like pitting. The only obvious sign is the brown film of precipitated iron hydroxide.
This is an example of a differential aeration cell, where the upper solution is an oxygenated cathode, and the lower solution contains the anodic portion of the nail. http://www.corrosion-doctors.org/Corrosion-Factors-Cells/corrosion-cells-aeration.htm
We consider this corrosion in acetic acid to be an electrolytic reaction, and also the reaction of the nail with $HCl$, but the scale of cathodes and anodes in the case of $HCl$ is so small that we can't detect them. And in $HCl$ the $H_2$ is generated so fast that it bubbles off and doesn't need to be depolarized by dissolved oxygen.
Adding a picture suggesting the state of the nail:

https://www.ddcoatings.co.uk/2343/what-is-differential-aeration-corrosion.


A Curious (definitely wrong) result of Absolute Entropy

The Gibbs free energy change is zero in the case of reversible processes carried out at constant temperature and pressure, but that isn't the case if these conditions are not observed.
As a demonstration consider an isothermal reversible expansion of an ideal gas.
Since the temperature is constant, the free energy change is given by
$$\Delta G = \Delta H - T \Delta S$$
Because the gas is ideal and T constant, $\Delta U = 0$, and from the second law of thermodynamics it follows that
$$\Delta S = \frac{q_\text{rev}}{T}= -\frac{w_\text{rev}}{T} = \frac{\int_{V_1}^{V_2}p dV}{T} = \frac{\int_{V_1}^{V_2} \frac{nRT}{V} dV}{T} = nR\log \left( \frac{V_2}{V_1} \right)$$
It also follows that
$$\Delta H = \Delta U + \Delta (pV) = 0 + \Delta (nRT) = RT \Delta n = 0$$
since the system is closed.
But then
$$\Delta G = - nRT\log \left( \frac{V_2}{V_1} \right) \neq 0$$
The free energy change is not zero because the pressure changed during the process.
For the reversible adiabatic expansion of an ideal gas
$$\begin{align}  \Delta S &= 0  \\ \Delta H &= \int_{T_1}^{T_2} C_p dT  = C_p \Delta T \\ \Delta G &= \int_{P_1}^{P_2} V dP - \int_{T_1}^{T_2} S dT =  (V_1P_1^{C_V/C_P})\left(\frac{C_P}{C_P-C_V}\right)( P_2^{1-C_V/C_P}-P_1^{1-C_V/C_P}) - S\Delta T  \end{align}$$
The integral over P in the expression for $\Delta G$ can be evaluated after substituting the following expression for the volume in terms of pressure:
$$V=(V_1P_1^{C_V/C_P}) P^{-C_V/C_P}$$


Spectrochemical series of TMs: why does the field-splitting parameter NOT steadily increase along a period?

A greater octahedral field split would be associated with a shorter $\ce{L\bond{->}M}$ bond length. Bond length is a function of:

size of both partners
strength of the bond

In particular, as we are examining metals in $\mathrm{+II}$ oxidation state, the metal ion’s size should increase somewhat steadily along the period while the ligand size should stay the same. This means that we need to be looking at bond lengths to understand the discrepancy.

For reference, here is a typical octahedral $\ce{[ML6]}$ complex ignoring π effects.

Figure 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.

We should immediately be able to see how field split is generated: it is the difference between the $\mathrm{t_{2g}}$ orbitals and the $\mathrm{e_g^*}$ orbitals; the latter of which are antibonding. Thus, having any electrons in $\mathrm{e_g^*}$ will weaken the $\ce{L\bond{->}M}$ bonds. This corresponds nicely with vanadium(II), a $\mathrm{d^3}$ ion, having the largest field split, and manganese(II) ($\mathrm{d^5}$) having the lowest.

However, the final value is dependent on very subtle differences and not easily predicted a priori.


Why do Grignard reagents act as sources of nucleophilic R− and not Br−?

In all likelihood the halide ion does attack the substrate.  But, the product such a reaction mode would form does not accumulate whereas the alkylated product does.
Recall that in a nucleophilic reaction halide ions are good leaving groups, at least if you stay away from fluorides which are rarely used for Grignard reagents.  So while the halide ion may add to the substrate, it may also be displaced if an alkyl-anion moiety from another Grignard molecule comes in, or possibly the halide ion may just launch on its own to attach to a magnesium atom (which, in a Grignard reagent, acts as an electrophilic center, for example by adding to the oxygen end of a carbonyl group).
In contrast, when the alkyl-anion moiety attacks, the carbon-carbon bond it forms is strongly favored thermodynamically and a lot harder to break; here we do not have a good leaving group.  So the alkylated product rather than the halogenated product is what accumulates.


What are the resonance structures for chlorite anion?

Disclaimer
The here used methods are also only approximations and I do not claim that the provided numbers are accurate. However, they are used here to make the point, that bonding situations involving resonance are not easily explained.

Oxygen is slightly more electronegative than chlorine, therefore the bonds will always be slightly polarised towards the oxygen.

The following resonance scheme is based on an NBO analysis on the DF-BP86/def2-SVP level of theory. The percentage value gives an estimate how close the given conformation is to an idealised Lewis structure. From this you could also derive how much such a structure would contribute to a resonance situation (in terms of Valence Bond theory).
What we do see here is, that the electronic structure is better described with more formal charges. The contribution of a π bond is best described by an ionic donor-acceptor structure, rather than breaking the octet rule at chlorine.
Noteworthy is also that the doubly ionic donor-acceptor configuration is as equally good a lewis structure than any that breaks the octet rule.
The worst representation is the one with a four electron three centre bond (which is also not part of the Lewis bonding picture anyway).

Videos like the one mentioned in the Wikipedia article are easy to follow, but often they are too simplified to be correct.


Why do we only consider relative motion of hydrogenic atoms?

We can do that because the variables have separated. The Hamiltonian of relative motion depends on the relative coordinates. The translational part depends on the coordinates of the center of mass. The two never mix. They don't interact; they don't even care about each other.

Separation of variables in various guises is an immensely powerful technique of solving PDE. You will encounter it time and again, including this very chapter of whatever textbook you are using, just a few lines below, when you'll start to actually solve the equation for hydrogen atom by splitting it into radial and angular parts.

So it goes.


Crystal Field Explanation of Colors of Coordination Compounds

Your assumption that the electron emits a photon when transitioning back to the ground state is not entirely correct.

The excited electron can also transition back by transferring the excitation energy to heat energy. The compound then warms up ever so slightly and the electron is buzzing away happily in the lower $d$ orbitals.

As to your P.S.-ed question, I'll quote Wikipedia:

  The main factors that cause broadening of the spectral line into an absorption band of a molecular solid are the distributions of vibrational and rotational energies of the molecules in the sample (and also those of their excited states). [...] In gas phase spectroscopy, the fine structure afforded by these factors can be discerned, but in solution-state spectroscopy, the differences in molecular micro environments further broaden the structure to give smooth bands.


Why isn't the scandium(2+) cation isoelectronic with potassium?

The neutral $\ce{Sc}$ atom has the electron configuration of $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^1 4s^2}$, and the $\ce{Sc^2+}$ ion has electron configuration $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^1}$.
The neutral potassium atom has the electron configuration of $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 4s^1}$. While both $\ce{Sc^2+}$ and $\ce{K}$ have the same number of electrons, their electron configurations are different and therefore they are not isoelectronic to one another. (Isoelectronic implies same number of electrons as well as the same structure)


What is the necessary volume of concentrated HCl to prepare a solution with a pH of 1.8?

Unfortunately no, you're not right. As mentioned in a now deleted comment by Bruno, your answer gives a $\mathrm{pH}$ of $-0.74$, which is not correct. You didn't give detailed steps of your calculations, and your first step doesn't seem like anything we should do.

We can solve this by making the calculations step by step. Let's start with the concentration of $\ce{H+}$, which will be the same as the concentration of $\ce{HCl}$, we will achieve:

$$\ce{[H+]} = 10^{-\mathrm{pH}} = \pu{0.01584893 mol//L}$$

Since we're going to prepare 5 liters, this means the amount of substance of $\ce{H+}$ should be

$$5 \times \pu{0.01584893 mol} = \pu{0.07924466 mol}$$

To transform this into amount of $\ce{HCl}$, we need the acid's molar mass ($\pu{36.46 g//mol}$) to convert from amount of substance to mass:

$$\require{cancel}%
g_{\ce{HCl}}
= \mathrm{\frac{36.46\ g}{\cancel{mol}}}
  \times 0.07924466\ \mathrm{\cancel{mol}}
= 2.88926 \ \mathrm{g}$$

Now that we have the amount of substance of $\ce{HCl}$, we calculate the volume, which is where the density comes in:

$$V_{\mathrm{HCl}}
= \mathrm{\frac{2.88926\ \cancel{g}}{1.179\ \cancel{g}\ mL^{-1}}}
= 2.450602 \ \mathrm{mL}$$

Now the last step: Our stock solution isn't pure $\ce{HCl}$, it's diluted in water. So actually the volume we need is:

$$V_{\mathrm{HCl}}
= 2.450602 \times \frac{100}{36}
= \pu{6.807229 mL}
\approx \pu{6.8 mL}$$

This might look that it is not much, but remember $\ce{HCl}$ is a very strong acid. For example, to prepare $\pu{1000 mL}$ of a $\pu{0.001 M}$ solution you only need $\pu{0.086 mL}$ of concentrated $\ce{HCl}$!

You can check this result by imputing the given density, weight percentage, final volume and $\ce{H+}$ concentration (first step) into Sigma-Aldrich's Molarity Calculator.


The process that transforms sugars (e.g. the ones in grape juice) into ethyl alcohol and then this latter into acetic acid is a biological one, that is, performed by micro-organisms. The first step is usually performed by yeasts such as Saccharomyces Cerevisiae, whereas the second step is performed by bacteria of the Acetobacter genus. Thus, your grape juice won't change to vinegar if you mix both liquids. All you will get is a sweeter vinegar, or a sour grape juice.


Lewis structures and skeletal structures show which atoms are connected to which other atoms, and the type of bonds (single, double, triple). They do not show the shape (geometry) of the molecule.

For example, the two skeletal structures below refer to the same molecule, butane:

When there is ambiguity (i.e. multiple molecules with the same skeletal structure), you can add information about the stereochemistry by adding dashes bonds and wedges. In your case, the molecule is achiral, so that does not apply.

To figure out the geometry, remember that carbon atoms with four single bonds have a tetrahedral geometry. You get other geometries when there is a double bond (trigonal planar) or a triple bond (linear) or other, more unusual features associated with a carbon atom (lone pair, non-zero formal charge, radicals, two double bonds).


What would be the enthalpy change for a isothermal expansion?

You have already mentioned that the change of internal energy is zero $(\Delta U=0)$ since, for an ideal gas, the internal energy $U$ only depends on amount of substance $n$ and temperature $T$, and  in a closed system $n$ is constant $(\Delta n=0)$ and during an isothermal process also $T$ remains constant $(\Delta T=0)$.

Enthalpy $H$ is defined as

$$H=U+pV$$

and the ideal gas law states that

$$pV=nRT$$

Thus

$$H=U+nRT$$

Since $n$ (closed system) and $T$ (isothermal process) are constant, the product $nRT$ is constant, and therefore, according to the ideal gas law, also the product $pV$ is constant.

Furthermore, since $U$ is constant during the given process, the sum $H=U+nRT$ as well as  $H=U+pV$ has to remain unchanged (answer d).

Note that your assumption

$$\Delta H = \Delta U + W$$

(where $W=p\Delta V$ is the pressure-volume work) only applies to processes under constant pressure $p$, since enthalpy is defined as

$$H=U+pV$$

and thus

$$\begin{align}
\mathrm dH&=\mathrm dU+\mathrm d(pV)\\
&=\mathrm dU+V\mathrm dp+p\mathrm dV
\end{align}$$

which simplifies to

$$\mathrm dH=\mathrm dU+p\mathrm dV$$

at constant pressure $(\mathrm dp=0)$.

However, the pressure does not remain constant during the process given in the question.


Which is more likely to form: silver chloride or silver phosphate?

The key concept here is solubility products which are a form of chemical equilibria. Once you have understood that, you will realise that it depends on the ions’ concentrations and relative solubilities which one will precipitate first.

In general, you have to consider the following principal equilibria:

$$\ce{Ag3PO4 v + 3 Cl- (aq) <=>[$k_1$] 3 Ag+ (aq) + 3 Cl- (aq) + PO4^3- <=>[$k_2$] 3 AgCl v + PO4^3-}\tag{1}$$

and futher ones that derive from the acidity equilibrium of the phosphate anions as well as the formation of silver hydroxide at higher $\mathrm{pH}$ values. I will choose to ignore these.

Equilibrium $k_1$ is actually only the precipitation of silver phosphate; we might instead write it as:

$$\ce{3 Ag+ (aq) + PO4^3- (aq) <=>[$k_1$] Ag3PO4 v}\tag{2}$$

The equilibrium constant is thus:

$$K_1 = \frac{a(\ce{Ag3PO4})}{a(\ce{Ag+})^3 \times a(\ce{PO4^3-})}\tag{3}$$

We are dealing with activities here; the activity of a solid is, by definition, always $1$.
This equilibrium is usually given in the inverse form in which the dissolved ions are the product and the solid salt is the reactant; in that case $K_1{}^{-1}$ is what you might know as the solubility product.

The same discussion can be made for the other equilibrium:

$$\begin{gather}\ce{Ag+ (aq) + Cl- (aq) <=> AgCl v}\tag{4}\\[1em]
K_2 = \frac{a(\ce{AgCl})}{a(\ce{Ag+}) \times a(\ce{Cl-})}\tag{5}\end{gather}$$

Again, the more common form is $K_2{}^{-1}$ which is also known as the solubility constant.

Knowing the values of both $K$ as well as the initial concentrations of all ions will allow you to calculate in which direction the reaction goes although the calculation may be tedious.


You need to show that $$\left(\frac{\partial U}{\partial V}\right)_T=-P+T\left(\frac{\partial P}{\partial T}\right)_V$$


Difference in acid strength of oxalic acid and malonic acid

Linear saturated dicarboxylic acids
$n$ = number of $\ce{-CH_2}-$ groups in chain
 n                        pKa1     pKa2   Ka1/Ka2
 0  Oxalic acid           1.25     4.14   776
 1  Malonic acid          2.83     5.69   724
 2  Succinic acid         4.21     5.41    15.8
 3  Glutaric acid         4.34     5.41    11.7
 4  Adipic acid           4.41     5.41    10.0
 5  Pimelic acid          4.50     5.43     8.51
 6  Suberic acid          4.526    5.498    9.4
12  Dodecanedioic acid    4.45     5.05     4.0

Dodecanedioic acid
Aromatic carboxylic acids
                                   pka1     pka2    pka3
 benzoic acid                      4.202    ----    ----
 ortho-phthalic acid               2.89     5.51    ----
 meta-phthalic acid                3.46     4.46    ----
 para-phthalic acid                3.51     4.82    ----
 o,o'-bibenzoic acid
 p,p'-bibenzoic acid
 benzene-1,3,5-tricarboxylic acid  3.12     3.89    4.70

Linear unsaturated dicarboxylic acids
$n$ = number of $\ce{-CH=CH}-$ trans groups in chain
n                            pka1      pka2
0  Oxalic acid               1.25      4.14
1  Fumaric acid              3.02      4.38
2  trans,trans-Muconic acid  3.87
3

Straight-chain, saturated carboxylic acids
$n =$ total number of carbon atoms in molecule ($n \geqslant 2$ has terminal methyl group).
n                     pKa
1  Formic acid         3.77
2  Acetic acid         4.76
3  Propionic acid      4.88
4  Butyric acid        4.82
5  Valeric acid        4.82
6  Caproic acid        4.88
15 Pentadecanoic acid  4.8

most $\mathrm{p}K_\mathrm{a}$ data from Wikipedia.

How does the inductive effect plays different roles in both acids?

First think of what happens when the carboxylic acid group of a straight-chain, saturated carboxylic acids ionizes to form the carboxylate ion. The $\ce{C=O}$ and $\ce{-C-O^{-}}$ bonds of course hybridize. This hybridization occurs faster than hydrogen bonds to the solvent form. The net result is effectively two oxygen atoms with 1/2 a charge each. This increases the ability of the  carboxylate ion to solvate, which increases its acidity.
In malonic acid the $\ce{C-C}$ bond "shares" some of that hybridization. This is the "inductive effect." The additional delocalization of the electrons from the carbon atoms makes the carboxylate ions more electronegative than they would be in a long linear saturated dicarboxylic acid. So the first ionization ,  the second ionization


How small is the smallest known carbon ring containing only double bonds?

TL;DR: Isolated derivatives of cyclo-$\ce{C18}$ most likely have the smallest isolated cumulenic 18-membered ring, though upon direct complexation of transition metal in/outside the ring the strain as well as its size can be reduced further.

In the past two decades a research group of Prof. Dr. François Diederich performed numerous studies on cyclo[n]carbons and monocyclic $\mathrm{sp}$-hybridized C-rings. He also published several reviews, e.g. see chapters "Cyclic and Linear Acetylenic Molecular Scaffolding" in [1, p. 43], and more recent "Oligoacetylenes" in [2, p. 443] devoted to versatile chemistry of cumulenes.
As @R.M. referred in the comment, early theoretical work [3] predicted special Hückel-aromatic cyclo-$\ce{C18}$ compound stabilized via two orthogonal $(4n + 2)$ $\pi$-electron systems, but later computational studies predict somewhat controversial electronic structures (from [2]):

Self-consistent field (SCF) calculations with a 3-21G or larger basis set predicted that the cyclic acetylenic $D_{\mathrm{9h}}$ structure 1a with alternating bond lengths represents the ground-state geometry[4].
But optimizations at the Møller-Plesset second-order perturbation theory (MP2) level including valence electron correlations as well as density functional theory calculations favor the cumulenic $D_{\mathrm{18h}}$ structure 1b as the most stable planar monocyclic geometry [5].

More from [1, p. 46]:

Accordingly, cyclo-$\ce{C10}$ is predicted to prefer a cumulenic $D_{\mathrm{5h}}$ structure, cyclo-$\ce{C14}$ is borderline, and the energies of its cumulenic $D_{\mathrm{7h}}$ and polyynic $C_{\mathrm{7h}}$ structures are nearly degenerate. It is clear that this fascinating theoretical controversy over the past years can only be definitively solved with the synthesis and characterization of cyclo-$\ce{C18}$.

cyclo-$\ce{C18}$ appeared to be a very elusive compound to isolate (on macroscoping scale typical products are anthracene and polymers), therefore only its derivatives were synthesized and characterized. Synthetic methods include the following paths: Retro-Diels-Alder; 3-Cyclobutene-l,2-dione route;  transition metal complexation [2]:

Generally, cyclic cumulenes have two types of strains: distortion from linearity imposed by $\mathrm{sp}$-hybridization and distortion from orthogonality of adjacent double bonds (cyclo[2n]carbons). [6]
Attempts to liberate the cyclocarbons strain in cyclic cumulenes is similarly relieved by complexing one of the double bonds directly to a transition metal. For example, like in this zirconocene-hexapentaene complex, where $\ce{C=C=C}$ angle is reduced to approx. $130^\circ$ by complexation with $\ce{Zr}$ [7]:

Unfortunally, up to date there are only mononuclear metal-cumulene complexes isolated in a large scale, and none of them are cyclo[n]carbons, but theoretically polynuclear transition metal complexation can significantly reduce the ring size.
References

Carbon Rich Compounds II, Macrocyclic Oligoacetylenes and Other Linearly Conjugated Systems; de Meijere, A., Ed.; Topics in Current Chemistry; Springer Berlin Heidelberg: Berlin, Heidelberg, 1999; Vol. 201; ISBN 3540653015.
Stang, P. J.; Diederich, F. Modern Acetylene Chemistry; Wiley-VCH: Weinheim, 2008; ISBN 9783527615261.
Hoffmann, R. Tetrahedron 1966, 22 (2), 521–538. DOI 10.1016/0040-4020(66)80020-0.
Diederich, F.; Rubin, Y.; Knobler, C. B.; Whetten, R. L.; Schriver, K. E.; Houk, K. N.; Li, Y. Science 1989, 245 (4922), 1088–1090. DOI 10.1126/science.245.4922.1088.
Hutter, J.; Luethi, H. P.; Diederich, F. J. Am. Chem. Soc. 1994, 116 (2), 750–756. DOI 10.1021/ja00081a041.
Stone, F. G. A.; West, R. Advances in organometallic chemistry; 1998; Vol. 42. ISBN 0120311429.
Suzuki, N.; Hashizume, D.; Yoshida, H.; Tezuka, M.; Ida, K.; Nagashima, S.; Chihara, T. J. Am. Chem. Soc. 2009, 131 (6), 2050–2051. DOI 10.1021/ja8077472.


Can potassium carbonate be used to calibrate hygrometers?

Note that all salts, used in humidity calibration, dissociate to their respective hydrated cations and anions when dissolved.
From the solution crystallizes the sesquihydrate $\ce{K2CO3·\frac 32 H2O}$ ("potash hydrate"). Heating this solid above 200 °C gives the anhydrous salt (Wikipedia).
It seems you have anhydrous carbonate, that hydrates itself exothermically. You can still use it, but you need to prepare rather the saturated solution with excess of carbonate, that would integrate the crystal water into itself.
Or, let the prepared saturated solution partially evaporate and crystalize.

What is important is maintained equilibrium.
$\text{Air 43% humidity }\ce{ <=> K2CO3} \text{ saturated solution } \ce{<=> K2CO3 . 3/2 H2O}$
Anhydrous $\ce{K2CO3}$, wetted by water, rather just absorbs water into its crystall lattice. If you add too little water, no solution would left, similarly what would happen to the plaster of Paris.
You need the carbonate in hydrate form, together with its saturated solution. Take some water, add there carbonate to dissolve as much of as it can, and then some. Let is settle.
If you had the hydrate, you could just moisten it. But you do not have it.
There is the third option - moistening  it enough so some liquid phase always remains. For $\pu{100 g }\ce{K2CO3}$, you need about $\ce{20 mL}$ of water just for the crystal water, so it becomes the part of the solid.
The carbonate solubility is $\pu{110.3 g/100 mL}$, so I would carefully mix some $$\pu{100 g}\  \ce{K2CO3} + \pu{50 mL}\ \ce{H2O}$$
or recalculated to  more suitable amounts.


Will PAEK polymers hydrolyze in concentrated sulfuric acid?

I resist concluding this but nothing else I can think of makes sense. The only way the polymer solution to become less viscous is through a cleavage of bonds reducing the MW of the polymer or dilution. Cleavage at the ether linkages would require reduction in an oxidizing environment or nucleophilic aromatic substitution which this polymer lacks the electron-withdrawing groups to effect, so those mechanisms should be eliminated (I concur with your colleague). Thus we can deduce that the cleavage must be occurring at the carbonyl groups.

To cleave the polymer at the carbonyl groups, the PAEK is likely undergoing an electrophilic aromatic substitution at the carbonyl groups. Though not commonly utilized, it is possible to substitute any electrophile on an aromatic ring in including acyl groups (especially with sulfuric acid which provides a high concentration of $\ce{H+}$ electrophiles) as shown in Figure 1 — essentially Friedel–Crafts acylation in reverse.

Figure 1. Electrophilic Aromatic Substitution Removing an Acyl Group.


Note that like most of the colleagues on this site with practical lab experience, I've used activated charcoal once in a while, but I never bothered to make it myself.

Supposed that you live on the countryside with enough space around, you can make your own charcoal in a earth-covered pile or an oven - unless local environmental regulations ban that.

Activation, even when starting from ordinary charcoal is different. The gas activation requires heating (and partial burning) of the charcoal at up to 1000 °C in a stream of air, carbon dioxide or water vapour. I'd rather not do it in my backyard.


Most reactive site for electrophilic aromatic substitution

As you know, in deciding which position undergoes aromatic electrophilic substitution, three elements must be taken in account: steric effect, mesomeric effect and inductive effect.

Let's call the ring containing atoms 1 and 2 "A", and the one containing atoms 3 and 4 "B".

Before deciding the importance of any of those effects, some considerations about the geometry of the molecule should be taken in account. It could be guessed, by thinking about the spatial arrangement of such molecule, that the high steric effect between rings A and B prevents their rotation, thus keeping them in a constantly "skewed" position in respect to the nitrated ring.

A representation of the molecule, with the Van Der Waals radius of the elements, follows:

Steric Effects: Because of the presence of  a nearby phenyl (the nitrated one) to any of the ortho positions of rings A and B, and because of the presence of the other phenyl in the middle of the possible electrophile's path, all the ortho positions might be considered as non favourable.

Mesomeric effect: One might guess that the resonance structures of the nitrated ring could extend to ring B, while leaving ring A untouched.
Actually, such a resonance structure requires a "double bond behaviour" for the aromatic carbons between ring B and the nitrated ring. But the molecule is extremely unlikely to behave that way, since the copresence of two aromatic rings prevents their own rotation, and locks them in a skewed position. This conformation, in other words, renders the pi system "less conjugated", and the mesomeric effect much less relevant.

Inductive effect: This is the final and, in this specific molecule, most important consideration. Simply, everything boils down to the proximity of the nitro EWG: this makes ring A more electron-poor, and ring B more electron-rich.

The answer is, hence: the meta position of ring B is attacked, or the "4" position in your image.


Is the reasoning of spin contamination correct?

Commuting operators indeed admit a set of simultaneous eigenfunctions, and since the non-relativistic electronic Hamiltonian commutes with total spin operators $\hat{S}_{z}$ and $\hat{S}^{2}$, the exact non-relativistic electronic wave function (which is an eigenfunction of the electronic Hamiltonian) is also an eigenfunction of the total spin operators. And this requirement can be also imposed on an approximate non-relativistic electronic wave function.

This is certainly true that the non-relativistic electronic Hamiltonian commutes not only with total spin operators $\hat{S}_{z}$ and $\hat{S}^{2}$, but with operators of the individual sources of spin $\hat{S}_{iz}$ and $\hat{S}_{i}^{2}$ as well, or to be more precise, with their trivial extensions on the state space of a many particle system. Thus, this is indeed true that one could adopt the uncoupled picture as well, rather then the usual coupled one, but what is the usefulness of such picture? In the absolute majority of cases the most important characteristic of a system is a property which describes it as a whole, rather that its components individually. For this particular case, recall, for instance, that the total spin of a molecular system is a very important characteristic of it. For instance, it is heavily used in electronic spectroscopy: important terms like term symbols, spin-allowed and spin-forbidden transitions, fluorescence, phosphorescence, etc. are all directly formulated in terms of total spin angular momentum of electronic states.

Now if you decide to use the uncoupled picture in which the total spin is unspecified you basically loose all the body of knowledge of electronic spectroscopy. But what do you gain instead? What is the point of using the uncoupled picture? You can specify the spin of each electron, yes, but but so what?


I am looking for any substance that would be unstable or dangerous when it is cooler or not on heat, and is stable when it is being boiled

Well, in theory, a few compounds could become explosive when crystallizing out of a solution (which is promoted by low temperatures) though I do not think that "turning off burners" would be enough. The "best" candidate I can imagine would be "azidoazide azide" which can detonate for... just any reason! But nobody uses it, just because of that.


Why do some periodic tables have 18 groups while others have 8?

Unfortunately, life sometimes is more complicated than school :)

Using 8 groups works well to teach the basic principle, but higher shells have space for more electrons. (And lower shells for fewer - ever wondered why there are only two elements in the first period?)

Hence, you'll get 18 electrons in the next shell (the "transition metals"), then 32 electrons ("Lanthanides/Actinides") - the general rule is $2n^2$, where $n$ is the number of the period. This is unfortunately made even more complex, as the energies of the shells are not quite what one would expect and they are not filled "in order." One probably should stop speaking of "shells" at this point and move towards "orbitals."

Anyway, the community has pretty much settled on using 18 groups nowadays. The idea behind the periodic system is to find regularities (hence the name) between elements.

It turns out, that individual groups and trends within them are similar in the 8 "main" groups (as you have learned). In the transition metals, there are still a fair few regularities, that justify this arrangement - but they are already a lot more similar to each other than say alkali-metals and halides. The Lantanides/Actinides finally are so similar to one another, that they are "grouped" (although they in fact are part of a period) together - they in fact so similar, that they usually only occur in mixtures in nature, and separating them is quite hard.

Finally: no heavier atoms have been discovered so far that would have to go into an even bigger "shell"/orbital. Hence, one could justify up to 32 groups, but for practical reasons, 18 are conventionally used.


How many atoms can be reduced in a single redox reaction?

I'm not sure if I understand your question correctly, but consider the reaction

$$\ce{CH3-SH + 3 O2 -> SO2 + CO2 + 2 H2O}$$

where methanethiol is oxidized with oxygen to sulfur dioxide, carbon dioxide, and water.

Thereby the oxidation states of sulfur and carbon change from $-2$ to $+4$ and the oxidation state of oxygen from $\pm0$ to $-2$. So sulfur and carbon atoms are both oxidized while oxygen is reduced.

I hope this answers your question.


The effects of acetic acid on porous concrete brick and mortar joints for someone who has no understanding of chemistry

Per Wikipedia on Fired Bricks, to quote their composition;

Normally, bricks contain the following ingredients:[21]
Silica (sand) – 50% to 60% by weight
Alumina (clay) – 20% to 30% by weight
Lime – 2 to 5% by weight
Iron oxide – ≤ 7% by weight
Magnesia – less than 1% by weight

I would expect slowly, with time, that the acetic acid could attack all of the above with the exception of Silica.
Flushing with water would reduce the magnitude of the attack, but I would not categorically claim that the brick structure is completely sound.


How to know which ionic compound have the highest lattice energy without knowing the values of the radii?

The Lattice Energy of Ionic Compounds is directly proportion to the Charge Density. As Ca, S have 2 units of charge while K, Cl have one and there is not much of a difference in size ( atleast not enough to overcome the doubling of the charge), the charge density of Ca and S ions would be more. Consequently, the lattice strength of CaS would be more.


Identifying protons in NMR using the J values

The proton(s) in question absorbs at 5.95 ppm, the olefinic region - so it cannot be D which is expected to absorb upfield in the aliphatic region.  B is the correct answer.  Proton B would be split by proton C into a doublet with a 15.5 Hz separation (the coupling constant $\ce{J_{BC}}$).  Each line in the doublet would be further split into another doublet with a J=8 Hz spacing by proton A.  So now we have 4 lines, a doublet of doublets.  Each of these four lines will be further split into a triplet with 1.5 Hz spacing by the two (hence the triplet) D protons.

Using the chemical shift of a proton and the observed splitting (e.g. the coupling constants, or J values) can go a long way in helping to determine a structure, particularly in relatively simple molecules.  As molecules get more complicated you can do advanced experiments to help with the structure determination.  One example would be spin decoupling experiments.  For example, if we "know" which signal is due to the D protons, we could irradiate them with energy during our nmr experiment.  If they are coupled to proton B (they are), then the triplet in the B signal should diappear and we would just see our doublet of doublets.  This type of experiment confirms which protons are coupled to which other protons.  There are many other advanced nmr techniques that can be used to help in structure elucidation.


What equipment is used for two-dimensional IR spectroscopy and how can the molecular structure of a compound be resolved?

You can find a schematic drawing of the box-CARS 2D IR setup in the publication
Enhancing signal detection and completely eliminating scattering using quasi-phase-cycling in 2D IR experiments, Robbert Bloem, Sean Garrett-Roe, Halina Strzalka, Peter Hamm, Paul Donaldson

https://pdfs.semanticscholar.org/e134/cdbb98a8d899a17848f8242c8160d4913f0f.pdf


I don't think that's how Sudan IV works. Sudan IV is a highly conjugated aromatic diazo compound, so it is already highly colored. In the lipid, the Sudan IV is basically hydrophobic so it will stain any lipid droplets suspended in water. The color is due purely to the Sudan IV, not a reaction.


Why does radical chlorination and bromination of propane occur at different positions?

The Hammond Postulate (HP): Exothermic reactions have early transition states (TS). Endothermic reactions have late TS's. [The HP may easily be demonstrated with a retractable, metal tape measure. Open several feet of the tape with the left hand holding the case and the other hand at the end of the tape. Bend the tape upward so that it looks like a TS. Raise your left hand above the right. This is an exothermic reaction. Note how the high point (TS) moves to the left. Make an endothermic reaction. Watch the TS move to the right.]
To compare the radical chlorination and bromination of propane, the following chart is instructive. Only the first propagation steps need to be examined because this is where the 1o vs. 2o choice is made. Chlorination (A & B) is exothermic (earlier TS) for both C-H bonds with the secondary more favorable by virtue of the weaker 2o C-H bond. Bromination (C & D) is endothermic (later TS) and more reversible than chlorination. The difference in the heat of reaction in both chlorination and bromination is 2 kcal/mol with the secondary radical formed in favor of the primary radical. But this equal difference does not explain the difference in the greater selectivity in the bromination vs. chlorination. The difference in activation energy (Ea) in the TS between secondary C-H bond cleavage vs. primary is greater in the bromination than in chlorination. [ADDENDUM: The ΔΔH‡ (ΔEa) for chlorination is 1 kcal/mol; for bromination, 3 kcal/mol. Therefore, bromination is more selective.][2] [continued]

The data for methane in the chart below is instructive. Fluorination has a very low Ea and it is more exothermic than chlorination. A fluorine atom is virtually indiscriminate in choosing among primary, secondary and tertiary C-H bonds. The less reactive chlorine atom has a higher Ea and shows greater discrimination. While these two halogenations are readily conducted at ambient temperature, bromination is conducted at elevated temperature to obtain useful rates of reaction. With a high Ea selectivity is the greatest with bromination.
To put a folksy spin on the discussion, a bromine atom has more "time" to make choices as to what C-H bonds it is capable of breaking while a fluorine atom is virtually a bull in a china shop.

[2]: Streitwieser and Heathcock, Introduction to Organic Chemistry, 4th ed., pg. 114.


Why does the reduction of a Weinreb amide give an aldehyde instead of an amine?

Reaction of acyl carbonyl compounds with strong nucleophiles (hydrides, organometallics) is a very common and useful reaction, although it has some limitations. One of the simplest cases is the reaction of an ester with a strong nucleophile. As one would expect, the strong nucleophile attacks the electrophilic carbonyl carbon, giving a tetrahedral intermediate. In most cases, the tetrahedral intermediate breaks down readily to give a new carbonyl compound. In this case, that carbonyl intermediate will be a ketone (from a Grignard, e.g.) or an aldehyde (from LiAlH4). Both of those functional groups are more reactive than the ester starting material, so the next equivalent of nucleophile will react faster with the aldehyde/ketone, eventually giving a primary (from LiAlH4) or tertiary (from a Grignard) alcohol product.

The situation at the tetrahedral intermediate changes slightly when an amide is the electrophile. As cfcief points out, the less electronegative nitrogen will kick out the more electronegative oxygen (stabilized by aluminum) to give an iminium intermediate. Again, this intermediate is more reactive than the starting material, so a second nucleophilc attack takes place.

The key feature of the Weinreb amide is that after the nucleophilic attack at the carbonyl, the tetrahedral intermediate is stabilized as a five-membered cyclic chelate. Without decomposition of the tetrahedral intermediate, the only active electrophile is more of the Weinreb amide starting material. On workup, the strong nucleophile is quenched, the metal washed out, and the tetrahedral intermediate is allowed to decompose to the aldehyde/ketone product.

Although the N-methoxy-N-methylamide was specifically designed by Weinreb to behave this way (ref), it wasn't until more than 20 years later that mechanistic evidence was observed (ref).


Dow's Process for the manufacture of phenol

The mechanism of this reaction is that $\ce{NaOH}$ being a strong base with some extreme conditions abstracts  H from C adjacent to C containing Cl while Cl also leaves.

This creates benzyne as an intermediate.

Subsequent addition of $\ce{NaOH}$ forms the phenoxide ion.


Why Aren't Chlorides Of The Noble Gases As Prevalent As Their Fluorides?

You know that the bigger the noble gas atom, the easier it is to ionize it. By the same token it is also easier for a Xe atom than for a He atom to share some electron density in a covalent bond.  That is, Xe is more reactive (a very little bit) than He (not at all).

For halogens, the tendency is in some sense the opposite: The smaller the halogen atom the more eager it is to gain electron density -- that is, the more electronegative it is.  Fluorine is so electronegative that it can persuade Xe, Kr and even Ar (with difficulty) to give up some electron density to form a covalent bond.  Cl is not quite so electronegative, and Br and I even less.  Basically you need the least "noble" of the noble gases and the greediest halogen of all to even have a chance at a bond.

One can justify this handwaving with thermodynamic calculations and molecular orbital arguments (though the latter are often handwavy too).


What are stereospecific and stereoselective reaction?

If you have a question, ask it. You say you've read the defination[sic]. What it is that you fail to understand?

IUPAC Gold Book

stereospecificity (stereospecific)

A reaction is termed stereospecific if starting materials differing only in their configuration are converted into stereoisomeric products. According to this definition, a stereospecific process is necessarily stereoselective but not all stereoselective processes are stereospecific. Stereospecificity may be total (100%) or partial. The term is also applied to situations where reaction can be performed with only one stereoisomer. For example, the exclusive formation of trans-1,2-dibromocyclohexane upon bromination of cyclohexene is a stereospecific process, although the analogous reaction with (E)-cyclohexene has not been performed.
The term has also been applied to describe a reaction of very high stereoselectivity, but this usage is unnecessary and is discouraged.

stereoselectivity
(Also contains definition of: diastereoselectivity)

The preferential formation in a chemical reaction of one stereoisomer over another. When the stereoisomers are enantiomers, the phenomenon is called enantioselectivity and is quantitatively expressed by the enantiomer excess; when they are diastereoisomers, it is called diastereoselectivity and is quantitatively expressed by the diastereoisomer excess.


Why does orbital optimization remove spin contamination?

Upon further thought, I've summarized it as follows. If anyone comes along with a better explanation, they'll get the cake!

Spin contamination is a fictitious effect arising from not achieving the full CI solution to the Schrödinger equation. For, had we found the exact solution, then the solution we found would be an exact eigenfunction of the ${\hat{S}}{}^2$ operator, as $[\hat H, {\hat{S}}{}^2]=0$.

In a general sense, orbital optimization can be viewed as simply allowing for more correlation to be recovered on top of whatever other improvements you've applied to your initial Hartree-Fock reference. So, by recovering more correlation, you've brought your solution closer to FCI. In doing so, you've brought your solution closer to a solution that contains no spin contamination.

However, there seems to be a stronger statement that could be made. This type of correlation seems especially potent in removing spin contamination.


Is a combustion necessarily a redox-reaction or can there be exceptions?

Yes, every combustion reaction is a redox reaction. Combustion is an exothermic reaction of a fuel and an oxidant (that's a big clue). Most commonly, the oxidant is oxygen, which is reduced to water in the process. The other products will depend on the fuel. Organic compounds will have carbon atoms of various oxidation states that will be oxidized to the +4 state as carbon dioxide product.


Why does buffer capacity of tartrate buffer generally decrease as the buffer pH increases?

Generally, the maximum buffer capacity is at $\ce{pK_a}$ . The tartaric acid is somewhat special for 2 reasons:

It is a diprotic acid with both $\ce{pK_a}$ very close, with the $\ce{pK_{a1}}$ rather low, being affected by the reason 2. :

$$\ce{pK_{a1}}=2.89,\ce{pK_{a2}}= 4.40 (L+)$$
The solution buffer capacity (not limited to presence of specific buffer substances) generally increases toward $\ce{pH}=0$. It means, the buffer capacity is not given by concentration of  of conjugated acid and base, but also by concentration of $\ce{H+}$ itself.
That in large extent masks the $\ce{pK_a}$ maximum. It means, at a slope, a local peak must be big enough to be a peak.


Does removing solution affect a supersaturated solution

When a solution is saturated, the remaining solid is in equilibrium with the solvent:
$$\ce{AB<=>[solv.]A+_{(solv)} + B-_{(solv)}}$$

That means that there is as much of your solid solved. If you take out some of the solution then the concentration remains the same, in your original flask as well as in the portion you withdraw.

If you remove only solvent, i.e. by heating your solution or diffusion over time. In this withdrawn portion crystals will form. The concentration will not change, only the amount of substance $n$ in your solution will change as the volume shrinks ($c=\frac{n}{V}$).


Briefly: valence orbitals on adjacent metal atoms interfere to form "bands" of closely spaced molecular orbitals that spread throughout the metal.

If you have N metal atoms with occupied s orbitals, they interfere to form an "s band" with N molecular orbitals. The lowest energy orbital is fully bonding; the orbital on the top of the band is fully antibonding. The spread of the band will be finite, and since N is a huge number, the molecular orbital energies will be very closely spaced, practically continuous.

If you have two valence s electrons in your metal, the s band will be full (each orbital can hold up to 2 electrons, you have 2N electrons, and N orbitals in the band). If you have only one valence s electron, the s band will be half full (the first N/2 orbitals will be filled, and the remaining orbitals will be empty).

Above the s band you'll have a similar p band. In gold, the p band is unoccupied, while the s band is half full (gold has 6s$^1$). However, IF the bottom of the p band overlaps with the top of the s band, the s electrons can easily move into the bonding MO's at the bottom of the p band. They can also move into any vacant orbitals in the s band that are just above the highest occupied orbital.  In either case, the s electron will be delocalized.

To figure out whether the valence electrons will delocalize, then, you'll have to decide what bands are present, where the highest occupied molecular orbital is, and what the width and the spacing of the bands are (does the unoccupied band overlap with the occupied band, or is there a gap between the two?)


Gold plating glass could probably be 99.99% complete. It would make a nice mirror and would probably stay unoxidized for a long time.

However, if I wanted to remove the gold, I think I would rinse the glass with HF solution, or perhaps NaOH solution, either of which would find those teeny tiny gaps and undercut the whole gold plate.

I don't think there would be any strong chemical bonding, and no mechanical strength to a gold film. Glass is just a reactive substrate, and protecting it would require a finite thickness, with mechanical strength.

An example of plating not working well is chromium plating (although it can be done very well, it requires much attention to detail). Automobile bumpers used to be chrome plated, but developed rust spots after a few years because of porosity in the chromium film. I believe nickel plating to reduce corrosion tendency (but Ni is not shiny enough) followed by chrome plating works reasonably well. But then there is hexavalent chromium toxicity... So now we just paint the bumpers or make them out of plastic.

Perhaps you could coat the glass with teflon or polypropylene.


Why are aprotic solvents unable to solvate negative ions well?

This can be attributed to the lewis acid/base properties of the solute/solvent. Anions (negative ions) and aprotic solvents both function as lewis bases. They 'repel' each other. However, a cation (positive ion) functions as a lewis acid and is 'attracted', resulting in greater dissolution.
How well a compound dissolves in a given solvent depends on a multitude of factors. This includes the two very important ones discusses here, polarity and lewis acid/base character.


Let me abstract your question to “Why is the $\style{font-weight:bold}{\ce{-CH(CH3)2}}$ group sometimes named ‘isopropyl’ but almost never ‘sec-propyl’?”

iso alkyls

‘Iso’, obviously as in ‘isomer’, in IUPAC organic nomenclature (last one is from 2013) evolved to non-preferred, retained names for several smaller alkyls, that are linear but specific single branching at the carbon next to the terminal carbon, by additional methyl.

These are (intentionally skipping our dour discussed $\ce{C3}$ branched alkyl):

isobutyl    $\ce{-CH2-CH(CH3)2}$
isopentyl   $\ce{-CH2-CH2-CH(CH3)2}$
(isohexyl    $\ce{-CH2-CH2-CH2-CH(CH3)2}$  , only in old IUPAC 1979 nomenclature, see here)

(However, in general naming of bulk chemical products, larger alkyls with different branching are named isoalkyl, e.g. ‘isooctyl’ $\ce{-CH2-CH(CH2CH3)-CH2-CH2-CH2-CH3}$.)

sec- alkyls

Obviously ‘sec-’ is for 'secondary’, meaning there's a branching into 2 carbons at the first carbon atom. The list of retained names has only one item:

sec-butyl $\ce{-CH(CH3)-CH2-CH3}$

but our candidate would fit in here as well.

Why only sec-butyl? A major reason could be that the next “sec-pentyl” would represent two different structures:

$\ce{-CH(CH3)-CH2-CH2-CH3}$
or $\ce{-CH(CH2CH3)2}$ ?

(Now one might ask, why there is not some naming system for alkyls with methyl branch at the first atom, similar to iso alkyls with the branching at the next-to-last carbon. I don't know.)

Why not sec-propyl? The $\ce{-CH(CH3)2}$ is a special case that can fall into both isoalkyl and sec-alkyl class. The better defined one, also resulting with simpler spoken name and simpler typography (no dash and italic) – isoalkyl, was once chosen. That's my humble hypothesis.

(And please note that the preferred IUPAC name for this group is propan-2-yl. Alternate name is 1-methylethyl. Isopropyl is still retained.)


Why does a catalyst increase the forward and reverse reactions equally

The equilibrium is unaffected because the barrier to both forwards and back reaction is reduced, and even though the amount is constant it works out that this effect cancels out. This is a consequence of the fact that the population of energy levels is exponentially distributed from the Boltzmann distribution. This also leads to Arrhenius type rate expressions.

If the energy in your figure from reactants to transition state is $E_a$ and the difference in products to reactants $G$ and we lower the energy $E_a$ by $E_0$ then using Arrhenius type rate expressions the equilibrium constant is $K_e=k_{forward}/k_{back}$

$\displaystyle K_e=\frac{\exp(-E_a/RT)}{\exp(-(E_a+G)/RT)}$,  and for the catalyst  $\displaystyle K_e^C= \frac{\exp(-(E_a-E_0)/RT)}{\exp(-(E_a-E_0+G)/RT)}$  which is the same as $K_e$if you expand the terms in the exponentials.


According to the international standard ISO 80000 Quantities and units – Part 9: Physical chemistry and molecular physics, the symbol $\mathrm p$ is interpreted as an operator $(\mathrm px=-\lg x)$. It is used in particular (but not only) in the definition of the quantity $\mathrm{pH}$:

$$\mathrm{pH}=\mathrm pa_{\ce{H+}}=-\lg\left( a_{\ce{H+}}\right)=-\lg\left(m_{\ce{H+}}\gamma_{\mathrm m,\ce{H+}}/m^\circ\right).$$

where $a_{\ce{H+}}$ is the activity of $\ce{H+}$ in solution and $\gamma_{\mathrm m,\ce{H+}}$ is the activity coefficient of $\ce{H+}$ on the molality basis at molality $m_{\ce{H+}}$. The standard molality $m^\circ$ is chosen to be equal to $1\ \mathrm{mol\cdot kg^{-1}}$.

The definition of $\mathrm p$ (and $\mathrm{pH}$) given by ISO is actually quoted from IUPAC Quantities, Units and Symbols in Physical Chemistry (Green Book).

In its current form, this definition can be traced back to a recommendation given in
  R. G. Bates and E. A. Guggenheim
  Report on the standardization of pH and related terminology
  Pure Appl. Chem., 1960, Vol. 1, No. 1, pp. 163–168
  http://dx.doi.org/10.1351/pac196001010163
which reads

  There already exists international agreement that $\mathrm{pH}$ should be written and printed on line in roman type. We recommend that, with the unique exception of $\mathrm{pH}$, the operator $\mathrm p$ (printed in roman) should denote $-\log_{10}$.
  For example
  $\mathrm p m_\mathrm H$ means $-\log_{10}m_\mathrm H$
  (…)
  where $m$ denotes molality (…)


What is the reaction between a copper anode and a zinc cathode?

Well the standard galvanic cell with $\ce{Zn}$ and $\ce{Cu}$ would usually look something like this, assuming there's a salt bridge.

and the reaction could be written as

\begin{align}
\ce{Cu^{2+}_(aq) +2e^{-} &<=> Cu^0} & E^\circ &=0.34\\
\ce{Zn^0 (s) &<=> Zn^{2+} (aq) +2e^-}& E^\circ &=+0.76\\
\ce{Cu^{2+}(aq) + Zn^0 (s) &<=> Cu^0 (s) + Zn^{2+}(aq)} & E^\circ &= +1.10
\end{align}

As for your measured emf, that can be calculated using the Nernst Equation.