What the plot shows is the reciprocal path length needed for a given absorbance to apply, for example slightly less than 100 m in the visible to $10^{-8}$ m in the uv, as you point out, but this is for $55.345$ molar water which is  very, very concentrated compared to most solutions from which an absorption spectrum is obtained, typically $<< 1 $ molar.

As you also point out the absorbance is $A=\log_{10}(I_0/I)=\epsilon[C]l$ with $\epsilon$ as the extinction coefficient usually in units $\pu{dm^3mol^{-1}cm^{-1}}$ and which depends upon wavelength, and $[C]=55.345 \pu{ mol}$ for pure water and l the path length in cm. What seems to be plotted is
$$\frac{1}{l}=\frac{\epsilon[C]}{A}$$
so is still proportional to the extinction coefficient and so still indicates regions where water absorbs greatly or otherwise. But it is not clear what value for A is used. It may be understood that $A=1$ always applies or that the value is taken at $1/e$ absorption. The values may also be based on $\log_e$ not $10$ so a factor of $2.303$ may be involved. As I'm not familiar with this much detail I can't help any more other than to suggest that  you will have to check by finding a true absorption spectrum or  extinction coefficient and work backwards and converting units as necessary.

In general the amount of absorption a molecule has can vary a huge amount. In water in the visible part of the spectrum absorption is low as there are only weak overtone bands from vibrational transitions and nothing else. At shorter wavelengths electronic absorption begins and these can have extinction coefficients of tens to hundreds of thousands in units of $\pu{dm^3mol^{-1}cm^{-1}}$.

A  concentrated solution of a dye  at, say, $1$ molar with a typical extinction coefficient of $10^4$ would have an optical density of $4$ in a $1$ cm cell or a reciprocal length of $10^6$ $\pu{m^{-1}}$ if $A=1$. Hope this helps.


Does K2CO3 react in minute amounts with glassware at higher temperatures?

In agriculture, potassium silicate ($\ce{K2SiO3}$) is used as foliar fertilizer ($\ce{K2SiO3}$ is water soluble). The major synthetic method of the preparation of $\ce{K2SiO3}$ is heating $\ce{SiO2}$ and $\ce{K2CO3}$ at high temperatures $(\pu{600-850 ^\circ C})$ in various mole ratios (e.g., Ref.1):
$$\ce{SiO2 + K2CO3 ->[\Delta] K2SiO3 + CO2} \tag1$$
According to Ref.1, alkali metal carbonates such as $\ce{Na2CO3, K2SiO3, Rb2CO3,}$ and $\ce{Cs2CO3}$, all give corresponding metal silicates in this reaction (e.g., silica to metal carbonate mole ratios used in these preparations are $1:2$ to $3:1$).
Keep in mind that the temperatures used in these preparations are below the melting points of both reactants, $\ce{SiO2}$ ($\pu{1713 ^\circ C}$) and $\ce{K2CO3}$ ($\pu{891 ^\circ C}$), thus, the reaction is essentially a solid state one.
Potassium silicate fertilizer grades were successfully produced by direct fusion of silica ($\ce{SiO2}$) and potassium salt ($\ce{KOH}$ and $\ce{K2CO3}$) in furnaces at temperatures up to melting point of the mixtures (melt temperatures are as high as $\pu{1350 ^\circ C}$). The range of the weight ratio of silicon dioxide/potasium solid used in this method is $1:1$ to $5:1$. The reaction between $\ce{SiO2}$ and $\ce{KOH}$ is being:
$$\ce{SiO2 + 2KOH ->[\Delta] K2SiO3 + H2O} \tag2$$
According to the Ref.1, the other salts (e.g., halides and sulfates) of the metal carbonate catalyze the reaction. For example, in preparing sodium silicate, 6.12 parts by weight sodium carbonate, 3.47 parts by weight silica of minus 325 mesh particle size, and 0.41 part by weight sodium sulfate were ground together (which is $\ce{SiO2 + Na2CO3 + 0.05Na2SO4}$). The ground mixture was heated at a temperature of approximately $\pu{700 ^\circ C}$. for $\pu{4.5 h}$, after which time, the reaction was found to be 51% complete (c.f., without catalyst under same condition, completion was ~$23\%$).
Based on these date, it is safe to say that when silica made contact with an alkali carbonate or hydroxide at elevated temperatures, we can expect a reaction between two chemicals, regardless of they are been in solid form or melt. Once washed, the formed metal silicate would be washed away by water, leaving silica with damaged surface. According to Wikipedia:

The most familiar, and historically the oldest, types of manufactured glass are "silicate glasses" based on the chemical compound silica (silicon dioxide, or quartz), the primary constituent of sand. Soda-lime glass, containing around 70% silica, account for around 90% of manufactured glass.

Therefore, it is no wonder when you applied alkali hydroxide solution (e.g., $\ce{NaOH}$ or $\ce{KOH}$) solution to glass surface for a long time and found the surface become roughed. The alkali carbonated would do the same in higher temperatures.
Note: $\ce{Li2SiO3}$ is only partially soluble in water. However, its $\ce{Li4SiO4}$ version has an advantage of being $\ce{CO2}$ absorber (Ref.3) according to the reaction:
$$\ce{Li4SiO4 + CO2 ->[\Delta] Li2SiO3 + Li2CO3} \tag3$$
Unlike other carbonates, two $\ce{Li2CO3}$ molecules react with one $\ce{SiO32}$ molecule to give $\ce{Li4SiO4}$:
$$\ce{2Li2CO3 + SiO2 ->[\Delta] Li4SiO4 + 2CO2} \tag4$$
References:

Isadore Mockrin, “Production of Silicates,” US Patent 1958, 2,823,098 (PDF).
Srie Muljani, Bambang Wahyudi, Ketut Sumada, Suprihatin, “Potassium silicate foliar fertilizer grade from geothermal sludge and pyrophylite,” MATEC Web of Conferences 2016, 58, 01021 (DOI: 10.1051/matecconf/20165801021) (Proceedings of the 3rd Bali International Seminar on Science & Technology)(PDF).
Xianyao Yan, Yingjie Li, Xiaotong Ma, Jianli Zhao, Zeyan Wang, “Performance of $\ce{Li4SiO4}$ Material for CO2 Capture: A Review,” Int. J. Mol. Sci. 2019, 20, 928 (22 pages) (DOI: 10.3390/ijms20040928).


Why heat of reaction under neither constant pressure or constant volume is considered (in my textbook, or probably somewhere else)?

The constant pressure and constant volume scenarios are presented when you first study thermodynamics because they present an easy way to simplify the calculation of energy changes ($\Delta U,\ \Delta H,$ etc.) based on the change of one easy to measure phenomenon.

  Internal Energy $U$

So, the change in internal energy must equal the sum of the work done $w$ and the heat transferred $q$. This is not random, it is a consequence of the Law of Conservation of Energy. If the internal energy of a system has changed then energy must be exchanged with the surroundings. There are only a few ways this can happen (heat, work, electromagnetic radiation, ...), and heat and work are the most important for most thermodynamic scenarios.

Thus, $\Delta U = q + w$. We need to be able to relate $q$ and $w$ to things we can measure. We calculate $q$ based on the temperature change of the system $\Delta T$ and the heat capacity of the system $c$. Thus, $q=c\Delta T$. Work is defined (in one way) as the change in volume of the system times the pressure. The negative sign is present to make sure that an expanding system is doing work on the surroundings (and thus losing energy): $w=-P\Delta V$.

Now we have $\Delta U = c\Delta T -P\Delta V$

  Constant Volume

If volume is constant $\Delta V = 0$, which means that $w=0$, regardless of pressure. Thus, $\Delta U = q = c\Delta T$.

  Constant Pressure

This one is trickier, but we can do it for a gaseous system.

We do not need to measure volume, since $PV=nRT$, but it is easier to measure volume than $n$, which may also be changing. If $n$ is constant, then, the equation for change in internal energy becomes:

$$\Delta U = c\Delta T - nR\Delta T$$

If $n$ is not constant, then we need to measure that change in volume and temperature.

If the process is in another state, like in solution, $\Delta V$ might be very challenging to measure. Let's hope we can measure $\Delta n$.

  Enthalpy $H$

Because of the challenge of measuring volume changes in solution, we invented a new thermodynamic potential $H = U + PV$. Total differentiation and whatnot gives us:
$$\Delta H = \Delta U +P\Delta V + V\Delta P = c\Delta T - p\Delta V + P\Delta V + V\Delta P$$
$$\Delta H = c\Delta T + V\Delta P$$

Thus, at constant pressure, $\Delta H = q = c\Delta T$, which is easy to measure. Enthalpy is similarly challenging at constant volume as internal energy is at constant pressure.

  Constant pressure and volume.

Under these conditions, $\Delta U = \Delta H = q = c\Delta T$.

  Pressure and volume are not constant

Okay, so if $w=-P\Delta V$, how do we calculate work? If pressure is changing, which value of $P$ do we use? It turns out we use all of them and we need some calculus to do it. This case is often left out because many introductory textbooks consider the likelihood that many students beginning their study of chemistry may not have taken calculus.

Here we go. We need to replace the total change in $V$ with the infinitessimal change in $V$ at each moment $\Delta V \rightarrow  dV$. $\Delta V$ is the sum of all of the infitinessimal $dV$ values. Now, we can multiply each $dV$ with the pressure at that moment and add them up to get work (which involves integration):

$$w = \sum_{j=1}^n P_j dV_j=\int_{V_i}^{V_f}PdV$$

However, we cannot leave well enough alone, because $P$ is changing, so we need to substitute for $P=\frac{nRT}{V}$.

$$w=\int_{V_i}^{V_f} \dfrac{nRT}{V}dV=nRT\dfrac{\ln V_f}{\ln V_i}$$

Now the above maths only work if temperature is constant. What if it's not?


A thorough benchmark of DFT functionals without dispersion corrections

There is a benchmark of the GMTKN30 database with and without  the D3 dispersion correction. See Figure 4 here: "Effect of the damping function in dispersion corrected density functional theory"

However, your statement

  I quickly run few calculations including dispersion corrections and found no noticeable difference.

is not supported by this.

Another example is this paper: "Improving intermolecular interactions in DFTB3 using extended polarization from chemical-potential equalization" where we compare PBE to PBE-D3 for 11 different datasets (see the SI). For many dispersion-dominated complexes, DFT (no D3) is qualitatively wrong, and in many other cases, the error is comparable to the mean interaction energy.

In general, I would strongly recommend against DFT without an empirical dispersion correction unless you are only focusing on electronic properties, but then you wouldn't be very interested in GMTKN30 anyway.

Since you say you are using G3/G4 you might also be interested in these recent benchmark papers:

"Benchmarking Compound Methods (CBS-QB3, CBS-APNO, G3, G4, W1BD) against the Active Thermochemical Tables: Formation Enthalpies of Radicals"
"Accurate reaction barrier heights of pericyclic reactions: Surprisingly large deviations for the CBS-QB3 composite method and their consequences in DFT benchmark studies"


What's the pKa (range) of a carbonyl functional group?

As per comments you are asking for the $\mathrm{p}K_\mathrm{a}$ value of a hydrogen directly attached to a carbonyl carbon. This means, you are asking only for non-enolisable aldehydes as ketones have no hydrogens on the carbonyl carbon. Aldehydes that have a hydrogen in α-position will be deprotonated there generating an enolate ($\mathrm{p}K_\mathrm{a} \approx 20$) — more acidic than a carbonyl hydrogen. So all we have left is 2,2-dimethylpropanal (or pivaloylaldehyde). Its $\mathrm{p}K_\mathrm{a}$ is not given on SciFinder, but should be somewhere between $45$ (an alkane) and $25$ (an alkyne), probably just slightly higher than an alkene ($40$).


Relationship between the partial pressure $p_J = x_J p$ and the perfect gas law $pV = nRT$

Partial pressure is defined as the pressure a component of a gaseous mixture would exert if it alone were to occupy the container at the same temperature as that of the mixture.
If the gases in the mixture are ideal, you can apply the ideal gas equation for it $$p_A V = n_ART$$where $p_A$ and $n_A$ represents the partial pressure and number of moles of a component A. Also $$p_tV = nRT$$ where $p_t$ and $n$ represents the total pressure and number of moles of the gaseous mixture. Dividing these equations you will get $$\frac{p_A}{p_t}=x_A$$ This equation will only be true for ideal gas mixture.
Also, partial pressure is a notional concept. It is not a the actual pressure exerted by the components. It was introduced to explain and form physical properties of gaseous mixtures.


Is melting of Tar for road construction a reversible change?

Tar
 is mainly made up of a collection of very heavy hydrocarbons, with
rather high melting temperatures. Commonly, it is mixed with some stones
 or gravel to make a harder surface for roads. It is possible to heat
this up to soften or melt this; you can try it yourself by heating it in
 a metal can over a stove (don't use a container that you plan to keep
clean) or in a wood or (char)coal fire.

EDIT: As MaxW pointed out, it would be a bad idea to do this inside the house because it would create a strong smell, similar to that created by whatever machine they used to make the road.


The actual drying agent is calcium sulfate semihydrate. It has low drying capacity, which decreases already after the uptake of $6.6\%$ water (theoretical maximum is $15.7\%$). Within these limits, however, it is possible, for example, to dry air at room temperature down to $0.0004\ mg$ of water per liter of air. It is suitable for drying almost all liquids and gases. It can be regenerated by heating to $190-230^{\circ}C$. Above $300^{\circ}C$ it loses its drying properties. The data comes from an old Merck's handbook on drying agents.


How do I control the negative temperature in an experiment?

For temperatures below room temperature cooling baths are used.
These baths consist of a solvent + something that has a low temperature already (ice, dry ice, or liquid nitrogen). A comprehensive list is given below:

Although all of these temperatures are technically achievable, in practice we rarely prepare most of them due to either cost (acetonitrile is quite expensive to be using for a cooling bath), toxicity (a benzene/CO2 bath is not ideal) or difficulty (a methanol/LN2 bath forms a very thick slurry which is difficult to work with).
Most reactions below room temperature are therefore conducted at 0 (ice/water) or at -78 (dry-ice/acetone), with in-between temperatures achieved by adding dry ice slowly to acetone until the desired temperature is reached.
In addition to using this kind of cooling baths, there are machines that will allow you to dial up a given temperature and they will cool the solvent to within a few degrees. These are incredibly useful but their size and cost restricts them from everyday use (most labs don't have one, and those that do won't have enough for everyone to use one everytime they need cryogenic conditions)


What is the mathematical foundation for the probabilistic nature of chemistry?

Actually, a huge amount of the probablistic nature of chemistry is simply the Law of Large Numbers (LLN). A mole of any substance is approximately $6\times10^{23}$ molecules, which is 600 sextillion molecules! The LLN kicks in at quantities much smaller than the sextillion level. Put simply, there are so many interactions occurring in any macro-level reaction that these are inevitably going to converge to the theoretical mean by the LLN.

Now your question adds in the issue of activation energy, but I still think it fits this model, because activation energy tends to produce a sharp difference: too low, and little reaction happens; but high enough, and we mostly go to completion. Once activation energy is high enough, the LLN applies.


Assigning Re and Si face for Carboxylic group

The carbonyl oxygen gets higher priority because it has a real oxygen bonded via $\sigma$ bond and also the double bond, but we count that as the oxygen bonding back to the carbon. We don't treat it like a real carbon; it's a "ghost" or "phantom" carbon with nothing attached to it. This carbon counts as a carbon but is lower priority than any other real carbon. That means that -O-(C) outranks -O-H.


Mechanism for simultaneous reduction of nitro group (into aniline) and aromatic substitution (into benzonitrile) by Ethyl cyanoacetate?

This reaction was conducted using a 3- to 5-fold excess of ethyl cyanoacetate (ECA) and KOH over the arylnitro compounds in dimethylformamide (DMF). It is also important to realize that most KOH is 15% water, which allows for a source of protons in this basic medium. Steps 1 --> 3 (Scheme 1) illustrate a possible route to nitrone 4. In so doing the ECA moiety is being oxidized as the arylnitro compound is being reduced. Addition of hydroxide to nitrone 4 forms 5, which can collapse in two ways. One pathway (another will be mentioned later) leads to reactive ethyl oxocyanoacetate 6 and an aryl hydroxylamine. Base deprotonates the hydroxylamine which adds to 6 to afford species 7 and liberates cyanide anion. Addition of cyanide to the aromatic ring of 7 affords arylamine 8 with loss of the half ester of oxalic acid. A second equivalent of 6 reacts with amine 8 to produce target compound 9.

Condensation (Scheme 2) of the anion 2 of ECA with ethyl oxocyanoacetate 6 provides ketodiester 10 which condenses with arylamine 8 to form the second targeted product 11. Scheme 3 provides another mode of collapse of intermediate 5 which leads to 9. The hydrolysis of 9 and 11 is routine. There are undoubtedly other variants of these Schemes that may apply.

Update: The reference provided by the OP was not the original paper in the series and the picture provided referred to reference 5. The original[1] reactions with 6-nitroquinoline used KCN as the base providing yields of 9 (IIa) in the 50% range and trace amounts of 11 (IIIa) in reaction times up to 5 hours.  After 24 hours the yield of 9 is greatly reduced with the formation of the oxamic acid derived from ester 9 by an unspecified mechanism of formation. The appearance of 16.5% of 11 should got be construed that 11 is derived from 9. Significantly, the use of KOH as a base produces 9 in 66.5% yield without the formation of 11.

Y. Tomioka, A. Mochiike, J. Himeno and M. Yamazaki, Chem. Pharm. Bull., 1981, 29(5), 1286.


How to separate HCl and NaOH from salt and water solution?

Unless time travel is an option, you could

Electrolyze the $\ce{NaCl}$ solution to obtain a solution of $\ce{NaOH}$, and $\ce{H2}$ and $\ce{Cl2}$ as gases.
Collect the gases and photolyze them. The dissociation energy of $\ce{Cl2}$ is $243\, \mathrm{kJ \cdot mol^{-1}}$, irradiation  at $\lambda$ < 490 nm will cleave $\ce{Cl2}$ to chlorine radicals and initiate the chain reaction to yield $\ce{HCl}$ gas.

$$\ce{Cl2 ->[h\nu] 2 Cl*}$$

$$\ce{Cl* +\  H2 -> HCl + H*}$$
$$\ce{H* +\  Cl2 -> HCl + Cl*}$$

EDIT 1

Unless you know exactly about the risk of both processes, don't do it!
This video gives a nice impression of how violently hydrogen and chlorine react upon irradiation! Note that it has nothing to do with the laser involved. A flashlight would give the same dramatic effect!

EDIT 2

The answer above, giving practical advices on how to get $\ce{HCl}$ and $\ce{NaOH}$ back apparently suggests that you can not simply revert the neutralisation reaction.

The neutralisation is irreversible.

In order to find out why this is the case, you might want to have a look at thermodynamics of this exothermic reaction, particularly at the entropy changes.


Does mixing calcium dichloride and magnesium sulfate induce a reaction, and how does it affect the hygroscopy of the sample?

You are correct.

In the system $\ce{Ca^2+*Mg^2+*Cl^-*SO4^2-*H2O}$, the least soluble phase is $\ce{CaSO4*2H2O}$.

Once you're mixing $\ce{CaCl2*2H2O}$ and $\ce{MgSO4*7H2O}$, your $\ce{CaCl2*2H2O}$ starts absorbing atmospheric moisture and then dissolving in it. Then you have a brine which dissolves the $\ce{MgSO4*7H2O}$, speeding up the kinetics of the thing. The solid that you precipitate by the recation has less $\ce{H2O}$ per mole of salt, so the excess $\ce{H2O}$ just ends up in the brine (the "water"). Pressing it speeds it up because there is more surface area of the two salts in contact, which is where the reaction happens.


Find ratio of number of atoms of $A$ and $B$

There are two radio nuclei A and B. A is α emitter and B is β
  emitter and their decay constant are in ratio 1:2

So B decays twice as fast as A.  If we want the actual number of decay events from A to equal the actual number of decay events from B at t=0, then we must have twice as much A present as B at t=0.

We are given that $\ce{k_{B} = 2k_{A}}$. For the decay events at T=0 to be equal for A and B, then $$\ce{[B]_{o} * k_{B}~=~[A]_{o} * k_{A}}$$
$$\ce{[B]_{o} * 2k_{A}~=~[A]_{o} * k_{A}}$$
or $$\ce{\frac{2}{1}~=~\frac{[A]_{o}}{[B]_{o}}}$$
At t=0, the concentration of A must be twice the concentration of B in order for equal decay events to occur from A and B


In general, a functional $F$ is a mapping from an arbitrary set $\mathcal{X}$ of functions to the set of complex numbers $\mathbb{C}$ or the set of real numbers $\mathbb{R}$:
$$
F : \mathcal{X} \mapsto \mathbb{R}.
$$
or
$$
F : \mathcal{X} \mapsto \mathbb{C}.
$$

For example, if you consider $\mathcal{X}$ as the set of polynomials with real coefficients, you can define a functional $F$ as
$$
F[f] = \int_0^1f(x)\,dx
$$
i.e. your functional $F$ takes a polynomial function $f\in\mathcal{X}$ (for example $f(x)=3x+1/2$) as an argument and returns a scalar (2 for $f(x)=3x+1/2$, as you can easily verify).

A density functional is simply a functional $F[f]$ where the argument $f$ is the electron density $\rho(\vec{r})$ (i.e. a density functional is a functional of the electron density). For example Hohenberg and Kohn showed that the energy $\epsilon$ of a quantum system is a functional of the density
$$
\epsilon=E[\rho]
$$
This means that when you plug the electron density of your system $\rho(\vec{r})$ into the energy functional $E[\rho]$ you get a number $\epsilon$, which is the energy of your system. The whole energy functional is not known explicitly, but some of its components are known. For example for the external potential energy we have
$$
V[\rho] = \int v(\vec{r})\rho(\vec{r})d\vec{r}
$$
and for the Coulomb interaction between electrons we have
$$
J[\rho] = \frac{1}{2}\iint \frac{\rho(\vec{r})\rho(\vec{r}')}{|\vec{r}-\vec{r}'|}\,d\vec{r}d\vec{r}'
$$
which are clearly functionals of the electron density.


Example of a similar tertiary ammonium reaction with mechanism

The thing is, it does react but there will be no cyclic product, in the end, I think that is the reason it is not the answer here.


Calculating the pH of a saturated calcium fluoride solution

In homework land you are right, but not in real life.

The $K_\mathrm b$ of $\ce{Ca^2+}$ ($\mathrm pK_\mathrm b = 2.43$ according to some sources) is within 1 log unit of the $K_\mathrm a$ of $\ce{HF}$.  In other words, near neutral pH, consideration of the $\ce{Ca^2+/CaOH+}$ equilibrium is almost as important as the $\ce{F-/HF}$ equilibrium.


When a sample of $\ce{CuSO4⋅5H2O}$ is dissolved in water, the $5$ molecules water are mixed with the water used as solvant, and cannot be distinguished any more from the other molecules of water. It does not really matter, as if you want to calculate the molarity, you never dissolve your sample of $\ce{CuSO4·5H2O}$ in exactly $1$ litre water. No ! You dissolve it in a volume of water big enough to get one liter solution, whatever the exact amount of water used for this dissolution. It should be a little less than one liter water. The only important parameter is the final volume of the solution, namely one liter.


When is there hybridization and when is there no hybridization?

As Ivan already mentioned, hybridisation is not a physical phenomenon. Physically, the most accepted theory at the moment is that you have nuclei and the electrons roam freely in between them, forming standing waves (orbitals). These standing waves generally spread across the entire molecule unless it is a standing wave of a core electron.

However, it is non-trivial for us chemists to come up with a proper description of these molecular orbitals. Hence, we use a certain set of assumptions. For example, the LCAO approach for molecular orbitals used in some calculations assumes a set of hydrogen-like orbitals on every atom and linear combines them to create the most stable set of molecular orbitals. Since that is still rather hard to do, chemists have also derived a pen-and-paper only approach that does this on a per-atom basis (called hybridisation) and then checks what kind of orbitals would form which bonds — so that’s two steps of linear combinations following each other.

But all of this is only our thought experiment to better describe the physical picture of a molecule. Nature does not need hybridisation or linear combinations. Nature just ‘does it’ and comes up with a final solution. So whenever hybridisation ‘is involved’, it is always man thinking about how to look at the molecule.

Once you have taken that for granted, the question may remain:

  But when do I need to take hybridisation into account for orbital considerations?

There’s a set of simple rules:

Elements of the third and higher periods will remain mostly unhybridised unless they require hybridisation for any specific issue ($\ce{SiH4}$ cannot be explained without $\mathrm{sp^3}$ hybridisation on silicon).
Halogens and noble gases will never be hybridised unless they are hepta-/ octacoordinated.
Generally assume a hybridisation for all elements of the second period. Use the one that best explains the geometry. For oxygen, assume at least one unhybridised p-orbital.
For a nitrogen with three single bonds, remember that the hybridisation will always switch between $\mathrm{sp^3}$ and $\mathrm{sp^2 + p}$. This is the nitrogen inversion.
Hydrogen needs no hybridisation.


Since soda[-lime] glass is specified, then Al will react with the alkali forming an aluminate, e.g. $\ce{Na2Al2O4}$. This may or may not dissolve in the molten glass, depending on concentration of the aluminate(s), temperature and cooling rate. If it precipitates, the glass would have a "frosted" appearance.


Differences between Urushibara nickel and Raney nickel

You can find information about the preparation of Urushibara nickel (U-Ni) in the original paper by Urushibara[1].

Regarding the differences with Raney nickel (R-Ni), surface characterization by means of XPS[2] had shown that R-Ni catalysts contain metallic aluminium in the activated phase as well as oxidized aluminium, while U-Ni catalysts contain metallic zinc in the activated phase in addition to oxidized zinc. Also, U-Ni catalysts can be described as supported nickel catalysts, while R-Ni catalysts are shown to have the well known skeletal structure.

The activity of U-Ni catalysts is very similar to R-Ni catalysts, both in nature of the reaction catalysed and in the rate of catalysis. One of the advantage of U-Ni is their simplicity of preparation, while R-Ni catalytic synthesis requires a long time. Despite this, U-Ni catalysts are also not very sensitive to impurities, so they can be prepared from commercial chemicals of ordinary grade, and they can be stored for a long time without decreasing the activity. Another advantage is the safer handling, since U-Ni catalysts are not inflammable upon contact with air, in contrast to Raney catalysts which are pyrophoric. This allows the catalyst to be recovered and regenerated. However, it should be noted that U-Ni catalyst durability is not satisfactory during the reaction, but this is also the case of R-Ni catalysts, so there is not really a big difference when it comes to practical use, maybe slightly supporting the use of U-Ni vs R-Ni as the former is less expensive and easy to produce and regenerate. The only real disadvantage of U-Ni catalysts is that for some particular reductions, they show less activity than R-Ni catalysts, favouring one reaction over another, for example U-Ni catalysts particularly favours the partial hydrogenation of acetylenic compounds to ethylenes[3].

     1 Urushibara, Yoshiyuki. "A New Method of Catalytic Hydrogenation". Bulletin of the Chemical Society of Japan 25 (4): 280–280
     2 Y. Okamoto, Y. Nitta, T. Imanaka and S. Teranishi, J. Chem. Soc., Faraday Trans. 1, 1979, 75, 2027
     3 Kazuo Hata, “New Hydrogenation Catalysts: Urushibara Catalysts” (1971), pp 235-237


Is poison still poisonous after its 'expiration date'?

It depends on what the poison is.
If we take the colloquial use of the word and include toxins and venoms, many are things like proteins that will certainly denature or otherwise degrade, eventually becoming harmless. e.g. tetrodotoxin, ricin, botulinum, etc.
I would expect that type of poison to have the shortest shelf-life as they are relatively fragile.

Many other poisons are small organic molecules.
These can often be degraded by oxidation in air, exposure to UV, hydrolysis etc. and would include things like nicotine and nerve agents like sarin and VX.
Many nerve agents, have shelf lives of a few years and research has actually been done to extend them for use in munitions.

Several metals are known to be poisonous (like lead, mercury, and cadmium) and are problematic because they are toxic in not only their elemental forms, but also in inorganic and organic compounds.
There may be a great difference in toxicity of the different forms, (see elemental mercury vs methylmercury), but most forms remain at least somewhat toxic.
These may last a very long time because reactions likely to occur under normal conditions may not render them safe, e.g. a chunk of cinnabar ($\ce{HgS}$ mineral) sitting on your desk will not undergo any significant change to render it safe, even on a geological timescale.


Need a pair of compounds that is similar in formula but different in structure

The classical example is dimethyl ester $\ce{CH3-O-CH3}$ and etanol $\ce{CH3-CH2-OH}$ pair. They have exactly same brutto-formula, but different structure and very different properties. The first is a gas, the second is a liquid easily reacting with sodium metal.

Carbon is also interesting example. It has two common forms with exactly same chemical formula $\ce{C}$, but very different properties: a very hard and transparent diamond and soft black graphite.


From the abstract of Solubility of Polymers and Copolymers in Supercritical $\ce{CO2}$

[…] data to temperatures of 270 °C and 3000 bar […]
Over the same range of conditions, $\ce{CO2}$ cannot dissolve polyethylene, poly(acrylic acid), poly(methyl methacrylate), poly(ethyl methacrylate), polystyrene, poly(vinyl fluoride), or poly(vinylidene fluoride) […]


What does "/2" in the structure of calcium benzoate mean?

It means half of calcium.

The organic acid is $\rm{COO}^-$, whereas calcium is $\rm{Ca}^{2+}$. For the structure to be fully correct, there are two benzoic acids per one calcium. Other way round this would be $\rm{(C_6H_5COO)_2Ca}$. Full drawing of this will be way to messy, so it is simpler to put half calcium to one acid.


"Don't pour oil into lye, pour lye into oil"

Lye is denser and tends to pour through the lighter oil, improving mixing (and possibly cutting down on the splattering of lye).  The same is true of concentrated mineral acids added to water.

In any case the hazards are minimized by also adding slowly and with stirring.


General Rule #1: Most elements use only s and p orbitals to form bonds, only transition elements and heavier elements use d, f, etc. orbitals in bonding.

General Rule #2: The more s-character in a bond the shorter the bond (reference).  For example

a $\ce{C(sp^3)-C(sp^3)}$ single bond length is ~ 1.54 $\mathrm{\mathring{A}}$
a $\ce{C(sp^2)-C(sp^3)}$ single bond length is ~ 1.50 $\mathrm{\mathring{A}}$
a $\ce{C(sp)-C(sp^3)}$ single bond length is ~ 1.46 $\mathrm{\mathring{A}}$

In an earlier SE Chem post, the structure and bonding in the trigonal bipyramid molecule $\ce{PCl_5}$ was discussed.  The molecular hybridization is pictured below.  The axial bonds are hypervalent (this concept is explained in the earlier post) and they are constructed from p orbitals.  The equatorial bonds are constructed from $\ce{sp^2}$ orbitals.  General Rule #2 suggests that the equatorial bonds will be shorter than the axial bonds since the equatorial bonds contain more s-character.

As the following structural diagram shows, this is indeed the case.

Now let's turn our attention to the pentagonal bipyramid structure as found in $\ce{IF_7}$ for example.  Here is the structure for this molecule

As you noted in your question, the equatorial bonds are now longer than the axial bonds.  Based on what we've discussed thus far, we might make an initial "guess" that in $\ce{IF_7}$ the axial bonds have more s-character than the equatorial bonds.  As it turns out, this is correct.

Here is a link to an article discussing the structure and bonding in $\ce{IF_7}$.  Only the abstract is freely available.  For completeness, I'll mention that there was an earlier SE Chem question related to $\ce{IF_7}$, but it doesn't seem to add much to the conversation.  Returning to the published abstract, the key part of the abstract states,

  "These features can be explained, however, by a bonding scheme
  involving a planar, delocalized pxy hybrid on the central atom for the
  formation of five equatorial, semi-ionic, 6-center 10-electron bonds
  and an sp hybrid for the formation of two mainly covalent axial
  bonds."

The abstract tells us that the equatorial bonds are formed from the $\ce{p_{x}}$ and $\ce{p_{y}}$ orbitals; the axial bonds are formed from the $\ce{sp_{z}}$ hybrid orbitals (after using two p orbitals for equatorial bonding we are left with one $\ce{p_{z}}$ orbital and one s orbital on the central iodine, hence, when they combine two sp hybridized orbitals result).  That is the answer to the question.  As we guessed, there is more s-character in the axial bonds than the equatorial bonds.  Hence, the axial bonds are now shorter than the equatorial bonds.

For completeness

Someone, after reading that last paragraph, is going to notice that there are 5 equatorial fluorines, but only 4 bonding sites with the two p orbitals - what gives?  First off, note that the $\ce{p_{x}}$ and $\ce{p_{y}}$ orbitals on iodine form hypervalent bonds with the p orbital on fluorine and so yes, there are only 4 equatorial points of attachment.  But note that the abstract also mentioned that the equatorial bonds are semi-ionic.  This suggests that we have a resonance structure where we have an [$\ce{IF_{6}^{+}}$][$\ce{F^{-}}$] ionic contributor.  We can draw 4 more resonance structures just like this for the other 4 equatorial $\ce{I-F}$ bonds.  This means that mixed in with our hypervalent p-p bonds is some ionic character - this allows us to make 5 equivalent equatorial bonds.  Said differently, our 5 equatorial bonds are all identical resonance hybrids obtained by mixing 2 hypervalent p bonds (each one providing bonding for 2 fluorines) with 1 ionic bond (providing bonding for 1 fluorine).


Why do Electrons leave the Zinc in a Galvanic Cell

First, let's think about what happens if you dip a single zinc electrode into some electrolyte solution (for now, it's not going to have copper or anything in it). This situation is subject to this equilibrium: $$\ce{Zn_{(s)}<=>Zn^2+_{(aq)} +2e-}$$

At first, there is no barrier to some of the zinc at the surface dissolving. The standard reduction potential is −0.7618 V (this is for the reverse of the reaction written above) so the oxidation process is energetically favourable. However, though the zinc ion can diffuse through the solution, there is nowhere for the two electrons to go, so they are trapped within the electrode. This excess charge opposes further oxidation—it becomes more and more difficult to force more charge into the electrode. For this reason, zinc electrodes do basically nothing in neutral solutions. N.B. It is possible for these electrons to be used to reduce the hydronium ion into hydrogen gas, but zinc adsorbs hydronium very poorly so this process is rather slow except in concentrated acid.

But what happens if we give these electrons somewhere to go? If we connect the zinc electrode to another electrode that is less easily oxidized than zinc (copper is often used because it's cheap, but it could be siver, gold, etc.), these electrons can flow into the other electrode and participate in a reduction reaction. In the case of a copper electrode, because there are no copper ions in our electrolyte, the reduction reaction that occurs (unless there is a more easily reduced substance in the electrolyte) is: $$\ce{2H3O+ + 2e- <=> H_{2(g)} +2H2O}$$
The two reactions balance each other out so we have no charge buildup to oppose the dissolution of the zinc.

To go back to the core of your question,

I'm just not sure what prompts them to leave in the first place if the positive copper
ions are not directly in contact with the zinc.

If there were any copper ions in the solution, electrons could be transferred directly from the zinc to the copper ions. If you dip a piece of zinc into a copper solution, this is exactly what happens—the zinc dissolves and small beads of copper are reduced on its surface, no second electrode required. Trying to make a zinc-copper cell with copper ion in the solution just makes the cell work worse as a good part of the zinc is lost through direct reduction of copper ions at the surface. The way to prevent this is to separate the two half-reactions with a salt bridge.

The reason the electrons leave in the first place (why the oxidation reaction above occurs) is that it is more energetically favourable for the zinc to be oxidized in the solution than to remain in its metallic state. Another way to think about is that zinc is more easily oxidized than copper so it must have electrons that are more easily lost (have a greater electrochemical potential) and these electrons will flow towards a lower electrochemical potential. Therefore, when you connect the two electrodes together, electrons flow from zinc to copper until an equilibrium is reached. Copper's electrons are not normally high enough energy to reduce the hydronium ion, but because of the electrons coming from the zinc, its electrochemical potential is raised enough for the reaction to occur.

This paper is a great reference for learning more about the lemon battery and other single solution cells (if you have ACS journal access, unfortunately).


Is ozone an electrophile in the ozonolysis reaction?

Electrons are concentrated on the two terminal oxygens in ozone but remember, ozone exhibits resonance. The electrons can be delocalized across each oxygen through p-orbitals.

So yes, at first if you think about the reaction in purely Coulombic terms then it doesn't make much sense. Why would electrons go to a region that already has a relative surfeit of electrons?

But what happens is that by attacking the terminal oxygen, the double bond effectively displaces electrons to the central, electron-deficient oxygen atom.

I'm sure someone can explain this in much greater detail with reference to molecular orbital theory but I'll leave that to someone else.


Carbon doesn't always form 4 bonds. Consider neutral methylene, $\ce{H2C:}$ in which it forms two leaving two unshared electrons (:). Now suppose we protonate that (with a super acid?). We'd have $\ce{H2CH+}$ or $\ce{H3C+}$, a cation just as when we protonate $\ce{H3N:}$ we get the ammonium cation $\ce{NH4+}$. In both cases the bond to the proton is made from the unpaired electrons on C or N and not by sharing an electron from C or N with one from hydrogen.

In computing formal charge we assume that all bond electrons are equally shared. The formula for formal charge is V - N - B/2 in which V is the number of valence electrons in the neutral atom (4 for carbon), N is the number of valence electrons not participating in a bond (0 in the case of methenium) and B the number of electrons in bonds (6 in the case of methenium) so that the formal charge is 4 - 0 - 3 = +1.

It's a different way of indicating charge distribution. If we write methenium as $\ce{H2CH+}$ it implies that the carbon retained the electrons for the most part so that the bond to the proton is largely ionic. In the formal charge scheme we assume that this bond is totally covalent, that the + charge of the proton has transferred to the carbon because the proton got a half share in two electrons thus neutralizing its charge and write it $\ce{C^\oplus H3}$. It's a formality.


I understand why a HS text might call it "electrolytic" , but in the real world it is called "cathodic protection" or CP. That is what to look for on the net. It is a multibillion dollar industry. Essentially all pipelines ( oil, gas, products, water , etc.) use it . A large majority off offshore steel structures ( oil  platforms, ships, etc) use it. And many steel products are galvanized . There are two basic types ; sacrificial coatings and impressed current. Sacrificial coatings are represented by galvanized steel , also aluminum and cadmium coatings. Impressed current, CP is also  two categories : Sacrificial anodes , usually aluminum ( which do not use rectifiers) , and inert anodes , usually titanium or graphite. These are buried in soils made more conductive with graphite powder (typically) and electrically connected to rectifiers  which are then connected to the steel to be protected . The voltage is adjusted and monitored along the protected steel , eg. pipeline. Pipelines are coated with an electrically insulating coating to reduce the amount of current required to maintain a protective voltage , normally lower than - 0.65 V ( Cu - Cu SO4 half cell) Too low a voltage can cause hydrogen gas bubbles under the coating and hydrogen embrittlement of certain steels.There is also anodic protection but it has very specific limited applications. So, there is such a thing as "electrolytic" protection.


Can I express the rate of a reaction as percent per second (%/s)?

According to IUPAC Recommendations for chemical kinetics [1], rates $\nu$ are defined for reactant $\ce{B}$ and product $\ce{Y}$ as a time $(t)$ derivative of the amount $n$ (in general) or per unit volume $V$ (for the closed system):

$$
\begin{array}{lll}
\hline
\text{System} & \text{Consumption} & \text{Formation} \\
\hline
\text{Open} & \displaystyle\nu(n_\ce{B}) = -\frac{\mathrm dn_\ce{B}}{\mathrm dt} & \displaystyle\nu(n_\ce{Y}) = \frac{\mathrm dn_\ce{Y}}{\mathrm dt} \\
\hline
\text{Closed} & \displaystyle\nu(c_\ce{B}) = -\frac{1}{V}\frac{\mathrm dn_\ce{B}}{\mathrm dt} & \displaystyle\nu(c_\ce{Y}) = \frac{1}{V}\frac{\mathrm dn_\ce{Y}}{\mathrm dt} \\
  & \displaystyle\nu(c_\ce{B})|_{V = \mathrm{const}} = -\frac{\mathrm d[\ce{B}]}{\mathrm dt} & \displaystyle\nu(c_\ce{Y})|_{V = \mathrm{const}} = \frac{\mathrm d[\ce{Y}]}{\mathrm dt} \\
\hline
\end{array}
$$

Rates of consumption and formation that are measured in $\pu{mol L-1 s-1}$ are defined for the closed system at constant volume, and molar concentration arises from its definition.
Whether you can or cannot substitute molar concentration $c_i$ with fractions can be answered by looking at the relationship between the two.

For example, molar concentration $c_i$ expressed in terms of mass fraction $\omega_i$ via the density of solution $\rho$ and the molar mass $M_i$ of the $i$th component:

$$c_i = \frac{\omega_i \rho}{M_i}\tag{1}$$

Similarly, molar concentration $c_i$ expressed in terms of mole fraction $x_i$ via the density of solution $\rho$ and the average molar mass $\bar{M}$ of the solution:

$$c_i = \frac{x_i\rho}{\bar{M}}\tag{2}$$

As you can see, there is a problem: as the reaction progresses, not only the amounts (or fractions) would change, but also the density and the molar mass of the solution may and will change in most cases, which leads to inapplicability of percent fractions as time derivatives.
All in all, in chemistry it's just easier to manipulate with the quantities based on amount of substance.

References

Laidler, K. J. A Glossary of Terms Used in Chemical Kinetics, Including Reaction Dynamics (IUPAC Recommendations 1996). Pure and Applied Chemistry 2009, 68 (1), 149–192. DOI: 10.1351/pac199668010149. (Free Access)


Is tetraammine dichloro cobalt (III) an optically active compound?

$\ce{Co[(NH3)_4Cl2]}$ can exist as cis and trans isomers as shown in the following figure.

image source

If a molecule contains a plane of symmetry it cannot be optically active or chiral.

In the cis isomer there is a plane of symmetry containing the $\ce{Cl-Co-Cl}$ atoms, therefore this isomer cannot be optically active.  With the trans isomer there are several planes of symmetry (one contains the $\ce{Co}$ atom and the four $\ce{NH3}$ groups, and there are four more planes of symmetry containing the $\ce{Cl-Co-Cl}$ atoms and either containing or bisecting the pairs of amino groups), therefore the trans isomer is also optically inactive.


Predicting products of chemical equations?

One would not "know" what the products of the reaction are, a priori. As mentioned by @IvanNerenten, the way you know is by studying chemistry. In fact, the reaction is different if you use cold NaOH vs hot NaOH. The reaction as written is with cold NaOH and is a disproportionation reaction.This is where the chlorine is both oxidized and reduced in the same reaction.


Can a bond be electrophilic or nucleophilic?

Bonds can indeed be considered to be nucleophilic. For example, in the electrophilic aromatic substitution mechanism an electrophile interacts with benzene’s π system. By analogy, we can also see this as a nucleophilic electron pair (situated in the HOMO) interacting with the electrophile’s LUMO.

However, since bonds are considered to be the bonding electrons which by definition carry a negative charge, bonds can only be nucleophiles or non-nucleophilic; never electrophiles.

Bonds also cannot be electron-deficient. A bond in a Lewis structure is made up of two electrons, full stop. In MO theory, we may no longer consider the bond localised but we still have an electron pair in a certain bonding orbital and thus never electron-deficient.

I suspect the book made an error here and wanted to point at the carbon atom so the text should be as you suggest.


All animals (including humans) consume significant amount of proteins. Proteins consist of aminoacids, which include amine nitrogen. This nitrogen traces back to mineral fertilizers humans use in agriculture. As proteins in our bodies are destroyed, the waste is utilized, formine urea (carbamide), so ammonia didn't cause too much trouble. However, urea is slowly hydrolized in water back into ammonia, so old urine stinks of (among other things) ammonia.


Models for diastereoselectivity in nitronate additions

Dave Evans taught me that you should generally analyze these types of problems by drawing out allyl conformations.

I've drawn the nitronate to minimize 1,3-allylic strain. Now, the top face of the nitronate is largely blocked by the t-butyl carboxylate. I strongly suspect that having a bulking alkyl group there is important to getting good selectivity.

This means that the electrophile will react on the bottom face, and that will lead to the desired stereochemistry depicted in 7 with the nitro group down on the ring.


Phase of reaction between glucose and potassium cyanide to form amygdalin

The glycoside amygdalin is a product of enzymatic biosynthesis. Its hydrolysis produces 2 molecules of glucose, benzaldehyde and hydrogen cyanide. Mixing potassium cyanide and sugar ( = sucrose = combined glucose + fructose) does not produce amygdalin. Neither you would get it, mixing glucose, benzaldehyde and hydrogen cyanide.


Why do chloroform and benzene exhibit negative deviation from Raoult's law?

This is because when these liquids are mixed, H-bonding type interactions are formed between hydrogen atom of chloroform (partial positive charge due to 3 Cl atoms) and 'pi' electron cloud of benzene ring.
Thus, chloroform-benzene interactions are stronger than chloroform-chloroform and benzene-benzene interactions.


Effect of impurities on Enthalpy of vapourization

Adding impurities(which mostly have low heat capacity, like salt) decreases the enthalpy of vaporization. This is because the impurities lower the heat capacity of the solution as a whole, making the enthalpy of vaporization lesser than the original solvent.

This is why even though the boiling point of a solution increases when impurities are added it is heated up faster(boils faster) than the one without impurities. For example, saltwater (salt is a non-volatile solute) boils faster than pure water because adding salt lowered the heat capacity of water as a result of which the enthalpy of vaporization is lesser.

Sources:

https://www.studocu.com/en/document/western-governors-university/natural-science-lab/mandatory-assignments/task-1-boiling-salt-water/5667142/view
https://www.thoughtco.com/adding-salt-to-boiling-water-607427

The addition of impurities lowers the heat capacity since the solvent molecules bind to the solute particles in a specific manner and hence require less heat to vaporize than they'd require when they're randomly moving.

Source:

https://socratic.org/questions/how-does-salt-change-the-specific-heat-capacity-of-water#143950


Which preposition to use for the reducing agent?

Either is fine.  In the most literal interpretation of the statement I would say by (as it does the actual action of reduction), however I personally would likely use with, to a scientist there is little difference though.  If you were filing a patent, the term might might matter a bit more, which a quick survey of the patent literature suggests with.  Again though this is minutia and nit-picky.

I would suggest phrasing your sentence more as:

  ...is reduced to sodium phenylselenolate with a solution of NaBH4 in alkaline ethanol .


Why amino acids (Zwitterion) become either negative or positive at low and high pH solutions?

Firstly a note about terminology. The word "terminus" is reserved for the N- or C-termini of a polypeptide chain. For a free amino acid, you should refer to the carboxyl and amino groups as the $\alpha$-$\ce{COOH}$ and $\alpha$-$\ce{NH2}$ groups respectively.

Anyway, the -$\ce{COOH}$ group is acidic; above a certain pH, typically around 2, it can be deprotonated to form a negatively charged -$\ce{COO-}$ group. On the other hand, the -$\ce{NH2}$ group is basic; below a certain pH, typically around 9, it can be protonated to form a positively charged -$\ce{NH3+}$ group. I made a short table describing the protonation states of the two groups.

$$\begin{array}{c|c|c|c}
\textbf{pH} & \textbf{Carboxyl group exists as} & \textbf{Amino group exists as} & \textbf{Net charge} \\
\hline
<2 & \ce{-COOH}\text{ (neutral)} & \ce{-NH3+}\text{ (positively charged)} & +1 \\
>2 & \ce{-COO^-}\text{ (negatively charged)} & \ce{-NH3+}\text{ (positively charged)} & 0 \\
>9 & \ce{-COO^-}\text{ (negatively charged)} & \ce{-NH2}\text{ (neutral)} & -1 \\
\end{array}
$$

For more information you can read Wikipedia or any biochemistry textbook - the first few chapters will often be devoted to discussing chemistry like this.


Possible Products in the Addition To a Double Bond in a Ring

why should the plane in which both the hydrogen atoms are present even
  matter?

In this case it doesn't.  The author of the example is laying the groundwork for the future.

The area above and below the double bond is often referred to as a "face". In your molecule, the two faces are not equivalent. If you build two models of your compound and in both color the top of the double bond red and the bottom blue, you'll find there is no way to superimpose one model onto the other where the red face is superimposed on the blue face.  These two faces are therefore not equivalent, they are said to be enantiotopic (mirror images) in this molecule.

In this case, addition of (an achiral reagent such as) hydrogen to the two faces produces the same molecule.  In more complex molecules, or when using chiral reagents this will not be the case.

Consider the case where there is a methyl group attached to the methylene group adjacent to the $\ce{C-Br}$ carbon in your molecule.  Now the faces are different (diastereotopic) and addition of hydrogen to the top and bottom faces will produce two different products (methyl and bromine on the same side or opposite sides of the molecule).

The concept of faces is important in chemistry.  Always consider if the faces are equivalent, enantiotopic or diastereotopic. Depending on what you find, the products from attack at the two faces may be the same, enantiomers or diastereomers.  The concept of faces is introduced in your problem.


Just as a quick answer, because apparently some people are in doubt so let me try to clean up the confusion.
The 'physics part'
Surface tension that people normally talk about is the liquid-air surface tension, in this case water-air, and I am pretty sure this is the surface tension you are referring to in your question. This water-air surface tension is a property of the water and air and has absolutely nothing to do with the physical shape or material of the container.
For completeness, I have to mention that the contact with the walls that have different chemical properties will change the shape of the water-air interface a bit, which you might misinterpret as a change in water-air surface tension but this is caused by the water-container and container-air surface tensions which indeed DO depend on the container.
The 'chemistry part'
What could * change the water-air surface tension is the exchange of some material out of the container into the water or the other way around. This would mean that you are not really looking at the water-air surface tension anymore, but rather at the (water$+$some solute)-air surface tension. Which might indeed be different, typically lower.
If this exchange is the mechanism then the time over which the surface tension will change is dependent on the kinetics of that process and without knowing what is being exchanged it is impossible to tell you what the time span will be, it could be seconds, it could be hours. Note that the explanation with stuff leaking out of or into the lota would mean that (assuming there will be equilibrium at some point) storing water in the lota and then drinking it from the glass would be the same as drinking it directly from the lota.

*I'm not saying it does I have no clue whether this happens with a lota


Derivation of formula for the degree of unsaturation

Well, that is simple.

A Carbon (C) makes 4 bonds, a Hydrogen (H) or a halogen (X) makes 1 bond and Nitrogen (N) makes three bonds. So in a molecule with $c$ Carbons, $h$ Hydrogens, $x$ halogens and $n$ Nitrogens, we will have $4c+h+x+3n$ "bonding capacity".

Assuming there are no rings, no double/triple bonds, we must have $c+h+x+n-1$ bonds in between these atoms [think like how many dashes you can draw between $m$ $\ce{O}$'s: $\ce{O-O-O}$ two dashes for three $\ce{O}$'s apparently].

Since each bond uses one "bonding capacity" from each of the two atoms its binding, $c+h+x+n-1$ bonds are using $2(c+h+x+n-1)$ bonding capacity. So the remaining capacity is:

$$(4c+h+x+3n)-2(c+h+x+n-1)=2c-h-x+n+2$$

If this is larger than zero, it means our molecule is unsaturated, i.e. still has a capacity to make bonds. And our assumption that "there are no rings, no double/triple bonds" is wrong. This bonding capacity is used in forming new bonds either in rings or in pi-bonds.  Since each bond uses two capacity one from each side we should have $c-h/2-x/2+n/2+1$ "unsaturated" bonds.


"Asymmetric molecules are necessarily polar"

Let's start by defining a polar molecule as a molecule with a dipole moment.

  Is a molecule with no symmetry necessarily polar?

Yes, a molecule with no symmetry belongs to point group $\ce{C_1}$ (e.g. the only symmetry element it contains is a $\ce{C_1}$ axis, rotation of the molecule 360/1 degrees converts the molecule back into itself).  The molecule is asymmetric and polar.

  Can a polar molecule still have some elements of symmetry (i.e. mirror
  image is the same as itself)? I think so ... because we can have
  achiral tetrahedral carbon molecules that are polar - i.e. methylene
  chloride.

Yes, methylene chloride, vinyl chloride, methanol, chloroform, etc. all contain planes of symmetry, but are still polar.

  So, is it proper to say that a molecule that lacks symmetry is
  necessarily polar?

If I understand correctly, this is a repeat of your first question.

  Or can a more nuanced statement based on symmetry be made? Is there a
  "minimum" level of symmetry needed to achieve non-polarity?

Yes

All asymmetric molecules are polar.  Such molecules belong to point group $\ce{C_1}$
Molecules belonging to point group $\ce{C_{n}}$ with n>1 are polar. 1,3-dimethylallene is an example (it has $\ce{C_2}$ symmetry, rotate it by 360/2 degrees about its $\ce{C_{2}}$ axis and it converts back to its original representation)
Molecules belonging to point group $\ce{C_{s}}$ are polar.  A sigma plane either contains or bisects molecules in this point group, and this plane is the only symmetry element.  Methanol, bromocyclopropane, vinyl chloride, etc. belong to this point group and are polar.
Molecules belonging to point group ${C_{nv}}$ are polar.  Formaldehyde and methylene chloride (both ${C_{2v}}$), chloroform (${C_{3v}}$) and hydrogen chloride (${C_{\infty v}}$) are examples.  This point group contains a ${C_{n}}$ axis and n sigma planes containing the axis.

If the molecule is not in one of these point groups, then it is not polar and does not have a dipole moment (or said differently, molecules in all other point groups [${D_n, S_n, C_{nh}, D_{nd}, D_{nh}, T_d, O_h}$ and higher] are not polar and do not have dipole moments).


How is charge density a useful measure of polarising power and polarisability?

What does higher charge density of a cation means? It means that the positive charge on a cation is concentrated in a small volume. Small, concentrated charge can cause greater distortion in the anion and therefore they have greater polarising power, or greater power to cause attract the negative electron cloud of the anion and hence greater distortion. The greater the positive charge on a cation, the larger the charge density will be (it is directly proportional to $Z$). Greater positive charge will cause greater distortion in the electron cloud of anion.

How do small and concentrated charge cause greater distortion? What will affect you more? A force applied on a very very small area of your body, or the same amount of force applied on whole of your body? Of course, the former one will affect us more, i.e., a small and concentrated force.

What about the anion? When the electron cloud of the anion is small, it is closer to the nucleus and experiences a greater attraction from the nucleus, therefore it is not easy to distort it. But, if the electron cloud is large, the inward force experienced by it from the nucleus is less, hence it can be polarised much easily. The inference is that the greater the radius, the larger the polarisability of the anion will be. What does greater radius mean? The greater the radius, the smaller the charge density will be (it's inversely proportional to $r$, or proportional to $1/r$).


Is plane of symmetry a necessary and sufficient condition for optical inactivity in coordination compounds?

The number $\pi$ for bowhead whales† is still 3.1415... Same thing here.
Chirality of coordination compounds is the same thing as chirality of organic compounds. If all your symmetry elements are just rotations, you have it; if some are improper rotations, then you don't. It is as simple as that, and is not going to be any simpler.
Can you just check for the planes and call it a day? Well, you kinda can, and that will do, until you come across a compound which doesn't have a plane, but still is achiral because of some other symmetry element.
Are there such compounds? Well, that's like asking whether there is a place on Earth where three bricks stand on top of each other. You may google long and hard, or you may apply for a travel grant to visit countries with rich traditions of brickwork masonry... or you may just take three bricks and put them on top of each other. That's it; problem solved.
Here are my bricks. I claim that the tetrahedral $\ce{[Zn(NH3)4}]^{2+}$ has an inversion axis $\bar4$ and hence is achiral. You may point out that it has a plane as well. Fine. Let's give it some substituents: change each ligand to $\ce{N(CH3)3}$. What? Does it still have a plane, given that everything which rotates is rotated to the appropriate position? Fine. Let's give it different substituents. Change each ligand to $\ce{CH3NHC2H5}$. Where is the plane now?
So it goes.

† There was an old anecdote about an even more old paper on biophysics which was said to include some volume calculations and some crude approximations of a whale as a cylinder, then of course the number $\pi$ showed up, followed by the hilarious phrase along the lines of "The number $\pi$ for bowhead whales is about 3.14".


What's the difference between "extent of reaction" and "position of equilibrium"?

Two differences. Firstly, the extent of reaction $\xi$ can be mathematically defined, whereas the "position of equilibrium" is for the most part qualitative, in the sense that we only say it "moves left" or "moves right", but not "the position of equilibrium is $x$".

More importantly, the extent of reaction does not actually tell you anything about the equilibrium. In this respect, it is more analogous to the reaction quotient $Q$ rather than the equilibrium constant $K$. If you measure the extent of reaction $\xi$ when the system is at equilibrium, then it does provide a measure of the equilibrium position. This is just like how if you measure $Q$ at equilibrium, it is equal to $K$. However, in principle you can measure $\xi$ (or $Q$) at any stage of the reaction, not just when it is at equilibrium.

Both $\xi$ and $Q$ are mathematical ways of describing how much products there are versus how much reactants, but they are not equivalent, they have different forms. See Wikipedia for more information.

So, I don't agree with the statement that "the extent of reaction is indicated by the value of the equilibrium constant" on two counts. Firstly, the extent of reaction is not directly "indicated" by $Q$ – they both measure the same thing, but not in the same way. Secondly, and crucially, $Q$ is not the equilibrium constant $K$: it is only equal to $K$ at equilibrium, which this sentence doesn't capture.


Is electron density a good measure of similarity between molecules?

TLDR: Defining the similarity between two molecules is an open field of research, but is fundamentally a subjective question which depends on your context. If you are interested in the similarity of molecular structures, then a graph-based measure might be a logical choice. If you're interested in similarities of reactivity, then a measure incorporating the electron density or valence electron density might make sense. Ultimately, only you can know which measure makes the most sense for the problem you are trying to solve.

This is a good question and one which does not (and likely never will) have a definitive answer. That is, there will never be a physically correct way to answer how similar two different molecules should be, or even what type of mathematical object that answer should be (i.e. a number, vector, matrix, etc.). Nonetheless, some measure of the similarity between systems is clearly a useful thing to know about, and it is becoming more practical as machine learning becomes more common in chemistry.

From what I have seen recently, many of the most commonly used similarity measures are derived from a representation of a molecule as a graph. I think there are two main reasons for this. First, most people who are interested in molecular similarity are also interested in machine learning. Hence, they are working with a relatively large dataset. Computing a graph representation of a molecule and performing operations on this graph is almost always significantly cheaper than computing some kind of quantum mechanical property for every molecule in your dataset. Second, machine learning projects in chemistry tend to be very collaborative and there many applied mathematicians and computer scientists that know a lot about graph theory and are hence eager to apply this knowledge in a new context.

I have found a review from 2003 which discusses some more chemically motivated methods of defining molecular similarity and some methods based on a graph representation of the molecule[1]. They also discuss a similarity measure based on the overlap a field. The most natural choice for the field being the electron density, though one could choose something like an electrostatic potential or even some kind of magnetic vector potential. What you choose would just depend on what type of similarity you're thinking of.

As an example, you propose using the electron density due to valence electrons. I would be surprised if this hasn't been done before, but I couldn't find any references where this is done. This would be very natural if one is interested in similarities with regards to reactivity. On the other hand, if you were interested in predicting how similar the x-ray absorption spectra of various compounds would be, then it would be stupid to leave out the core electrons from the density as those are the electrons which are most relevant. This further illustrates that the similarity between two molecules is unanswerable because it is not a well-defined question. That is, this similarity will inevitably be connected to some way in which two molecules produce a similar outcome in some physical situation. If you're interested in predicting a molecule which will behave similarly to some other molecule in a reaction, then valence electron density could be a great descriptor, but this descriptor might be entirely useless if you're actually interested in finding molecules with similar x-ray absorption spectra.

Now, chemists have a serious habit of trying to make a connection between everything a molecule does and the structure of that molecule. Very often, this works out extraordinarily well, so chemists keep doing it. I think this is another reason why graph based measures of similarity have become so popular. Virtually all methods of determining molecular similarity based on a graph begin by computing the adjacency matrix of the molecule based on connectivity along covalent bonds (or maybe hydrogen bonds). Then, some operation is performed on a pair of graphs which results in a number or vector or somethings else which is interpreted as a measure of similarity. Or, an operation is performed on each graph individually, and then some measure of distance is used to define the similarity of these graphs. These operations are usually called kernels. A review of some of the kernels which have been used to define graph similarity can be found here[2].

There are all sorts of tricks which can be played with the adjacency matrix to give some more physical meaning to the measure of similarity. For instance, you might weight each of the edges by the atom-atom distance defining that connection. You might also weight each of the nodes with an atomic charge or polarizability and incorporate this into the adjacency matrix. You might also choose to do this and then work with the laplacian of the graph rather than the adjacency matrix. Anything at all can be done really. You just have to choose what makes sense for your context.

More recently, people have come up with measures that also incorporate similarities of the subgraphs of two graphs[3]. This seems like it has some potential advantages if functional groups, etc. are important to you.

References:

[1]: Nikolova, N., & Jaworska, J. (2003). Approaches to measure chemical similarity–a review. QSAR & Combinatorial Science, 22(9‐10), 1006-1026.

[2]: Rupp, M., & Schneider, G. (2010). Graph kernels for molecular similarity. Molecular Informatics, 29(4), 266-273.

[3]: Kondor, R., & Pan, H. (2016). The multiscale laplacian graph kernel. In Advances in Neural Information Processing Systems (pp. 2990-2998).


Thermal decomposition of mercury thiocyanate

The reaction is the decomposition of mercury(II) thiocyanate:

  $\ce{2Hg(SCN)2 → 2HgS + CS2 + C3N4}$

Where $\ce {C3N4} $ is the compound that makes up the "snakes," and grow due to the evolution of gases produced by the decomposition of the other products:

  $\ce{CS2 + 3O2 → CO2 + 2SO2} $

  $\ce {2C3N4 → 3(CN)2 + N2} $

  $\ce {HgS + O2 → Hg + SO2}$

This is also known as the "Pharaoh's Serpent" in pyrotechnics. The resulting solid can range from dark graphite grey to light tan in color with the inside generally much darker than the outside.
Click here and here for reference.


I checked the literature, it has been reported that fluorine ions replace hydroxyl ions in osseous tissue (bone) so by making a measurement of the flourine content of bone it is possible to estimate the age of the bones.

One early paper on the subject is Cook, S. F.; Ezra-cohn, H. C. An Evaluation of the Fluorine Dating Method. Southwestern Journal of Anthropology 1959, 15 (3), 276–290. DOI: 10.1086/soutjanth.15.3.3628977.

One method would be to use neutron activation anaylsis on bones to measure the flourine-19, using the n,gamma reaction. This method does have a problem as sodium-23 can form flourine-20 by a n,alpha reaction. As a result a correction will have to be made for the sodium content of the bones.

This is discussed in Cheng, T. P.; Anderson, H. D.; Mills, D. S.; Spate, V. L.; Baskett, C. K.; Morris, J. S. Determination of the fluoride distribution in rabbit bone using instrumental neutron activation analysis. J. Radioanal. Nucl. Chem. 1997, 217 (2), 171–174. DOI: 10.1007/BF02034437.


How to decide acidic or basic nature of oxides

In general, the electropositive character of the oxide's central atom will determine whether the oxide will be acidic or basic. The more electropositive the central atom, the more basic the oxide. The more electronegative the central atom, the more acidic the oxide. Electropositive character increases from right to left across the periodic table and increases down the column.
The trend of acid-base behaviour is from strongly basic oxides on the left-hand side to strongly acidic ones on the right, via an amphoteric oxide (aluminium oxide) in the middle. An amphoteric oxide is one that shows both acidic and basic properties.
This trend applies only to the oxides of the individual elements in the highest oxidation states for those elements. The pattern is less clear for other oxides.

We define non-metal oxide acidity in terms of the acidic solutions formed in reactions with water. For example, sulfur trioxide reacts with water to forms sulfuric acid.
In sum, acidic oxides are oxides of non-metals, and basic oxides are oxides of metals.

There are three non-metal oxides from the upper right portion of the periodic table, $\ce{CO}$, $\ce{NO}$, and $\ce{N2O}$, which have such low oxidation numbers for the central atom that they give neutral aqueous solutions.

Since the acidity of a cation rises rapidly with charge, d-block elements that exhibit a wide variety of oxidation numbers may have one or more oxides that exhibit only basic properties and one or more oxides that exhibit only acidic properties. The higher the oxidation number, the more acidic the corresponding oxide. Chromium is an example of such an element. $\ce{CrO}$ is basic, $\ce{Cr2O3}$ is amphoteric and  $\ce{CrO3}$ is acidic.


How can I measure amount of ozone byproduct coming out of hydroxyl generator

Ozone is very reactive, so unless you have an industrial scale hydroxyl generator, the ozone will readily react with organic matter in the air and surrounding surfaces before it builds up to dangerous levels. There are commercial portable instruments that are sold to monitor ozone concentration, they will be in the \$500 to $1500 price range for the instrument. You can also look at this simple experimental apparatus described in the Journal of Chemical Education for a cheaper alternative.


How to apply the Heisenberg Uncertainty Principle in calculations on the movement of bacteria?

$\newcommand{\upmu}{{\large\unicode[Times]{x3BC}}}%Remove this line when upright greek characters are implemented by mathjax$
(Hover over the boxes to see the results.)
The velocity of the  bacteria is $\mathbf{v} = 9~\mathrm{\upmu m/s} = 9\times10^{-6}~\mathrm{m/s}$, with an uncertainty of $9.00\%$ you can calculate the absolute uncertainty in the velocity $\Delta \mathbf{v}$ as the difference of the upper bound $(\mathbf{v} + 0.09\mathbf{v})$ and the lower bound $(\mathbf{v} - 0.09\mathbf{v})$.

   $$\Delta \mathbf{v} = (\mathbf{v} + 0.09\mathbf{v}) - (\mathbf{v} - 0.09\mathbf{v}) = 0.18\mathbf{v} = 1.62\times10^{-6}~\mathrm{m\,s^{-1}}$$

The momentum is defined as
$$\mathbf{p} = m\mathbf{v},$$
hence the absolute uncertainty in the momentum is
$$\Delta\mathbf{p} = m\Delta\mathbf{v}.$$
The mass is $m = 0.100~\mathrm{fg} = 1.00\times10^{-19}~\mathrm{kg}$.

   $$\Delta\mathbf{p} = m\Delta\mathbf{v} = 1.62\times10^{-25}~\mathrm{kg\,m\,s^{-1}}$$

Now you can use this to calculate the uncertainty in position $\Delta\mathbf{x}$.
$$\Delta\mathbf{x}\Delta\mathbf{p} \geq \frac{h}{2}$$
Planck's constant is $h = 6.62607004\times 10^{-34}~\mathrm{m^2\,kg\,s^{-1}}$.

   $$\Delta\mathbf{x} \geq \frac{h}{2\Delta\mathbf{p}}\\\Delta\mathbf{x} \geq 2.07\times10^{-9}~\mathrm{m}$$

Since you have not included any of the steps how you arrived at your result, I cannot help you finding your mistake.


Defining sensitivity of TOF Mass Spec - Does the detector play a role?

The term sensitivity has been misused in analytical chemistry a lot (even in German scientific literature). Be careful if you read articles especially the older ones. Sensitivity, in a true sense, refers to the slope of the calibration curve. Thankfully IUPAC sorted this out and defined it as "The slope of the calibration curve. If the curve is in fact a 'curve', rather than a straight line, then of course sensitivity will be a function of analyte concentration or amount. If sensitivity is to be a unique performance characteristic, it must depend only on the chemical measurement process, not upon scale factors."

When we wish to refer to the minimum amount that can be detected, it is called the limit of detection (LOD). Hope this clarifies the terminology. Note that sensitivity and limit of detection are related. Higher the sensitivity (= slope of calibration curve), lower the detection limit, provided that noise does not increase.

Coming to the detectors: The limit of detection of all detectors which are based on counting (basically we count ions indirectly in mass spectrometry) is limited by statistics rather not by electronics! This is the fundamental limit put by mother Nature. It is called shot noise. Whenever you count discrete random events, in case of a mass spec. random impinging of ions, there is always an error associated with it which is proportional to square root of N, where N is the number of events.

You can read the article on shot noise (for educational purposes).The Ultimate Limit in Measurements by Instrumental Analysis: An Interesting Account of Schroteffekt and Shot Noise, https://pubs.acs.org/doi/abs/10.1021/acs.jchemed.8b00202


How can aqueous KOH give alcohol by SN2 of alkyl halide?

When you are considering a particular SN2 reaction, you have to consider the nucleophilicity as well as the nature of the leaving group of both the groups.

Here as you mentioned, $\ce{OH-}$ and $\ce{I-}$  (and other halide ions, except $\ce{F-}$, which is a very bad nucleophile) have comparable nucleophilicity in protic solvents. But, as for their leaving nature, $\ce{I-}$ (as well as other halides) is far more better leaving group than $\ce{OH-}$, because we know weak bases are better leaving groups. That is why the reaction is feasible.


Determine vmax and enzyme concentration (Michaelis–Menten)

Michaelis-Menten kinetics is given by
$$V = V_{\max} \frac{S}{K_M + S}$$
Taking the inverse of both sides to linearise the equation,
$$\frac{1}{V} = \frac{1}{V_\max} +\frac{K_M}{V_{\max} S}$$
Taking $y = 1/V$ and $x = 1/S$ in $y = A + Bx$ leads by linear regression to $A = 0.4767$, $B = 4.4754$ (not including units). Therefore,
$$V_\max = \frac{1}{A} = \pu{2.10 mMs^{-1}}$$
$$K_M = B V_\max = \pu{9.39 mM}$$
For the new situation with different $V$ and $S$ and necessarily different $[E_0]$ , $V_{\max}$ would change because $V_{\max} = k_b [E_0]$ but $K_M$ remains constant, thus
$$V_{\max} = V \frac{K_M + S}{S} = \pu{0.101mM s^{-1}} \frac{(\pu{9.39 mM} + \pu{24 mM})}{\pu{24 mM}} = \pu{0.141 mM s^{-1}}$$
You can then calculate $k_b$ by $k_b = V_{\max}/E_0$.


What does the reaction of hydrogen and iodine to hydrogen iodide look like?

To show an example for equilibrium reactions to your students, you can use a reaction which has visual change as shifting the equilibrium position such as color. My best bet is, use of cobalt(II) chloride equilibrium with water:
$$\ce{[CoCl4]^2-(aq) + 6H2O(l) <=> [Co(H2O)6]^2+(aq) + 4Cl-(aq)}$$

The beauty her is The ion, $\ce{[Co(H2O)6]^2-(aq)}$ is pink while $\ce{[CoCl4]^2-(aq)}$ is blue.  According to the Le Chatelier’s principle, at relatively low chloride concentrations (e.g., diluting the solution or adding $\ce{AgNO3(aq)}$ to remove $\ce{Cl-}$ ions), the equilibrium shifted to the right and, therefore, the solution becomes pink. However, if there is a large concentration of excess chloride (e.g., adding $\ce{Cl-}$ ions by means of adding $\pu{12 M}~\ce{HCl}$ drop wise or adding $\ce{CaCl2}$ beads), the equilibrium tends to shift to the left, making the solution blue.

In addition, similar to its sensitivity to the concentration of solutes, the equilibrium is also sensitive to temperature as well (See picture below). At colder temperatures, the equilibrium tends to shift to the right(pink color at LHS of the picture) while at warmer temperatures, it lies to the left (blue color at RHS of the picture):

Reading:

http://www.rsc.org/learn-chemistry/resource/res00000001/the-equilibrium-between-two-coloured-cobalt-species?cmpid=CMP00005957
https://eic.rsc.org/cpd/understanding-equilibrium-a-delicate-balance/2000012.article
http://research.chem.psu.edu/mallouk/ilab/Equilibrium_and_Le_Chatelier.htm


Why does more heat transfer take place in a reversible process than in a irreversible process?

In the Clausius inequality, the temperature in the denominator is supposed to be the temperature at the boundary of the system where the heat transfer is taking place.  This important point is often omitted from many thermodynamics textbooks.  In a reversible process, the temperature at the boundary is essentially equal to the temperature of the system, because the system temperature is uniform throughout.  But, in an irreversible process, the temperature of the system is not uniform, and the temperature at the boundary can differ substantially from even the average temperature of the system.  So the Clausius inequality does not always require that $\delta Q_{rev}\geq \delta Q$.


What can be a metal cleanser for aluminum?

Really, ASTM procedures must be followed as written for certified testing. However, if this is for your edification or for a school project, you might use a commercial cleaner, such as those below, or just one ingredient such as phosphoric acid.

There are a number of commercial aluminum corrosion cleaners, such as Loctite Duro Aluminum Jelly and CRC MaryKate Aluminex Clean and Shine.

If you want to make your own:

Ingredients listed in the Aluminum Jelly Material Safety Data Sheet include phosphoric acid, diethyleneglycol monoethyl ether, nonionic surfactant, isopropyl alcohol , polysaccharide and sulfuric acid. [Similar formulaions may have contained hydrofluoric acid to break the highly-adherent $\ce{Al2O3}$ surface film and corrosion, but due to the hazards of $\ce{HF}$, I believe it is no longer sold in most places, for good reason.]

Aluminex Clean and Shine's MSDS includes ethoxylated tallow alkyl amines, sodium decylglucosides, hydroxypropyl phosphate and potassium iodide.


Effect of pressure on viscosity of liquids and gases

One of the best websites for all advanced water properties is a website maintained by Dr. Martin Chaplin http://www1.lsbu.ac.uk/water/physical_anomalies.html#Pvisc

Keep in mind that viscosity is a really complicated subject and water is among the unique solvents. In general viscosity of liquids increases with pressure. At pressure extremes, let us say, 100,000 psi, many liquids become solids. Water also becomes a solid at really really high pressures. So when you talk about viscosity, you should talk about what pressure and what temperature.

From the website:

  Water's pressure-viscosity behavior [534, 2890] can be explained by
  the increased pressure (up to about 100-200 MPa) causing deformation,
  so reducing the strength of the hydrogen-bonded network, which is also
  partially responsible for the viscosity. This reduction in
  cohesiveness more than compensates for the reduced void volume. It is
  thus a direct consequence of the balance between hydrogen-bonding
  effects and the van der Waals dispersion forces [558] in water; with
  hydrogen-bonding prevailing at lower temperatures and pressures. At
  higher pressures (and densities), the balance between hydrogen-bonding
  effects and the van der Waals dispersion forces is tipped in favor of
  the dispersion forces and the remaining hydrogen bonds are stronger
  due to the close proximity of the contributing oxygen atoms [655].
  Viscosity, then, increases with pressure.

Second question on the viscosity of gases

The textbooks seems to be talking about a limited range of P. Viscosity of gases is dependent on very low or very high pressures. However, for ideal gases and normal pressures, it is only dependent on the temperature. Yes, it is quite counterintuitive. Also viscosity of gases increases with temperature. This is another surprise. https://en.wikipedia.org/wiki/Viscosity#Pure_gases

You can try posting this second query in Physics forum. I don't know the derivation of that Wikipedia expression.


How to predict which complex is formed and which are soluble?

Please be mindful, in my opinion, that your source material may not be of the highest quality.
For example, NH4OH is more accurately represented as aqueous ammonia (that is, NH3(aq), see, for example, this source's comment).
Second, Al(OH)3 does not form a complex with ammonia or, as your reference claims, is 'partially soluble' with aqueous ammonia to any significant extent, and if at all, likely due to the presence of OH-. See, for example, this ebook, Basic Analytical Chemistry
 discussing various metal compounds interactions with aqueous ammonia.
Third, a more accurate depiction of the action of ammonia on Cu(OH)2, for example, is the formation of a complex of the form, $\ce{[Cu(NH3)_x(H2O)y](+)}$, where x + y = 6, with also a possible charge of 2+ for cupric, where the concentration of ammonia influences the value of 'x' (supplying guidance on your question as to which possible complex may form).
Lastly, high strength ammonia is needed to react with some of the listed metal hydroxides and is characteristic of the particular metal. See my suggested source, that may address your question on which salts are (or, are not) soluble.


Are all enolised molecules called enolates?

There are two questions in here.

First enols/enolates. It is correct that all enolates are enolised carbonyl compounds but not all enolised carbonyl compounds are enolates. Take for example pentan-2,4-dione. While in protic solvents this will indeed have two $\ce{C=O}$ groups, if dissolved in unpolar solvents it will enolise to allow for a cyclic intramolecular hydrogen bond for stabilisation (see scheme).

Obviously, the resulting compound is an enol, as the functional group $\ce{R2C=CR-OH}$ (a hydroxy group bound directly to a vinylic double bound) is present. However, it is not an anion so it is not an enolate.

Most carbonyl compounds prefer the keto-configuration but if there are electronic factors that favour enolisation (e.g. enolisation results in a sufficiently large conjugated system) then an enol will be preferred.

Enols can be easily deprotonated at the oxygen so the generation of an enolate is straightforward. However, since ketone and enols are tautomers of another with the key limiting step being hydrogen transfer ([1,3]-suprafacial hydrogen migration is forbidden), once the proton is removed the difference becomes purely electronic and thus deprotonated ketones are routinely referred to as enolates whether or not the cation seems more attached to the oxygen or the carbon. (Reaction mechanisms differ; sometimes only one attack (often carbon) results in a productive mechanism so it seems as if the other had not reacted or be not present but this is often a hindsight fallacy.)

The term enolate is not limited strictly to ketones. For one of the more synthetically relevant examples, see the Evans enolate used in Evans alkylation or Evans aldol reactions or the Myers enolate used in Myers alkylation reactions; both are amide enolates.

However, these compounds are still characterised by the presence of a $\ce{(C-)C=O}$ bond which upon enolisation becomes a $\ce{(C=)C-OH}$ functionality. In the case of a nitrile or an imine, the oxygen is not present so even though a similar reaction may take place it is not correct to call the resulting product enol. For imines, the compound is known as an enamine and happens to be the standard outcome if a secondary amine is reacted with a carbonyl group.

For nitriles, I am not aware of a similar shorthand and would probably simply call it a nitrile anion (as does Wikipedia). (It is worth pointing out that nitrile anions are somewhat different from enamides/enolates as they are essentially an aza-ketene, not an aza-allyl anion.)


Applying vacuum at the end of a flash chromatography column

All chromatography needs to work is some pressure differential, no matter how it's established. Gravitational chromatography uses hydrostatic pressure from the weight of the solvent, and a typical gravitational column can have a roughly ~20-50 mbar pressure difference between the top and bottom of the column; flash chromatography also has the hydrostatic pressure, but has an additional positive gas pressure at the top from e.g. pumped air, and can typically reach ~100 mbar pressure differences.

The principle of chromatography is to force compounds in a mixture to stay at the interface of a stationary phase (e.g. silica) and a mobile phase. The slight differences in how molecules behave in this boundary are added up over time, creating the observed separation.

Since this is a surface process, having a stationary phase with higher specific surface area and better packing effectively means molecules spend more time undergoing separation at the interface than being dissolved in the bulk of the eluent, where no separation happens. However, a more tightly packed stationary phase also restricts eluent flow. Under a simplified set of assumptions, decreasing the stationary phase particle size by half (and therefore increasing specific surface area by a factor of 4) requires 4 times the pressure differential to produce the same flow rate. So getting better separation in reasonable timeframes requires larger pressure differentials.

This is where vacuum chromatography techniques come in. Vacuum pumps are a staple laboratory instrument, and can easily achieve negative pressures of ~900 mbar with common glassware. Therefore, without too much hassle, it is possible to obtain good flow rates using finer stationary phase particle sizes and get improved separation.

Of course, larger pressure differentials can be achieved with positive pressure at the top of the column. Medium, high and ultra-high pressure chromatography (MPLC, HPLC, UHPLC) make use of stationary phases with very fine particles (from ~10 μm to ~1 μm diameter) which start to look more like silt or clay. The only way to force liquids through these stationary phases is with very large pressure differentials, from ~5 bar to as much as ~1000 bar, but they can achieve incredible resolution of compounds during separation. The reason UHPLC isn't used for every single separation is cost and scalability; it's an engineering challenge to control such enormous pressures safely. Glass is very resilient under compression (which is why glassware can commonly hold a vacuum), but weak under expansion. This means such positive pressures need much more rigid containers, such as thick steel columns. This also limits their feasible sizes and therefore the amount of stationary phase/eluent/compound they can handle.

As an example of a vacuum chromatography technique, I point towards the excellent video by Daniel Pedersen detailing dry column vacuum chromatography (DCVC). There is also a preceding publication which goes into further details. For what it's worth, I've found this vacuum chromatography technique to be far superior to flash chromatography.


It is more acurate to say that they aren't very far from being electrically neutral. If you take a charged sphere and drop it into a beaker of some electrolyte sitting on an insulating counter, the result is a charged solution, albeit slightly.

Also, in spite of your amusing ceiling analogy (it made me chuckle), the argument about electrostatic attraction and repulsion does justify why solutions (and solids) can't be too far from being electrically neutral.  The flaw in the analogy is that humans don't need ceilings to exist; but you can' t have two charged beakers without a force between them. If each has a charge of 1 C -- which is a mere $10^{-5}$ times the charge of a mole of electrons -- and they're sitting 1m apart the force between them is about a million tons -- clearly this situation won't last very long!


Are carbocation intermediates in equilibrium?

Are the two carbocations in an equilibrium or are they formed in specific, fixed quantities?

$\ce{R-CH=CH2 + H^+ <=> R-CH2-CH2^+ + RCH^+-CH3 <=> alcohol~products}$

Yes, the two carbocations formed are in equilibrium. Initially the two carbocations will be formed in a ratio dependent upon the difference in activation energies leading to them (kinetic control).  However, since all of the steps in this reaction involve equilibria, after a while when equilibrium is established several things can happen

many carbocations will revert back to starting material which can then be reprotonated and reform the carbocations
or alcoholic product can be reprotonated, eliminate water and reform the carbocations
or the carbocations themselves can equilibrate directly by hydrogen shift

Once the equilibrium is established, the ratio of carbocations formed will be dependent upon the energy difference between the carbocations themselves (thermodynamic control).  Often, but not always, the relative stabilities of the transition states leading to the carbocations matches the relative stabilities of the carbocation intermediates themselves.

  If they are in equilibrium, i.e., the primary and secondary carbocation, why and how would the secondary carbocation revert back
  to being a less stable primary carbocation?

why - that's what happens in equibria; even though phenol is much, much more stable than its enol counterpart, there will be an exceedingly small amount of the enol present at equilibrium

how - by any of the 3 routes mentioned in the bullets above

  Also, is the hydride shift that converts the primary to a secondary carbocation the only shift that can occur; even if after the shift, a
  more stable carbocation can be formed (depending on the attached alkyl
  group)?

Other groups like alkyl or phenyl can shift too, but hydrogen shift is usually faster than the others.


Abundant substances which I may use like coupling agents

If you don'y want to use maleated polyolefin (the best coupling agent for this composite), you can use oxidized polyethylene, ethylene propylene rubber, ethylene-co-glycidyl methacrylate or polyolefin elastomer.


Why does using NaI as a nucleophile and acetone as solvent lead to an SN2 reaction mechanism instead of SN1? Is I- not considered a weak nucleophile?

As Jan wrote, there isn't a hard and fast rule encompassing every mechanism, so there may not be a definitive answer to the question. Nevertheless, I found an interesting site that talked a little bit about the Finkelstein reaction:

$$\ce{NaI + RX -> NaX + RI}\qquad\ce{X} = \ce{Cl}, \ce{Br}$$

This is in acetone solvent (I don't know how to format it along the arrow).

If you go by Henry Rzepa's blog, the solvation of sodium halides by acetone leads to a solvent shell that isn't well "fit" for a chloride or bromide anion. The writer discusses how the gaps between solvated sodium ions is just large enough to fit an iodide anion, which means the anion has minimal distance to the sodium atoms around it, and therefore a minimized potential energy. Bromide and chloride are too small, leaving a gap of empty space around them, and the distance to sodium cations is larger than optimal. It becomes energetically more favorable for the sodium and halide ions to simply fall out of solution than arrange themselves in a solvated cage. This drives the reaction forward in accordance with Le Chatelier's principle.

While this relates to the solubility of the halide salts, maybe it can also serve to speculate on the mechanism of the Finkelstein reaction. In the vicinity of the solvated $\ce{NaI}$ cage, there isn't a favorable space to release bromide or chloride into, so the iodine attack is forced to proceed through $\mathrm{S_N2}$.

This is far from rigorous research, so I'd take it with a grain of "salt" as a correct answer, but thought it might be interesting to think about.


Did you consider PMMA?

A 3 mm Plexiglas G UVT layer seems to have a transmittance > 80% at $\lambda = 300~\mathrm{nm}$ and above.


For molecules to be enantiomers, do the groups have to be non-superimposable mirror images even when rotated around C-C single bonds?

Real molecules surely do rotate around their single C-C bonds all the time(*). So yes, if two models can be superimposed after an internal rotation of such sort, this means they represent the same molecule, as is the case with your B and C.

This, BTW, is the reason why you should orient all substituents in a standard manner when drawing your Fischer projections. Otherwise any molecule would have just too many variants, and you'll get bogged down in them. Think of rotating the upper group the third way (the first two being B and C). Think of rotating the lower group too...

Then again, I've rarely seen Fischer projections used for anything other than sugars. Typically people just determine the absolute configuration (R or S) and go with that. It gives you a definitive answer, after all.

(*) That is, most of them do. When you have bulky substituents, rotation may pretty well get blocked, and then you may be able to separate the isomers that are (nominally) produced by this rotation - they are no longer equivalent. Look at BINOL or BINAP, for example.


Keeping an open, partially empty soda bottle bubbly longer

The fizz is dissolved carbon dioxide (or carbonic acid with which it is in equilibrium) which is put into the bottle under quite some pressure (around $\pu{120 kPa}$ at $\pu{4 ^\circ C}$ or $\pu{250 kPa}$ at $\pu{20 ^\circ C}$; from a first glance on the google results).
$$\ce{CO2 (g) + H2O (aq) <=> H2CO3 (aq)}$$

Because the bottle (in first approximation) is a closed system, there is an established equilibrium between the "air" in the bottle (gaseous $\ce{CO2}$) and the dissolved $\ce{CO2}$. Once the bottle is opened the equilibrium is disrupted and the dissolved carbon dioxide is no longer as soluble as before.

If you close the bottle again and use the little pump you essentially add pressure again, shifting the equilibrium back to where more dissolved carbon dioxide is stable. It won't be as effective as the factory produced bottle, but it will slow down loosing the fizz.

The volumes also play some role here, ideally the aqueous phase should be much larger than the gaseous phase, as you need more molecules producing the pressure if you have more space. An half empty bottle therefore is not ideal to start with.

Squeezing the air out of the bottle would in the worst case result in the opposite effect that you want, in the best case it won't change much. As you correctly remarked, the bottle reforms which means that you would have created lower pressure, essentially dragging out more carbon dioxide molecules. If that was not the case, you would reduce the volume of the gaseous phase, which would probably help a little. Without adding high pressure though it will still loose most of the fizz over time, until equilibrium is reestablished.

In principle there is not really very much you can do. Keep the bottles refrigerated (as carbon dioxide is more soluble at lower temperatures) and try to finish them quickly. Alternatively, get yourself a soda maker and enjoy all the fizzy drinks in all the customised quantities whenever you want. (I guess, this is not a plug for soda maker machines, I personally prefer non fizzy drinks.)


How many empty orbitals do elements have?

In my view there is an important difference between empty orbitals and empty valence orbital.  There are an infinite number of orbitals, a finite number of filled orbitals, but only one valence orbital.

The definition of a valence electron is usually something like, "The electrons in the outermost shell are the valence electrons--the electrons on an atom that can be gained or lost in a chemical reaction."  Therefore, a valence orbital would be the orbital holding those electrons.  If the valence orbital is full, it is full; there are no other valence orbitals.

Based on this understanding, it would be correct to say "the fluoride anion has no empty valence orbitals."