Why is ortho-hydroxybenzoic acid more acidic than its para-isomer?

I don't think that in the first link it is stated that the intermolecular $\ce{H}$-bonding is stronger than the intramolecular one. On the contrary, the accepted answer starts as:

"I don't think there is much of a difference between the strengths of intramolecular or intermolecular hydrogen bonds."  With which I agree.

In that post they discuss the effect of intermolecular and intramolecular $\ce{H}$-bonding on the melting and boiling points of the compounds.

The answer by @AllYouCanEat86 that follows in the same post I think is more relevant to your question.

Therefore in our case there are intermolecular $\ce{H}$-bonds in p-hydroxybenzoic acid but they don't last for long due to the thermal motion and higher degrees of freedom of the $\ce{COO-}$ and $\ce{OH}$ groups. In contrast, in the ortho-isomer the $\ce{COO-}$ and $\ce{OH}$ groups are locked in close proximity due to the aromatic ring and therefore are forced to maintain the $\ce{H}$-bond for much longer. This could be thought as an equivalent to the chelation effect in metal complexes. Lastly, one could also argue that in the case of o-hydroxybenzoinc acid, there is both the option of intermolecular and intramolecular $\ce{H}$-bonding compared to the para where there is only intermolecular $\ce{H}$-bonding.


Which mutated enzymes act as better drugs than their non-mutated form?

There are mutants of clotting factor VIII that have longer half lives and better activity than natural Factor VIII. Factor VIII is one of the missing proteins commonly involved in hemophilia. Hemophiliacs usually use injections of Factor VIII to prevent bleeding. The mutant form should last longer in the blood, with better activity, resulting in fewer injections.

source: https://encrypted.google.com/patents/US7211559

Since this is a 2007 patent, I don't know if this product is actually being used to treat patients, drug approval takes a long time.


When considering Van der Waals forces, why do dipoles form?

on average they form dipoles.

Not quite. On average they form the nice (electrically neutral) electron orbitals we know and love. However, when they happen to be in a dipole state, there is an attraction between the two atoms, and that is evidently more than enough to overcome the randomly occurring opposing dipole states. That's just mathematics.

  why is the potential energy of a dipole lower than a non-dipole

It isn't. But quantum mechanics allows electrons to visit higher energy states, with a probability dependent on the energy difference.

  What determines which electrons are going to move to the middle of the atoms to form a dipole, and which ones move to the outside of an atom

Nothing. The  "movement" of electrons is just a random statistical fluctuation in the location of the electrons' centers of charge. Quantum-mechanical events like this are inherently nondeterministic.  And strictly speaking, you can't tell electrons apart anyway. There is no "which electron," at one point in time you have one electron here and another there, the next they may have switched places. Or not, you can never know.

  (I assume that a dipole is formed like this $e−..ion...........e−..ion$ for the centre of positive charge shifted rightwards of the centre of negative charge)

Don't think of those as being electrons, just the statistical average of the negative charge distribution.

  given that the jitters are instantaneous, and photons travel only at the speed of light, how are dipoles formed quickly enough before the next jitter

First, They're not quite instantaneous (that would have them basically everywhere possible at once), and not quite jitters.

In summary, don't think of electrons as particles with a distinct position moving around, just think of them as waves, with a peak that occasionally forms, and that peak causes the electric arrangement of two nearby atoms to, on average, be mildly attractive.


Which one has the highest boiling point?

Boiling point elevations proportional to the molar concentration of dissolved particles.  So that's what you have to figure out, being careful to (1) properly convert mass to moles and (2) recognizing that some of the solutes can form ions so you can get more than one particle per "molecule".

Let's look at glucose.  If you work out the molecular mass it's about 180 grams per mole.  Divide that into 10 grams and you get about 0.056 moles of glucose molecules in your one liter of solution.

Now try sodium sulfate.  For this one you have a "molecular" mass of 142 grams per mole, so you seem to get about 0.070 mole of particles.  Except ... Sodium sulfate forms ions, each "molecule" gives one sulfate ion and two sodium ions.  Thus three particles overall, and you have to multiply your 0.070 figure by that factor of 3.  Thus the sodium sulfate solution will have 0.21 mole of ions in one liter.  The sodium sulfate solution will therefore give a higher boiling point than the glucose solution.

Work it out now with the other two choices and pick the one with the most particles in solution including ion formation.  Be sure to use reasonably accurate atomic masses because this set of choices could throw you for a loop.


Do secondary alcohols prefer to undergo SN1 or SN2 reactions when reacting with a strong acid?

Actually, secondary alcohols undergoes $\mathrm{S_N1}$ or $\mathrm{S_N2}$ depending upon the substrate you are using. For e.g. In your case(as provided above) the substrate has steric hindrance nearby alcoholic group and the carbocation will be stable due to the migratory aptitude of the methyl group. Hence, here should be $\mathrm{S_N1}$ and you have to see the structure of substrate to decide $\mathrm{S_N1}$ or $\mathrm{S_N2}$ for secondary alcohols.
Also, alcohols are not good leaving groups also! But after protonation the water is a very good leaving group.


After mixing, which species has the dominant affect on the pH of the solution that remains?

CsOH is a strong base.

0.5 L X 1.5 M = 0.75 moles of hydroxide.

Similarly, there are 0.5 moles of monoprotic acid.

After mixing there will be 0.25 moles of hydroxide in 1 liter of solution.

In other words [OH-] = 0.25M

pH is approximately 14 + log (0.25) = 13.4


Why does electronegativity generally increase across a period?

Electronegativity is the tendency of atom to attract the electron cloud towards itself in a covalent bond. Across the period, since the electrons are filled in the same shell(which means it should be of the same size without any electron) and the amount of positive charge increases and as a consequence the size of the atom also decreases(Each electron is pulled more strongly by a nucleus with higher positive charge). The electrons are hence more tightly held as you go across the period.

  I thought though it was because as you go across a period, an atom
  will slowly fill up its outer shell, and hence reduce its need to gain
  or hog electrons?? Or is this incorrect?

A need to gain or lose electron is dictated by the nearest configuration of greatest stability(Usually Octet). And this is not closely related to Electronegativity either. Electronegativity only has to do with how strongly can an atom pull the electron cloud towards itself. Actually, as you go across the period, the tendency to gain electron increases(Indicated by increase in Electron Affinity) and not decrease as you thought.


For Gibbs Free Energy, the standard equation is ΔG=ΔH−TΔS. The overall change in Gibbs Free Energy will depend, at least in part, on the enthalpy of dissolution for the specific substance. As for entropy, increasing entropy by breaking up the crystal decreases Gibbs Free Energy. The saturation point is where ΔG for the dissolution equals zero: where ΔH = TΔS. As ΔH and ΔS vary depending on the particulars of the solute and solvent, the saturation point also varies.
The equation you cite is specifically for when you have two solutions, so the dissolution of the solid is not described by it. Since the saturation point depends on the enthalpy of dissolution, you can't use the log equation to find it.


Giving advice on preventing sunburns may fall under health advice, which is not permitted on this site. For advice, you should ask a medical professional.
The following information is from The Melanoma Foundation.
What is the difference between sunscreen and sunblock?

Sunscreens can be classified into two major types: chemical and physical. Chemical sunscreens contain special ingredients that act as filters and reduce ultraviolet radiation penetration to the skin. These sunscreens often are colorless and maintain a thin visible film on the skin. These sunscreens usually contain UVB absorbing chemicals and more recently contain UVA absorbers as well.
Physical Sunscreens, most often referred to as sunblocks, are products containing ingredients such a titanium dioxide ($\ce{TiO2}$) and zinc oxide ($\ce{ZnO}$) which physically block ultraviolet radiation (UVR). Sunblocks provide broad protection against both UVB and UVA light. They can be cosmetically unacceptable to many people, because they are often messy, visible and do not easily wash off. However, some new zinc oxide products are available in brightly colored preparations which are popular with young people. The amount of sun protection these sunblocks provide, while potentially high, cannot be quantified in the same manner as sunscreen SPFs. Physical sunscreen is recommended for individuals who have unusual sensitivity to UVR. Most recently on the sun protection scene is sun-protective clothing designed to block UVA and UVB radiation. The effective SPF is greater that 30.

When should you use a sunscreen?

Sunscreens should be used daily if you are going to be in the sun for more than 20 minutes. Most people will receive this amount of sun exposure while performing routine activities. They can be applied under makeup. There are many cosmetic products available today that contain sunscreens for daily use because sun protection is the principal means of preventing premature aging and skin cancer. Sunscreen used on a regular basis actually allows some repair of damaged skin. Because the sun's reflective powers are great - 17 percent on sand and 80 percent on snow - don't reserve the use of these products for only sunny summer days. Even on a cloudy day 80 percent of the sun's ultraviolet rays pass through the clouds. Skiers beware, ultraviolet radiation increases 4 percent for every 1,000-foot increase in altitude.

To add to this last part, sunblock manufacturers usually include instructions with their products on the proper way to use them.


To my understanding, only $\ce{NaC#CH}$ and $\ce{LiCH3}$ are organometallic compounds.

According to the IUPAC definition of organometallic compounds, these must contain

  at least one metal-carbon bond between an organic molecule, ion, or radical and a metal.


Identifying lone pairs when looking for formal charges

Drawing common electron pairs like they appear in most organic molecules is only then relevant, when

they play a keyrole in the reaction process
they should explain a special molecule structure
they are of an unusual quantity and thus resulting in a (intermolecular) formal charge

Otherwise especially molecules with a lot of atoms with unpaired electrons (e. g. saccharides) can get confusing. And in addition: they take more time to draw ;)
Of course when learning organic chemistry its good to think of them a lot, because of their high reaction potential.

You can be a 100 % sure that when they would not be there in compound 1 that it would be highlighted somehow, because that would result in a very unstable state.


Why do unbonded electrons exist in pairs?

A single unpaired electron is called a radical, and usually aren't stable. Each electron orbital can hold two electrons, one spin up, other spin down. So a lone pair is just a full electron orbital, and cannot accept another electron. An orbital with 1 electron will readily react to get another electron and form a pair.

Check this: Free Radical

The first reaction they show is UV light being used to break $\mathrm{Cl}_2$ gas into $2\mathrm{Cl}$ radicals, each with a single lone electron, forming free radicals. These radicals could be used as free radical polymerization initiators, or for other reactions.

In the structure you show, the molecule forms a trigonal planar configuration, with the lone pair forming the third arm. If you could stably achieve two free radicals instead of the lone pair, it would have tetrahedral configuration, with the radicals forming 2 of the 4 arms.

The theory behind it is more quantum mechanics than I can remember, but I know it involves the Pauli exclusion principle and Hund's rule.


Application of Nernst Equation in Potentiometric titration of Fe2+ with Ce4+

It is an interesting question. Daniel Harris is revising his book with my former mentor. Hope he clarifies this section in the revised version. Your point number 1 is misleading. The reason is that before the titration, theoretically there is no Fe(III). So Nernst equation should not be used- electrode potential is infinite (log 0 is undefined). When you start the titration (Ce (IV) in the burette), the moment you add Ce(IV), it is immediately reduced by Fe(II) to Ce(III). You do not have the Ce(IV/III) couple in the solution. Before the endpoint, Fe(III/II) couple controls the potential of the indicator electrode.

Do you remember what that indicator electrode is made of? It is just a platinum wire and the reference electrode is usually silver-silver chloride electrode, all enclosed in a single unit. The platinum wire is oblivious to iron or cerium. All it "senses" is the charge. Imagine a single Pt electrode dipped in an iron solution and forget about the presence of cerium. Imagine that a metal is a source (sea) of electrons.

The moment you dip the electrode, instantaneous exchange of electrons takes place between the metal and the reducible or oxidizable ions. Electrons can be released by Fe(II) to form Fe(III) and similarly electrons can be gained by Fe(III) to form Fe(II). This exchange rate is dependent on the concentration ratios of Fe(II) and Fe(III).

As a result of this exchange, the Pt metal develops a charge. It is very surprising that you only need to remove a very very very small number of electrons to develop a measurable surface charge. Hence, there is no perceptible change in the concentration of Fe ions. If platinum electrode develops a charge, the interface (solution electrode boundary) develops an opposite charge until equilibrium is achieved.

Now given the fact that Pt electrode is charged i.e., if we connect Pt to another electrode, we can sense the potential difference between the Pt electrode and another electrode, which can be any other electrode. Now that you wish to have a stable. For that purpose you need an electrode whose potential is fixed i.e. a reference electrode.


s orbitals have spherical symmetry, and p orbitals have a dumb-bell shape.  When these orbitals overlap, then they form the bonding orbitals (and corresponding antibonding orbitals).  The p-orbitals are aligned along Cartesian (xyz) axes.  If all three p-orbitals overlap, then the s-orbitals can't get 'close enough' to overlap and form a fourth bond.  If we start mixing s and p orbitals, the geometry always puts electron density away from the other bonding atom.

Concisely: there are no arrangements which allow overlap of atomic orbitals that create four bonding molecular orbitals with the proper orientation.

C2 as a molecule can exist, it will simply only have 3 bonds plus two unpaired electrons confined to each carbon.

If you want a more elaborate answer we have to unpack the heavy quantum mechanics, and you won't be seeing that until your junior year.


What is the correct lewis structure of SCH⁻?

This model seems reasonable, as you can see from the formal charges carbon has the -1 value while the rest are 0. One thing that you have forgot is to add brackets around the structure to show that it is a ion, include its overall charge.

$$\ce{[\overset{..}{\underset{..}{S}}=\overset{..}{C}-H]-}$$

Although, I suppose a structure of $\ce{[\overset{..}{C}#\overset{..}{S}-H]-}$ would work with the same formal charge, except that sulfur has the expanded octet, which is possible but I am not sure if that is more reasonable or not, it probably is not.

It is a very odd ion, to say the least, probably very unstable. Perhaps you have mistaken the hydrogen with nitrogen? It would be more reasonable to be asked to draw the lewis structure for $\ce{[SCN]-}$, thiocyanate ion. I am not even know how to pronounce or name the $\ce{[SCH]-}$ion you are required to draw, or if it even exists. Maybe that is why you need to draw it though.


What is the lethal dose ( LD50 ) of coffee?

According to one reference:

LD$_{50}$ for coffee = 118 servings at 240 ml each; and

LD$_{50}$ for espresso = 175 shots (volume not specified),

for a 75 kg person.

A much more comprehensive quantitative description is given here for the LD$_{50}$ of coffee for an 80 kg person (113 cups at 250 mL each, 135 mg caffeine per serving). They account for half-life of elimination of caffeine, and arrive a the conclusion that one can't actually reach the LD$_{50}$ (15,360 mg) for caffeine due to metabolism, elimination, and the limit on how much water can be processed by the kidneys.

Finally, as noted in the last reference and hinted at above, the amount of water a person would imbibe on the way to reaching the 113 cup total (about 30 L) would likely kill the person (water intoxication).


If you consider the carbon with as star (C*), it bears a C(CH3)(=CH2) group and an H group (not shown). But it is also part of a ring and in that case, we actually follow the exact same rules!

From C*, if we go to the left or the right, we get a CH2 group. So let's get further: on the left we get a CH group with a double bond to another carbon, while on the right, we get a C atom with a double bond to an oxygen and a single bond to a carbon atom.

Because you get different groups if you follow the ring from C* to the left or the right, it means that C* is chiral!


What is the formula for theoretical buffer capacity for a diprotic buffer system?

The logic is the same used for a monoprotic weak acid, considering the two ionization steps, as indicated by @MaxW.

For a diprotic weak acid we have:

The first ionization

\begin{align*}
\ce{H2A + H2O &<=> H3O+ + HA-} &K_{\ce{a}1}=\frac{\ce{[H3O+][HA-]}}{\ce{[H2A]}}
\end{align*}

and the second ionization

\begin{align*}
\ce{HA- + H2O &<=> H3O+ + A^{2-}} &K_{\ce{a}2}=\frac{\ce{[H3O+][A^{2-}]}}{\ce{[HA-]}}
\end{align*}

To make further calculations easier let's assume that the strong base added is monoprotic:

\begin{align*}
\ce{BOH &-> B+ + HO-}
\end{align*}

Charge balance of the solution is given by the equation

$$\ce{[H3O+]} + \ce{[B+]} = \ce{[OH-]} + \ce{[HA-]} + \ce{2[A^{2-}]}$$

When combined give us formula for the amount of the strong base (for short $\ce{[H3O+]}$ was written as $\ce{[H+]}$)

$$C_{\ce{B}} =
\frac{K_{\ce{w}}}{\ce{[H+]}}-
\ce{[H+]}+
\frac{C_{\ce{H2A}}K_{\ce{a}1}\left(\ce{[H+]}+2K_{\ce{a}2}\right)}
{\ce{[H+]}^2 + K_{\ce{a}1}\ce{[H+]} + K_{\ce{a}1}K_{\ce{a}2}}$$

when $K_\ce{w}$ is the water ionization equilibrium constant, $C_{\ce{H2A}}$ is the concentration of diprotic weak acid, $K_{\ce{a}1}$ and $K_{\ce{a}2}$ are the acid dissociation constants, and $C_{\ce{B}}$ is the concentration of strong base added.

The buffer capacity of a diprotic weak acid-conjugate bases buffer is defined as the
maximum amount strong base that can be added before a significant change in the pH will occur.

\begin{align*}
\beta &= \frac{dC_{\ce{B}}}{d\ce{pH}} = \frac{dC_{\ce{B}}}{d\ce{[H3O+]}}\frac{d\ce{[H3O+]}}{d\ce{pH}}\\
\end{align*}

Differentiating the equation we should then find that

\begin{align*}
\beta &= \left[-\frac{K_{\ce{w}}}{\ce{[H+]}^2}-1 -
\frac{C_{\ce{H2A}}K_{\ce{a}1} \left(\ce{[H+]}^2 + 4K_{\ce{a}2}\ce{[H+]} + K_{\ce{a}1}K_{\ce{a}2} \right)}
{\left(\ce{[H+]}^2 + K_{\ce{a}1}\ce{[H+]} + K_{\ce{a}1}K_{\ce{a}2}\right)^2}\right]
\left(-\ce{[H+]}\ln{10}\right)
\end{align*}

So finally buffer capacity is given by

\begin{align*}
\beta &= 2.303\left[\frac{K_{\ce{w}}}{\ce{[H+]}}+\ce{[H+]}+
\frac{C_{\ce{H2A}}K_{\ce{a}1}\ce{[H+]}\left(\ce{[H+]}^2 + 4K_{\ce{a}2}\ce{[H+]} + K_{\ce{a}1}K_{\ce{a}2} \right)}
{\left(\ce{[H+]}^2 + K_{\ce{a}1}\ce{[H+]} + K_{\ce{a}1}K_{\ce{a}2}\right)^2}\right]
\end{align*}

you can then plot $\beta$ vs $\ce{pH}$ and from this you should be able to find what you want.

Look a example based on this question. I hope it be useful.


Reaction of 2-ethyl-5-methyl-3-phenylfuran under acidic conditions

Your answer looks perfectly reasonable. Acid hydrolysis of furans is a well established route to 1,4 diketones. The heat is required to overcome the resonance stabilisation of the furan ring.


Why is potassium permanganate used to kill mollusks?

The provided link (Lab Tutorial at Washington University in St. Louis)  states the following:

Another problem for water-treatment plants in the Great Lakes and Mississippi Valley is zebra mussels, which can "congregate" at the water intake and clog the screen, eventually sealing it off. Potassium permanganate, KMnO4, can be used to kill these mollusks.

Zebra mussels are a problematic invasive species in the midwestern USA. Permanganate has been found to inhibit their proliferation. The main effect is apparently oxidation. From a report released by the US military (ref [1.]) on use of oxidizing molluscicides including permanganates:

In general they have similar modes of action based on the oxidation of organic matter,which leads to toxic and lethal effects. They are suitable for use in preventative treatment, where they are added to a system throughout the breeding season at from 0.1 to 0.5 mg/L (ppm) total residual oxidant (TRO) to prevent settling. For reactive treatments,continuous application of 0.5 to 1.0 mg/L TRO for 2 to 4 weeks can eliminate established adult colonies, but concentration and contact time required depends on temperature, water chemistry, and physiological state of the zebra mussels. Mussels do detect oxidants, and shell closure for up to 2 weeks may reduce efficacy in adults(Claudi and Mackie1994)

See also ref [2] cited in ref [1.].
Ref [3] compares the toxicity of various molluscicides, noting that

An  LC50 value  could  not  be  estimated  for antimycin,  Bulab  6002,  Bultab  6009.  Calgon DM -DACC,  KML  V2.  KML  V54.  potassium  chloride, and  potassium  permanganate  due  to  insufficient mortality  of  the  test  organisms.

In addition, potassium can also have an effect on mollusks. For instance, from a thesis (Ref [4]):

Potassium can reach increased levels that are toxic to aquatic organisms, such as mollusks, in the environment (Romano and Zeng 2007). For example, Fisher et al. (1991) reported that increased potassium concentrations were toxic to Dreissena polymorpha, causing asphyxiation by damaging gill epithelium. In addition, Imlay (1973) suggested that high potassium concentrations could be the cause of absent mussel populations or lower diversity within Upper Midwest mussel assemblages. Any river or stream with a potassium concentration of 7 mg/L or higher lacked mussels, but mussels could be found in rivers with a concentration lower than 4 mg/L.

Note responses vary across organisms. Discussion of that is bound to take us increasingly away from chemistry.
References

Zebra mussel chemical control guide (ERDC/ELTR-00-1) by Susan L Sprecher, Kurt D. Getsinger; prepared for U.S. Army Corps of Engineers (2000).

Klerks,P.L., Fraleigh, P.C,and Stevenson, R.C.(1993)."Controlling zebra mussel {Dreissena polymorpha) veligers with three oxidizing chemicals: Chlorine, permanganate, and peroxide+iron."Zebra mussels: Biology, impacts,  and control. T.F.Nalepa and D.W.Schloesser, eds., Lewis Publishers, Boca Raton,FL,621-641.

Diane L. Waller, Jeffrey J. Rach, W. Gregory Cope, and Leif L. Marking, Susan W. Fisher, Henrycka Dabrowska. Toxicity of Candidate Molluscicides to Zebra Mussels (Dreissena polymorpha) and Selected Nontarget Organisms. J.  Great  Lakes  Res.  19(4):695-702 Internat.  Assoc.  Great  Lakes  Res.,  1993.

Gibson, Kesley. (2015). ACUTE TOXICITY TESTING ON FRESHWATER MUSSELS (BIVALVIA: UNIONIDAE) AND FRESHWATER SNAILS (GASTROPODA: CAENOGASTROPODA).


What is a good (small) model system for ligand metal charge transfer practises?

AFAIK, permanganate ion is colored because of intense line, corresponding to oxygene-manganese charge transfer. This is noted in wikipedia article on charge-transer complex, but, unfortunately, without quote, and I'm unable to come with one right now

Edit:
Aha. Here there is a MO diagram on $\ce{MnO4-}$ ion and extensive discussion of (mis)understanding of its spectra, including statement I made above.


Cauchy-Schwarz inequality for integral derivatives

This is indeed a way that pre-screening of two-electron integrals is done, and if you thought of this yourself, then I'm very impressed! Though I doubt that is too much of a compliment.

Unfortunately, Tyberius is correct that you cannot be sure this inequality will hold after taking the derivative. Fortunately, the Cauchy-Schwartz inequality holds for any arbitrary vectors which have an inner-product. This means that you can simply switch the order you are doing things and it will work.

For what comes below, I will basically replicate some work from a paper by Reinhart Alrichs et al., and will use their notation as well [1].

In what follows, a superscipt $i$ represents a derivative in any of the cartesian $x$, $y$, or $z$ directions.

The Cauchy-Schwarz inequality states that,
$$\tag{1}\label{eq:1}
|(ab|cd)|\le Q_{ab}Q_{cd}
$$
where,
$$\tag{2}\label{eq:2}
Q_{ab}=(ab|ab)^{\frac{1}{2}}
$$
and the same for $Q_{cd}$.

If we take the derivative in the $i$ direction of $(\ref{eq:1})$, we have,
$$\tag{3}\label{eq:3}
(ab|cd)^i=(a^ib|cd)+(ab^i|cd)+(ab|c^id)+(ab|cd^i)
$$
which is contained in the work you have done, but with some extra stuff from the original inequality.

Each of these terms, however, are still a part of the original space and hence still have a well-defined inner-product. This means we can apply the Cauchy-Schwarz inequality directly to these derivatives as a means of screening the integrals.

That is,
$$\tag{4}\label{eq:4}
|(a^ib|cd)|\le Q_{a^ib}Q_{cd}
$$
and so one for the other derivatives.

The advantage of this is that one can give an upper bound for the size of these integrals based on calculations which have to be done for the other integrals anyways, and some simple to compute auxiliary quantities. See [1] for details on these auxiliary quantities (eqs. 10-19). I do not fully understand this part of the procedure, so I'm not restating it here.

Next, you asked where one should do the screening. The authors of [1] advocate what they call a "global screening" methodology. This means that all of the quantities required for calculating the upper-bound are pre-computed (before any integral loops but after any checks for symmetry), and compared to some pre-specified threshold, say $10^{-10}$. Although we do have to compute some new things, these are much cheaper than the integrals themselves and thus save quite a bit of time.

Ref. [1] also goes into detail about some test cases as well as doing the same screening for second-derivatives and hence the application of this pre-screening to Hessians.

A quite different approach which is very interesting to read is given in ref. [2], which came after and improves on [1]. This method, quite naturally I think, takes advantage of a generalization of Cauchy-Schwarz known as Holder's Inequality. For our purposes, the relevant form of Holder's inequality is (see the section titled Lebesgue Measure or ref. [2]),
$$\tag{5}
\int|f(x)g(x)|dx\le\left(\int|f(x)|^mdx\right)^{\frac{1}{m}} \left(\int|g(x)|^n dx\right)^{\frac{1}{n}}
$$
for $\frac{1}{m}+\frac{1}{n}=1$.

Notice, however, that while this strengthens the inequality by giving a class of possibilities for different $m$ and $n$, it weakens the inequality in that we may only use one variable $x$. Thus, to apply this to 2-electron integrals, the authors of ref. [2] first integrate out one of the variables, which gives us the same inequality for our physical situation as,
$$\tag{6}
\int|(ab|cd)|\le\left(\int|V_{ab}|^md\textbf{r}\right)^{\frac{1}{m}} \left(\int|(cd|cd)|^n d\textbf{r}\right)^{\frac{1}{n}}
$$
where $V_{ab}$ is the potential due to electron 1. Thus, in this case the screening must always take place after the first coordinate is integrated-out. This method, however, is very flexible in that it apparently gives tighter bounds (a smaller maximum), and also has many possible solutions. Ref. [2] discusses four of these possible solutions, and compares to the Schwarz pre-screening method, which is a special case of what they derive. Again, some special quantities must be pre-computed, but this is not such a big deal for the time it saves.

Ref. [2] does not discuss derivatives, though I assume the same method ought to possible as in the first part, only things might be complicated by the fact we need we need to integrate out a coordinate. Maybe not in the first derivative case, but in the second derivative case, this would present a problem as far as I can see.

[1] Horn, H., Weiß, H., Haser, M., Ehrig, M., & Ahlrichs, R. (1991). Prescreening of two‐electron integral derivatives in SCF gradient and Hessian calculations. Journal of computational chemistry, 12(9), 1058-1064.

[2] Gill, P. M., Johnson, B. G., & Pople, J. A. (1994). A simple yet powerful upper bound for Coulomb integrals. Chemical physics letters, 217(1-2), 65-68.


What is the reason of the color difference for the cis/trans isomers of tetraamminedichloridocobalt(III) chloride?

If you search a library for a copy of Inorganic Chemistry by Miessler and Tarr, they have an excellent diagram showing which explains the phenomenon.

$\ce{Cl-}$ ligands have lone electrons that can donate into the metal, as you said yourself. The trick is to recognize what effect this has on the $t_\mathrm{2g}^*$ and $t_\mathrm{2g}$ orbitals. Remember that the five frontier orbitals in an octahedral configuration are the $t_\mathrm{2g}$ and $e_\mathrm{g}^*$ orbitals.

When a $\ce{Cl-}$ ligand pi-donates into the metal center, it can only interact with the $\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, $\mathrm{d}_{yz}$ $(t_\mathrm{2g})$ orbitals because the $\mathrm{d}_{x^2-y^2}$, $\mathrm{d}_{z^2}$ $(e_{g}^*)$ orbitals have only a sigma-type interaction with the ligand (for them, there is no node, draw this to ensure yourself). When these ligands like $\ce{Cl-}$ pi-donate, they fill the $t_\mathrm{2g}$ bonding orbitals, widening the gap between them and the open the $T_\mathrm{2g}^*$ antibonding orbitals. These antibonding orbitals become much closer in energy to the $e_{g}^*$ orbitals, so you reduce the octahedral field splitting energy $\Delta_\mathrm{O}$ which redshifts the absorbance of the compound.

The reason the trans isomer is more redshifted (green as opposed to purple) is due to the fact that in the trans isomer, the ligand-metal pi overlapping does not clash. On the other hand, if you draw the ligand-metal pi overlapping generated by two cis $\ce{Cl-}$ ligands, you will find that there is always one orbital that contains a lobe where the ligand overlap does clash. When this clash does exist, it becomes less favorable for the $\ce{Cl-}$ ligands to pi-donate into the metal via that clashing orbital -- so, the $t_\mathrm{2g}^*$ antibonding energy level is closer in energy to the $t_\mathrm{2g}$ bonding energy level, and further in energy from the $e_\mathrm{g}^*$ energy level. In the cis compound, field splitting $\Delta_\mathrm{O}$ is higher and the color should be purple instead of green.

If you are having trouble seeing the clash, draw a $\mathrm{d}_{xy}$ orbital on a metal $\ce{M}$, and then draw two $\ce{Cl-}$ ligands such that $\ce{Cl-M-Cl}$ is a right angle (cis ligands). When you try to show pi-donation from both ligands, you will find that there is one lobe of the $\mathrm{d}_{xy}$ orbital that is used by both; that is the clash that is disfavored.


Hydrolysing a compound by repeated methyl and hydride shifts

The reaction will stop after first methyl rearrangement as it leads to tertiary allylic carbocation which is highly stable, so there will be no further rearrangement, as it does not lead to more stable carbocation intermediate. Thus one will get tertiary allylic alcohol.


Standard hydrogen electrode, why hydrogen?

There are two separate issues here. The first one is defining a zero for potential. In theoretical physics, it is the energy of two charges separated by infinity distance.

In electrochemistry, this is chosen to be the potential of the hydrogen electrode since that is something that can be made reproducibly and hydrogen was the de facto 'zero' for a number of other quantities.

These days, there are much more practical and easy to use reference electrodes like the silver/silver chloride (as chipbuster mentioned) or the saturated calomel or the mercury/mercurous sulfate. Here is a good list of them from a manufacturer of such electrodes: http://www.koslow.com/select_reference_probe

Full disclosure: I used to work for a company that resells electrodes from Koslow.


Water, as simple as it might appear, has quite a few extraordinary things to offer. Most does not seem to be as it appears.

Before diving deeper, a few cautionary words about hybridisation. Hybridisation is an often misconceived concept. It only is a mathematical interpretation, which explains a certain bonding situation (in an intuitive fashion). In a molecule the equilibrium geometry will result from various factors, such as steric and electronic interactions, and furthermore interactions with the surroundings like a solvent or external field. The geometric arrangement will not be formed because a molecule is hybridised in a certain way, it is the other way around, i.e. a result of the geometry or more precise and interpretation of the wave function for the given molecular arrangement.

In molecular orbital theory linear combinations of all available (atomic) orbitals will form molecular orbitals (MO). These are spread over the whole molecule, or delocalised, and in a quantum chemical interpretation they are called canonical orbitals. Such a solution (approximation) of the wave function can be unitary transformed form localised molecular orbitals (LMO). The solution (the energy) does not change due to this transformation. These can then be used to interpret a bonding situation in a simpler theory.

Each LMO can be expressed as a linear combination of the atomic orbitals, hence it is possible to determine the coefficients of the atomic orbitals and describe these also as hybrid orbitals. It is absolutely wrong to assume that there are only three types of spx hybrid orbitals.
Therefore it is very well possible, that there are multiple different types of orbitals involved in bonding for a certain atom. For more on this, read about Bent's rule on the network.[1]

Let's look at water, Wikipedia is so kind to provide us with a schematic drawing:

The bonding angle is quite close to the ideal tetrahedral angle, so one would assume, that the involved orbitals are sp3 hybridised. There is also a connection between bond angle and hybridisation, called Coulson's theorem, which lets you approximate hybridisation.[2] In this case the orbitals involved in the bonds would be sp4 hybridised. (Close enough.)

Let us also consider the symmetry of the molecule. The point group of water is C2v. Because there are mirror planes, in the canonical bonding picture π-type orbitals[3] are necessary. We have an orbital with appropriate symmetry, which is the p-orbital sticking out of the bonding plane. This interpretation is not only valid it is one that comes as the solution of the Schrödinger equation.[4] That leaves for the other orbital a hybridisation of sp(2/3).
If we make the reasonable assumption, that the oxygen hydrogen bonds are sp3 hybridised, and the out-of-plane lone pair is a p orbital, then the maths is a bit easier and the in-plane lone pair is sp hybridised.[5]

A calculation on the MO6/def2-QZVPP level of theory gives us the following canonical molecular orbitals:

(Orbital symmetries: $2\mathrm{A}_1$, $1\mathrm{B}_2$, $3\mathrm{A}_1$, $1\mathrm{B}_1$)[6,7]

Since the interpretation with hybrid orbitals is equivalent, I used the natural bond orbital theory to interpret the results. This method transforms the canonical orbitals into localised orbitals for easier interpretation.

Here is an excerpt of the output (core orbital and polarisation functions omitted) giving us the calculated hybridisations:

     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
   2. (1.99797) LP ( 1) O  1            s( 53.05%)p 0.88( 46.76%)d 0.00(  0.19%)
   3. (1.99770) LP ( 2) O  1            s(  0.00%)p 1.00( 99.69%)d 0.00(  0.28%)
   4. (1.99953) BD ( 1) O  1- H  2
               ( 73.49%)   0.8573* O  1 s( 23.41%)p 3.26( 76.25%)d 0.01(  0.31%)
               ( 26.51%)   0.5149* H  2 s( 99.65%)p 0.00(  0.32%)d 0.00(  0.02%)
   5. (1.99955) BD ( 1) O  1- H  3
               ( 73.48%)   0.8572* O  1 s( 23.41%)p 3.26( 76.27%)d 0.01(  0.30%)
               ( 26.52%)   0.5150* H  3 s( 99.65%)p 0.00(  0.32%)d 0.00(  0.02%)
 -------------------------------------------------------------------------------

As we can see, that pretty much matches the assumption of sp3 oxygen hydrogen bonds, a p lone pair, and a sp lone pair.

Does that mean that the lone pairs are non-equivalent?
Well, that is at least one interpretation. And we only deduced all that from a gas phase point of view. When we go towards condensed phase, things will certainly change. Hydrogen bonds will break the symmetry, dynamics will play an important role and in the end, both will probably behave quite similarly or even identical.

Now let's get to the juicy part:

  Second, if so, does this have any significance in actual physical systems (i.e. is it a measurable phenomenon), and what is the approximate energy difference between the pairs of electrons?

Well the first part is a bit tricky to answer, because that is dependent on a lot more conditions. But the part in parentheses is easy. It is measurable with photoelectron spectroscopy. There is a nice orbital scheme correlated to the orbital ionisation potential on the homepage of Michael K. Denk for water.[8] Unfortunately I cannot find license information, or a reference to reproduce, hence I am hesitant to post it here.

However, I found a nice little publication on the photoelectron spectroscopy of water in the bonding region.[9] I'll quote some relevant data from the article.

  $\ce{H2O}$ is a non-linear, triatomic molecule consisting of an oxygen atom  covalently bonded to two hydrogen atoms. The ground state of the $\ce{H2O}$ molecule is classified as belonging to the $C_\mathrm{2v}$ point group and so the electronic states of water are described using the irreducible representations $\mathrm{A}_1$, $\mathrm{A}_2$, $\mathrm{B}_1$, $\mathrm{B}_2$. The electronic configuration of the ground state of the $\ce{H2O}$ molecule is described by five doubly occupied molecular orbitals:
  $$\begin{align}
\underbrace{(1\mathrm{a}_1)^2}_{\text{core}}&&
\underbrace{(2\mathrm{a}_1)^2}_{\text{inner-valence orbital}}&&
\underbrace{
  (1\mathrm{b}_2)^2 (3\mathrm{a}_1)^2 (1\mathrm{b}_1)^2
       }_{\text{outer-valence orbital}}&&
\mathrm{X~^1A_1}
\end{align}$$

  [..]

  In addition to the three band systems observed in HeI PES of $\ce{H2O}$, a fourth band system in the TPE spectrum close to 32 eV is also observed. As indicated in Fig. 1, these band systems correspond to the removal of a valence electron from each of the molecular orbitals $(1\mathrm{b}_1)^{-1}$, $(3\mathrm{a}_1)^{-1}$, $(1\mathrm{b}_2)^{-1}$ and $(2\mathrm{a}_1)^{-1}$ of $\ce{H2O}$.

As you can see, it fits quite nicely with the calculated data. From the image I would say that the difference between $(1\mathrm{b}_1)^{-1}$ and $(3\mathrm{a}_1)^{-1}$ is about 1-2 eV.

TL;DR
As you see your hunch paid off quite well. Photoelectron spectroscopy of water in the gas phase confirms that the lone pairs are non-equivalent. Conclusions for condensed phases might be different, but that is a story for another day.

Notes and References

What is Bent's rule?
Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?

Formal theory of Bent's rule, derivation of Coulson's theorem (Wikipedia).
Worked example for cyclo propane by ron.

A π orbital has one nodal plane collinear with the bonding axis, it is asymmetric with respect to this plane. A bit more explanation in my question What would follow in the series sigma, pi and delta bonds?
With in the approximation that molecular orbitals are a linear combination of atomic orbitals (MO = LCAO).
The terminology we use for hybridisation actually is just an abbreviation:
$$\mathrm{sp}^{x} = \mathrm{s}^{\frac{1}{x+1}}\mathrm{p}^{\frac{x}{x+1}}$$
In theory $x$ can have any value; since it is just a unitary transformation the representation does not change, hence
\begin{align}
 1\times\mathrm{s}, 3\times\mathrm{p}
   &\leadsto 4\times\mathrm{sp}^3 \\
   &\leadsto 3\times\mathrm{sp}^2, 1\times\mathrm{p} \\
   &\leadsto 2\times\mathrm{sp}, 2\times\mathrm{p} \\
   &\leadsto 2\times\mathrm{sp}^3, 1\times\mathrm{sp}, 1\times\mathrm{p} \\
   &\leadsto \text{etc. pp.}\\
   &\leadsto 2\times\mathrm{sp}^4, 1\times\mathrm{p}, 1\times\mathrm{sp}^{(2/3)}
\end{align}
There are virtually infinite possibilities of combination.
This and the next footnote address a couple of points that were raised in a comment by DavePhD. While I already extensively answered that there, I want to include a few more clarifying points here. (If I do it right, the comments become obsolete.)

  What is the reason for concluding 2 lone pairs versus 1 or 3? For example Mulliken has in table V the b1 orbital being a definite lone pair (no H population) but the two a1 orbitals both have about 0.3e population on H. Would it be wrong to say only one of the PES energy levels corresponds to a lone pair, and the other 3 has some significant population on hydrogen? Are Mulliken's calculations still valid? – DavePhD

The article Dave refers to is R. S. Mulliken, J. Chem. Phys. 1955, 23, 1833., which introduces Mulliken population analysis. In this paper Mulliken analyses wave functions on the SCF-LCAO-MO level of theory. This is essentially Hartree Fock with a minimal basis set. (I will address this in the next footnote.) We have to understand that this was state-of-the-art computational chemistry back then. What we take for granted nowadays, calculating the same thing in a few seconds, was revolutionary back then. Today we have a lot fancier methods. I used density functional theory with a very large basis set. The main difference between these approaches is that the level I use recovers a lot more of electron correlation than the method of Mulliken. However, if you look closely at the results it is quite impressive how well these early approximations perform.
On the M06/def2-QZVPP level of theory the geometry of the molecule is optimised to have an oxygen hydrogen distance of 95.61 pm and a bond angle of 105.003°. This is quite close to the experimental results.
The contribution to the orbitals are given as follows. I include the orbital energies (OE), too. The contributions of the atomic orbitals are given to 1.00 being the total for each molecular orbital. Because the basis set has polarisation functions the missing parts are attributed to this. The threshold for printing is 3%. (I also rearranged the Gaussian Output for better readability.)

Atomic contributions to molecular orbitals:
2:  2A1  OE=-1.039   is   O1-s=0.81 O1-p=0.03   H2-s=0.07   H3-s=0.07
3:  1B2  OE=-0.547   is   O1-p=0.63             H2-s=0.18   H3-s=0.18
4:  3A1  OE=-0.406   is   O1-s=0.12 O1-p=0.74   H2-s=0.06   H3-s=0.06
5:  1B1  OE=-0.332   is   O1-p=0.95

We can see that there is indeed some contribution by the hydrogens to the in-plane lone pair of oxygen. On the other hand we see that there is only one orbital where there is a large contribution by hydrogen. One could here easily come up with the theory of one or three lone pairs of oxygen, depending on your own point of view. Mulliken's analysis is based on the canonical orbitals, which are delocalised, so we will never have a pure lone pair orbital. When we refer to orbitals as being of a certain type, then we imply that this is the largest contribution. Often we also use visual aides like pictures of these orbitals to decide if they are of bonding or anti-bonding nature, or if their contribution is on the bonding axis.
All these analyses are highly biased by your point of view. There is no right or wrong when it comes to separation schemes. There is no hard evidence for any of these obtainable. These are mathematical interpretations that do in the best case help us understand bonding better. Thus deciding whether water has one, two or three (or even four) lone pairs is somewhat playing with numbers until something seems to fit. Bonding is too difficult to transform it in easy pictures. (That's why I am not an advocate for cautiously using Lewis structures.)
The NBO analysis is another separation scheme. One that aims to transform the obtained canonical orbitals into a Lewis like picture for a better understanding. This transformation does not change the wave function and in this way is as equally a representation as other approaches. What you loose by this approach are the orbital energies, since you break the symmetry of the wave function, but this is going much too far to explain. In a nutshell, the localisation scheme aims to transform the delocalised orbitals into orbitals that correspond to bonds.
From a quite general point of view, Mulliken's calculations (he actually only interpreted the results of others) and conclusion hold up to a certain point. Nowadays we know that his population analysis has severe problems, but within the minimal basis they still produce justifiable results. The popularity of this method comes mainly because it is very easy to perform. See also: Which one, Mulliken charge distribution and NBO, is more reliable?
Mulliken used a SCF-LCAO-MO calculation by Ellison and Shull and was so kind to include the main results into his paper. The oxygen hydrogen bond distance is 95.8 pm and the bond angle is 105°. I performed a calculation on the same geometry on the HF/STO-3G level of theory for comparison. It obviously does not match perfectly, but well enough for a little bit of further discussion.

NO  SYM    HF/STO-3G : N(O)     N(H2)      |      Mulliken : N(O)     N(H2)
1   1A1    -550.79     2.0014   -0.0014    |      -557.3     2.0007   -0.0005
2   2A1     -34.49     1.6113    0.3887    |       -36.2     1.688     0.309
3   1B2     -16.82     1.0700    0.9300    |       -18.6     0.918     1.080
4   3A1     -12.29     1.6837    0.3163    |       -13.2     1.743     0.257
5   1B1     -10.63     2.0000    0.0000    |       -11.8     2.000

As an off-side note: I completely was unable to read the Mulliken analysis by Gaussian. I used MultiWFN instead. It is also not an equivalent approach because they expressed the hydrogen atoms with group orbitals.
The results don't differ by much. The basic approach of Mulliken is to split the overlap population to the orbitals symmetric between the elements. That is a principal problem of the method as the contributions to that MO can be quite different. Resulting problematic points are occupation values larger than two or smaller than zero, which have clearly no physical meaning. The analysis is especially ruined for diffuse functions.
At the time Mulliken certainly did not know about anything we are able to do today, and under which conditions his approach will break down, it still is funny to read such sentences today.

  Actually, very small negative values occasionally occur [...]. [...] ideally to the population of the AO [...] should never exceed the number 2.00 of electrons in a closed atomic sub-shell. Actually, [the orbital population] in some instances does very slightly exceed 2.00 [...]. The reason why these slight but only slight imperfections exist is obscure. But since they are only slight, it appears that the gross atomic populations calculated using Eq. (6') may be taken as representing rather accurately the "true" populations in various AOs for an atom in a molecule. It should be realized, of course, that fundamentally there is no such thing as an atom in a molecule except in an approximate sense.

For much more on this I found an explanation of the Gaussian output along with the reference to F. Martin, H. Zipse, J. Comp. Chem. 2005, 26, 97 - 105, available as a copy. I have not read it though.
Scroll down until the bottom of the page for the image, read for more information: CHEM 2070, Michael K. Denk: UV-Vis & PES. (University of Guelph) If dead: Wayback Machine
S.Y. Truong, A.J. Yencha, A.M. Juarez, S.J. Cavanagh, P. Bolognesi, G.C. King, Chemical Physics 2009, 355 (2–3), 183-193. Or try this mirror.


Are metallic/ionic bonds weaker than covalent bonds?

Quartz and diamond are stronger substances because their molecules form network covalent structures. These structures form a lattice-like structure, much the same as ionic compounds.

This molecular network is also the reason that diamond and quartz form a crystalline structures, just like you'd see in ionic substances such as NaCl. Some other structures you might want to look into are Graphite and Graphene, which are both allotropes of carbon (allotropes are, simply put, different molecular arrangements of an element).

The network structure combines to make the substance stronger than normal covalent bonded substances.

So to answer your question, substances with standard covalent bonds seem to be weaker than those with ionic bonds because the ionic bonds tend to form a lattice structure, that makes them much stronger. You can see this in the fact that the boiling points of ionic salts are much higher than that of a covalent substance like water. However, when covalent bonds form network covalent structures, atoms combine to form a singular macromolecule that is much stronger than singular covalent bonds.


Electrochemical half cell without redox couple

Nernst's equation

The Nernst Equation can be derived from the Gibbs energy $\Delta G = nFE $. The free energy for a cell reaction:

$$\Delta G = \Delta G^\circ - RT \ln Q$$

can be written as:

$$E = E^\circ - \frac {RT}{nF} \ln Q$$

for reaction

$$\ce{O + $n$e -> R}$$

$$E = E^\circ -  \frac{RT}{nF} \ln\frac{a_\mathrm R}{a_\mathrm O}$$

where $a_i$ is activity of species $i$.

Activity
In a solution, the activity of $i$ is defined: $a_i=\gamma_i \frac {C_i}{C^0}$ where $C_i$ is the concentration of solute, $C^\circ$ is standard concentration ($1\ \mathrm M$), and $\gamma_i$ is unitless activity coefficient.

The equation can be simplified in:

$$E = E^{\circ'} -  \frac{RT}{nF} \ln\frac{[\ce{R}]}{[\ce{O}]}$$

But for the reaction

$$\ce{Cu^2+ + 2e- -> Cu}$$

the Nernst's equation will be:

$$
E = E^{\circ'} -  \frac{RT}{nF} \ln\frac{[\ce{R}]=1}{[\ce{O}]}
$$

because activity of solid (deposited copper) is 1.

If you try to calculate the potential of $0.01\ \mathrm M$ solution you will obtain the difference between standard concentration and the potential at $0.01\ \mathrm M$.

$$\begin{align}
E &= E^{\circ'} -  \frac{0.02559\ \mathrm V}{2} \ln\frac{1}{0.63}\\[6pt]
&= 0.34\ \mathrm V -  0.0059\ \mathrm V
\end{align}$$

The computed potential ($0.0059\ \mathrm V$) is lower than the standard potential and they are probably neglecting it.


How to determine at temperature and pressure to keep Cl2 a gas?

What you need to know is the vapor pressure of $\ce{Cl2}$ as a function of temperature.  Just ensure that the release pressure of the chlorine is lower than the vapor pressure, and it will be a vapor.

I found one web site that gives a graph (and a literature reference) of $\ce{Cl2}$ vapor pressure as a function of temperature from -30 °F to 220 °F.  The references they give are:

  Reprinted from Chlorine Manual, the Chlorine Institute, 3rd ed. 1959. Data for graph from R. Kapoor and J.J. Martin, Thermodynamic Properties of Chlorine, Eng. Res. Inst. Univ. of Michigan, 1957.

At -30 °F they report $\ce{Cl2}$ vapor pressure is "0" psi (which has to be an approximation; the vapor pressure of all materials is non-zero at any non-zero temperature) and at 220 °F it is 587 psi.


Why are Ka and Kb only used for weak acids and bases, respectively?

This is absolutely not true. $\mathrm{p}K_\mathrm{a}$ values are frequently quoted amoung organic chemists for very strong bases and sometimes even for very strong acids. For reference, check out the Evans $\mathrm{p}K_\mathrm{a}$ table and the values quoted therein.


Are ethyl iodophenylundecylate and Ethyl-10-(Iodophenylundecanoate) the same compound?

The "structures" you are trying to post and their respective names do not match, as I cannot spot a phenyl group in there. (Also it is hard to tell from this diagram.) But based on what you gave us, I believe these are the same compounds, where ethyl 10-(4-iodophenyl)undecanoate is the correct IUPAC name and Ethyl iodophenylundecylate is a shortened synonym. I also found trivial names for this compound, for example Pantopaque and Iofendylate.
You can access this molecules record on ChemSpider and you will find a couple more names attached to it. From the same page is the following structure taken:

(source: chemspider.com)


Why does 9-fluorenone dissolve in hexane?

Firstly, as a general rule, organic compounds with more then 6 carbon have higher London forces then polar forces.  This is a very general rule and doesn't work in all circumstances. Thus, this molecule is not as polar.

Secondly, cyclic compounds have much weaker London forces then molecules that have straight chains. Thus, this makes the London forces not so high(relative).

Thirdly, to have something dissolve in hexane it must have weak intermolecular forces, but not weaker then hexane itself, as hexane only has to suspend the solute molecules in solution and prevent the solute from clumping up and falling to the bottom.

So, quite simply, what happens is that the hexane molecules surround the 9-Fluorenone molecules, which diffuse through the solution.


What is the typical resistance of a piece of 1 cm^2 of carbon paper?

If there is no published resistance for that carbon sheet, then the best choice is to do exactly what you have done, i.e., measure the resistance per square. However, that specification is in given in the Strem catalog! It might be difficult o find if you do not spell the company's name correctly.
Two notes:

Resistance of a uniform, homogeneous sheet is measured per square, not per cm2. Whether the sheet is 1 mm on a side, or 100 km on a side, the resistance from one edge to the other would be the same; in your case, 3 Ω. [Consider that making the separation wider between edges is exactly compensated by the increased length of the conducting sheet.]
"Carbon paper" referred, in English, to paper with a mix of lampblack carbon and wax, though now it also applies carbon microfiber paper. For clarity, I suggest using the name Strem uses in it's catalog.


How to model 98% ethanol solution concentration graph over time

As Poutnik pointed out, if you boil a 98% ethanol solution (2% H$_2$O), then you will lose a few % of ethanol until you get to ~95.5% ethanol; then the rest of the solution will boil off as if it were a pure compound, for the composition will not change. Well, this is strictly true at the high temperature of boiling at 1 atm pressure. At about 0.1 atm, the azeotrope breaks, and so at that boiling temperature (about 35$^o$C, from a graph in https://wps.prenhall.com/wps/media/objects/3311/3391416/blb1105.html),

there is no azeotrope. I suspect that at lower temperatures than 35$^o$C, there may be no azeotrope. Nevertheless, if you start with a 98% ethanol solution, it won't change much until you have exported off most of the ethanol (like ~75% or so).
I would suggest cutting the concentration to about 50% ethanol, put it in a large beaker (~1 liter), stir with a small magnetic stirring bar at low speed to keep the contents somewhat homogeneous, put in a thermometer and a hydrometer with a range of 0.9 to 1.0 and engage a mild cross current of air across the beaker.
You will be able to determine the concentration of ethanol from a diagram like this:

from  https://www.engineeringtoolbox.com/ethanol-water-mixture-density-d_2162.html using the thermometer and hydrometer. Time is a variable, but is useless for determining any general scientific principle. The conditions of your experiment include air pressure (you can get that), air current (you can describe that - if you don't use any air current, it will take hours or days to do your experiment).
You can read off the volume of solution you have from the graduations on the beaker, and you can then plot the composition of the solution as its volume decreases by evaporation. The measurements are unfortunately not very accurate, but you will be able to make a graph and then spend several pages describing your lab work and potential errors and look up graphs and tables like the ones I have used. A likely error will be that the temperature will not be constant, the liquid level will drop so that the air current will have different strengths.
And as MaxW pointed out, the modeling may not be very "realistic". But I think people will be impressed with your use of scientific equipment and your analysis (which will be of the highest caliber, I'm sure).
While 50% ethanol is a lot less flammable than 98%, it still is a fire hazard, especially in quantities as big as a liter. Not only the liquid, but also the vapor is hazardous, so make sure that there are no flames or sparking devices within about 20 feet of your experiment. And have good overall ventilation: ethanol vapor can put you to sleep or make you dizzy. For extra security, you should plan to have a fire extinguisher handy (and know how to use it).


When calculating a dihedral angle for a chemical bond, how is the direction defined? What are the most common convention?

The IUPAC Gold Book states that,

  In a Newman projection the torsion angle is the angle (having an absolute value between 0° and 180°) between bonds to two specified (fiducial) groups, one from the atom nearer (proximal) to the observer and the other from the further (distal) atom. The torsion angle between groups A and D is then considered to be positive if the bond A-B is rotated in a clockwise direction through less than 180° in order that it may eclipse the bond C-D: a negative torsion angle requires rotation in the opposite sense.

This implies a range of -180 to 180, which is by far the most typical range used, but I certainly have seen reports ranging from 0..360, e.g., 225 degree dihedrals.


You're looking for the point where the normalized absorption and emission spectra cross each other.

The UV/Vis absorption spectra is almost always higher in energy (shorter wavelengths) than the emission spectra, because the molecule goes from the electronic and vibrational ground state to the electronic excited state. Usually this also means some vibrational modes are also excited. Since the molecule is typically in either the vibrational ground state or a low-energy vibrational state, the absorption process can go up several vibrational levels (i.e, the energy absorbed is higher than just the basic electronic excitation).

For emission, the molecule often loses some vibrational energy, so the emission comes from the ground vibrational state of that electronic excited state. The molecule ends up in the ground electronic state, but a vibrationally excited state (i.e., it loses a bit less energy than the $E_{0-0}$ energy difference).

So if you normalize the absorption and emission spectra, the crossing point should roughly correspond to the $E_{0-0}$ difference.


There are two ways I can think of to get rid of the remaining water:

As per LDC3's answer, flush the system with ethanol which will hopefully not have the problem of slow dry time.
Put the glassware in an oven, which will also nicely sterilize the system.


What's a minimal yet chemically-meaningful kinetic system for an oscillating reaction?

I'm not sure if this is what you're looking for (it may well still be too abstract), but oscillations can occur in the Gray-Scott system, which is
$$
A + 2B \to 3B\\
B \to P,
$$
where $P$ is an inert product, and the reaction is assumed to take place in a flow reactor that provides a supply of A, giving rise to the dynamics
$$\begin{align}
\frac{\mathrm da}{\mathrm dt} &= f(1-a) - ab^2;\\
\frac{\mathrm db}{\mathrm dt} &= ab^2 - (f+k)b,\\
\end{align}
$$
where $f$ is a rate determined by the flow reactor, $k$ is the rate of the $B\to P$ reaction, and the rate constant of the autocatalytic reaction has been set to 1 without loss of generality by scaling $f$ and $k$ relative to it.

With the appropriate choice of the parameters $f$ and $k$, oscillation can happen because $B$'s concentration increases autocatalytically, but then it overshoots its food source (i.e. the concentration of A), which then builds up again, allowing the cycle to repeat.

You might not like the trimolecular step, but I've found you generally get similar behaviour if you split it up into something like
$$
A + B \to C + B\\
B + C \to 2B\\
B \to P.
$$
(Simply using $A+B\to 2B$ doesn't work, because its kinetics don't have the right sort of nonlinearity.)

I would say that this has an advantage over the Ball model that you posted in that it obeys the laws of thermodynamics. (At least from what you showed in your question, it looks to me that Ball's model only oscillates because the reverse reactions have been neglected, and if they weren't then it would have to go to equilibrium, because it's a closed system.) It makes explicit that you need a power source (the supply of $A$) for oscillation, and illustrates the connection between oscillatory behaviour and autocatalytic kinetics.


Please go through papers by Prof. Ralf Ludwig. You will certainly gain a lucid understanding of the subject both at experimental end as well as computational end. He advocates use of low frequency vibrational modes such as Far IR along with quantum chemistry for studying hydrogen bonding and allied weak interactions in very simplistic yet the most accurate way.

References:

"Hydrogen Bonding in protic ionic liquids: Reminiscent of Water." K. Fumino, A. Wulf, R. Ludwig., Angew. Chem., 2009, 121, 3230-3233. Angew. Chem. Int. Ed., 2009, 48, 3184-3186.
"Advancing into Water’s “No Man’s Land”: Two Liquid States?" D. Paschek, R. Ludwig, Angew. Chem., 2014, 126 (44), 11888-1890. Angew. Chem. Int. Ed., 2014, 53 (44), 11699–11701.


Are we wrong about Bakelite formation mechanism?

C-acylation : acylating agent (acyl chloride or anhydride) and $\ce{AlCl3}$
O-acylation : acylating agent (acyl chloride or anhydride).This is the one which you have talked about in the question.

Since the formation of bakelite is by Lederer Manasse reaction which proceeds through electrophilic aromatic substitution on phenol, its path is similar to that in C- acylation and not O-acylation.
Hence, the polymerisation of phenol and formaldehyde takes place either from two places:

For para substitution end:

For ortho substitution end:


What causes the difference in heat energy released by combustion for the butanol isomers?

The heat of combustion (or any reaction) is the energy that is set free when you make the products ($\ce{CO2}$ and $\ce{H2O}$ here, in their respective stochiomeric amounts) out of single atoms (this is the same for all isomers), minus the energy needed to break all bonds in the educts. The latter is different for the different isomers, because the energy in a chemical bond not only depends on the two partners but also on their neighbours (the "shape" of the molecules).

The usual procedure for this calculation is to not go over single atoms but the elements (and compounds) in their most stable form at standard conditions. These energies a called standard enthalpies of formation. There are long tables listing them for all possible compounds.


What is the usage of orbitals more complex than f orbitals?

Surprisingly, I learned that there are also usages for orbitals g,h,i
  and even j.

Actually, the letter "j" is not used, so it is s, p, d, f, g, h, i, k, l, etc.

The higher angular momentum orbitals do enter the domain of science, due to excited states of atoms. Transitions to and from excited states are observable through atomic spectroscopy.

For example there is the article Microwave spectroscopy of Al I atoms in Rydberg states: D and G terms


Quantum mechanical (or QFT) description of mesomeric effect

As far as I understand you search a text which provides you with the information about how to quantify the mesomeric and inductive properties of parts of an three dimensional organic molecule. The quantification should account for symmetries in the underlying potential.

From my point of view, any modern quantum chemical program provides you exactly this. So a good way to go are the usual books from Szabo and Ostlund or (a bit more modern and complete) from Helgaker, Jorgensen and Olsen. Both start from a quantum description of fermionic systems and later on introduce the potentials arising due to the physical nature of the system. However, geometrical symmetries are not covered in these books (maybe apart from solving the Schrödinger equation for a particle in spherically symmetric field - $\ce{H}$-like systems). If you're interested in general in molecular symmetries, then the book of Bunker and Jensen might be of interest.

However, those texts usually use the linear combination of atomic orbitals for molecular orbitals (LCAO-MO) approach, i.e., a certain number of atomic orbitals is combined to form molecular orbitals (MOs). Effects like "mesomerization" are obtained directly as a result because MOs are "spread over the whole molecule", i.e., the probability density to find an electron at position $\mathbf{r}$ is roughly constant along the bonds of the mesomeric system.

Your question states that

  [m]ost texts argue in terms of atomic orbitals that are somehow bent and
  combined and hybridized [...]

I agree that the usual chemistry starter books do this. And as you state, the bendedness of such an orbital (like the $sp^3$) hybridization arises from the symmetry of the underlying potential (in the $sp^3$-case the tetrahedral [$T_d$] potential of the four $\ce{H}$ atoms around $\ce{C}$ in methane). In LCAO-MO theory, however, symmetry is not necessary, i.e., you can place your atomic orbitals with their angular momenta at the atoms, and you would still get the correct answer. But by symmetry adaption of your basis (basis, because the atomic orbitals form the basis within which the MOs are spanned), you can obtain basis functions that already have a symmetry corresponding to the molecular geometry. This step is not entirely complicated but requires a bit of knowledge about group theory. If you are interested particularly in this part, I would recommend a book about Hückel MO (HMO) theory, like, e.g., Elektronenstruktur organischer Moleküle by Martin Klessinger or the book of Heilbronner and Bock. HMO theory is a simplified LCAO-MO theory in order to treat organic molecules that consist mainly of $\pi$-conjugated systems. In these books they explain how to symmetry-adapt a basis of $p_z$ orbitals e.g. for the calculation of the MOs of $D_{6h}$ symmetric molecules like benzene. While for the non-symmetry adapted basis you need to solve a $6\times6$ eigensystem, it simplifies very much in a symmetry adapted basis.

Lastly, I would like to mention that inductive as well as mesomeric a mere qualitative models in order to quickly come up with an idea how a system changes its properties under substitution of specific parts. None of those quantities is are actual calculable properties (there is no operator that returns the expectation value for this property when applied on a wave function). In this way a quantum chemical theory cannot teach you directly about

  3d dependences of the inductive effect[.]

But you can calculate your system and its properties (absorption spectra, electron distribution, electronic/charge density, electric dipole moents, ...) and do the same for a substituted system. From both data you can come up with a hypothesis about the source and change of mesomeric and inductive effects. The same arguments hold, of course, for aromaticity. There, it suits to cite a professor:

  Aromaticity is like a unicorn - something everybody knows, but nobody has ever seen.


How does the pressure change in the dimerization of nitrogen dioxide?

If you assume ideal gas in a constant volume this works, but x is not y. 'a' and 'p' are related via the ideal gas law(a = n): P = n * R*T/V .

Take in mind that the pressure actually increases as the reaction progesses as more particles are formed.


if a balloon full of warm air was placed in a refrigerator, what is the result?

The balloon will shrink and get smaller as the gas inside it gets cooler, the pressure inside the balloon falls. You might also have seen that a bottle partially filled with warm water kept in a refrigerator squeezes a bit because of the air pressure inside the bottle falls. This result directly follows from the gas laws.


There are different notions of atomic radius; the one you're using seems to be the metallic radius, which is half the distance between nearest neighbors in the metal.  This notion is very sensitive to the number of electrons per atom involved in bonding.  Scandium has only 3 valence electrons, while $\ce{Ti}$ has 4.  These all participate, in some extent, in the "electron soup" that holds metals together.  I haven't been able to figure out exactly to what extent, but  It's fair to say though that the 4 valence electrons of Ti bind the nuclei together significantly more tightly than the 3 of $\ce{Sc}$. As a result, the $\ce{Ti}$ atoms get significantly closer together.  (An analogous situation is the covalent radius of $\ce{F2}$, of around $\pu{70 pm}$, versus that of $\ce{O2}$, of around $\pu{60 pm}$; although the covalent radius tends to decrease across a period, it grows from $\ce{O}$ to $\ce{F}$ because $\ce{F2}$ has a single bond while $\ce{O2}$ has a double one.)

As you progress further along the transition metals the delocalization of d electrons in the metal goes down.  That is, although there are more d electrons in $\ce{Co}$ than in $\ce{Fe}$, their effectiveness in binding atoms together isn't really any larger.  As a result, the distance between neighbors (hence the metallic radius) is the same for both.


IUPAC name of aldehyde with two functional groups

You have correctly identified the principal characteristic group as aldehyde.
According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), simple mono- and dialdehydes are named substitutively using the suffix ‘al’.

P-66.6.1.1.1 Mono- and dialdehydes derived from alkanes are named substitutively using the suffix ‘al’ added to the name of the parent hydride with elision of the final letter ‘e’ of the parent hydride before ‘a’.

For example: pentanal and pentanedial, respectively. Thus, the compound that is given in the question is a butanedial.
The terminal locants of aldehydes, i.e. ‘1,4’ in case of butanedial, are not cited.

P-14.3.4.1 Terminal locants are not cited in names for mono- and dicarboxylic acids derived from acyclic hydrocarbons and their corresponding acyl halides, amides, hydrazides, nitriles, aldehydes, amidines, amidrazones, hydrazidines, and amidoximes, when unsubstituted or substituted on carbon atoms.

Also the locant ‘2’ of 2-methylbutanedial is not cited.

P-14.3.4.3 The locant is omitted in monosubstituted symmetrical parent hydrides or parent compounds where there is only one kind of substitutable hydrogen.

Therefore, the preferred IUPAC name is methylbutanedial.

By way of comparison, the Blue Book contained the similar example ‘2-chlorobutanedioic acid’, which was later corrected to ‘chlorobutanedioic acid’.


What are analytical expressions for the isothermal change of Gibbs energy?

With regard to the Van der Waals equation, have you considered integrating by parts:  $V\mathrm dp=\mathrm d(pV)-p\mathrm dV$.  That can certainly be integrated analytically.

The answer will be in terms of $V$, but that’s not a big limitation.


Does the ammonium ion hydrogen bond with water?

There is a detailed consideration of this issue in  Hydrogen Bonding of the Ammonium Ion Journal of Molecular Structure 345 (1995) 77-81.

First, the try to be clear about what they mean by "hydrogen bond":

  For the purposes of the discussion, a hydrogen
  bond is taken to be a specific short-range interaction
  in which the N-H bonds within the
  ammonium ion are significantly affected by the
  surrounding ions, and not simply electrostatic
  attraction.

They conclude that ammonium ions rotate too freely in water for significant hydrogen bonding to exist.

In gas phase calculations, they find that one water molecule hydrogen bonds to one ammonium ion (in other words, the dimer is hydrogen bonded), based upon the lengthening of the N-H covalent bond, but as you add 2, 3, and 4 water molecules, the N-H bond length returns to the free ammonium N-H bond length and the O --- H length increases.


Determine which orbitals will form hybrids with one another

Let's take your questions in reverse order.

  what is the rule in general to determine which orbitals will form
  hybrids?

The "general" rule is that you mix one atomic orbital for each bond you want to form.  In methane, carbon has 4 valence electrons and is forming 4 bonds so it mixes 4 atomic orbitals (1 2s and 3 2p) generating 4 $\ce{sp^3}$ hybrid orbitals, and it places 1 electron in each orbital.

In the case of aluminum trihydride, aluminum has 3 valence electrons and will form 3 bonds, one to each of the 3 hydrogen atoms.  Therefore, aluminum must mix 3 atomic orbitals (1 3s and 2 3p) generating 3 $\ce{sp^2}$ orbitals and places 1 of its electrons in each of these orbitals.

  why do the lower orbitals form a hybrid?

A 2s orbital is relatively close in energy to a 2p orbital and the same applies for a 3s and 3p orbital.  The closer in energy 2 orbitals are, the more effective is their mixing.  Therefore, 3s and 3p orbitals can mix (form new hybrid orbitals) effectively.  The reason orbitals mix in the first place is to form new orbitals that have greater directionality and can therefore form stronger bonds.  After all, we could form 3 bonds to aluminum trihydride by just bonding to the 3s orbital and separately to 2 3p orbitals.  Two of these bonds would be oriented 90 degrees from one another and the third bond would be fluxional, moving around the surface of a (3s) sphere. But hybridizing the aluminum so that we can form 3 $\ce{sp^2}$ orbitals and use them for bonding with hydrogen produces a more stable molecule and one with the observed geometry (3 bonds in a plane with a 120 degree angle between them).  If hybridization doesn't produce stronger bonds and a molecule that is more stable, then the molecule wouldn't hybridize. $\ce{PH3}$ is an example of such a molecule, the central phosphorous atom remains unhybridized.


Resonance structures of 4‐aminobut‐3‐en‐2‐ylium

Just shift the pi-bond from the double bond to the right side. The carbocation then shifts to the left. That's the second resonance structure. The first one is already shown, making a total of two different structures (D).

You might be wondering about the allylic lone pair on the Nitrogen atom. Well, that could have resulted in another resonance pair, but if you shift the lone pair to the adjacent C-C bond on the right, there will be a total of 3 charges on the resonance structure. This is a highly unstable form, so it's not a significant resonance structure. Thus, it can be neglected.


I suspect that you would have both these molecular equations:

\begin{align}
\ce{2KCl(aq) + Na2SO4(aq) &-> K2SO4(aq) + 2NaCl(aq)}\\
\ce{Ca(NO3)2(aq) + Na2SO4(aq) &-> 2NaNO3(aq) + CaSO4(aq)}
\end{align}

To answer your question with "What is a complete reaction?":

  A complete reaction is one where all of at least one of the available reactants is used up and converted into products.

So for example $\ce{KCl}$ and $\ce{Na2SO4}$: In this equation you will need $\pu{2 M}$ $\ce{KCl}$ and $\pu{1 M}$ $\ce{Na2SO4}$ in order for all the reactants to react. However, you only have $\pu{1 M}$ $\ce{KCl}$ (so $\pu{1 M}$ chloride ions). This means that, as you mentioned, you will need the double concentration for the $\ce{KCl}$ solution to achieve a complete reaction.

So in this case the only way for all $\ce{Na2SO4}$ molecules to react is to add the double amount of $\ce{KCl}$ solution. Then this is a complete reaction : at least one reactant is used up in the reaction.

But say we have a situation in which we have a $\pu{2 M}$ $\ce{KCl}$ solution, but a $\pu{2 M}$ $\ce{Na2SO4}$ solution, then all the $\ce{KCl}$ would react, but still \pu{1 M} of $\ce{Na2SO4}$ will not have reacted.If in your lab assignment it is the meaning that you react all $\ce{Na2SO4}$, then in this case you don't have a complete reaction. Only when a two times larger volume of the $\ce{KCl}$ is added (to 2M $\ce{Na2SO4}$), then you will once again achieve a complete reaction. Furthermore you would just get a solution of solvated ions.

In the case of $\ce{Ca(NO3)2}$ and $\ce{Na2SO4}$: In this equation you will need $\pu{1 M}$ $\ce{Ca(NO3)2}$ and $\pu{1 M}$ $\ce{Na2SO4}$ in order for all the reactants to react. You have $\pu{10 ml}$ of both compounds, each $\pu{1 M}$ in concentration: so in this case there is no need to use a larger volume of aqueous $\ce{Ca(NO3)2}$.


You do need the electromotive series or reactivity series to answer that question. If you will not have access to it during such exams, you'll need to memorize the order of some common elements, and to understand the arrangement of the Periodic Table.

The most reactive metals are at the bottom left of the table, low in Group I. The "noble" or coinage metals are clustered near the center, with reactivity generally decreasing within a column. E.G., Au is less reactive than Ag, which is less reactive than Cu.

Later, you may learn about the electron configuration of atoms and how that determines reactivity, but at this point, knowing the reactivity series would get you though the test.


Show which group the element with Z=31 belongs

This is the old periodic table in which are still used these groups. You should avoid its use, but if your teacher needs you to learn them anyway, here they are:


Visual explanation between Molecule vs Compound vs Element vs Atom vs Substance?

Here's a cone view.

Everything from large trees to atoms are viewed by our eyes in the cone view (just my perceptions...). This is a very interesting concept. It's like shooting an electron (into or out) of an element, that changes an physical & chemical properties of the atom, compound and substance etc..

Think about a pencil, and what it's made of.

This may be quite far from a deep chemistry expertise. So take this as a simple representation to accommodate the diagrams that you began with and looking from a different perspective. We aren't trying to fit a facts/theory into our understanding and visualisation but other way around.

An important distinction is the clarity of molecule, compound and mixture.
Molecules are constructed of atoms of same species. A compound is formed by atoms of different species. Compounds are of 2 different types mainly,

Covalent
Ionic
Intermetallic
Coordinate

Each of this bonds, origin goes back to a single atom and involvement of its orbitals energy and electrons.

For an example,

The water that we drink is a molecule and also a compound, made of two H atoms and oxygen atom covalently bonding. Isn't it odd to read in your paper or an article, "the water compound..." or wouldn't it feel fit to read as "a water molecule...".

The salt we add to balance flavour is a compound, made of one sodium atom (+charged ion) and chloride atom (-charged ion) ionically bonding.

The air we breath is a mixture of various molecules : $H_2O, H_2, O_2, N_2, CO_2$ etc (not atoms).

  Put it this way, all compounds could be molecules, but (not) vice
  versa.

Here's the top view (transverse plane from the very top) of the cone.

In physical sciences the matter is classified in pyramids like the following. Reference is in the image itself.

There are more advanced and slightly different version of discussions on the same topic here on pubchemblog : "difference between xyz..."..

The starting cone isn't pretty as your images. But I will make them more user-friendly and eye-candy if you can comment. This almost feels like a drawing session now, apart from the conceptual outlook. I am happy to read more about intermetallic and coordinate bonding as well to address that in the answer. It is not something I come across on daily basis as I am not in electronics or soldering/alloy analytical chemistry. Apart from the cone, other picture references are taken from here.


Which is most likely the acidic proton in ascorbyl palmitate?

The other H (hydrogen) will be more acidic. The ascorbate anion is stabilized by electron delocalization, as I have show below in terms of resonance between two canonical forms.


How to neutralize the stink of decomposed fish?

Plants like fully decomposed (composted) organic materials just like humans prefer cooked food. The solution which you are preparing, is going to hurt the plants rather badly. For example, cow dung is a good organic fertilizer but fresh manure will burn the plants. Why don't you bury this fish waste in the ground or soil, mix it with old leaves and let it compost for 1-2 months? No smell should arise because soil bacteria and perhaps worms/insects will take care of it. Above all everything will be well covered with soil.
Do not add vinegar or lemon juice to your liquid (which should go to waste anyway). Vinegar has salt and plants don't grow well with salty water.


Reaction Pathways: What to do when IRC geometries are relaxed (optimized) and the energy changes significantly

IRC is a gradient following algorithm that tries to follow the imaginary mode. Check the algorithm (several one implemented, so really depends on which you are talking about) and you will see that unlike in optimisation algorithms, there is a big emphasis on following the real gradient (and not just jump to the endpoint as fast as possible).

It works well around the TS where the gradient is large and has a well define direction. However getting closer to the optimised geometry, the requirements of following the exact gradient and having an effective optimization algorithm are contradicting. In analytical sense it may be not obvious, but when you have numeric gradients, small second derivatives (these are almost always very soft modes), the direction of the gradient may have big errors and you should keep a small stepsize to check it thoroughly.

In summary, the situation you wrote down is more or less common at some extent in almost all the IRC calculations due to the numerical nature of the problem. Harder imaginary frequencies may help.


Can a NOESY NMR experiment be halted partway through and give meaningful results?

There are two big differences between the STD experiment and a NOESY that matter when aborting the experiment early. The STD is usually run as a 1D experiment, and it's using the phase cycle to substract the on- and off-resonance spectra from each other.

When you acquire a 1D spectrum with multiple scans, the scans are not saved individually but directly added up to one FID, at least on Bruker spectrometers. So if you abort a 1D experiment (you need to execute tr to do that in most cases, as the FID is only written to disk after all scans are finished in most pulse sequences), you can't just throw away the last incomplete phasy cycle. And with experiments like the STD, your actual signal is the difference of two larger signals, so one incomplete substraction can ruin your spectrum.

In a 2D spectrum the increments for the indirect dimension are measured and written to disk sequentially, so you can throw away the last incomplete increment if you would need to. In most cases the incomplete increment will not be saved at all to disk, so you'll end up with one less data point in the indirect dimension, but no incomplete ones that were aborted in the middle of the phase cycle. This depends on how exactly the pulse program is written, but I suspect that most standard sequences handle it this way.

I've stopped HSQC and NOESY spectra before they were finished pretty often, and never observed any artifacts that could be traced to the abortion of the experiment.


White powder observed after boiling water in electric kettle for many weeks

It's likely that your home is supplied with "hard" water (that is, water with a high mineral content). These dissolved minerals typically include calcium or magnesium ions (from dissolved calcium bicarbonate and magnesium bicarbonate, respectively). Heating or boiling your "hard" water in a kettle will soften it by precipitating the calcium carbonate as its solubility decreases with increasing temperature.

This (calcium carbonate) could be the white powder you have observed after evaporation. Hard water should not pose any health risks, and in fact can add some extra calcium and magnesium to your diet. If it's still concerning to you, there are hundreds of products available that can soften your water before you use it, just google "water softener" and you'll be overwhelmed.

http://en.wikipedia.org/wiki/Hard_water


In your first reaction, $\ce{HIO2}$ and $\ce{HIO3}$ react to form $\ce{2IO2.}$ and $\ce{H2O}$.

In your second reaction, $\ce{IO2\cdot}$ is being reduced back to $\ce{HIO2}$ by $\ce{Mn^{2+}}$. Since the first reaction produced $\ce{2IO2.}$, this second reaction needs to happen twice for each time the first reaction occurs. The pair of reactions is really:

$$\ce{ IO3- + HIO2 + H+ -> 2IO2. +H2O }$$
$$2\left(\ce{IO2. +Mn^{2+} +H2O-> HIO2 + Mn(OH)^{2+}} \right)$$

When you combine these reactions into an overall reaction, the instances of $\ce{IO2.}$ cancel and you are left with a net gain of $\ce{HIO2}$.

$$\ce{IO3- + H+ +2Mn^{2+} + H2O-> HIO2 + 2Mn(OH)^{2+}}$$

Of course, the net reaction does not hint at the autocatalytic behavior. To emphasize that the reaction is autocatalytic, you could write the overall reaction in a not quite simplest form:

$$\ce{IO3- + HIO2 +  H+ +2Mn^{2+} + H2O-> 2HIO2 + 2Mn(OH)^{2+}}$$


Which one has higher priority according to Cahn–Ingold–Prelog rules?

You are correct. In both cases $\ce{CH2OH}$ has higher priority. the Cahn-Ingold-Prelog system uses atomic number as its primary determinant of priority. However, a popular misconception is that the CIP system uses something related to "size", like mass, number of carbon atoms, length of chain, etc.

Here is a way that I use when I teach the subject. In the following compound, the chirality center has two bonds to carbon atoms. The carbon atom on the left is bonded two an oxygen atom and two hydrogen atoms. The carbon atom on the right has two bonds to another carbon atom and one bond to a hydrogen atom. Write these connections vertically below the carbon atoms with highest atomic number at the top. Compare each row, and the first difference is where you judge. In this case, the first difference is the first row: oxygen has a higher atomic number than carbon. $\ce{CH2OH}$ has higher priority than $\ce{CH=CH2}$.

You can do the same analysis for $\ce{CH2OH}$ vs. $\ce{C(CH3)3}$.


Decomposition of CaCO3 in a sealed vessel

You are right that at $1100~\mathrm{K}$ there will not be any decomposition of $\ce{CaCO3}$ in the presence of $\ce{CO2}$.

But, as the temperature increases further to $1120~\mathrm{K}$, $\ce{CaCO3}$ starts decomposing because at this temperature $Q_\mathrm p < K_\mathrm p$.

As the reaction is taking place in a closed container, by viewing the reaction mixture as an ideal gas mixture, we can calculate the amount of $\ce{CaCO3}$ decomposed at $1120~\mathrm{K}$:

$$\mathrm P(\ce{CO2}) \times V = 1 \times R \times 1100 \tag1$$
$$\mathrm P(\ce{CO2}) = 0.80$$
$$\mathrm P(\ce{CO2}) \times V = (1+n) \times R \times 1120 \tag2$$
$$\mathrm P(\ce{CO2}) = 1.14$$

On solving, $n= 0.399$.


In the graph provided, which of the substances is the most efficient in acting as an antacid?

Normally, the efficacy of a drug is determined by the area under the curve of the entire time course of the drug in the body.  The graphs here only show the initial rate of the change in pH but not the entire time course after the peak pH is reached.  I assumed that these results were generated by an acid-base titration experiment in the laboratory and not generated in vivo.  Therefore I cannot answer which antacid is more effective and/or clinically safe.

Although calcium carbonate shows a rapid initial rate of pH neutralization, it may not be suitable for all patients.  Too rapid a pH rise in the digestive tract may cause nausea and vomiting and may trigger an “acid rebound”, meaning that the body tries to compensate by secreting more acid to maintain the normal gastric pH for food digestion, making the heartburn even more unbearable.

Some patients take calcium carbonate as a source of calcium, but one also needs vitamin D in order for the calcium to be absorbed.  Besides, calcium carbonate yields carbon dioxide gas with acid which is still acidic, and bloating.  For patients with compromised parathyroid function, too much calcium in the blood (hypercalcemia) after the bones are already saturated with calcium may start to deposit in the heart and in the kidneys to form calcium oxalate kidney stones (different from uric acid stone).  Therefore calcium antacid may not be the best antacid for all patients.

If calcium antacid is not tolerable, one can use aluminum hydroxide (Aludrox®) which is a good alternative.  However aluminum hydroxide can cause constipation.  Magnesium hydroxide (milk of magnesia) is another good alternative, but magnesium hydroxide can cause diarrhea because magnesium is a smooth muscle relaxant.  Therefore the product Maalox® or Mylanta® which combines aluminum hydroxide with magnesium hydroxide will be easier on the stomach.  Also, there are products that contain simethicone (which is an anti-flatulent) as an extra ingredient to help with bloating.

The choices for an antacid are out there, make a good, sound one.


What is the difference between "molecular mass", "average atomic mass" and "molar mass"?

Atomic mass refers to the average mass of an atom. This has dimensions of mass, so you can express this in terms of daltons, grams, kilograms, pounds (if you really wanted to), or any other unit of mass. Anyway, as you said, this is an average of the masses of the isotopes, weighted by their relative abundance. For example, the atomic mass of $\ce{O}$ is $15.9994~\mathrm{u}$. $\mathrm{u}$ is short for unified atomic mass unit and 1 u is equivalent to $1.661 \times 10^{-24}~\mathrm{g}$. It is exactly the same as the dalton, but from what I've seen, the term dalton is used more when discussing polymers, biomolecules, or mass spectra.
Molecular mass refers to the average mass of a molecule. Again, this has dimensions of mass. It's just the sum of the atomic masses of the atoms in a molecule. For example, the molecular mass of $\ce{O2}$ is $2(15.9994~\mathrm{u}) = 31.9988~\mathrm{u}$. You don't need to calculate the relative isotopic abundance or anything for this because it's already accounted for in the atomic masses that you are using.
The term molar mass refers to the mass per mole of substance - the name kind of implies this. This substance can be anything - an element like $\ce{O}$, or a molecule like $\ce{O2}$. The molar mass has units of $\mathrm{g~mol^{-1}}$, but numerically it is equivalent to the two above. So the molar mass of $\ce{O}$ is $15.9994~\mathrm{g~mol^{-1}}$ and the molar mass of $\ce{O2}$ is $31.9988~\mathrm{g~mol^{-1}}$.
Sometimes you may come across the terms relative atomic mass ($A_\mathrm{r}$) or relative molecular mass ($M_\mathrm{r}$). These are defined as the ratio of the average mass of one particle (an atom or a molecule) to one-twelfth of the mass of a carbon-12 atom. By definition, the carbon-12 atom has a weight of exactly $12~\mathrm{u}$. This is probably clearer with an example. Let's talk about the relative atomic mass of hydrogen, which has an atomic mass of $1.008~\mathrm{u}$:
$$A_\mathrm{r}(\ce{H}) = \frac{1.008~\mathrm{u}}{\frac{1}{12} \times 12~\mathrm{u}} = 1.008$$
Note that this is a ratio of masses and as such it is dimensionless (it has no units attached to it). But, by definition, the denominator is always equal to $1~\mathrm{u}$ so the relative atomic/molecular mass is always numerically equal to the atomic/molecular mass - the only difference is the lack of units. For example, the relative atomic mass of $\ce{O}$ is 15.9994. The relative molecular mass of $\ce{O2}$ is 31.9988.
So in the end, everything is numerically the same - if you use the appropriate units - $\mathrm{u}$ and $\mathrm{g~mol^{-1}}$. There's nothing stopping you from using units of $\mathrm{oz~mmol^{-1}}$, it just won't be numerically equivalent anymore. Which quantity you use (mass/molar mass/relative mass) depends on what you are trying to calculate - dimensional analysis of your equation comes in very handy here.

Summary:

Atomic/molecular mass: units of mass
Molar mass: units of mass per amount
Relative atomic/molecular mass: no units

A small (and unessential) note about the definition of the $\text{u}$. It's defined by the $\ce{^{12}C}$ atom, which is defined to have a mass of exactly $12 \text{ u}$. Now the mole is also defined by the $\ce{^{12}C}$ atom: $12 \text{ g}$ of $\ce{^{12}C}$ is defined to contain exactly $1 \text{ mol}$ of $\ce{^{12}C}$. And we know that one mole of $\ce{^{12}C}$ contains $6.022 \times 10^{23}$ atoms - we call this number the Avogadro constant. That means that $12 \text{ u}$ must be exactly equal to $(12 \text{ g})/(6.022 \times 10^{23})$, and therefore,
$$1 \text{ u} = \frac{1 \text{ g}}{6.022 \times 10^{23}} = 1.661 \times 10^{-24} \text{ g.}$$


Assumptions:

Reaction of carbon dioxide with water is neglected.
Vapour pressure of water is negligible.
Volume of solution does not change on dissolution of carbon dioxide.
Moles of carbon dioxide dissolved in water is very less as compared to moles of water and thus, $X_{\ce{CO2_{(aq.)}}}≈\frac{n_{\ce{CO2(aq)}}}{n_{\ce{H2O(l)}}}$

Initially no $\ce{CO2}$ is dissolved in water and so initial moles of $\ce{H2O_{(l)}}$ and $\ce{CO2_{(g)}}$ can be calculated as follows:
$$
n_{{\ce{H2O_{(l)}}}_i} = \frac{\rho V}{M} =\pu{\frac{500 \times 0.9975}{18} mol}= \pu{27.71 mol}\\
n_{{\ce{CO2_{(g)}}}_i} = \frac{P_iV}{RT} = \pu{ \frac{50 \times 0.07}{0.0821 \times 283} mol}= \pu{0.15 mol}\\
$$
Now after the equilibrium is settled some amount of carbon dioxide will dissolve in water according to Henry's law.
Final pressure in the container can be written as:
$$
P_f= \frac{n_{{\ce{CO2_{(g)}}}_f}RT}{V}\\
$$
And by Henry's law we can say,
$$
P_f= K_H \times X_{\ce{CO2_{(aq.)}}}
$$
Thus from both these equations we can conclude:
$$
K_H \times  \frac{n_{\ce{CO2(aq)}}}{n_{\ce{H2O(l)}}} = \frac{n_{{\ce{CO2_{(g)}}}_f}RT}{V}\\
$$
We also known that,
$$
n_{{\ce{CO2_{(g)}}}_f} = n_{{\ce{CO2_{(g)}}}_i} - n_{{\ce{CO2_{(aq)}}}_f} \\
$$
So,
$$
K_H \times  \frac{n_{\ce{CO2(aq)}_f}}{n_{\ce{H2O(l)}}} = \frac{(n_{{\ce{CO2_{(g)}}}_i} - n_{{\ce{CO2_{(aq)}}}_f})RT}{V}\\
$$
Solving this we get,
$$
n_{\ce{CO2(aq)}_f} = \pu{0.135 mol}\\
$$
This correspondingly means:
$$
n_{{\ce{CO2_{(g)}}}_f}= \pu{0.015 mol}\\
P_f= \pu{5 atm}\\
$$
The final pressure is less than initial pressure as expected. The mistake you were making was implicitly considering $P_f=P_i$ when finding the mole fraction of carbon dioxide in water, which clearly isn't the case.


The answer is in front of you. $E_v$ is the Total translational  kinetic energy not the Average kinetic energy. If you want average translational kinetic energy per mol this is given by: $E_v/n$ or $3/2RT$.


Why is a supercritical fluid not a liquid?

Beyond the elementary school list of solid liquid and gas, there are many states of matter which exhibit peculiarly different properties. Wikipedia gives a list of the states of matter.

  My book says that because in a container above the critical temperature and critical pressure, a fluid of density similar to liquid exists uniformly in the container and there is no surface or meniscus that separates the liquid from the gas.

That is true.

  Does that mean that a liquid always has a meniscus which separates it from the liquid's vapour?

No, if the container is absolutely full of liquid then there is no gas phase.

  Does that mean that in free space without no gravity, a gas/vapour will try to expand and the liquid will try to contract and form spherical blobs due to surface tension?

Yes, but for the liquid there will also be evaporation. The rate of evaporation depends on the temperature of the liquid. As evaporation occurs the liquid will also cool. If there is enough liquid in the blob, then the matter can get cold enough to form a solid which will then sublime.

  Can I draw the conclusion from this that a liquid is the phase when attractive forces are dominant and repulsive forces are dominant in gas/vapour phase?

This is more or less true. I'd quibble about the nature of the "repulsive forces." I don't think that "repulsive forces" is the best way to model water in the form of steam. Rather the water molecules have too much kinetic energy to stick together.