Why is liquid oxygen more reactive than gaseous oxygen?

Since oxidation is generally strongly exothermic, a large part of the difference in reactivity can be explained by kinetics (rather than thermodynamics): the denser liquid brings oxidant (oxygen) into intimate contact with the oxidized substance throughout the reaction. Consumed oxygen is quickly replenished by surrounding liquid oxygen. In the gas the exothermic nature of combustion reactions accelerates reaction with oxygen gas by increasing the collision rate between oxygen molecules and the surface being oxidized. In the liquid this temperature effect is not as important because the oxidized surface is already bathed in oxygen molecules. Also, although the gas might be at a higher T, as assumed in the question, local heating of liquid oxygen around the oxidized surface due to the highly exothermic nature of most oxidation reactions can also add to the faster propagation rate of reactions in the liquid. The importance of thermal effects might also be reduced (if not trumped) by the acceleration of concerted events during the oxidation (such as sequential events at the same or neighboring sites or while in a short-lived activated state), since in the liquid collisions between surface and more than one oxygen molecule at neighboring sites will occur at rates that are orders of magnitude faster than in the gas.

This is admitedly an answer in need of some quantitative support.

Here's a "back-of-the-envelope" justification for the importance of density:

It turns out that if you use a model based on the Maxwell-Boltzmann kinetic theory then the collision frequency with a surface is proportional to $p/T^{1/2}$. Therefore, if you ignore the thermal activation energy for reaction, you can assume that the reaction rate is proportional to $p/T^{1/2} = \rho _m RT^{1/2} = RT^{1/2}/V_m$. This means we will have the same collision frequency for two different set of values of T/V when

$$\left(\frac{V_{m2}}{V_{m1}}\right)^2=\left(\frac{T_{2}}{T_{1}}\right)$$

Therefore, we can counter the effect of a 200 K drop in T by increasing the density (reducing the molar volume) by a factor of ~14. Now at STP the molar volume of a gas is ~20 L/mol, while for a compressed (liquified) gas the molar volume is 100 times smaller, say 0.2 L/mol. Therefore the increase in density more than makes up for the drop in T, at least within this simplified model.

In more general terms, we can explain away much of the loss of the kinetic advantage of higher temperature by the higher density in the liquid. So based on this simplified argument oxygen at a liquid-like density can be more reactive than a much warmer gas.

Note that the activation energy for reaction of $\ce{O2}$ is not negligible due to the relatively stable structure of the oxygen molecule. Therefore kinetic effects involving rapid heating and thermal activation of oxygen (due for instance to local accumulation of thermal energy) are probably also important to explain the reactivity of LOX.


How can I generate 250 GPa to create metallic hydrogen?

According to The Compressive Strength of a Perfect Diamond 410-480 GPa is possible.  So, yes, the diamond experiences the applied pressure, but it is strong enough to sustain it.

For more recent work, see Implementation of micro-ball nanodiamond anvils for high-pressure studies above 6 Mbar


According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the names of esters are usually formed by placing the alcoholic component in front of the name as a separate word. The particular subsections concerning esters from a single alcoholic component and multiple acid components read as follows:

  P-65.6.3.3.3 Polyesters formed from a single ‘alcoholic’ component

  Esters derived from a single ‘polyhydroxylic’ component are named by placing the name of the ‘polyhydroxylic’ component denoted by a multivalent organyl group (alkyl, aryl, etc.) in front of the name(s) of the acid component denoted by the anion names derived from the appropriate acid(s) (…).

  P-65.6.3.3.3.1 When anions are identical, functional class multiplicative nomenclature is used. Names are formed by citing the multivalent group, the multiplicative prefix, and the multiplied anionic component name. Multiplicative prefixes ‘di’, ‘tri’, etc. are used when anions are unsubstituted; when substituted, prefixes ‘bis’, ‘tris’, etc. are used.

In this case, the polyhydroxylic component is denoted by the multivalent organyl group ‘butane-1,3-diyl’. The acid component is denoted by the anion name ‘hexanoate’, which is derived from the corresponding acid ‘hexanoic acid’. Since both acid components are identical and unsubstituted, the multiplicative prefix ‘di’ is used in functional class multiplicative nomenclature. Therefore, the preferred IUPAC name for the compound given in the question is (3R)-butane-1,3-diyl dihexanoate.


Derivation and application of the Henderson-Hasselbalch equation

The reaction taking place is the quantitative reaction:
$$\ce{AH + OH- -> A- + H2O}$$

The initial concentrations are:
 $$\ce{[OH- ]_0}=\frac{0.001}{0.1}=0.01  \mathrm{M}$$  $$\ce{[AH ]_0}=\frac{0.002}{0.1}=0.02 \mathrm{M}$$
The equilibrium concentrations:
 $$\ce{[AH ]}=\frac{0.002-0.001}{0.1}=0.01 \mathrm{M}$$
 $$\ce{[OH- ]}=10^{-8}  \mathrm{M}$$
 $$\ce{[A- ]}=0.02-0.01=0.01 \mathrm{M}$$
Using the equation of question (ii):

$\ce{pH = p}K_a=6$


The equation for $\Delta H$ for a van der Waals gas should be:
$$\Delta H=C_p^{IG}\Delta T-2a\left(\frac{1}{V_2}-\frac{1}{V_1}\right)+b\left(\frac{RT_2}{V_2-b}-\frac{RT_1}{V_1-b}\right)$$where $C_p^{IG}$ is the heat capacity of the gas in the Ideal Gas (IG) region (low pressures and high specific volumes).  For constant temperature ($T_1=T_2=T$), this equation reduces to:$$\Delta H=-2a\left(\frac{1}{V_2}-\frac{1}{V_1}\right)+bRT\left(\frac{1}{V_2-b}-\frac{1}{V_1-b}\right)$$

Explanation:
In the link https://physics.stackexchange.com/questions/321968/internal-work-done-by-a-gas/322170?noredirect=1#comment725750_322170, I showed that, for a van der Waals gas, the internal energy U is given by:  $$U(T,V)=U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-\frac{a}{V}$$
where $U(T_{ref},\infty)$ refers to the internal energy in the ideal gas reference state $T=T_{ref}$ and $V\rightarrow \infty$ (i.e., a reference state in which the gas exhibits ideal gas behavior).

The enthalpy H of the gas is equal to U + PV:
$$H(T,V)=\left[U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-\frac{a}{V}\right]+\left[-\frac{a}{V^2}+\frac{RT}{V-b}\right]V$$
Combining the terms on the right hand side gives:
$$H(T,V)=U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-2\frac{a}{V}+\frac{RTV}{V-b}$$
But, $$\frac{V}{V-b}=\left[1+\frac{b}{V-b}\right]$$
So, substituting into the equation for H gives:
$$H(T,V)=U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-2\frac{a}{V}+\frac{RTb}{V-b}+R(T-T_{ref})+RT_{ref}$$
But, $$U(T_{ref},\infty)+RT_{ref}=H(T_{ref},V)$$and$$\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}+R(T-T_{ref})=\int_{T_{ref}}^{T}{C_p^{IG}(T')dT'}$$where $C_p^{IG}(T)$ is the ideal gas heat capacity at constant pressure.  So, finally, we obtain:$$H(T,V)=H(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_p^{IG}(T')dT'}-2\frac{a}{V}+\frac{RTb}{V-b}$$


Is it possible to produce sodium bicarbonate from potassium bicarbonate?

Boiling off some water is a decent plan. Perhaps it should be done at reduced pressure so as to keep the temperature low. Magnesium and calcium bicarbonates decompose at temperatures above 60C in water to yield insoluble carbonates, and several data sheets for sodium bicarbonate suggest the same. One said decomposition was complete at 100C. I suppose you could consider that any H2CO3 in equilibrium with NaHCO3 could decompose to CO2 + H2O. Heat would just drive the equilibrium to completion. I suppose that's why I could find no solubility data for NaHCO3 or KHCO3 at temperatures higher than 60C.

One way to go about the preparation is to dissolve 58.5 grams (1 mole) of NaCl in 260 grams H2O, and 100 grams of KHCO3 (1 mole) in 260 grams H2O. Heat both to 60C; they will completely dissolve. Then pour one into the other (it may make a difference which way you do it) and stir. The amount of water is sufficient to dissolve 1 mole (84 grams) of NaHCO3 at 60C (but the other salts could and probably will affect this). So the mixed solution will be clear; all salts will be soluble.

Cooling to 20C will (should!) ppt 34.1 grams of NaHCO3, leaving 49.9 grams in solution (again, ignoring the effect of the KCl). Cooling to OC should ppt another 14 grams of NaHCO3, leaving 35.9 grams in solution. You could then boil off about half of the water before the solubility limit of the KCl is reached, but because of the possible loss of CO2, you may not get much more NaHCO3.


How does neutralization of free radicals by $\beta$-carotene work?

$\beta$-Carotene is a highly delocalised $\pi$ system.
According to George Britton there are numerous mechanisms to achieve this.

Oxidation
Oxidising radicals can remove an electron from the carotene molecule.
$$\ce{CAR + .R+ -> .CAR+ + :\!R}$$
Reduction
Analogous it can accept an additional electron.
$$\ce{CAR + .R- -> .CAR- + R}$$
Hydrogen Abstraction
It is possible to abstract a hydrogen atom from a saturated carbon in allylic position.
$$\ce{H-CAR + .R -> .CAR + HR}$$
Addition
You are correct in assuming, that a free radical may form a bond with that system. This is most common for hydroxyl radicals.
$$\ce{CAR + .OH -> .CAR-OH}$$

You are also correct in assuming, that another radical will always be created. However, the single electron will be very delocalised (resonance stabilised) throughout the whole system and not nearly as reactive as the free radical before.

The remaining molecule can then of course attract more free radicals to form other singlet species, which the body can digest or excrete.


Why does adding an acid speed up electrolysis of water?

Primarily due to

  Pure water conducts an electric current very poorly and, for this reason, is difficult to electrolyze.

(Source: Water Electrolysis)

and that salt and acids are able to act as an electrolyte catalyst as most salt and acids dissolve (hence dissociate) in water, specifically, according to the Wikipedia page,

  If a water-soluble electrolyte is added, the conductivity of the water rises considerably. The electrolyte disassociates into cations and anions; the anions rush towards the anode and neutralize the buildup of positively charged $\ce{H+}$ there; similarly, the cations rush towards the cathode and neutralize the buildup of negatively charged $\ce{OH-}$ there. This allows the continued flow of electricity.

A diagram of the electrolysis of water with dissolved salt/acid ions is shown below:

Image source

Further information can be found in the article A Review of Water Electrolysis (Zoulias et al).


Why does C=O have a larger dipole moment than C-O?

According to Wikipedia, bond dipole moment depends on:

Distance between atoms and
Overall charge difference, not just electronegativity difference.

Resonance tells us that there is some amount of charge separation in $\ce{C=O}$ bonds because of the $\ce{C+-O-}$ contributor. This difference in charge, in addition to the electronegativity difference, is more significant than the decrease in distance between atoms, hence the larger dipole moment for $\ce{C=O}$.


Calculating amount of base needed to achieve a specific pH in a mixed ethanol-water solvent

I'm looking for some way to calculate how much base I will need to add to an unbuffered solvent consisting of a mixture of ethanol and water in order to basify it to a specific reading (10) on an a pH probe calibrated with an aqueous standard.

I think you have summarized the main issue yourself. The reading of a pH probe in a basic alcoholic solution is just a positive real number. It has no significance whatsoever. The value of 10 means nothing. The pH probe has no idea that it is being dipped in an aqueous organic solution. I am sure the probe will take a very long time to equilibrate to give a stable "real number" output. Since those numbers such as pHe =10 have no theoretical basis it is hard to calculate anything. This is confirmed by a quick look at the standard method.

The ASTM D6423 method says " Readings are taken at exactly 30 s because the meter reading will change throughout the analysis due to glass-electrode voltage drift caused by dehydration of the glass-electrode membrane. Because alcohol dehydrates the glass-electrode membrane, time spent soaking in alcohol solution should be minimized.
4.3 The electrode is soaked in water-based pH 7 buffer between readings to prepare it for the next sample. This rehydrates the glass electrode, a necessary step to preserve the electrode's response characteristics.
4.4 The electrodes are cleaned/rehydrated at least every ten samples by alternately soaking several times in 1 mol/L NaOH solution and 1 mol/L H2SO4 (or 1 mol/L HCl) for about 30 s each. New electrodes are treated by this procedure before first use."

The practical way to "calculate" perhaps is to do a acid base titration. Calculate the titer for each batch with a suitable indicator and add the amount of base accordingly.


Electronic model with highest prediction rate

The so called gold standard of quantum chemical calculations of ground state energies and properties is usually considered to be Coupled Cluster Singles Doubles (with perturbative) Triples as a method. This method is based on ab initio molecular orbital theory and it is size consistent, but not variational. A basis set of at least a triple zeta quality with additional valence and polarisation functions should be employed, good examples for this are Dunning's (augmented) correlation consistent basis sets, e.g. (aug-)cc-pV{T,Q,5,6}Z, or Weigend's def2-basis set family, def2-TZVPP, def2-QZVP, but there are various others. (If you like to have some citations, drop a note in the comments.) This level of theory is usually denoted as CCSD(T)/TZVP, where TZVP is a placeholder for one of the various basis sets.

In principle CC theory is exact at the complete basis set limit, but in practise, this is not solvable. Another theory that is in principle exact, because they become equivalent, is Full Configuration Interaction, F-CI. The truncated variants CISD, CISD(T), CISDT, ... are variational, but not size consistent. For chemical systems it is much more important, that the calculations are size consistent, therefore these methods lost their importance. An attempt to solve this was to introduce Quadratic Configuration Interaction, QCI, but this is almost the same as coupled cluster theory, so it was not successful either.

The importance of Many Body Perturbation Theory, MBPT, of high orders, like Møller Plesset Perturbation Theory of fourth order, MP4, has very much decreased because of it inefficiency.

There are some cases, which cannot be accurately described by these models. Those are cases, where the ground state cannot be well defined by one determinante, the initial guess wave function obtained from Hartree-Fock, HF. Here you have to use a different description. A Multi Configuration, MC, or Multi Reference, MR, ansatz is necessary. The qualitative equivalent to HF is Complete Active Space Self Consistent Field, CASSCF. Based on this you can use Perturbation Theory to include more correlation effects. The best possible description so far (judging accuracy versus computational cost) is second order perturbation theory on a CASSCF wave function, for short CASPT2. It is also very reliable for spectroscopic calculations. There are also various developments on Multi Reference Configuration Interaction as another theory. (I have limited expertise on that field.)

Another way of treating chemical systems is Valence Bond theory, or with the application to modern computational aspects. Here you propose the mixing of various Lewis like structures, to obtain the ground state energy and properties. The formalism can get quite complicated, maybe this is why its performance for excited states properties is not very accurate at the moment. It is still a very active field of research, by a few scientific groups. (I am not one of them.)

However, affordable spectroscopic calculations have still huge errors and sometimes the numbers can vary quite a bit. Currently most used is a variety of methods derived from Time Dependent Density Functional Theory, TDDFT. An ab initio method of great importance in this field is Configuration Interaction Singles. Also fairly well implemented are approximate coupled cluster techniques, e.g. CC2. The struggle for finding an accurate, robust and computationally cheap method are one of the main goals of modern quantum chemistry. (I can apply these methods fairly well, but I have no deeper insight how they work.)

For extended systems like surfaces and solids there are various approaches. Since they get very demanding very soon, the most commonly used practical approaches are based on plain or hybrid density functionals. They often also employ plane wave basis sets with periodic boundary condition. (Unfortunately I am not an expert on that field either.)

As you can see, there is no definite answer to your question and the development of new and efficient methods is still a very active field. Depending on what type of description you want or need, there are many possibilities. The most advanced models are obviously for ground state energies and properties, since they have been researched for about a century now.

I hope this helps you, I apologise for the chatty answer. If something remains unclear, drop me a note in the comments.


On the mechanism of the nitrogen-removal step of the Wharton olefin synthesis

The Wharton rearrangement presents an opportunity to convert one α,β-unsaturated ketone to its regioisomeric allylic alcohol. To this end Ohloff, et al.,1,2 explored the transformation of α-ionone 1 into α-damascone 2. Epoxide 3 was exposed to hydrazine in methanol to afford bicyclic allylic alcohol 4 (30%) and the uncyclized allylic alcohols 5 and 6 (20%, ~1:1). The proposed mechanism (3 $\rightarrow$ 7 $\rightarrow$ 8 $\rightarrow$ 9) goes through the vinyldiazene 8, which loses nitrogen to form reactive species. Ohloff proposed a mechanism invoking the vinyl anion 9, which is either protonated by methanol to afford acyclic allylic alcohols 5 and 6 or undergoes cyclization with the endocyclic double bond followed protonation of the resultant secondary anion. The bothersome aspect of this mechanism is, can a vinyl anion in a sea of methanol survive long enough to avoid protonation and undergo cyclization? The same point can be raised about  the traditional Wolff-Kishner reaction run at elevated temperature in protic ethylene glycol.

Stork and Williard3 explored the scope of the cyclization under Ohloff conditions finding that both 5- and 6-membered rings can be formed. Cyclization of epoxyketones 10a and 10b formed 11a (85%) and 11b (60%), respectively, as mixture of stereoisomers. In the former example, little of the uncyclized product was formed. These authors suggested a radical-based mechanism because in a typical reaction, the ratio of cyclized-to-uncyclization product was not markedly affected by employing the more acidic β,β,β-trifluoroethanol or the less acidic tert-butyl alcohol as a solvent. The authors argue that if a vinyl anion were involved, the more acidic solvent would be expected to increase the amount of uncyclized product at the expense of cyclized material. An intense yellow color developed which faded with time upon mixing the epoxide and hydrazine independent of whether or not cyclization occurred. A strong UV absorption at $\pu{232 nm}$ and a weak one at $\pu{409 nm}$ were attributed to the vinyldiazene. A concerted vinyldiazene cyclization was considered but with less enthusiasm.

Taber and Stachel4 designed a set of experiments to resolve this dichotomy based on the known stereochemistry of radical and anionic cyclizations. Tri-n-butylstannane-initiated cyclization of bromide 12a gave an excess of cis-cyclopentane 13a over trans-14a in accord with the cyclization of bromide 12b to 13b and 14b (13b>14b).5 Alternatively, metal-halogen exchange of iodide 15a with t-BuLi gave more of trans-cyclopentane 14a over the cis-isomer 13a in an anionic cyclization in accord with the results of Bailey6 for iodide 15b. The Wolff-Kishner reduction of ketone 16a in triethyleneglycol (TEG) afforded an 11:1 ratio of trans-14a over cis-13a. This result led the authors to conclude that the Wolff-Kishner cyclization is anionic as opposed to radical in nature.

While it is not necessary that the Wharton rearrangement and the Wolff-Kishner reduction must proceed via the same mechanism, in my opinion there is ample opportunity for more definitive results. Should an occasion arise where only the radical or anionic mechanism would apply, then the definitive experiment has been conducted.

References:

G. Ohloff, G. Uhde, "Über eine ungewöhnliche Cyclisationsreaktion bei der Umsetzung von (+)‐Epoxy‐α‐dihydrojonon mit Hydrazinhydrat (English translation: About an unusual cyclization reaction in the reaction of (+)-epoxy‐α-dihydrojonone with hydrazine hydrate)," Helv. Chim. Acta, 1970, 53(3), 531-541 (DOI:  https://doi.org/10.1002/hlca.19700530309).
K. H. Schulte-Elte, V. Rautenstrauch and G. Ohloff, "Synthesen der stereoisomeren (±)‐γ‐Damascone (English translation: Syntheses of the stereoisomeric (±)‐γ-damascones)," Helv. Chim. Acta, 1971, 54(7), 1805-1812 (DOI:  https://doi.org/10.1002/hlca.19710540709).
Gilbert Stork, Paul G. Williard, "Five- and six-membered-ring formation from olefinic $\alpha, \beta$-epoxy ketones and hydrazine," J. Am. Chem. Soc., 1977, 99(21), 7067-7068 (DOI: https://doi.org/10.1021/ja00463a053).
Douglass F. Taber, Shawn J. Stachel, "On the mechanism of the Wolff-Kishner reduction," Tetrahedron Lett. 1992, 33(7), 903-906 (DOI: https://doi.org/10.1016/S0040-4039(00)91571-5).
A. L. J. Beckwith, I. Blair, G. Phillipou, "Preferential cis cyclization of 6-hepten-2-yl and related radicals. Example of orbital symmetry control," J. Am. Chem. Soc., 1974, 96(5), 1613-1614 (DOI: https://doi.org/10.1021/ja00812a066).
William F. Bailey, Timo T. Nurmi, Jeffrey J. Patricia, Wei Wang, "Preparation and regiospecific cyclization of alkenyllithiums," J. Am. Chem. Soc., 1987, 109(8), 2442-2448 (DOI: https://doi.org/10.1021/ja00242a032).


How to calculate the aqueous solubility of CO2 vs Pressure

The acidity of HCl is so great compared to H2CO3 that you have no equilibrium between Na2CO3 and HCl; the equilibrium is only between CO2 and water and the vapor space. H2CO3 is one of the forms of CO2 in water, but it is unnecessary to delve into this.

After your reaction, you will have a solution of NaCl in which some of your CO2 is dissolved; the rest of the CO2 will be in the vapor space. We will assume that the solubility of CO2 in this solution is the same as in pure water. Data from CRC Handbook, per 100 cc H2O at 1 atm pressure: 171 cm^3 at 0C, 90 cm^3 at 20C. In grams: 0.348 at 0C, 0.145 at 25C, 0.097 at 40C, 0.058 at 60C.

That's the chemistry. Now the mathematics is more complicated, and I'll leave that up to you. But a few points to help you get started: if you have a container with a volume of 100 cc and put 100 mL of water into it, you could also dissolve 171 cm^3 (0.348 grams) of CO2 into that same volume at 1 atm pressure (OK, maybe a slight volume increase - ignore it for simplicity). If you put only 50 mL H2O in that 100 cc volume and added 171 cm^3 of CO2, the pressure in the vessel would be greater than 1 atm (but less than 1.71 atm, which is the pressure you would have with no water in it at all).


Δ symbolizes difference. For chemical reactions, it's usually difference between energy that reactants already have and energy needed for reaction. Then Δ = $E_a$ - activation energy of the reaction. Activation energy is like a "potential barrier" for reaction to happen. And Δ means that we've added this amount of energy so energy of reactants changed by Δ (difference).


What is the geometric configuration of the four fluorine atoms during the synthesis of xenon tetrafluoride?

The answer appears to be right there in the linked synthesis details - the nickel is non-catalytic and is only there to generate a passivated surface that won't degrade in the high-temperature fluorine atmosphere.

[Edit]

The plot thickens! this paper suggests that xenon may in fact abstract fluorine off the passivated nickel fluoride surface due to observed zeroth order kinetics, which lends some serious credence to jonsca's original inference of Ni being catalytic. This will take some more investigation...


Something denser than lead but with lead like properties?

Here's a link to a listing of the elements in the periodic table sorted by density.  The density of lead is 11.35 g/cm^3.  If you wanted higher density, no toxicity and to really let your pinky live in style, then gold with a density of 19.32 would be a possibility.  Other relatively available elements that are denser than lead and without toxicity concerns include palladium (12.02), rhodium (12.41), tungsten (19.35) and platinum (21.45). If you're willing to be a bit less dense than lead, then you can reduce price by using copper (8.96), molybdenum (10.22) or silver (10.5).  Alloys of some of these elements are also available at high density, but the cost (due to fabrication costs) is likely to be as high or higher than that the element itself.


First of all, I'd ask: what do you admit as "chemistry"? You mentioned thermodynamics as being a field where you have "models to predict results". But thermodynamics is extremely important in chemistry; it wouldn't be right if we classified it as being solely physics. There is a large amount of chemistry that can be predicted very well from first principles, especially using quantum mechanics. As of the time of writing, I work in spectroscopy, which is a field that is pretty well described by QM. Although there is a certain degree of overlap with physics, we again can't dismiss these as not being chemistry.
But, I guess, you are probably asking about chemical reactivity.
There are several different answers to this depending on what angle you want to approach it from. All of these rely on the fact that the fundamental theory that underlies the behaviour of atoms and molecules is quantum mechanics, i.e. the Schrödinger equation.*
Addendum: please also look at the other answers, as each of them bring up different excellent points and perspectives.
(1) It's too difficult to do QM predictions on a large scale
Now, the Schrödinger equation cannot be solved on real-life scales.† Recall that Avogadro's number, which relates molecular scales to real-life scales, is ~$10^{23}$. If you have a beaker full of molecules, it's quite literally impossible to quantum mechanically simulate all of them, as well as all the possible things that they could do. "Large"-ish systems (still nowhere near real-life scales, mind you — let's say ~$10^3$ to $10^5$) can be simulated using approximate laws, such as classical mechanics. But then you lose out on the quantum mechanical behaviour.
So, fundamentally, it is not possible to predict chemistry from first principles simply because of the scale that would be needed.
(2) Small-scale QM predictions are not accurate enough to be trusted on their own
That is not entirely true: we are getting better and better at simulating things, and so often there's a reasonable chance that if you simulate a tiny bunch of molecules, their behaviour accurately matches real-life molecules.
However, we are not at the stage where people would take this for granted. Therefore, the ultimate test of whether a prediction is correct or wrong is to do the experiment in the lab. If the computation matches experiment, great: if not, then the computation is wrong. (Obviously, in this hypothetical and idealised discussion, we exclude unimportant considerations such as "the experimentalist messed up the reaction").
In a way, that means that you "can't predict chemistry": even if you could, it "doesn't count", because you'd have to then verify it by doing it in the lab.
(3) Whatever predictions we can make are too specific
There's another problem that is a bit more philosophical, but perhaps the most important. Let's say that we design a superquantum computer which allowed you to QM-simulate a gigantic bunch of molecules to predict how they would react. This simulation would give you an equally gigantic bunch of numbers: positions, velocities, orbital energies, etc. How would you distil all of this into a "principle" that is intuitive to a human reader, but at the same time doesn't compromise on any of the theoretical purity?
In fact, this is already pretty tough or even impossible for the things that we can simulate. There are plenty of papers out there that do QM calculations on very specific reactions, and they can tell you that so-and-so reacts with so-and-so because of this transition state and that orbital. But these are highly specialised analyses: they don't necessarily work for any of the billions of different molecules that may exist.
Now, the best you can do is to find a bunch of trends that work for a bunch of related molecules. For example, you could study a bunch of ketones and a bunch of Grignards, and you might realise a pattern in that they are pretty likely to form alcohols. You could even come up with an explanation in terms of the frontier orbitals: the C=O π* and the Grignard C–Mg σ.
But what we gain in simplicity, we lose in generality. That means that your heuristic cannot cover all of chemistry. What are we left with? A bunch of assorted rules for different use cases. And that's exactly what chemistry is. It just so happens that many of these things were discovered empirically before we could simulate them. As we find new theoretical tools, and as we expand our use of the tools we have, we continually find better and more solid explanations for these empirical observations.
Conclusion
Let me be clear: it is not true that chemistry is solely based on empirical data. There are plenty of well-founded theories (usually rooted in QM) that are capable of explaining a wide range of chemical reactivity: the Woodward–Hoffmann rules, for example. In fact, pretty much everything that you would learn in a chemistry degree can already be explained by some sort of theory, and indeed you would be taught these in a degree.
But, there is no (human-understandable) master principle in the same way that Newton's laws exist for classical mechanics, or Maxwell's equations for electromagnetism. The master principle is the Schrödinger equation, and in theory, all chemical reactivity stems from it. But due to the various issues discussed above, it cannot be used in any realistic sense to "predict" all of chemistry.

* Technically, this should be its relativistic cousins, such as the Dirac equation. But, let's keep it simple for now.
† In theory it cannot be solved for anything harder than a hydrogen atom, but in the last few decades or so we have made a lot of progress in finding approximate solutions to it, and that is what "solving" it refers to in this text.


Simple procedure for showing presence of milk proteins in clarified butter (ghee)

A classic test for protein is Biuret solution, using readily available reagents:  sodium hydroxide, copper sulfate and potassium sodium tartrate (Rochelle salt). The test is not specific for milk protein, but what other protein would be in pure butter?

Use caution with the chemicals -- the first listed is quite caustic, and the second is poisonous. BTW, save some leftover Rochelle salt if you'd like to make a piezoelectric crystal.


Let's reword it:

Traces of oil and grease are distilled off at temperatures below their boiling points.
The temperature in the frying pan is above 100 °C.
The food does contain significant amounts of water.

This sounds like the conditions for steam distillation, except that you're collecting the mixture of water and water-insoluble material everywhere in the kitchen and not in a flask.


Is an explosion merely an exothermic reaction?

The key parts to the definition of an explosion are "rapid" and "increase in volume".

Exothermic reactions do not necessarily meet these requirements. Exothermic reactions need only release energy to the surroundings by converting chemical potential energy into kinetic energy.

For example, the combustion of methane is very exothermic, with a standard enthalpy change of -890 kJ/mol.

$\ce{CH4(g) + 2O2(g)->CO2(g) +2H2O(g)} \space  \space \space \space \Delta_c H^o=-890 \text{ kJ/mol}$

However, at room temperature this reaction is very slow. Mixtures of oxygen and methane do not spontaneously ignite without a spark source. Also, there is no change in pressure with this reaction, since there are 3 moles of gas on both sides of the equation.

On the other hand, consider nitroglycerin. Yes, it's combustion is significantly exothermic. More importantly, the combustion happens rapidly, needs only the heat of friction or the energy of impact to ignite, and produces significantly more gas than it consumes:

$\ce{4C3H5O9}(l) \ce{+1O2(g)->12CO2(g) +10H2O(g) +6N2(g)} \space  \space \space \space \Delta_c H^o=-5660 \text{ kJ/mol}$

All thermodynamic data were mined from the NIST Chemistry Webbook.


Will Proving or Disproving of any of the following have effects on Chemistry in general?

All right, let's sum it up.

P vs NP, if solved the way that pretty much everybody expects (that is, $\rm P\ne NP$), will have no far-reaching consequences, because that's what we were thinking for quite a while now. If it would miraculously happen to be otherwise (that is, $\rm P=NP$), that would be quite a shock to many fields, especially to computer science, and by extension, to computational chemistry. But I don't believe in miracles.
Navier–Stokes (no matter how it will turn out) will have consequences in fluid dynamics, and by extension, maybe in some relatively narrow areas of chemical technology.
Yang–Mills will cause repercussions in the areas of physics which are the most distant from chemistry.
The rest of the problems, as far as I understand, have no bearing whatsoever on our material world.


The effect of the groups at para position on SN2 reactivity

This is what I found from your reference to Clayden. You just had to read the next paragraph ( I have highlighted the point ). You can see that March and Clayden have agreed on the same point, that electron withdrawl from the phenyl/allyl group increases the effective conjugation in the transition state.
So I hope You can understand what will be the correct order and why.

P.S. I know this question is old, but it was unanswered, and someone may benefit from an answer.

Reference: Jonathan Clayden, Nick Greeves, S.Warren, P.Wothers -Organic Chemistry (2000) - Pg 422.


How come uranium's relative atomic mass is 238.03 when it only contains isotopes with a mass number of 238 or less?

Approximately 99.3% of uranium on Earth is the $\mathrm{^{238}U}$ isotope, and this specific isotope has an atomic mass of $\mathrm{238.05\ u}$, where $\mathrm{u}$ is the atomic mass unit, equivalent to 1/12 the mass of a $\mathrm{^{12}C}$ atom. Including the other isotopes to obtain the average atomic mass drags the value down a little, but it still ends up being slightly larger than $\mathrm{238\ u}$.

It is interesting to point out why the atomic mass is slightly different from the mass number. The masses of a free proton or neutron are both slightly larger than $\mathrm{1\ u}$ ($\mathrm{1.0073\ u}$ and $\mathrm{1.0087\ u}$, respectively). Some of this mass is converted to energy and lost when neutrons and protons (collectively called nucleons) bind to form nuclei, which means bound nuclei will always be lighter than the sum of the masses of their component nucleons. Different nuclei release different amounts of energy per nucleon depending on how many protons and neutrons are inside the nucleus, and on their relative proportions. Very interestingly, the curve for nuclear binding energy per nucleon versus mass number starts off at zero for $\mathrm{^{1}H}$ (there can be no attractive nuclear forces between nucleons if there's only one nucleon!), then rises until it shows a maximum around nickel (specifically for the isotope $\mathrm{^{62}Ni}$) and from there on slowly decreases.

Since the atomic mass unit is standardized with respect to the $\mathrm{^{12}C}$ isotope of carbon, then in approximate terms, the nuclei with a lower binding energy per nucleon than $\mathrm{^{12}C}$ will have less of the initial free proton/neutron mass lost to energy (and thus an isotope will have an atomic mass slightly higher than its mass number), while for nuclei with a higher binding energy per nucleon than $\mathrm{^{12}C}$, more mass will be lost to energy and the resulting isotope will have an atomic mass slightly lower than its mass number.

Taking a horizontal line through the $\mathrm{^{12}C}$ isotope in this graph and looking at its intersections with the nuclear binding energy curve will show the regions where this flip happens; isotopes for atoms below carbon will tend to have higher atomic mass than mass number, isotopes for atoms between carbon and thorium tend to have a lower atomic mass than mass number, and then isotopes for atoms above thorium (such as $\mathrm{^{238}U}$) will again tend to have higher atomic masses than mass numbers. This is only an approximate analysis, as protons and neutrons don't have the same mass, so isotopes very close to the transition region don't behave perfectly smoothly.

Edit: The flipping effect I mention can be seen very nicely in the list of isotopes shown in this site. Look at the (atomic) masses for the isotopes, and see how the fractional part of the mass starts out as circa $.01$ (above the isotope's mass number), then decreases to $.001$, reaches exactly zero (by definition) for $\mathrm{^{12}C}$, then soon after flips under to $.999$ and keeps going lower all the way to $.90$ around zirconium and tin isotopes before slowly rising up to $.98$ for bismuth $\mathrm{^{209}Bi}$ and then flipping back over to $.04$ for thorium and $.05$ for uranium.

Now here's a question: I said the maximum of nuclear binding energy per nucleon happened at $\mathrm{^{62}Ni}$, so why does the fractional part of the masses in the table keep decreasing all the way to tin $\mathrm{^{116}Sn}$? It's because these masses are not normalized with respect to how many protons and neutrons there are in the nucleus. If you were to take all the isotopic masses and divide them by their respective mass numbers, you would see the minimum around $\mathrm{^{62}Ni}$, as expected (actually, the minimum ratio of isotope mass to mass number happens at $\mathrm{^{56}Fe}$ because it has a slightly higher proportion of protons to neutrons. As I said, the small difference between proton and neutron masses creates slight irregularities which I did not take into account for simplicity).


How to calculate n factor for O2 and H2 in this reaction?

Here is the unbalanced reaction
$H_2 +O_2----> H_20$
I hope you know how to calculate oxidation number
 On product side
In $H_20$ oxidation number of $H$ is  $+1$ and $O$ is $-2$
On reactant side
oxidation number of $H$ and $O$ is  $0$
So n-factor is magnitude in change of oxidation number
So n factor of $H$ is $1-0$ i.e $1$ and $O$ is $-2-0$ i.e $2$(magnitude)
Now to balance the reaction (for 2 reactant rex) you just interchange the n factors
You get
$1O_2 +2H_2 -----> 2H_20$
Now to calculate equivalent weight formula is
$E.w=molar mass/N.F$


Someday I was wondering as you are doing but  I think that now my confusion has lessened considerably after talking with experts in the field and enrolling in a master program of pharmaceutical chemistry and drug design.

Different kinds of simulations can be called modelling of cellular processes. The first are the rules of systems biology and the second are the rules of chemistry or more fundamentally, quantum chemistry.

An example of systems biology is the simulation of the heart muscle fiber. a single cell, contractility as a function of the different ions concentrations, K, Ca, Cl, Na, etc.

The rules of quantum mechanics when applied to systems of atoms and molecules yields quantum chemistry. Quantum chemistry can calculate the tendency of a certain chemical reaction a function of the electron density and energy in certain regions. Molecular Mechanics and Molecular dynamics are more convenient when simulating bigger systems, larger number of atoms and molecules. With large number of molecules statistical mechanics must be taken into account to realistically simulate the relative population of molecules that have a certain energy.

In you question, I don't think that analytical chemistry has any thing to with computer simulation of cellular processes. From the the disciplines you outlined I thank that you are more interested in Quantum chemistry/Molecular mechanics/Molecular dynamics than that of systems biology. In my opinion you have to focus on: Physical chemistry, quantum chemistry, statistical thermodynamics,  Drug discovery and design. Also, In order to understand molecular dynamics you must have a background in motion laws of newton.

Occasionally you will need to gain a background in some specific points but, however, mastering the basics of the disciplines I mentioned will give you a comfortable platform from which you can move to your target easily.

Regards.


The test for $\ce{Cr(III)}$ can be performed in the solid state using an oxidative melt. You will need

chromium(III) oxide (obviously)
sodium carbonate
potassium nitrate
a Bunsen or Teclu burner
a magnesia trough or a crucible
suitable pliers to hold the trough

Mix the green pigment with up to the threefold amount of sodium carbonate and potassium nitrate, place the mixture on the magnesia trough and melt it.

A yellow melt indicates the formation of sodium chromate according to

$\ce{Cr2O3 + 3KNO3 + 2Na2CO3 -> 2Na2CrO4 + 3KNO2 + 2CO2 ^}$


Software to assist drawing of complex, three-dimensional skeletal formula from scratch or from existing crystal structure

Thanks to vapid and their comment

  There are editors that have 3D mode and they can display a 3D molecule in 2D manner. MarvinSketch is first that comes to my mind, but there are probably more.

I figured a crude solution, which I hope might as well serve as a starting point for other answers.

First of all, I completely forgot that MarvinSketch (free (registration required), available for Windows, Linux, macOS) has support for 3D.
Drawing in 3D from scratch with MarvinSketch is probably not the best idea as even though it has a support for optimizing (cleaning) 3D structures, it sometimes gives weird results for organics and practically not suitable for inorganic complexes and especially clusters.

Arguably, the best and the most affordable third-party tool for creating such structures is Avogadro (open source, available for Windows, Linux, macOS).

If there is a crystal structure at hand (e.g. CIF data), then one must do two things first:

crop/grow the fragment of interest;
assign proper bond orders and export the structure in one of the formats that preserve bond orders (such as MOL, MOL2, PDB or CML).

I believe Avogadro can do both, but I personally like to do part 1 in Olex2 (free (registration required), available for Windows, Linux, macOS) by force of habit.

As a proof of concept, I took the crystal structure for the trefoil knot, allowed it to grow from assymetric unit to a complete molecule as well as cut out solvent and guest molecules, so I was left with the initial shape:

which I immediately exported to the MOL file by running file trefoil-knot.mol.

Next, I opened this MOL file with Avogadro, which automatically assigned bond orders:

and the file was exported to MOL once again, and subsequently opened in MarvinSketch (I reduced fog in settings, made implicit hydrogens hidden and used F7 button to align the molecule on screen):

And the final result came out to look like this:

The image is, however, not quite as perfect as the one in the paper.
For instance, there are some artifacts about inflection points as if the program was confused how to place double bonds.
Also, if you look closer, the grayed out part which was supposed to be further from the viewer, actually appears in front, which I attribute to the fact that Marvin was designed as a 2D molecular editor first and foremost, so it doesn't always handles depth (or the aforementioned z-buffer) adequately.


What concentration is phenolphthalein indicator ? 1% or 0.1%

During titration of small amounts of acids, the molar amount of the indicator in 1-2 drops of $\pu{1 \%}$ indicator solution may not be negligible compared to the acid molar amount, affecting the result.
So for that cases, $\pu{0.1 \%}$ solution is used, to be able to dose smaller indicator amounts.
As the phenolphalein molar mass is about $M=\pu{318 g/mol}$, 1 drop of $\pu{1 \%}$ solution is equivalent about 3 drops of $\pu{0.01 M}\ \ce{NaOH}$.


Compound having the highest enol content will be?

If you can form an enol where the carbon-carbon double bond is conjugated to one or more other double bonds, the enol is more stable and the equilubrium mixture will therefore have more.  With that in mind, try seeing what happens to (3) if you tautomerize with the proton on the bridging carbon between the two carbonyl groups.

For a further exercise: render the structure for cyclohexa-2,4-dien-1-one where the carbonyl group is conjugated to two carbon-carbon double bonds in the ring.  Now tautomerize to the enol and see what "enol" you get.  When the conjugation of the enolic double bond forms an aromatic ring, the stabilization effect is so strong that the enol takes over completely.


Why isn't this resonance possible in an imidazole ring?

You need to understand the basic reason for the aromatic nature of materials. A molecule is said to be aromatic only if:

A ring in it is planar,
Continuous delocalisation is possible throughout the concerned cyclic ring, and
This delocalisation stabilises the ring.

Now, for continuous delocalisation to be possible, the $\mathrm{\pi}$ orbitals of all the elements in the ring need to be parallel to each other ( They need to overlap sideways for formation of  $\mathrm{\pi}$ bonds).

In your example, consider the carbons in the ring. All of their $\mathrm{\pi}$ orbitals are parallel to each other and PERPENDICULAR to the plane of paper. On the other hand, the imine nitrogen is $\mathrm{sp^2}$ hybrid. The lone pair of electrons is in an  $\mathrm{sp^2}$ hybrid orbital, i.e, IN the plane of the paper. And mind you, it's fixed (can't rotate without distorting the geometrical shape of the ring, that is, which would be too much to expect from the poor guy.)

So the imine Nitrogen cannot give the lone pair anywhere. it can only play around with the  $\mathrm{\pi}$ bond already on it.

Moreover, as soon as the amine Nitrogen takes up a proton, it loses its ability to use the lone pair for delocalisation. The condition for planarity itself is not fulfilled ($\ce{(R)2-N+-(H)2}$ has tetrahedral geometry), so there's no use of discussing aromaticity. Before the proton uptake, the ring was aromatic because the amine Nitrogen was using the lone pair for continuous delocalisation, effectively attaining planar geometry. But now... It can no longer be done.

Conclusion: Resonance is still possible(between all atoms of the ring except the amine Nitrogen), but Continuous Delocalisation is not. So, the system loses its aromaticity.


Disubstituted cyclic alcohols - looking for IUPAC rule

For preferred IUPAC names, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), explicitly stipulates when locants are omitted. Most of these rules apply to the omission of the locant ‘1’.

In particular:

  P-14.3.4.2 The locant ‘1’ is omitted:

  (…)

  (c) in monosubstituted homogeneous monocyclic rings;

  (…)

One example for this rule given in the Blue Book is cyclohexanethiol. Likewise, cyclopentanol is the preferred IUPAC name for the first compound that is given in the question.

This rule, however, is not applicable to the other two compounds that are given in the question since they are not monosubstituted.

Furthermore, we can find a general rule that is relevant in this case:

  P-14.3.3 Citation of locants

  In preferred IUPAC names, if any locants are essential for defining the structure of the parent structure or of a unit of structure as defined by its appropriate enclosing marks, then all locants must be cited for the parent structure or that structural unit. For example, the omission of the locant ‘1’ in 2-chloroethanol, while permissible in general usage, is not allowed in preferred IUPAC names, thus the name 2-chloroethan-1-ol is the PIN. (…)

Therefore, the preferred IUPAC names are cyclopentane-1,1-diol and 1-methylcyclopentan-1-ol, respectively. The omission of the locant ‘1’ is not permissible for the preferred IUPAC name.

Nevertheless, the name ‘1-methylcyclopentanol’ describes the correct compound and may be used in general nomenclature.


Do ions also form special electron configurations? (just as Copper and Chromium do)

Check out the NIST database for electronic configurations for atoms and ions in the gas phase. If you play around a bit, you will find that yes, some ionizations will not follow the expected trends from the aufbau principle. For example, search for rhenium and compare the configurations between $\ce{Re^{11+}}$ and $\ce{Re^{17+}}$, and you'll see some of them are unexpected (and rather hard to explain). Osmium shows an anomaly going from $\ce{Os^{+}}$ to $\ce{Os^{2+}}$. There are other elements with ions containing aufbau anomalies, but I can't remember many from the top of my head.


Should double or triple bonds have preference for low locants in IUPAC nomenclature?

The appropriate passage in the 2013 Blue Book is as follows:

P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride [...] is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.

Therefore, the preferred IUPAC name for the above compound is hexa-1,3-dien-5-yne, as the double bonds receive the lower locants (1,3) than in the alternative name hexa-3,5-dien-1-yne.
Note that this preference for double bonds over triple bonds is only relevant when there is no difference between the locant set for the multiple bonds. In the following example, the locant set (1,4) is lower than (2,5) and hence hex-4-en-1-yne is the preferred name, even though it has a higher locant for the double bond:


The exercise in question is a prediction of the appearance energy of the methyl cation fragment $\ce{CH3+}$. The authors predict that value by estimating the enthalpy for the reaction of methane ionization followed by loss of a hydrogen atom. The enthalpy change estimation is based on published values for heats of formation and bond dissociation energies that have been determined experimentally.

The values presented, however, are rounded to the nearest integer or tenth, and the authors continue to round values during the explanation. For example, 12.6 eV is converted to the closest integer value of 1216 kJ/mol. All values in Fig 2.7 are presented as integers.

There seem to be two possible explanations for the final value being given as 14.35 rather than your calculated value of 14.365. Most likely, the authors have chosen to round to the nearest multiple of 0.05 rather than the nearest tenth or hundredth. Alternatively, they could have used unrounded values throughout the calculation rather than the rounded numbers presented in the step-by-step description. Either way, the difference is insignificant relative to the (im)precision with which the value can be predicted or experimentally determined.


Yes, but.

Yes, in the conventional low-level models, one would consider the two carbons in benzyne you mentioned as sp-hybridised. And that does mean that their orbitals seem to be pointing the wrong way.

However, a better picture would be to use an orbital which is a lot closer to sp²-hybridisation, and an even better picture would calculate orbitals computationally (I can’t do that, but Orthocresol can).

The thing is, all these pictures show, that the triple bond in benzyne is rather different from those in linear molecules or large cycloalkynes such as cyclodecyne ($\ce{C10H16}$). It contains a large angular strain, not unlike very small saturated rings such as cyclopropane. The strain causes the triple bond to be rather weak and easily cleaved/attacked — which paves the way for all reactions in which benzyne is an intermediate.

Larger cycloalkynes with eight or nine carbons have been isolated, but for benzyne only in situ studies have been made according to my quick check on SciFinder two seconds ago. In layman’s terms, you can think of the angle strain by saying ‘I am a triple bond, I want a $180°$ angle, but all I can have is $120°$ so I’m rather unhappy!’ A straight-up $180°$ angle is not possible, however, because of the other two atoms in the benzyne ring.

Kukolich, McCarthy and Thaddeus published two papers (one together with Tanjaroon) on microwave-spectrum measurements to determine the structure and compare it to in silico structures. Their results give a $126°$ angle along $\ce{C^1-C^6-C^5}$ (your numbering scheme). This is more obtuse than in benzene ($120°$) but still far away from $180°$. However, the major constraint is the angle $\ce{C^6-C^1-C^2}$, which is compressed to a mere $111°$.[1,2]

Before the objection is raised and as ron pointed out in the comments: The triple bond is indeed a bond and not a diradical as can be derived from its behaviour. E.g, it concertedly and stereospecificly adds to dienes.

References
[1] S. G. Kukolich, C. Tanjaroon, M. C. McCarthy, P. Thaddeus, J. Chem. Phys. 2003, 119, 4353. DOI: 10.1063/1.1593015
[2] S. G. Kukolich, C. Tanjaroon, M. C. McCarthy, P. Thaddeus, J. Phys. Chem. A 2004, 108, 2645. DOI: 10.1021/jp031344p


Estimate the concentration of HCl and volume of 6 M HCl required to complete precipitate each of the given salts

Your calculations are looking good. I didn’t check the numerical values, I’ll just assume you’re able to do your algebra and use a calculator correctly.

What you got are two different concentrations required to start to precipitate all the cations in your solution. However, you are asked to completely precipitate the cations. Well — that is a nigh impossible task unless you define a concentration threshold which you want to reach. Your threshold could be $1\,\mathrm{mM}$, $1\,\mathrm{µM}$, $1\,\mathrm{nM}$ or whichever you see fit. Use this desired concentration to recalculate the concentration of chloride ions required for both cations. Then continue on.

I recalculated the value in the spoiler tag for $1\,\mathrm{µM}$ residual cation concentration.

   $4.12\,\mathrm{M}$ chloride, equivalent to $0.687\,\mathrm{mL}$ or 18 drops.
Backwards calculation tells me, that you had a residual lead concentration of ca. $9.58\,\mathrm{µM}$, explaining your second result.


Would Schweizer's reagent dissolve nitrocellulose?

Schweizer's reagent dissolves cellulose because it forms a complex with hydroxyl groups of glucose moieties. Therefore, a completely nitrated cellulose will not dissolve in Schweizer's reagent. However, the reagent is quite basic and it will cause a hydrolysis of the nitrocelulose, so it may dissolve eventually, but after precipitation it will be normal cellulose again. But there is also a good information for you: nitrocellulose easily dissolves in many organic solvents e.g., acetone, ethyl acetate, diethyl ether and so on. So once you have a nitrocellulose, you do not need to play with Schweizer's reagent.


Why is measurement in molarity preferred over molality?

Molality is used for those people trying to understand how the melting point or the boiling point of water or any other substance, changes when they add a given amount of salt or any other impurity to one kilogram of this pure substance. They don't care about the volume of the obtained solution. They will not keep the solution once obtained. As an example, they just want to check if it is correct to say that adding $1$ mole of a solute to $1$ kg water lowers the freezing point by $1.86°$C, whatever the nature of the solute.

Molarity is used for another group of people. They don't mind about the process of dissolution of a substance in a solvant. They don't care about the volume of water to be used to make $1$ liter of the solution. They just want to know the composition of the solution they have to use, i.e. how many moles (or grams) of a solute are present in a given volume of this solution. If for a given use, they must have a solution containing exactly $9$ g NaCl per liter, they don't care about the way of dissolving the salt. All they need to know is whether this solution can be used for making an injection, or any other use.


Are geminal disubstituted alkenes more stable than their cis/trans isomers?

This is a very good question and if popular books give conflicting answers, then it must be reasoned out. Unfortunately, Paula Bruice has given the wrong answer while the other two books have given no explanation for this comparison.

For the answer I assume that you know about hyperconjugation and the various contributing structures it involves.  This gives a basic idea of hyperconjugation:

Source: tutorcircle.com

Note that on hyperconjugation, the other carbon of the alkene gets a negative charge on it.

If the hyperconjugation structures of 2-methylprop-1-ene and trans-but-2-ene are drawn:

you will notice that:

$ \ce{C_1} $ of the former has a negative charge on it
$ \ce{C_2} $ of the latter has a negative charge on it
The latter's methyl group makes the negatve charge less stable due to +I (inductive effect)
The former has no such destablizing effect as the carbon containing the negative charge is just connected to hydrogens
As all the hyperconjugation structure have this effect applying on them, the net effect cant be ignored for the overall stability

Hence the overall stability of  2-methylprop-1-ene > trans-but-2-ene


Note: As stated in a comment to the question by Loong, the preferred IUPAC name according to the 2013 nomenclature recommendations is:

  3-(10,11-dihydro-5H-dibenzo[a,d][7]annulen-5-ylidene)-N,N-dimethylpropan-1-amine

The only change this makes to the analysis it that we can ignore the ‘incorrect’ cycloheptene, because [7]annulen is automatically a cycloheptatriene-type molecule.

First, let’s separate the name into its fragments. We have:

3-something-propane-1-amine
N,N-dimethyl
(10,11-dihydro-something)
(something-5H-dibenzo[a,d]cycloheptene-something)
(something-5-ylidene)

I’m going to start with item 4, since that is the main bicycle. Cycloheptene sounds like this molecule:

But the [a,d] and the dibenzo is telling us that we have a fused polycyclic unsaturated hydrocarbon, so we will need to add two more double bonds. (Skipping a picture of this step.) Dibenzo means that there are two fused benzene rings, and the letters in square brackets symbolise the attachment points. a stands for 1,2; d for 4,5 (analogously for other letters). With that, we get a tricyclic compound with many double bonds. However, it is still ambiguous how exactly to place the double bonds, we need the information 5H — it tells us that the ‘additional’ hydrogen must be at the 5-position. Beware: The 5-position does not reference the initial numbering we used for fusing the rings; the atoms are renumbered according to the rules for aromatic polycycles: Starting at the top right and going clockwise, ignoring all bridgehead atoms. So we get 5H-dibenzo[a,d]cycloheptene which looks like this:

Next up, item 3: 10,11-dihydro. This means that the double bond in the seven-membered ring is hydrated, i.e. there is a single bond instead.

And finally, item 5: 5-ylidene. We are attaching something to carbon 5 via a double bond. We’ll get back to that.

Item 1: 3-something-propane-1-amine. That’s nice and simple, we’ll first consider propane-1-amine:

N,N-dimethyl (item 2) tells us that there are two methyl groups attached to the nitrogen, giving us the following:

And now we need to add the entire tricyclo-blurb to the propane’s 3-position using a double bond:

And voilà, there is amitriptylene!


Is there a mathematical generalization for the number of isomers (structural and configurational) of a given compound?

Generally speaking, graph-theoretical enumeration aims at counting chemical compounds as graphs (2D structures).
In other words, it is concerned with constitutional
(or structural) isomers.

The most famous method for combinatorial enumeration of graphs is based on Polya's theorem: G. Polya and R. C. Read,
Combinatorial Enumeration of Groups, Graphs, and Chemical Compounds (Springer, 1987).
G. Polya, R. E. Tarjan, and D. R. Woods,

Notes on Introductory Combinatorics (Birkhauser, 1983) Chapter 6.
An introduction to the practical uses of Polya's theorem from a chemical viewpoint has appeared:   O. E. Polansky,

Polya's method for the Enumeration of Isomers
(MATCH Commun. Math. Comput. Chem. 1, 11--31 (1975)). free access.
Many examples of chemical application of Polya's theorem have been described in
the following book: A. T. Balaban (Ed.), Chemical Applications of Graph Theory (Academic Press, 1976).

To accomplish enumeration of 3D structures (concerned with configurational isomers or stereoisomers), in contrast, it is necessary to take chirality/achirality of substituents into explicit consideration.

Symmetry-itemized enumeration of chemical compounds as 3D structures can be accomplished by adopting Fujita's unit-subduced-cycle-index (USCI) approach: S. Fujita,
Symmetry and Combinatorial Enumeration in Chemistry (Springer 1991).
Gross enumeration without symmetry-itemization can be accomplished by
adopting Fujita's proligand method: S. Fujita,
Combinatorial Enumeration of Graphs, Three-Dimensional Structures, and Chemical
Compounds (University of Kragujevac, Kragujevac, 2013).
For the enumeration of alkanes ($\ce{C_nH_{2n+2}}$) and
monosubstituted alkane ($\ce{C_nH_{2n+1}X}$) as 3D structures,
I would like to introduce an account article, which is freely available: S. Fujita,
Numbers of Alkanes and Monosubstituted Alkanes. A Long-Standing Interdisciplinary Problem over 130 Years (Bull. Chem. Soc. Japan, 83, 1--18 (2010)).
The following article is also available freely: S. Fujita, Numbers of Achiral and Chiral Monosubstituted Alkanes Having a Given Carbon Content and Given Numbers of Asymmetric and Pseudoasymmetric Centers (Bull. Chem. Soc. Jpn., 81, 193--219 (2008)).

To grasp perspectives of the theoretical foundations of stereochemistry and of the state-of-the-art chemical enumeration, see also the following recent book: S. Fujita,
Mathematical Stereochemistry (De Gruyter, Berlin, 2015)


Volume changes when mixing DMSO and water

If we assume that the density of the mixture is just the average of the individual densities, would get 1 litre of a mixture with a density of 1.04635 kg/l.
However, measured density (1, table 1) is 1.0682.

We thus expect to see a volume that is 0.9795 times the added individual volumes.

In this case: 0.9795 litres.

LeBel, R. G.; I. Goring, D. A. Density, Viscosity, Refractive Index, and Hygroscopicity of Mixtures of Water and Dimethyl Sulfoxide. J. Chem. Eng. Data, 1962, 7 (1), 100–101


In the blue bottle experiment, will it eventually be "used up" after repeated use?

The blue bottle experiment, and other similar reactions like the "chemical traffic light", are not infinitely repeatable.  They are driven by the oxidization of glucose (or some other suitable reducing agent, such as ascorbic acid) by atmospheric oxygen, and will eventually stop working once the bottle runs out of either glucose or oxygen.  However, each cycle only consumes a rather small amount of glucose and oxygen, so the reaction can be repeated quite many times.

Also, the blue indicator dye (methylene blue) itself is not used up in the reaction.  Even if the reaction does "run out of fuel", it can be made to work again just by adding more glucose to the solution and/or simply by opening the bottle to let more oxygen in.  However, since the glucose is not getting fully oxidized all the way to water and carbon dioxide in this reaction, the various incomplete oxidization products will gradually accumulate in the solution.

I'd expect that, if you kept adding more and more glucose to the solution, eventually you'd end up with a sticky sugary alkaline goo full of partly oxidized glucose products that was too viscous to be swirled around the flask any more.  At that point, you'd really have no choice but to dispose of the experiment and start over with fresh ingredients.


Can water electrolysis detect the water impurities?

[…] can this method detect the impurities in water […]

In principle, yes. The electrolysis can, as an example oxidize iron ions ($\ce{Fe^2+ -> Fe^3+}$) The latter forms a brownish, voluminous oxide hydroxide, $\ce{FeO(OH)·n H2O}$.

The WHO suggests an upper limit of 0,2 mg/l iron in drinking water.

Am I right that your drinking water comes from a well? Did you perform any analyses on the the water quality or consult local authorities concerning possible health risks and their prevention?

  […] or is it just some dirty sales tactic […]

I can't comment on the dirty part, but salesmen need to sell. We don't know whether the reference sample was actually produced by reverse osmosis or just a sample of distilled water (which you would not want to drink over a longer time).

Even if reverse osmosis can reduce the mineral content in your drinking water, you might want to check whether the additional costs for operation are really worth it.

Again, try to obtain some information from local authorities.


Why is the iodide anion a good nucleophile but a poor base?

You should clarify whether the iodide anion is a good nucleophile in a polar protic or polar aprotic solvent.

Also let's note that nucleophilicity is a kinetic property, while acidity/basicity are thermodynamic properties. Iodide ion's lack of basicity in water reflects its conjugate acid's ($\ce{HI}$) lack of stability; it's relatively easy to ionize its proton due to the weak hydrogen-iodide bond.

On the other hand, the high nucleophilicity of iodide ion in water reflects the high rate of reaction between iodide ion and some electrophilic atom (other than hydrogen).

The relative lack of basicity of the iodide anion in water is precisely what makes it a good nucleophile. Its lack of affinity toward protons - i.e. the protons of water - physically frees $\ce{I^-}$ to attack electrophilic atoms (not limited to carbon). There is little hydrogen bonding between iodide ion and water.

On the other hand, the relative basicity and thus strong hydrogen-bond forming capabilities of $\ce{F^-}$ makes it a poorer nucleophile in water. Rates of reaction between fluoride anion and some electrophilic atom are slowed because the fluoride anion is trapped (strongly solvated) by water molecules. And there will also be a sizable number of fluoride anions that have completely abstracted a proton from water to make $\ce{HF}$, which isn't nucleophilic at all.

In a polar aprotic solvent, however, where there are no positively polarized hydrogens for a nucleophile to hook onto; $\ce{F^-}$ becomes a much stronger nucleophile. Actually,  $\ce{F^-}$ becomes a better nucleophile than $\ce{I^-}$, likely due to $\ce{F^-}$'s greater charge density.


Is a hydrogen molecule smaller than hydrogen atom?

You can define the size of atoms and molecules in various ways. You can, for instance, derive a measure of the size of a hydrogen molecule from the density of solid hydrogen:

  Solid hydrogen has a density of 0.086 g/cm3 making it one of the lowest-density solids.

From the above you can derive an effective radius for $\ce{H2}$ of $\pu{2.10 Å}$, which is eerily similar to $4 \times \pu{0.53 Å}= \pu{2.12 Å}$.

However this is a distraction from the main point of the statement in the book, which is simply that concentration of electron density between the two $\ce{H}$ nuclei results in an "end-to-end" distance (the distance, using some measure to determine the edges of the molecule, along the internuclear axis) that is less$^\dagger$ than $\pu{2.12 Å}$, and that the internuclear distance is less than $\pu{2\times\pu{0.53 Å}= 1.06 Å}$ (which it is: the H-H bond length is $\pu{0.74 Å}$). An internuclear distance shorter than twice the lone H radius is consistent with bonding, a process by which electrons and nuclei are allowed to occupy lower-energy regions of the potential than they would when the atoms are far appart.

But the answer to the question is no, a hydrogen molecule is still bigger than a hydrogen atom.

$\dagger$ But just barely. Define the edge of an atom or molecule as the point where the electron density equals that at a distance of $\pu{1 Bohr}$ ($\pu{0.529 Å}$) from the nucleus in a hydrogen atom. Next find the edges in a simple  MO model of $\ce{H2}$ ([3]). The edges are $\pu{1.056 Å}$ from the center of the molecule, that is, the molecule along the longest axis has a diameter of $\pu{2.11 Å}$, barely less than $4\times$ the Bohr radius.

[3]: Pearson, Ralph G., and Palke, William E.. Proc.Natl.Acad.Sci.USA, Vol. 77, No.4, pp.1725-1727, April 1980.


Can someone confirm and/or correct my logic regarding this mechanism?

The decarboxylation occurs specifically at that position, because that is a $\beta$-keto acid. In presence of slight amount of heat only the decarboxylation occurs due to the presence of keto group adjacent to it. Actually, in acidic medium only,it decaboxylates in presence of heat as follows,

But no keto group is present at the carboxylic acid group attached at the benzylic position so it cannot decaboxylate.


In practice, do polar molecules actually align in an electric field?

These are two entirely different and unrelated phenomena, to the point that you'd better unsee that other question and my answer to it, if you can. Ions are very much not like molecules. Ions are charged; molecules are not. Molecules don't form the double layer, nor do they even know what is Debye length. The electric field (though somewhat diminished, by a factor of dielectric constant) penetrates the whole volume of liquid, so all molecules feel it the same way; none is shielded.

Yes, polar molecules get oriented in the electric field, albeit to a very tiny degree. Think of it this way: the molecule still bounces and rotates around, but it spends a little bit longer in the field-aligned orientation than in any other. To orient all molecules in a truly stiff way, like soldiers in formation, the energy of alignment has to overtake the energy of thermal motion, which requires enormous electric fields and is ultimately impossible, because a breakdown would occur way sooner than you get there.


Reaction between carboxylic acids and amines

You have a false dichotomy.

Diamines and diacids can form salts with doubly charged anions and cations. Sadly, simply searching SigmaAldrich’s catalogue for the phrase diammonium did not turn up any example; however, I assume this is because these salts are unlikely to be used and thus wouldn’t sell. Chemically, proton transfer is one of the fastest reactions possible and thus simply mixing a diacid and a diamine in a suitable solvent would result in the diammonium diate salt as it would for an ammonium monoate salt.

Amide bond formation is a different reaction that typically requires activation or harsher conditions (e.g. heating). It is a condensation that results in the release of water. All that is required is the presence of an amino group and a carboxylic acid. So aminoethane and ethanoic acid can be heated in solution to produce N-ethylacetamide.

The one difference between the mono and the di systems is that the mono system has no of polymerising or cyclising. So the amide bond will always consist of one acid fragment and one amine fragment while in the case of a diamine and a diacid polymerisation or cyclisation can occur.


Order of Acidic strength: Phosphorus oxoacids

Acidic strength of the acids is determined on the relative tendency to donate one proton.

The source where you have seen then $\mathrm{p}K_\mathrm{a}$ value has considered the tendency of first deprotonation of the acids ($\ce{H3PO2}$ deprotonates once, $\ce{H3PO3}$ can deprotonate twice and $\ce{H3PO4}$ can deprotonate thrice).

The double bonded oxygen has negative inductive effect (-I). And this electron withdrawing effect is experienced by one $\ce{H}$ of the one $\ce{-OH}$ group in $\ce{H3PO2}$. Hence it has greatest acidic strength.

Similarly, the -I effect of $\ce{O}$ is experienced by two $\ce{H}$ of the two $\ce{-OH}$ groups in $\ce{H3PO3}$, hence electron withdrawing effect is reduced because it is getting distributed between two groups. Hence, it has lesser acidic strength. Similarly, in $\ce{H3PO4}$, it is distributed between 3 groups and so the acidic strength is least.

Thus the acidic strength, is:
$$\ce{H3PO2 > H3PO3 > H3PO4}$$
(I hope it's clear.)


Find the minimum temperature at which the reaction will become spontaneous

You are only partially correct: negative Gibbs free energy at $\pu{273 K}$ is a sign of a spontaneous reaction.
However, the question asks when the reaction will become spontaneous and not to select the minimal temperature value from the multiple choices list which corresponds to the spontaneous process.

So, you were supposed to find the equilibrium temperature, e.g. when $ΔG = 0$.
You have correctly started with the equation for the free Gibbs energy:

$$ΔG = ΔH - TΔS = 0  \quad\implies\quad T_\mathrm{eq} = \frac{ΔH}{ΔS}$$

Unknown entropy change $ΔS$ can be determined from the Gibbs free energy $ΔG_1$ at initial temperature $T_1$:

$$ΔS = \frac{ΔH - ΔG_1}{T_1}$$

Finally,

$$T_\mathrm{eq} = \frac{ΔH}{ΔS} = \frac{ΔH}{ΔH - ΔG_1}T_1 = \frac{\pu{-90 kJ mol-1}}{\pu{-90 kJ mol-1}-(\pu{-45 kJ mol-1})}\cdot\pu{273 K} = \pu{546 K}$$


Boric acid does not dissociate in aqueous solution, rather it is acidic due to its interaction with water molecules (i.e accepts $\ce{OH-}$) to form the tetrahydroxyborate ion:

$\ce{B(OH)3 + H2O -> [B(OH)4]- + H+}$

$\ce{B(OH)3}$ does not act as a proton donor but behaves as a Lewis acid i.e. it accepts a pair of electrons from $\ce{OH-}$ ion and in this process $\ce{H+}$ ions are furnished.

I understand monobasic means there should be 1 replaceable hydrogen atom in the molecule, but $\ce{B(OH)3}$ in itself actually doesn't contain any replaceable hydrogen atom; rather, on interaction with water (i.e after accepting a $\ce{OH-}$ ion) a hydrogen ion is liberated. But if you see the after effect, say adding a monobasic acid (which holds the definition correctly) like $\ce{HCl}$ to water, per molecule it liberates one $\ce{H+}$ ion, similarly each $\ce{B(OH)3}$ also liberates one $\ce{H+}$ after it accepts $\ce{OH-}$ from water. So only looking at the after effects/ final result, both $\ce{HCl}$ and $\ce{B(OH)3}$ liberate 1 hydrogen per molecule though the process through which it was liberated is different.

So considering the after effect we consider $\ce{B(OH)3}$ as an acid and as it finally liberates 1 electron per molecule (though after accepting $\ce{OH-}$), we still can consider it to be a monobasic Lewis acid. At least, Wikipedia calls it a monobasic Lewis acid.


Combustion of NaCl(aq) when testing conductivity

There is not "burning", one might see a spark. It is a well known phenomenon. Do you see lightening in the sky? Air is an insulator, yet when very high voltage is present it becomes a conductor. It is called dielectric barrier discharge. In the same way, discharge between two electrodes can occur in pure water (almost an insulator). Adding NaCl to water makes it a good conductor but still it is mostly water, and if the electrodes are close enough, electrical breakdown can occur. Instead of ions carrying the current as it should happen in an aqueous conducting solution, the current becomes electronic even in the presence of water. Apparently, the local heating between the electrodes is so high, that a gaseous plasma is formed between the electrodes and plasma are electrically conducting.
As suggested in the comment, always exercise cause caution and all experiments should be done under proper supervision by a trained person/teacher.


What does ozone smell when treating with household water?

Although the primary products of chloride oxidation resulting from the ozonation of saline waters are chlorate salts ("Photochemical oxidation of chloride ion by ozone in acid aqueous solution", Environ Sci Pollut Res (2015) 22:16554 – 16569), a portion of the chloride is also known to oxidized to molecular chlorine, $\ce{Cl2}$.

Human olfactory senses can detect $\ce{Cl2}$ at levels as low as 0.2 ppm in air. This World Health Organization publication reports that some non-seawater contaminated surface waters in the UK contain greater than 40 mg/L chloride, and some seawater-contaminated aquifers used for drinking water in both the US and Philippines contain chloride concentrations on the order of a few hundred mg/L.

The degree of oxidation of $\ce{Cl-}$ to $\ce{Cl2}$ by ozone depends on several factors including temperature, lighting conditions, pH of the well water in question, and of course chloride concentration. However, based on the information given above, low-percent level conversion rates of $\ce{Cl-}$ to $\ce{Cl2}$ by ozone oxidation should be feasible, which under the right conditions, would result in the production of a noticeable chlorine odor. It indeed seems that $\ce{Cl-}$ to $\ce{Cl2}$ conversion represent the most likely scenario for the chlorine odor detected from the treated well water.

NOTE: OP mentioned an "infra red lamp" as part of the disinfectant process for the drinking water.  I suspect this really meant, or was in addition to, a UV lamp, which would further promote the oxidation of chloride per my first reference.


Reflected light (gloss) is a maximum when the angle of observation is equal to the angle of incidence:  https://en.wikipedia.org/wiki/Gloss_(optics). The image shows reflection from a plane surface with some roughness that gives diffuse reflection.

At Brewster's angle, the reflected ray is strongly polarized. The image is intended to indicate a depth of penetration in transparent materials; the polarization is not relevant here.

In conductors, like metals, and strong absorbers, like iodine, there is essentially no transmission, and almost all the light is reflected from a plane surface, because the penetration depth is exceedingly small.

In a dielectric (i.e., a nonconductor like glass or fats), some light will be transmitted, and reflection occurs after a penetration of some depth - on the order of a wavelength of the light. Glass is atomically uniform on this scale, and  a fat molecule (e.g., glyceryl trioleate) would be a glob about 2 or 3 millimicrons in diameter, therefore essentially homogeneous on this scale. Any liquid volume larger than that would have no discernible inhomogeneity and reflection would be a uniform wavefront, specular (mirror-like) rather than diffuse.

But suspensions and runny emulsions and milk are not homogeneous on that scale. They have particles on the order of a wavelength of light (~500 millimicrons) in the liquid, so reflection occurs from sites of different composition within this penetration depth. These materials reflect similar to reflection from a roughened metal surface and give a diffuse reflection rather than a gloss.


How does LC50 translate to "real world" exposure for paint thinner?

The $\mathrm{LD_{50}}$ and $\mathrm{LC_{50}}$ (lethal dose and lethal concentration) values are measures for acute toxicity as you noted. Long-term exposure is generally not depicted well by these acute toxicity measures. A.K. pointed out, how improbable it is to have a serious risk of acute poisoning. (Also note that the vapour pressure is given as $2~\mathrm{mmHg}$, which is $2.7~\mathrm{mbar}$ in proper units, meaning that evaporation will be very slow at ambient temperature and pressure.)

Chronic toxicity data is typically given as workplace exposure limits, under the reasoning that if you work with something every day and its concentration is less than the harmfulness limit, it should be okay. Scanning your MSDS, I did not come across any such value, so there should be no serious chronic toxicity if you keep your room well vented and don’t forget to recap the bottle, etc.


What are the roles of pyridine and DCM in the acylation of an alcohol?

Dichloromethane (DCM) is a common organic solvent and this is its role in both reactions (and in most cases where it is used). The first reaction is indeed the conversion of a carboxylic acid to an acyl chloride. The DMF is a catalyst and the reactions proceeds through an imidoyl chloride intermediate, $\ce{Me2N+=CHCl}$.

Pyridine is a weak base, with a $\mathrm{p}K_\mathrm{aH}$ of 5.21 and it is often used to remove extra protons during the course of a reaction. It is also a good nucleophile for attacking carbonyl groups to activate them towards attack by other, less good nucleophiles (thanks to @orthocresol and @DGS for pointing this out). In this reaction it serves both these purposes.

An acid anhydride can also be used as the acylating agent. The principle is the same: pyridine acts as a nucleophilic catalyst and generates the same intermediate as above.


Change in concentration when acids are diluted

Let suppose $c, c_1, c_2$ are concentrations of weak acids, and $K_\mathrm{a}, K_\mathrm{a1}, K_\mathrm{a2}$ are their respective acidity(dissociation) constants.

The equations $\ce{[H+]}=\sqrt{K_\mathrm{a} \cdot c}$ for a weak acid, respectively for 2 acids $\ce{[H+]}=\sqrt{K_\mathrm{a1} \cdot c_1 + K_\mathrm{a2} \cdot c_2}$

are approximately valid, if $c, (c_1, c_2) \gg K_\mathrm{a}(, K_\mathrm{a1}, K_\mathrm{a2}) \gg 10^-7 = \sqrt{K_\mathrm{w} }$, what also implies $\ce{[H+]} \gg 10^{-7}$.

Then we can approximate all $\ce{H+}$ comes from the acid dissociation, but OTOH, all acid is in non-dissociated form, leading to $$K_\mathrm{a} = \frac{\ce{[H+][A-]} }{\ce{[HA]}} \simeq \frac{{[H+]}^2 }{ c}$$ and $$\ce{[H+]}=\sqrt{K_\mathrm{a} \cdot c}$$

If, OTOH, $c \ll 10^{-7}=\sqrt{K_\mathrm{w}}$  and $c \ll K_\mathrm{a}$ we can approximate all acid is dissociated.

Then $$K_\mathrm{w} =  \ce{[H+][OH-]}$$

$$\ce{[H+]} - \ce{[OH-]} = c_1 + c_2$$

$$\ce{[H+]} - \frac{K_\mathrm{w}}{\ce{[H+]}} = c_1 + c_2$$

$${\ce{[H+]}}^2 - (c_1 + c_2) \cdot \ce{[H+]} - K_\mathrm{w} = 0 $$

$$\ce{[H+]} = \frac{c_1 + c_2 \pm \sqrt {{(c_1 + c_2)}^2 + 4 \cdot K_\mathrm{w} }}{2}  $$

$$\ce{[H+]} = \frac{\pu{2.1e-9} \pm \sqrt {{(\pu{2.1e-9})}^2 + \pu{4e-14} }}{2} = \pu{1.0105e-7 mol/L }$$

$$\mathrm{pH} = 6.995$$

It is obvious, that if with full dissociation $\mathrm{pH}$ does not reach even the value $6.99$, the result of partial dissociation cannot be $6.7$


What happens when a semipermeable membrane allows only one solute to pass through?

I am afraid you are stuck because you considered glucose as osmotically active.
But you assumed that glucose can freely move across the membrane, which is the opposite of osmotic activity.
So, short-short answer : ignore the glucose, it is an osmolyte, but it is not osmotically active on the membrane you are thinking of; its concentration is going to equilibrate on each side of the membrane :
$$[\mathrm{C_6H_{12}O_6}] = \frac{1 \mathrm{mol}}{2 \mathrm{L}} = 0.5\, \mathrm{mol\, L^{-1}}$$
This solution of glucose is simply the solvent.
Also, you may have inverted terms in the equilibrium equation.

A longer answer :
First we assume that the volume and density of the solution are unchanged by the solutes added (ideal solutions).
We are left with $1 \,\mathrm{osmol\cdot L^{-1}}$ NaCl on one side of the membrane, and no osmolyte on the other side.

The equilibrium between hydrostatic and osmotic pressure ($i c R T $) then writes :
$$ 2 \rho g y = i \frac{n}{(h+y)A} RT  $$
Solving for $y$, we get
$$ 2y = \sqrt{\frac{2inRT}{\rho g A}+h^2}-h$$
Using the following values :  $i=2$, $n=0.5 \,\mathrm{mol}$, $R=8.3\,\mathrm{kg\cdot  m^2\cdot s^{-2}\cdot K^{-1}\cdot mol^{-1}} $, $T=3\cdot 10^2\,\mathrm{K}$, $\rho=10^3\,\mathrm{kg\cdot m^{-3}}$, $g=9.8\,\mathrm{m\cdot s^{-2}}$ ,  $A=10^{-2}\, \mathrm{m^{-2}} $ and $h = 10^{-1} \mathrm{m} $, we obtain :
$$ 2y \simeq 7\, \mathrm{m} $$
This value is actually unreachable given the volume of solvent and the geometry of your beaker. All the glucosed water would be drawn on one side, and equilibrium could not be reached.

The osmotic pressure of a solution containing $1\,\mathrm{osmol\cdot L^{-1}}=10^3 \,\mathrm{osmol\cdot m^{-3}} $ is approximately $\Pi = 2.5\, \mathrm{MPa} = 25 \, \mathrm{bar}$. To equilibrate such a pressure with hydrostatic pressure, you need a column of water of depth $z = \frac{\Pi}{\rho g} \simeq 260\,\mathrm{m}$.

The osmolarity of sea water is about $1\,\mathrm{osmol\cdot L^{-1}}$. In a desalination plant, sea water is purified by reverse osmosis using pressures from 50 to 80 bar. Osmotic equilibrium is of utmost importance in physiology, but this you must know better than me. Unless there is active transport, diffusion of solutes, or extraordinary high applied pressures (which may be relevant for deep-sea creatures), water is always going from the least to the most concentrated side of the membranes, so as to even out the osmolarities. Hope it helped.


Reaction between n-butylithium and either an aromatic bromide or an alkylbromide

It's quite a long time that I did myself, but I used BuLi to transmetalate. I.e. $\ce{MesBr + BuLi -> MesLi + BuBr}$


Why is a drop of chloroform stretched/pulled apart on a water surface?

As you pointed out, the surface tension of water-air interface is greater than the surface tension of chloroform-air interface. At $25~^\circ\mathrm{C}$, the surface tension of water-air interface is $71.99\pm0.05~\mathrm{mNm^{-1}}$ and that of chloroform-air interface is $26.67~\mathrm{mNm^{-1}}$ (source).
Now, let's understand why a drop of chloroform placed over water gets puller apart forming a layer. Surface tension is the force per unit length. More the surface tension, more is the force per unit length and vice versa. Surface tension is similar to tension in a stretched rubber sheet. As you place a drop of chloroform on water, the system looks as shown in the following diagram:

As we seen earlier, surface tension of water-air interface is greater than that of chloroform-air interface. So in the boundary separating chloroform and water there exists an imbalance of force. The force vectors are the red arrows in the diagram above and their strength is denoted by their length and weight (thickness).
Due to this, the drop of chloroform expands to form a layer on top of water. This can also be looked in the nature's tendency to lower energy. The surface energy of chloroform-air interface is lesser than that of water-air interface. So, a lower energy configuration is preferred.


The reaction is given here.

$$\ce{3PbI2 + 8HNO3 ->[\Delta] 3Pb(NO3)2 + 3I2 + 2NO + 4H2O}$$

  Lead(II) iodide react with nitric acid to produce  lead(II) nitrate,
  iodine, nitrogen monoxide and water. Nitric acid - 30% solution. The
  reaction takes place in a boiling solution.


The vapour pressure lowering is a consequence of Raoult's law with a non-volatile solute, i.e. a solvent that has zero vapour pressure (or at least, lower vapour pressure than the solvent).

Regarding the second part (“How can we derive, understand or quantify various thermodynamic parameters (ΔG,ΔS,ΔH etc) for the process of deliquescence?”), I am not sure exactly what you are looking for. This is a complex phenomenon, and there is a lot of research literature of the topic. To give only one example, see

  “Water-Solids Interactions: Deliquescence”, Lisa J. Mauer and Lynne S. Taylor, Ann. Rev. of Food Sci. Technol., 2010, 1, 41-63.


What is exactly meant by position of equilibrium?

The term "equilibrium shifting" is mostly used for a reaction that was already at equilibrium, but we changed something (in the simplest case adding one reactant or product to the equilibrium mixture so the reaction is no longer at equilibrium). The "equilibrium is shifting" is a shortcut for saying there is a net reaction to re-establish equilibrium.

The term is used both in cases where the equilibrium constant does not change (change in one of the concentrations) and in cases where the equilibrium constant does change (change in temperature, for example).

  Does it mean that the rate of forward reaction is now faster than the rate of backward reaction, which would mean that the reaction isn't at equilibrium anymore?

Yes, for some reason the reaction is out of equilibrium, otherwise there would be no net reaction (no "shift").

  I'm confused with the terms "position of equilibrium" and "equilibrium shifting towards left/right".

"Position of equilibrium" could be roughly equated with the reaction quotient Q or more generally with the set of all concentrations at equilibrium.

Taking a change in temperature as an example, you start with one set of equilibrium concentrations ("old position of equilibrium" corresponding to some reaction quotient $Q_1$). As you change the temperature, the reaction is no longer at equilibrium because the equilibrium constant changes (lets say from $K_{T_1}$ to $K_{T_2}$), so $Q_1$ is equal to the former equilibrium constant $K_{T_1}$, but not to the current equilibrium constant $K_{T_2}$). There is a net reaction until you reach the new equilibrium ("new position of equilibrium" corresponding to some other reaction quotient $Q_2$). At the first temperature $T_1$ after equilibrium is established, $Q_1 = K_{T_1}$. At the second temperature, $Q_2 = K_{T_2}$ once the equilibrium is re-established.

For adding or removing one of the species, you start with the old position of equilibrium, then the addition or removal of a species changes Q, which no longer equals K. As a consequence, there is a net reaction that changes Q back to K. Now, you have a new set of concentrations, but the reaction quotient is back to what it was at equilibrium. I'm not sure whether to call this a new position of equilibrium, or to say the reaction went back to equilibrium (this might be confusing because the equilibrium corresponds to a third set of concentrations - first set: original equilibrium concentrations; second set: concentrations right after the manipulation; third set: concentrations after the net reaction re-establishes equilibrium).


Find volume of ammonia gas required to prepare a solution of given pH

The $\mathrm{pH}$ of your final ammonia solution is $11.0$. When you dissolve ammonia in water it tends to ionize with water to give following equilibrium:
$$\ce{NH3 + H2O <=> NH4+ + OH-} \quad K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} = 1.8 \times 10^{-5}$$
Suppose final concentration of $\ce{NH3}$ is $c$, and equlibrium concentrations of $\ce{NH4+}$ and $\ce{NH4+}$ are $\alpha$. Thus:
$$K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} = \frac{\alpha \cdot \alpha}{c - \alpha} = 1.8 \times 10^{-5}$$
$$\alpha^2 + 1.8 \times 10^{-5}\alpha - 1.8 \times 10^{-5}c =0 \tag1$$
Since, $\mathrm{pOH}$ of the solution is $3.0$ ($14 - \mathrm{pH}$), $[\ce{OH-}] = \alpha = 1.00 \times 10^{-3}$, from equation $(1)$:
$$(1.00 \times 10^{-3})^2 + 1.8 \times 10^{-5}\times 1.00 \times 10^{-3} = 1.8 \times 10^{-5}c $$
$$1.8 \times 10^{-5}c = 1.00 \times 10^{-6} + 1.8 \times 10^{-8} = 1.018 \times 10^{-6} $$
$$\therefore \ \ c = \frac{1.018 \times 10^{-6}}{1.8 \times 10^{-5}} = \pu{5.66 \times 10^{-2} molL-1} $$
This is exactly what you got without showing your work. However, you made only $\pu{100 mL}$ of solution. Thus, amount of ammonia dissolved in $\pu{100 mL}$ of water is:
$$$\pu{0.100 L} \times \pu{5.66 \times 10^{-2} molL-1} = \pu{5.66 \times 10^{-3} mol}$
Assuming gaseous ammonia is behaving like ideal gas, you can use $PV = nRT$ here to find the volume of $\pu{5.66 \times 10^{-3} mol}$ of ammonia at STP:
$$V= \frac{nRT}{P} = \frac{(\pu{5.66 \times 10^{-3} mol})(\pu{0.082 LatmK-1mol-1})(\pu{298 K})}{\pu{1 atm}} = \pu{0.138 L}$$


Aromatic nomenclature, when to use numbers or prefixes?

1,4-dimethylbenzene and p-xylene are both equally correct nomenclatures. You could also say 4-methyl toluene if you wanted. I personally prefer the first example because it allows me to visualize the molecule immediately.

As for 2,4-diiodonitrobenzene, this is indicating three substituents. You have the nitro group, and two iodo groups - one para to the nitro, and one ortho to the nitro.


What is the pH at both equivalence points of titration between diprotic tartaric acid and NaOH?

For the 1st equivalence point, the tartaric acid $\ce{H2A}$ is titrated to $\ce{NaHA}$

For the solutions of acid salts, we can with advantage use the formula

$$\mathrm{pH}=\frac{\mathrm{p}K_\mathrm{a1}+\mathrm{p}K_\mathrm{a2}}{2}$$

that can be derived from equilibrium

$$\ce{2 HA- <=> H2A + A^2-}$$

For the 2nd equivalence point there is  $\ce{Na2A}$ solution, which is hydrolyzed with equilibrium

$$\ce{A^2- + H2O <=> HA- + OH-}$$

With neglecting OH- from water hydrolysis, $[\ce{HA^-}] = [\ce{OH^-}]$

$$\begin{align}
K_\mathrm{a2}&=\frac{[\ce{H+}][\ce{A^2-}]}{[\ce{HA-}]}\\
\mathrm{p}K_\mathrm{a2}&=\mathrm{pH} - \log{[\ce{A^2-}]} +\log {[\ce{HA-}]}\\
&=\mathrm{pH} - \log{[\ce{A^2-}]} +\log {[\ce{OH-}]}\\
&=\mathrm{pH} - \log{[\ce{A^2-}]} -\mathrm{pOH}\\
&=\mathrm{pH} - \log{[\ce{A^2-}]} -14 + \mathrm{pH}\\
\mathrm{pH}&=7+\frac12\left(\mathrm{p}K_\mathrm{a2} +\log{ c_\mathrm{\ce{Na2A}}}\right)\\
\end{align}$$

The molarity of $\ce{NaOH}$ solution $c_{\ce{NaOH}}=\frac{m}{M\times V}=\pu{0.09801  mol/L}$

The volume of $\ce{NaOH}$ solution for given volume of tartaric acid solution

$$V_{\ce{NaOH}}=V_{\ce{H2A}}\times \frac{2\times c_{\ce{H2A}}}{c_{\ce{NaOH}}}\\=0.0102\times V_{\ce{H2A}}$$

The final concentration

$$\begin{align}
[\ce{Na2A}]&=\frac{[\ce{H2A_\mathrm{init}}]\cdot V_{\ce{H2A}}}{V_{\ce{H2A}} +V_{\ce{NaOH}}}\\
&=\frac{[\ce{H2A_\mathrm{init}}]\cdot V_{\ce{H2A}}}{V_{\ce{H2A}} +
0.0102  \cdot  V_{\ce{H2A}}}\\
&=\frac{[\ce{H2A_\mathrm{init}}]}{1+ 0.0102}\\
&=5\times 10^{-4}/1.0102\\
&=\pu{4.9495\times10^{-4} mol/L}\\
&\overset{\mathrm{rounding}}=\pu{4.95\times 10^{-4} mol/L}\\
\end{align}$$


Difference between radial, planar, angular and spherical nodes

The wave function $\Psi(r,\theta,\phi)$ of a one-electron (hydrogen-like) atom is seperable as the product of a radial function $R(r)$ and an angular function $Y(\theta,\phi)$

$\Psi(r,\theta,\phi) =  R(r)Y(\theta,\phi)$

If $R(r_1) = 0$, there exists a radial node.  The radial node is a sphere with radius $r_1$. Therefore the terms "radial node" and "spherical node" are the same.

$Y(\theta,\phi)$ is further seperable as

$Y(\theta,\phi) = P(\theta)F(\phi)$

If either $P(\theta)$ or $F(\phi)$ is zero for a given respective angle value, there is an angular node.  However, a angular node is not necessarily a planar node.  An angular node could be a planar node or a conical node.

$F(\phi)$ being zero corresponds to a planar node.

$P(\theta)$ being zero corresponds to either a conical node or a planar node (some think of the planar case as a specical case of conical, with apex angle being 180 degrees)

Overall, there will be $n-1$ nodes.

$l$ nodes will be angular

$n-l-1$ nodes will be radial (spherical)


Before entering the kinetics realm, it is advisable to check the equilibrium. Besides the straightforward combustion reactions, you have two more, Boudouard reaction, which favors formation of $\ce{CO}$ from $\ce{CO_2}$ and Water-gas shift reaction.

The actual reactions which contribute to the equilibrium depend on the fuel. In case of pure carbon, you at least get rid of the Water-gas shift reaction.

You might need to solve the system iteratively, as the temperature of the equilibrium mixture is result of the combustion reactions, but also affects the position of equilibrium.